Online Education for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

Online Education for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

These Solutions are part of Online Education RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6

Other Exercises

Question 1.
Find the sum of the following arithmetic progressions :
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 1
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 2
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 3
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 4
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 5
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 6
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 7

Question 2.
Find the sum to n term of the A.P. 5, 2, – 1, 4, -7, …,
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 8

Question 3.
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 – 6n.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 9

Question 4.
Find the sum of last ten terms of the A.P.: 8, 10, 12, 14,…, 126. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 10

Question 5.
Find the sum of the first 15 terms of each of the following sequences having nth term as
(i) an = 3 + 4n
(ii) bn = 5 + 2n
(iii) xn = 6 – n
(iv) yn = 9 – 5n
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 11
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 12
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 13
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 14

Question 6.
Find the sum of first 20 terms of the sequence whose nth term is an = An + B.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 15

Question 7.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 – 3n. [CBSE 2004]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 16
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 17

Question 8.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 – 3n. [CBSE 2004]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 18

Question 9.
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, …, is 116. Find the last term.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 19
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 20

Question 10.
(i) How many terms of the sequence 18, 16, 14, … should be taken so that their sum is zero ?
(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?
(iii) How many terms of the A.P. 9, 17, 25,… must be taken so that their sum is 636 ? [NCERT]
(iv) How many terms of the A.P. 63, 60, 57, ……… must be taken so that their sum is 693 ? [CBSE 2005]
(v) How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero? [CBSE 2016]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 21
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 22
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 23
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 24
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 26

Question 11.
Find the sum of the first
(i) 11 terms of the A.P. : 2, 6, 10, 14,…
(ii) 13 terms of the A.P. : -6, 0, 6, 12,…
(iii) 51 terms of the A.P.: whose second term is 2 and fourth term is 8.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 27
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 28
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 29

Question 12.
Find the sum of
(i) the first 15 multiples of 8
(ii) the first 40 positive integers divisible by
(a) 3, (b) 5, (c) 6
(iii) all 3-digit natural numbers which are divisible by 13. [CBSE 2006C]
(iv) all 3-digit natural numbers, which are multiples of 11. [CBSE 2012]
(v) all 2-digit natural numbers divisible by 4. [CBSE 2017]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 30
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 31
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 32
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 33
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 34
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 35
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 36

Question 13.
Find the sum :
(i) 2 + 4 + 6 + ……….. + 200
(ii) 3 + 11 + 19 + ………. + 803
(iii) (-5) + (-8) + (-11) + ……. + (-230)
(iv) 1 + 3 + 5 + 7 + …….. + 199
(v) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + ……… + 84
(vi) 34 + 32 + 30 + ………. + 10
(vii) 25 + 28 + 31 + ……….. + 100 [CBSE 2006C]
(viii) 18 + 15\(\frac { 1 }{ 2 }\) + 13 + ……… + (-49\(\frac { 1 }{ 2 }\))
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 37
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 38
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 39
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 40
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 41
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 42

Question 14.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?
Solution:
First term of an A.P. (a) = 17
Last term (l) = 350
Common difference (d) = 9
Let n be the number of terms Then an = a + (n – 1) d
=> 350 = 17 + (n – 1) x 9
=> 350 = 17 + 9n – 9
=> 9n = 350 – 17 + 9 = 342
n = 38
Number of terms = 38
Now Sn = \(\frac { n }{ 2 }\) [a + l]
= \(\frac { 38 }{ 2 }\) [17 + 350] = 19 (367) = 6973

Question 15.
The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 43
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 44

Question 16.
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 45

Question 17.
If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms ?
Solution:
12th term of an A.P. = -13
Sum of first 4 terms = 24
Let a be the first term and d be the common difference
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 46

Question 18.
Find the sum of n terms of the series
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 47
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 48
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 49

Question 19.
In an A.P., if the first term is 22, the common difference is -4 and the sum to n terms is 64, find n.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 50

Question 20.
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms ?
Solution:
In an A.P.
5th term = 30
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 51

Question 21.
Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution:
In an A.P.
No. of terms = 51
Second term a2 = 14
and third term a3 = 18
Let a be the first term and d be the common
difference, then
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 52
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 53

Question 22.
If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Solution:
Let a be the first term and d be the common difference of an A.P.
Sum of 7 terms = 49
and sum of 17 terms = 289
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 54
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 55

Question 23.
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 56

Question 24.
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences. [CBSE 2008]
Solution:
In an A.P.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 57

Question 25.
In an A.P., the first term is 22, nth term is -11 and the sum to first n terms is 66. Find n and d, the common difference. [CBSE 2008]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 58
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 59

Question 26.
The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 60

Question 27.
The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 61

Question 28.
The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P. [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 62
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 63

Question 29.
If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its nth term. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 64
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 65

Question 30.
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 66
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 67

Question 31.
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 68
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 69

Question 32.
The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 70
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 71

Question 33.
In an A.P., the sum of first ten terms is -150 and the sum of its next ten terms is -550. Find the A.P. [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 72
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 73

Question 34.
Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term. [CBSE 2012]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 74

Question 35.
In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P. [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 75

Question 36.
The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ? [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 76

Question 37.
Find the number of terms of the A.P. -12, -9, -6,…, 21. If 1 is added to each term of this A.P., then find the sum ofi all terms of the A.P. thus obtained. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 77
= 6 x 9 = 54
If we add 1 to each term, then the new sum of so formed A.P.
= 54 + 1 x 12 = 54 + 12 = 66

Question 38.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 78

Question 39.
The sum of first n terms of an A.P. is 5n – n2. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 79

Question 40.
The sum of the first n terms of an A.P. is 4n2 + 2n. Find the nth term of this A.P. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 80

Question 41.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P. [CBSE 2013]
Solution:
Let a be the first term and d be common difference
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 81

Question 42.
The sum of first n terms of an A.P. is 5n2 + 3n. If its mth term is 168, find the value of m. Also, find the 20th term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 82

Question 43.
The sum of first q terms of an A.P. is 63q – 3q2. If its pth term is -60, find the value of p, Also, find the 11th term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 83

Question 44.
The sum of first m terms of an A.P. is 4m2 – m. If its nth term is 107, find the value of n. Also, find the 21st term of this A.P. [CBSE 2013]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 84

Question 45.
If the sum of the first n terms of an A.P. is 4n – n2, what is the first term ? What is the sum of first two terms ? What is the second term ? Similarly, find the third, the tenth and the nth terms. [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 85

Question 46.
If the sum of first n terms of an A.P. is \(\frac { 1 }{ 2 }\) (3n2 + 7n), then find its nth term. Hence write its 20th term. [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 86
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 87

Question 47.
In an A.P., the sum of first n terms is \(\frac { { 3n }^{ 2 } }{ 2 } +\frac { 13 }{ 2 } n\). Find its 25th term. [CBSE 2006C]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 88
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 89
a25 = 8 + (25 – 1) x 3 = 8 + 24 x 3 = 8 + 72 = 80
Hence 25th term = 80

Question 48.
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 90

Question 49.
Find the sum of first n odd natural numbers.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 91

Question 50.
Find the sum of all odd numbers between
(i) 0 and 50
(ii) 100 and 200
Solution:
(i) Odd numbers between 0 and 50 are = 1, 3, 5, 7, …, 49 in which
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 92

Question 51.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 93
Hence proved.

Question 52.
Find the sum of all integers between 84 and 719, which are multiples of 5.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 94

Question 53.
Find the sum of all integers between 50 and 500, which are divisible by 7.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 95

Question 54.
Find the sum of all even integers between 101 and 999.
Solution:
All integers which are even, between 101 and 999 are = 102, 104, 106, 108, … 998
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 96

Question 55.
(i) Find the sum of all integers between 100 and 550, which are divisible by 9.
(ii) all integers between 100 and 550 which are not divisible by 9.
(iii) all integers between 1 and 500 which are multiplies of 2 as well as of 5.
(iv) all integers from 1 to 500 which are multiplies 2 as well as of 5.
(v) all integers from 1 to 500 which are multiplies of 2 or 5.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 97
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 98
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 99
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 100
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 101
= 250 x 251 + 505 x 50 – 25 x 510
= 62750 + 25250 – 12750
= 88000 – 12750
= 75250

Question 56.
Let there be an A.P. with first term ‘a’, common difference d. If an denotes its nth term and S the sum of first n terms, find.
(i) n and S , if a = 5, d = 3 and an = 50.
(ii) n and a, if an = 4, d = 2 and Sn = -14.
(iii) d, if a = 3, n = 8 and Sn = 192.
(iv) a, if an = 28, Sn = 144 and n = 9.
(v) n and d, if a = 8, an = 62 and Sn = 210.
(vi) n and an, if a = 2, d = 8 and Sn = 90.
(vii) k, if Sn = 3n2 + 5n and ak = 164.
Solution:
In an A.P. a is the first term, d, the common difference a is the nth term and Sn is the sum of first n terms,
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 102
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 103
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 104
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 105
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 106
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 107

Question 57.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 – S4). [NCERT Exemplar, CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 108

Question 58.
A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5m/minute every succeeding minute. After how many minutes, the policeman will catch the thief? [CBSE 2016]
Solution:
Let total time be 22 minutes.
Total distance covered by thief in 22 minutes = Speed x Time
= 100 x n = 100n metres
Total distance covered by policeman
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 109

Question 59.
The sums of first n terms of three A.P.S are S1, S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2. [CBSE 2016]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 110
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 111

Question 60.
Resham wanted to save at least 76500 for sending her daughter to school next year (after 12 months). She saved ₹450 in the first month and raised her savings by ₹20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
Solution:
Given : Resham saved ₹450 in the first month and raised her saving by ₹20 every month and saved in next 12 months.
First term (a) = 450
Common difference (d) = 20
and No. of terms (n) = 12
We know sum of n terms is in A.P.
Sn = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
Sn = \(\frac { 12 }{ 2 }\) [2 x 450 + (12 – 1) x 20]
=> Sn = 6[900 + 240]
=> Sn = 6720
Here we can see that Resham saved ₹ 6720 which is more than ₹ 6500.
So, yes Resham shall be able to send her daughter to school.

Question 61.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 112

Question 62.
Ramkali would need ₹ 1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved ₹ 50 in the first month of this year and increased her monthly saving by ₹ 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school? [CBSE 2014]
Solution:
Admission fee and books etc. = ₹ 1800
First month’s savings = ₹ 50
Increase in monthly savings = ₹ 720
Period = 1 year = 12 months
Here a = 50, d = 20 and n = 12
S12 = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
= \(\frac { 12 }{ 2 }\) [2 x 50 + (12 – 1) x 20]
= 6[100 + 11 x 20]
= 6[100 + 220]
= 6 x 320 = ₹ 1920
Savings = ₹ 1920
Yes, she will be able to send her daughter.

Question 63.
A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100 more than he did in the preceding year. How much did he save in the first year ?
Solution:
Savings in 10 years = ₹ 16500
S10 = ₹ 16500 and d = 7100
Sn = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
16500= \(\frac { 10 }{ 2 }\) [2 x a + (10 – 1) x 100]
16500 = 5 (2a + 900)
16500 = 10a + 4500
=> 10a = 16500 – 4500 = 12000
a = 1200
Saving for the first year = ₹ 1200

Question 64.
A man saved ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.
Solution:
Savings for the first year = ₹ 32
For the second year = ₹ 36
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 113

Question 65.
A man arranges to pay off a debt of ₹ 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one – third of the debt unpaid, find the value of the first installment.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 114
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 115

Question 66.
There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
Solution:
Number of trees = 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 116
Distance between one to other tree = 5 m
Distance between first near and the well = 10 m
Now in order to water the first tree, the gardener has to cover 10m + 10m = 20m
and to water the second tree, the distance to covered is 15 + 15 = 30 m
To water the third tree, the distance to cover is = 20 + 20 = 40 m
The series will be 20, 30, 40, ……….
where a = 20, d = 30 – 20 = 10 and n = 25
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 117

Question 67.
A man is employed to count ₹ 10710. He counts at the rate of ₹ 180 per minute for half an hour. After this he counts at the rate of ₹ 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 118
=> (n – 59) (n – 60) = 0
Either n – 59 = 0, then n – 59 or n – 60 = 0, then n = 60
Total time = 59 + 30 = 89 minutes or = 60 + 30 = 90 minutes

Question 68.
A piece of equipment cost a certain factory ₹ 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost ?
Solution:
Cost of a piece of equipment = ₹ 600,000
Rate of depreciation for the first year = 15%
for the second year = 13.5%
for the third year = 12.0% and so on
The depreciation is in A.P.
whose first term (a) = 15
and common difference (d) = 13.5 – 15.0 = -1.5
Period (n) = 10
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 119

Question 69.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each prize.
Solution:
Total sum = ₹ 700
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 120

Question 70.
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3 (S20 – S10). [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 121

Question 71.
Solve the question: (-4) + (-1) + 2 + 5 + … + x = 437. [NCERT Exemplar]
Solution:
Given equation is,
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 122
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 123

Question 72.
Which term of the A.P. -2, -7, -12, … will be -77 ? Find the sum of this A.P. up to the term -77.
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 124

Question 73.
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n. [NCERT Exemplar]
Solution:
Given that, first term of the first A.P. (a) = 8
and common difference of the first A.P. (d) = 20
Let the number of terms in first A.P. be n
Sum of first n terms of an A.P., Sn
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 125
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 126

Question 74.
The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middle most flag Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag? [NCERT Exemplar]
Solution:
Given that, the students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school.
Given that, the number of flags = 27
and distance between each flag = 2 m.
Also, the flags are stored at the position of the middle most flag i. e., 14th flag and Ruchi was given the responsibility of placing the flags.
Ruchi kept her books, where the flags were stored i.e., 14th flag and she could carry only one flag at a time.
Let she placed 13 flags into her left position from middle most flag i.e., 14th flag.
For placing second flag and return his initial position distance travelled = 2 + 2 = 4 m.
Similarly, for placing third flag and return his initial position, distance travelled = 4 + 4 = 8 m.
For placing fourth flag and return his initial position, distance travelled = 6 + 6 = 12 m.
For placing fourteenth flag and return his initial position, distance travelled = 26 + 26 = 52 m.
Proceed same manner into her right position from middle most flag i.e., 14th flag.
Total distance travelled in that case = 52 m.
Also, when Ruchi placed the last flag she return her middle position and collect her books.
This distance also included in placed the last flag.
So, these distances from a series.
4 + 8 + 12 + 16 + … + 52 [for left]
and 4 + 8 + 12 + 16 + … + 52 [for right] .
Total distance covered by Ruchi for placing these flags
= 2 x (4 + 8 + 12 + … +52)
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 127
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 128
Hence, the required is 728 m in which she did cover in completing this job and returning back to collect her books.
Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position
= (2 + 2 + 2 + … + 13 times)
= 2 x 13 = 26 m
Hence, the required maximum distance she travelled carrying a flag is 26 m.

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Online Education NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist

In Online Education NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist are part of NCERT Solutions for Class 10 English. Here we have given NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist.

Board CBSE
Textbook NCERT
Class Class 10
Subject English Footprints Without Feet
Chapter Chapter 6
Chapter Name The Making of a Scientist
Category NCERT Solutions

Online Education NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist

TEXTUAL EXERCISES

Read and Find Out (Pages 32 & 34)

Question 1.
How did a book become a turning point in Richard Ebright’s life ?
Answer:
A book entitled ‘The Travels of Monarch X’ changed Richard Ebright’s life. It was because it told him how monarch butterflies migrate to Central America. It opened the wide world of science to this curious young collector.

Question 2.
How did his mother help him ?
Or
How did Richard’s mother help him to become a scientist ? (CBSE 2014)
Answer:
His mother was a source of great inspiration for him. She always encouraged him and provided constant company to him. In the day she brought friends for him. In the night she herself worked with him. This way she helped him.

Question 3.
What lesson does Ebright learn when he does not win anything at a science fair ? (Imp.) (CBSE 2011)
Answer:
Ebright learns a lesson that the real experiments would win him a prize. It was not simply a neat display of things. The real experiments showed the things clearly and proved the point. The subject he knew was about the insect work. He had been carrying this work in the past several years.

Question 4.
What experiments and projects does he then undertake ?
Answer:
He took up his eighth-grade project. It was to find the cause of a viral disease that killed all monarch caterpillars.

Next year his science fair project was to test the theory of the butterflies. It was that viceroy butterflies copied monarchs. His next theory was to see whether birds ate monarchs. He found that monarchs don’t taste good to birds.

Next he began the research. It led to his discovery of an unknown insect hormone.

Question 5.
What are the qualities that go into the making of a scientist ?
Answer:
These qualities are : first-rate mind, curiosity, will to win for the right reasons.

Think About It (Page 38)

Question 1.
How can one become a scientist, an economist, a historian … ? Does it simply involve reading many books on the subject ? Does it involve observing, thinking and doing experiments ? (V. Imp.)
Answer:
One can become a scientist, an economist or a historian. He can become so not by reading many books on the subject. First of all, one must have a first-rate mind. One must have the capacity and the will to win for the right reasons. With all these one must seek the result through experiments. This involves observing with a sharp mind, thinking and testing things through experiments.

Question 2.
You must have read about cells and DNA in your science books. Discuss Richard Ebright’s work in the light of what you have studied. If you get an opportunity to work like Richard Ebright on projects and experiments, which field would you like to work on and why ?
Answer:
I have read a bit about cells and DNA in the science books. Richard Ebright’s work really is very important. It shall let the scientists and others know the origins of diseases etc. This will be possible because the field of experiments in cells is vast. Study of cells and DNA let one have knowledge about heredity and what makes this. If ever I get a chance to work like Ebright, I shall adopt this field. It would help humanity in fighting various diseases.

Talk About It (Page 38)

Question 1.
You also must have wondered about certain things around you. Share these questions with your class, and try and answer them.
Answer:
Like others I have thought about certain changes around me. I have tried to think their answers. These questions are :

  1. Why do mango leaves change colours ?
  2. Where do days and nights go ?
  3. How pigeons return to their perches after long flights to distant places ?
  4. How do small pipal seeds hide in them the features of big pipal trees ?

Science teachers may have convincing answers.

We hope the NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist help you. If you have any query regarding NCERT Solutions for Class 10 English Footprints Without Feet Chapter 6 The Making of a Scientist drop a comment below and we will get back to you at the earliest.

The Sermon at Benares Extra Questions and Answers Class 10 English First Flight

Online Education for The Sermon at Benares Extra Questions and Answers Class 10 English First Flight

Here we are providing Online Education for The Sermon at Benares Extra Questions and Answers Class 10 English First Flight, Extra Questions for Class 10 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-english/

Online Education for The Sermon at Benares Extra Questions and Answers Class 10 English First Flight

The Sermon at Benares Extra Questions and Answers Very Short Answer Type

The Sermon At Benares Extra Questions And Answers Question 1.
What was the original name of the Buddha?
Answer:
The original name of the Buddha was Siddhartha Gautama.

Sermon At Benares Extra Questions Question 2.
What did Siddhartha study in school?
Answer:
Siddhartha was sent away to school to study the sacred Hindu scriptures.

The Sermon At Benares Extra Questions Question 3.
What did Siddhartha see at the age of twenty-five?
Answer:
At the age of twenty-five, he saw a sickman, an aged man, a funeral procession and a monk begging.

Sermon At Benares Class 10 Extra Questions Question 4.
Where did Siddhartha get enlightened?
Answer:
Siddhartha got enlightened under a peepal tree after seven days.

Extra Questions Of The Sermon At Benares Question 5.
Why was the tree renamed as‘Bodhi Tree’?
Answer:
The Buddha got wisdom under the tree. So it was renamed ‘Bodhi tree’.

Sermon At Benaras Extra Questions Question 6.
What was the problem of Gotami?
Answer:
Gotami’s son had died. She wanted him to live again.

Extra Questions Of Sermon At Benaras Question 7.
What did she see in the city?
Answer:
She watched the light of the city.

The Sermon At Benares Class 10 Extra Questions Question 8.
What did she learn from the flickering of lights?
Answer:
She learnt that the fate of men is just like the city lights that flickered and extinguished again and again.

The Sermon At Benaras Extra Questions Question 9.
Who is being addressed by the Buddha?
Answer:
The people of Benares were being addressed by the Buddha.

The Sermon At Benares Short Questions And Answers Question 10.
What did he say about life of mortals?
Answer:
Buddha said that all mortals have to die.

The Sermon At Benares Questions And Answers Question 11.
What is inevitable?
Answer:
Death is inevitable.

The Sermon At Benares Important Questions Question 12.
What have the life and death of the man been compared to?
Answer:
The life and death are being seen and compared with the ripe fruits which have to fall.

The Sermon at Benares Extra Questions and Answers Short Answer Type

Question 1.
How did Gautama came to be known as the Buddha?
Answer:
Siddhartha Gautama wandered for seven years and finally sat down under a Peepal tree till he got enlightenment. After seven days of enlightenment, he renamed the ‘Bodhi Tree’ and began to teach and to share his new understanding and came to be known as the Buddha.

Question 2.
Why did prince Siddhartha leave the palace and became a beggar?
Answer:
While out hunting prince Siddhartha chanced upon a sick man, an aged man and then a funeral procession and he also saw a monk begging. He realised that world is full of sorrow so he left the palace in the search of enlightenment.

Question 3.
How did Buddha get enlightenment?
Answer:
Siddhartha Gautama wandered for seven years and finally sat down under a Peepal tree till he got enlightenment. After seven days of enlightenment, he renamed the ‘Bodhi Tree’ and began to teach and to share his new understanding and came to be known as the Buddha.

Question 4.
Why did the Buddha choose Benares to preach his first sermon?
Answer:
The Buddha preached his first sermon at the city of Benares. This city is regarded as the most holy of the dipping places of the River Ganga. That sermon has been preserved. It reflects the Buddha’s wisdom about one kind of suffering which cannot be understood.

Question 5.
What did Kisa Gotami say to Buddha? What reply did he give to her?
Answer:
Kisa Gotami went to the Buddha and asked him to give her the medicine to make her son alive. The Buddha asked her to bring him a handful of mustard seed. It should be brought from a house where no one had died.

Question 6.
Why was Kisa Gotami sad? What did she do in her grief?
Answer:
Kisa Gotami’s only son died and she was full of grief. She carried the dead son to all her neighbours. She asked them for medicine so that her son can be alive again. The people remarked that she had lost her senses.

Question 7.
What did Kisa Gotami do after the Buddha had asked her for a handful of mustard seed?
Answer:
Kisa Gotami went from house to house to get a handful of mustard seed. People gave her the mustard
seed. But when she asked if anyone had died in their family, they regretfully told her that the livings were few, but the deads were many. Kisa Gotami found no house where someone had not died.

Question 8.
What did the Buddha want to explain to Kisa Gotami?
Answer:
The Buddha wanted to explain to Kisa Gotami that man cannot get peace of mind by grieving. On the contrary, his pain will be greater and his body will suffer. He will make himself sick and pale. A man who overcomes all sorrows will be free from sorrow and will be blessed.

Question 9.
Why did Kisa Gotami go from house to house?
Answer:
Kisa Gotami was a lady who lived in Benares. Her only son had died and she could not’bear it. She went from house to house to cure his dead son. Someone told him about the Buddha and she reached him to cure his dead son.

Question 10.
What was the basic idea of the Buddha’s preaching?
Answer:
The basic idea of the Buddha’s preaching was that death is the ultimate truth of life. It is that every living being has to die one day. No one can escape from death. Grief cannot console anyone. We must accept this universal truth.

The Sermon at Benares Extra Questions and Answers Long Answer Type

Question 1.
What is the nature of the life of the human beings according to the Buddha?
Answer:
The life of mortals in this world is troubled and brief. It is combined with pain. Nobody can avoid dying. After reaching old age, there is death. Such is the nature of human beings. Just as ripe fruits are in danger of falling; so mortals are in danger of death. As all earthen vessels after a certain period of time break, so is the life of mortals. All have to die. Only he can get peace of mind who does not lament, complain and grieve. He who has overcome sorrow will be free from sorrow, and be blessed.

Question 2.
How did Gautam Buddha get enlightenment? Explain with reference to the story.
Answer:
Siddhartha Gautama was a prince who was protected from the sufferings of the world. At the age of twenty-five, he came across a sick man, an aged man, a funeral procession and a monk begging for alms. These sights moved him so much that he went out into the world to seek enlightenment concerning the sorrows he had witnessed. Siddhartha wandered for seven years and finally sat down under a Peepal tree.

He vowed to stay there until enlightenment came. Being enlightened after seven days, he renamed the tree the Bodhi Tree, the Tree of Wisdom. Thereafter he began to teach and share his new understanding. At that point he became to be known as the Buddha. He preached his first sermon at the holy city of Benares.

Question 3.
Why and how did Siddhartha Gautama become the Buddha?
Answer:
Gautama Buddha was born as a prince named Siddhartha Gautama in northern India. At the age of twelve, he was sent away for learning the Hindu sacred scriptures. Four years later he returned home to marry a princess. They had a son and lived for ten years as befitted royalty. Upto the age of twenty-five, the prince was shielded from the sufferings of the world.

Then while going out for hunting, he came across by chance a sick man, an aged man, a funeral procession, and finally a monk begging for alms. These sights moved the prince so much that he went out into the world to seek a state of high spiritual knowledge concerning the sorrows of human beings. He wandered for seven years and finally sat down under a peepal tree. He got enlightenment after seven days. After that he came to be known as the Buddha.

Question 4.
How did Buddha make Kisa Gotami understand about the reality of death?
OR
What lesson on death and suffering did the Buddha teach Gotami in the chapter. The Sermon at Benaras’.
Answer:
Kisa Gotami’s only son had died. In her grief she carried the dead child to all her neighbours, asking them for medicine. Then a man suggested her to go to Sakyamuni, the Buddha. Kisa Gotami requested Gautama Buddha to give her the medicine that would cure her son.

The Buddha replied that he wanted a handful of mustard-seed which must be taken from a house where no one had lost a child, husband, parent or friend. Kisa Gotami went from house to house. People pitied her but she could not find any house where near and dear one had not died. She thought how selfish she had been in her grief. She realised that death is common to all human beings.

Question 5.
“The World is afflicted with death and decay, therefore the wise do not grieve, knowing the. terms of the world”. Explain with reference to the story “The Sermon at Benares”.
Answer:
According to Gautama Buddha, the world is affected by sufferings, disease or pain, death and decay. Therefore, the wise persons do not mourn, knowing the terms of the world. Nobody can get peace of mind from weeping or mourning. Rather his pain will be greater and his body will suffer. He will make himself sick and pale, yet the dead cannot be saved from his expression of sorrow. He who seeks peace should “draw out the arrow” of lamentation, complaint and grief. In this manner, she who has become composed will obtain peace of mind. He who has overcome all sorrow will become free from sorrow, and be blessed.

Question 6.
In ‘The Sermon at Benares’, the Buddha preached that death is inevitable and we need to overcome the suffering and pain that follows. Discuss.
Answer:
It is very painful to lose someone or something we love. When we lose someone it is a great emotional suffering. The more we grieve over death, the more painful it will be for us. Death is always unwelcome. We must realize that we are all mortals and death is common to all. We may lament and cry but we cannot bring dead back to life. All the riches of the world cannot bring life back. We must realize that death is common to all rich or poor. Those who have overcome sorrow will become free from sorrow and are blessed.

The Sermon at Benares Extra Questions and Answers Reference to Context

Read the following passages and answer the questions that follow:

Question 1.
Gautama Buddha (563 B.C – 483 B.C.) began life as a prince named Siddhartha Gautama, in northern India. At twelve, he was sent away for schooling in the Hindu sacred scriptures and four years later he returned home to marry a princess. They had a son and lived for ten years as befitted royalty. At about the age of twenty-five, the Prince, heretofore shielded from the sufferings of the world, while out hunting chanced upon a sick him, then an aged man, then a funeral procession, and finally a monk begging for alms. These sights so moved him that he at once went out into the world to seek enlightenment concerning the sorrows he had witnessed.
(i) How did Gautam Buddha begin his life?
(ii) For how long did he live as a prince?
(iii) How did it change his life?
(iv) Pick out the word from the passage that means same as—‘a person who belongs to the family of a king or queen’.
Answer:
(i) Gautam Buddha began his life as a prince, named Siddhartha Gautama.
(ii) He lived as a prince for about twenty five years.
(iii) He was moved deeply by the sorrow he had witnessed. He at once became a beggar and went out into the world to seek enlightenment.
(iv) ‘royal’.

Question 2.
He wandered for seven years and finally sat down under a peepal tree, where he vowed to stay until enlightenment came. Enlightened after seven days, he renamed the tree the Bodhi Tree (Tree of Wisdom) and began to teach and to share his new understandings. At that point he became known as the Buddha (the Awakened or the Enlightened). The Buddha preached his first sermon at the city of Benares, most holy of the dipping places on the River Ganges; that sermon has been preserved and is given here. It reflects the Buddha’s wisdom about one inscrutable kind of suffering.
(i) Why did he wander?
(ii) Where did he get enlightened?
(iii) Why was the tree renamed as “Bodhi Tree’?
(iv) Pick out the word from the passage that means the same as—“religious talk”.
Answer:
(i) He wandered for seven years in search of knowledge.
(ii) He got enlightened after seven days under a fig peepal.
(iii) The tree was renamed ‘Bodhi Tree’. ‘Bodhi tree’ stands for the tree of wisdom. The Buddha got wisdom under that tree.
(iv) ‘sermon’.

Question 3.
Kisa Gotami had an only son, and he died. In her grief she carried the dead child to all her neighbours, asking them for medicine, and the people said, “She has lost her senses. The boy is dead.”At length, Kisa Gotami met a man who replied to her request, “I cannot give thee medicine for thy child, but I know a physician who can.”And the girl said, “Pray tell me, sir; who is it?” And the man replied, “Go to Sakyamuni, the Buddha.” Kisa Gotami repaired to the Buddha and cried, “Lord and Master, give me the medicine that will cure my boy.” The Buddha answered, “I want a handful of mustard-seed.” And when the girl in her joy promised to procure it, the Buddha added. “The mustard-seed must be taken from a house where no one has lost a child, husband, parent or friend.”
(i) What had happened to Kisa Gotami?
(ii) What did the people remark?
(iii) What did Buddha ask the girl for?
(iv) Pick out the word from the passage that means the same as—‘went to’.
Answer:
(i) Kisa Gotami was grieved at the death of her only son.
(ii) The people remarked that she had lost her senses.
(iii) The Buddha asked the girl to procure mustard seed from a house where no one had died before.
(iv) ‘repaired’.

Question 4.
Poor Kisa Gotami now went from house to house, and the people pitied her and said, “Here is mustard- seed; take it!” But when she asked, “Did a son or daughter, a father or mother, die in your family?” they answered her, “Alas! the living are few, but the dead are many. Do not remind us of our deepest grief.” And there was no hopse but some beloved one had died in it.
(i) Why did Kisa Gotami go from house to house?
(ii) What response did she get?
(iii) Pick out the word from the passage that means same as—“pain”.
(iv) Was Kisa Gotami awe to bring mustard-seed from any house?
Answer:
(i) Kisa Gotami went from house to house to get the mustard seeds.
(ii) She got the negative response about the mustard seeds. She couldn’t find a house where no one had died.
(iii) ‘grief.
(iv) No, she could not bring mustard-seed.

Question 5.
Kisa Gotami became weary and hopeless, and sat down at the wayside watching the lights of the city, as they flickered up and were extinguished again. At last the darkness of the night reigned everywhere. And she considered the fate of men, that their lives flicker up and are extinguished again. And she thought to herself, “How selfish am I in my grief! Death is common to all; yet in this valley of desolation there is a path that leads him to immortality who has surrendered all selfishness.”
(i) Why was Kisa Gotami hopeless?
(ii) Why did she sit down at the wayside?
(iii) What did she see in the city?
(iv) Pick out the word from the passage that means the same as—an area which is filled with deep sorrow’.
Answer:
(i) Kisa Gotami was hopeless because the Buddha asked her to bring mustard-seed from a house where nobody had died before.
(ii) She sat down at the wayside because she had become weary and hopeless.
(iii) She watched the light of the city. The lights flickered up and extinguished again and again.
(iv) “valley of desolation’.

Question 6.
The Buddha said, “The life of mortals in this world is troubled and brief and combined with pain. For there is not any means by which those that have been born can avoid dying; after reaching old age there is death; of such a nature are living beings. As ripe fruits are early in danger of falling, so morals when born are always in danger of death. As all earthen vessels made by the potter end in being broken, so is the life of mortals. Both young and adult, both those who are fools and those who are wise, all fall into the power of death; all are subject to death. (Pages 134-135)
(i) What did the Buddha say about the life of mortals?
(ii) What cannot be avoided by those that have been born?
(iii) What is inevitable?
(iv) Find the word from the passage that means the same as—‘those bound to die’.
Answer:
(i) The Buddha said that the life of mortals in this world is short, full of troubles and combined with pain.
(ii) Death cannot be avoided by those who have been born.
(iii) Death is inevitable.
(iv) ‘mortals’.

Question 7.
“Of those who, overcome by death, depart from life, a father cannot save his son, nor kinsmen their relations. Mark! while relatives are looking on and lamenting deeply, one by one mortals are carried off, like an ox that is led to the slaughter. So the world is afflicted with death and decay, therefore the wise do not grieve, knowing the terms of the world.
(i) Why did the Buddha give this sermon to Kisa Gotami?
(ii) What is the fate of mortals in the world?
(iii) Why do the wise men not grieve?
(iv) Pick out the word from the passage that means the same as—‘the killing of animals for their meat’.
Answer:
(i) The Buddha gave this sermon to Kisa Gotami to console her because her son had died.
(ii) Mortals are carried off like an ox that is led to the slaughter.
(iii) Due to knowing the terms of the world, the wise men do not grieve.
(iv) ‘slaughter’.

Question 8.
“Not from weeping not from grieving will anyone obtain peace of mind; on the contrary, his pain will be the greater and his body will suffer. He will make himself sick and pale, yet the dead are not saved by his lamentation. He who seeks peace should draw out the arrow of lamentation, and complaint, and grief. He who has drawn out the arrow and has become composed will obtain peace of mind; he who has overcome all sorrow will become free from sorrow, and be blessed.”
(i) Who is the speaker in the above passage?
(ii) How does ‘weeping1 affect a person?
(iii) What should one do to obtain peace?
(iv) Pick out the word from the passage that means same as—‘expression of sorrow’.
Answer:
(i) The Buddha is the speaker in the above passage.
(ii) Weeping and grieving cause pain and suffering. A person who weeps does not obtain peace.
(iii) One should never lament, complain and indulge in grief to obtain peace of mind.
(iv) lamentation.

Online Education for RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS

Online Education for RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS

These Solutions are part of Online Education RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions.
Question 1.
Write the value of k for which the system of equations x + y – 4 = 0 and 2x + ky – 3 = 0 has no solution.
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 1

Question 2.
Write the value of k for which the system of equations 2x – y = 5 6x + ky = 15 has infinitely many solutions.
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 2

Question 3.
Write the value of k for which the system of equations 3x – 2y = 0 and kx + 5y = 0 has infinitely many solutions.
Solution:
3x – 2y = 0
kx + 5y = 0
Here a1 = 3, b1 = -2, c1 = 0
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 3

Question 4.
Write the value of k for which the system of equations x + ky = 0, 2x – y = 0 has unique solution.
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 4

Question 5.
Write the set of values of a and b for which the following system of equations has infinitely many solutions.
2x + 3y = 7
2ax + (a + b) y = 28
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 5
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 6

Question 6.
For what value of ft, the following pair of linear equations has infinitely many soutions.
10x + 5y – (k – 5) = 0
20x + 10y – k = 0
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 7

Question 7.
Write the number of solutions of the following pair of linear equations :
x + 2y – 8 = 0
2x + 4y = 16 (C.B.S.E. 2009)
Solution:
x + 2y – 8 = 0 => x + 2y = 8 ….(i)
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 8

Question 8.
Write the number of solutions of the following pair of linear equations :
x + 3y – 4 = 0
2x + 6y = 7
Solution:
RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS 9

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Online Education HOTS Questions for Class 10 Science Chapter 12 Electricity

Online Education HOTS Questions for Class 10 Science Chapter 12 Electricity

These Solutions are part of Online Education HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 12 Electricity

Question 1.
Following table gives the resistivity of three samples in (Ωm)
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 1
Which of them is a good conductor ? And which of them is an insulator ? Explain why ? (CBSE 2012)
Answer:
A material having low resistivity is a good conductor. Since, resistivity of sample A is the least among all other materials, so sample A is a good conductor. A material having high value of resistivity is an insulator. Therefore, sample C is an insulator.

More Resources

Question 2.
The electrical resistivity of few materials is given below in ohm-meter. Which of these materials can be used for making element of a heating device ?
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 2
(CBSE Sample Paper, CBSE 2012, Similar CBSE 2014)
Answer:
For making element of a heating device, we use alloy instead of pure metals.The resistivity of material D lies in the range of resistivities of alloys. Therefore, material D can be used for making element of a heating device.

Question 3.
Electrical resistivities of some substances at 20°C are given below :
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 3
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 4
Answer the following questions in relation to them :

  1. Among silver and copper, which one is a better conductor ? Why ?
  2. Which material would you advise to be used in electrical heating devices ? Why ? (CBSE 2012)

Answer:

  1. A material whose electrical resistivity is low is a good conductor of electricity. Since the electrical resistivity of silver is less than that of the copper, so silver is a better conductor than the copper.
  2. For making the elements of heating devices, alloy is used instead of a pure metal. This is because the resistivity of an alloy is more than that of a metal and alloy does not burn (or oxidise) even at higher temperature. Out of the given substances, nichrome is an alloy, so nichrome is used in electrical heating devices.

Question 4.
The following table gives the resistivity of three samples :
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 5
Which of them is suitable for heating elements of electrical appliances and why ? (CBSE, 2010, 2012)
Answer:
For making the heating elements of electrical appliances, alloy is used instead of a pure metal. This is because alloy does not burn even at higher temperature. The resistivity of sample C is of the order of an alloy, so sample C is suitable for heating elements of electrical appliances.

Question 5.
Two metallic wires A and B are connected in series. Wire A has length l and radius r, while wire B has length 2l and radius 2r. Find the ratio of the total resistance of series combination and the resistance of wire A, if both the wires are of same material
Answer:
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 6

Question 6.
Two metallic wires A and B of same material are connected in parallel. Wire A has length l and radius r and wire B has length 2l and radius 2r. Compute the ratio of the total resistance of parallel combination and the resistance of wire A.
(CBSE Sample Paper)
Answer:
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 7

Question 7.
Two students perform the experiments on series and parallel combinations of two given resistors Rand R2 and plot the following V-I graphs.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 8
Which of the graphs is (are) correctly labelled in terms of the words ‘series’ and ‘parallel’. Justify your answer.
(CBSE Sample Paper, 2012)
Answer:
In first graph, slope of I-V graph = Resistance.
Since in series combination, resistance is more than the resistance in parallel combination, therefore slope of I-V graph for series combination is more than the slope of resistance.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 9
Hence, graph is correctly labelled. In second graph, slope of resistance

Question 8.
Two electric circuits I and II are shown in figure.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 10
(i) Which of the two circuits has more resistance ?
(ii) Through which circuit, more current passes ?
(iii) In which circuit, the potential difference across each resistor is equal.
Answer:
(i) Equivalent resistance of series combination of resistors is more than the equivalent resistance of the parallel combination of resistors. So, the resistance of circuit I is more than the resistance of circuit II.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 11
So current in circuit II is more than the current in circuit I.
(iii) Potential difference across each resistor is equal in circuit II.

Question 9.
An electrician puts a fuse of rating 5 A in that part of domestic electrical circuit in which an electrical heater of rating 1.5 kW, 220 V is operating. What is likely to happen in this case and why ? What change, if any needs to be made ?
(CBSE Sample Paper)
Answer:
The fuse will melt and the circuit breaks if electric current more than the rating of fuse (i.e., 5 A) flows in the circuit. Electric current flowing in the circuit,
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 12
Since, current flowing in the circuit (6.82 A) is more than the rating of fuse (5 A), therefore, the fuse will melt and the electrical heater does not work. To operate the heater, fuse of rating 10 A is to be put in the circuit.

Question 10.
The electric power consumed by a device may be calculated by using either of the two expressions: P = I2R or P = V2/R . The first expression indicates that the power is directly proportional to R,whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained ?
(CBSE Sample Paper)
Answer:
P = I2R is used when current flowing in every component of the circuit is constant. This is the case of series combination of the devices in the circuit.
P = V2/R is used when potential difference (V) across every component of the circuit is constant. This expression is used in case of parallel combination in the circuit. In series combination, R is greater than the value of R in parallel combination.

Question 11.
Three V-I graphs are drawn individually for two resistors and their series combination. Out of A, B, C which one represents the graph for series combination of the other two. Give reason for y or ansiver. (CBSE 2011)
Answer:
Slope of V-I graph = resistance of a resistor.
When two resistors are connected in series,Volts the resistance of this combination (R = R1 + R2) is more than the resistance of both the resistors.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 13
Since, slope of C is greater than the slopes of A and B. Therefore, C represents the graph for series combination of the other two.

Question 12.
V-I graphs for the two wires A and B are shown in the figure. If we connect both the wires one by one to the same battery, which of the two will produce more heat per unit time ? Give justification for your answer (CBSE 2014, 2015)
Answer:
Heat produced per unit time = V2/R
Now slope of V-I graph = R (resistance of wire).
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 14
Since slope of V-I graph for wire A is greater than the slope of V-I graph for wire B, therefore, resistance of wire A is greater than the resistance of wire B. Hence, more heat will be produced per unit time in wire B than in wire A.

Question 13.
The resistivities of some substances are given below :
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 15
Answer the following questions in relation to them giving justification for each :
(i) Which material is best for making connecting cords ?
(ii) Which material do you suggest to be used in heater elements ?
(iii) You have two wires of same length and same thickness. One is made of material A and another of material D. If the resistance of wire made of A is 2 Ω, what is the resistance of the other wire ?
Answer:
(i) Material A is best for making connecting cords as its resistivity is the lowest one.
(ii) For heater elements, material of high resistivity is used. Therefore, the material E is to be used in heater elements.
HOTS Questions for Class 10 Science Chapter 12 Electricity image - 16

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Online Education for RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Online Education for RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Other Exercises

Question 1.
Mark the correct alternative in each of the following:
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 1

Question 2.
A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones is
(a) 63
(b) 126
(c) 21
(d) 130
Solution:
Radius of sphere (R) = 10.5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 2

Question 3.
A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is
(a) 1 : 3
(b) 1 : \(\sqrt { 3 } \)
(c) 1 : 1
(d) \(\sqrt { 3 } \)  = 1

Solution:
Surface area of hemispherical part = surface area of conical part
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 3

Question 4.
A solid sphere of radius r is melted and cast into the shape of a solid cone of height r, the radius of the base of the cone is
(a) 2r
(b) 3r
(c) r
(d) 4r
Solution:
Radius of solid sphere = r
Volume = \((\frac { 4 }{ 3 } )\) πr³
Now height of the cone so formed = r and
let radius = r1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 4

Question 5.
The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is 5 cm, then height of the cone is
(a) 10 cm
(b) 15 cm

(c) 18 cm
(d) 24 cm
Solution:
Let height of cone = h
and let r be its radius
∴ Volume =  \((\frac { 1 }{ 3 } )\) πr²h
Now radius of cylinder so formed = r
and height = 5 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 5

Question 6.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of the canvas required in m2 is
(a) 1760
(b) 2640
(c) 3960
(d) 7920
Solution:
Diameter of tent = 105 m
Height of the cylindrical part (h1) = 4 m
Slant height of conical part (l) = 40 m
and radius (r) =  \((\frac { 105 }{ 2 } )\) m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 6
surface area of the tent = curved surface area of conical part + curved surface area of cylindrical part =
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 7

Question 7.
The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 8

Question 8.
A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by
(a) 4.5 cm
(b) 3 cm
(c) 4 cm
(d) 2 cm
Solution:
Radius of sphere (r) = 6 cm
Volume = \((\frac { 1 }{ 3 } )\) πr³ = \((\frac { 4 }{ 3 } )\) π (6)³ cm³
= \((\frac { 4 }{ 3 } )\)  x216π = 4x 72π cm³ = 28871 cm³
Radius of vessel (r²) = 8 cm
Let height of water level = h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 9

Question 9.
If the radii of the circular ends of a bucket of height 40 cm are of lengths 35 cm and 14 cm, then the volume of the bucket in cubic centimeters, is .
(a) 60060
(b) 80080
(c) 70040
(d) 80160
Solution:
Height of the bucket (h) = 40 cm
Upper radius (r1)  = 35 cm
and lower radius (r2) = 14 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 10

Question 10.
If a cone is cut into two parts by a horizontal plane passing through the mid¬point of its axis, the ratio of the volumes of the upper part and the cone is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 6
(d) 1 : 8
Solution:
In the figure, C and D are the mid-points and CD || AB which divide the cone into two parts
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 11
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 12

Question 11.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be \((\frac { 1 }{ 27 } )\) of the volume of the given cone, then the height above the base at which the section has been made, is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm
Solution:
Height of given cone (h1) = 30 cm
Let r1 be its radius
Then volume of the larger cone = \((\frac { 1 }{ 3 } )\) πr1²h1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 13

Question 12.
A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 14
Solution:
Height of cone = h
Volume of solid = 3 x volume of cone
Let h be the height of the cylinder and r be its radius, then
Volume of cylinder = πr²h1
and volume of cone = \((\frac { 1 }{ 3 } )\) πr²h1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 15

Question 13.
A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m deep, then its capacity is
(a) 176 m3
(b) 196 m3
(c) 200 m3
(d) 110 m3
Solution:
A reservoir is a frustum in shape which Upper diameter = 8 m
and lower diameter = 4 m
∴ Upper radius = \((\frac { 8 }{ 2 } )\) = 4 m
and lower radius = \((\frac { 4 }{ 2 } )\) = 2 m
Height (h) = 6m
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 16

Question 14.
Water flows at the rate of 10 metre per minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm ?
(a) 48 minutes 15 sec
(b) 51 minutes 12 sec
(c) 52 minutes 1 sec
(d) 55 minutes
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 17

Question 15.
A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is
(a) 12 cm
(b) 24 cm

(c) 36 cm
(d) 48 cm
Solution:
Radius of a cylindrical vessel (r1) = 18 cm
and height (h1 ) = 32 cm
∴ Volume of sand filled in it = πr1²h1
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 18

Question 16.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(a) 607t cm2
(b) 6871 cm2
(c) 12071 cm2  
(d) 136TI cmc
Solution:
Diameter of base of a right circular cone = 16 cm
Radius (r) = \((\frac { 16 }{ 2 } )\) = 8 cm
and height (h) = 15 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 19

Question 17.
A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is
(a) 12π cm3
(b) 15π cm3
(c) 16π cm3
(d) 20π cm3
Solution:
A cone is formed be rotating the right angled triangle above the side 3 cm
Height of cone (h) = 3 cm
and radius (r) = 4 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 20

Question 18.
The curved surface area of a cylinder is 264 m2 and its volume is 924 m3 The ratio of its diameter to its height is
(a) 3 : 7
(b) 7 : 3
(c) 6 : 7
(d) 7 : 6
Solution:
Curved surface area of a cylinder = 264 m2
and its volume = 924 m3
Let r be its radius and h be its height, then 2πrh = 264
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 21

Question 19.
A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. The radius of the cone is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 22

Question 20.
The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 1 : 2
(b) 2 : 3
(c) 9 : 16
(d) 16 : 9
Solution:
Ratio in volumes of two spheres = 64 : 27
= (4)³ : (3)³
∵ Volume is in cubic units
∴ Length will be units while areas are in square units
∴ Areas will be in the ratio = (4)² : (3)² = 16:9       (d)

Question 21.
If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is
(a) 12 cm
(b) 24 cm
(c) 30 cm
(d) 36 cm
Solution:
Let radii of 3 metallic spheres are
r1= 6 cm
r2 = 8 cm
r= 10 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 23
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 24

Question 22.
The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 12 cm
Solution:
Diameter of cylinder = 12 cm
∴ Radius (r1) = \((\frac { 12 }{ 2 } )\) = 6 cm
and height (h) = 12 cm
∴  Surface area = 2πrh = 2π x 6 x 12 cm²
= 144π cm²
Now surface area of sphere = 1447c cm²
Let r2 be its radius, then

Question 23.
The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 25
Solution:
Radius of cylindrical log (r) = 1 cm
and height (h) = 5 cm
The radius of the greatest sphere cut off from the cylindrical log will be = radius of the log = 1 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 26

Question 24.
A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 27
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 28

Question 25.
12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is
(a) \(\sqrt { 3 } \) cm
(b) 2 cm

(c) 3 cm
(d) 4 cm

Solution:
Diameter of solid cylinder = 16 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 29

Question 26.
A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
Solution:
Diameter of a metallic sphere = 6 cm
∴  Radius = \((\frac { 6 }{ 2 } )\) = 3 cm
∴  Volume = \((\frac { 4 }{ 3 } )\) πr1³ = \((\frac { 4 }{ 3 } )\) π (3)³ cm³ = 36π cm³
∴  Volume of cone = 36π cm³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 30

Question 27.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Solution:
Internal diameter of a hollow sphere = 4 cm
and external diameter = 8 cm
∴ Internal radius (r) = \((\frac { 4 }{ 2 } )\) = 2 cm
and external radius (R) = \((\frac { 8 }{ 2 } )\) = 4 cm
∴  Volume of metal used = \((\frac { 4 }{ 3 } )\) π (R³ – r³)
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 31

Question 28.
A solid piece of iron of dimensions 49 x 33 x 24 cm is moulded into a sphere. The radius of the sphere is (a) 21 cm
(b) 28 cm
(c) 35 cm
(d) None of these
Solution:
Dimension of a solid piece = 49 x 33 x 24 cm
Volume = 49 x 33 x 24 cm³ = 38808 cm³
Volume of a sphere = 38808 cm³
Let r be its radius, their
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 32

Question 29.
The ratio of lateral surface area to the total surface area of a cylinder with base diameter 1.6 m and height 20 cm is
(a) 1 : 7
(b) 1 : 5
(c) 7 : 1
(d) 5 : 1
Solution:
Ratio in lateral surface area and total surface area
Base diameter = 1.6 m = 160 cm
Height (h) = 20 cm
∴ Radius = 80 cm
Now, lateral surface = 2 πrh = 2 π x 80 x 20 = 3200 π
and 2 πrh x 2 πr2 = 3200 π + 2 π (80)²
= 3200 π + 2 π x 6400
= (3200 + 12800) π = 16000 π
Ratio = 3200 π : 6000 π = 1.5 (b)

Question 30.
A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is h. If the total height of the solid is 3 times the volume of the cone, then the height of the cylinder is
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 33
Solution:
Let r be the radius of the solid = height of the conical part = h
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 34

Question 31.
The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is 
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 35
Solution:
Radius of cone = r
and height = r
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 36

Question 32.
The radii of two cylinders are in the ratio 3 : 5. If their heights are in the ratio 2 : 3, then the ratio of their curved surface areas is
(a) 2 : 5
(b) S : 2
(c) 2 : 3
(d) 3 : 5
Solution:
Ratio in radii of two cylinders = 3 : 5
and in their heights = 2 : 3
Let r1 = 3x, r2 = 5x
h1= 2y, h2 = 3y
∴ Curved surface area of first cylinder = 2πr1h1
= 2π x 3x x 2y = 12πxy
and curved surface area of second cylinder
= 2πr2h2 = 2π x 5x x 3y = 30πxy
∴ Ratio = 12πxy : 30πxy = 2 : 5 (a)

Question 33.
A right circular cylinder of radius r and height It (h = 2r) just enclose a spehre of diameter
(a) h
(b) r
(c) 2r
(d) 2h
Solution:
Radius of right cylinder = r
Height = h or 2r(∵ h = 2r)
Diameter of sphere encloses by the cylinder = 2r (c)

Question 34.
The radii of the circular ends of a frustum are 6 cm and 14 cm. If its slant height is 10 cm, then its vertical height is
(a) 6 cm
(b) 8 cm

(c) 4 cm
(d) 7 cm
Solution:
Radii of circular ends of frustum an 6 cm and then
∴ r1 = 14, r2 = 6
and slant height (l) = 10 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 37

Question 35.
The height and radius of the cone of which the frustum is a part are h1, and r1 respectively. If h2 and r2 are the heights and radius of the smaller base of the frustum respectively and h2 : h1 = 1 : 2, then r2 : r1 is equal to
(a) 1 : 3
(b) 1 : 2
(c) 2 : 1
(d) 3 : 1
Solution:
Height of cone = h1
and radius = r1
Height of frustum = h2
and radius = r2
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 38

Question 36.
The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10 cm, then its lateral surface area is
(a) 321π cm²
(b) 300π1 cm²
(c) 260π cm²
(d) 250π cm²
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 39

Question 37.
A solid frustum is of height 8 cm. If the radii of its lower and upper ends are 3 cm and 9 cm respectively, then its slant height is
(a) 15 cm
(b) 12 cm
(c) 10 cm
(d) 17 cm
Solution:
In the frustum,
Upper radius (r1) = 9 cm
and lower radius (r2) = 3 cm
and height (h) = 8 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 40

Question 38.
The radii of the ends of a bucket 16 cm high are 20 cm and 8 cm. The curved surface area of bucket is
(a) 1760 cm²
(b) 2240 cm²
(c) 880 cm²
(d) 3120 cm²
Solution:
Height of bucket (h) = 16 cm
Upper radius (r1) = 20 cm
and lower radius (r2) = 8 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 41

Question 39.
The diameters of the top and the bottom portions of a bucket are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm, then the cost of painting its outer surface at the rate of 50 paise/ cm² is
(a) Rs. 1582.50
(b) Rs. 1724.50

(c) Rs. 1683
(d) Rs. 1642
Solution:
Diameter of upper and lower portions of a bucket are 42 cm and 28 cm
and height (h) = 24 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 42
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 43

Question 40.
 If four times the sum of the areas of two circular faces of a cylinder of height 8 cm is equal to twice the curve surface area, then diameter of the cylinder is
(a) 4 cm 
(b) 8 cm
(c) 2 cm
(d) 6 cm
Solution:
Let r be the radius of the cylinder
Height of = 8 cm
Sum of areas of two circular faces = 2πr²
Curved surface area = 2πrh = 2πr x 8
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 44

Question 41.
If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of orginal cylinder is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1 (CBSE 2012)
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 45

Question 42.
A metalic solid cone is melted to form a solid cylinder of equal radius. If the height of the cylinder is 6 cm, then the height of the cone was
(a) 10 cm
(b) 12 cm
(c) 18 cm
(d) 24 cm   [CBSE 2014]
Solution:
Let r be the radius in each case = r
Height of cylinder = 6 cm
Volume of cylinder = Volume of cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 46

Question 43.
A rectangular sheet of paper 40 cm x 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is
(a) 3.5
(b) 7
(c) 80/7
(d) 5
Solution:
Length of rectangular sheet(l) = 40 cm
and width (b) = 22 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 47

Question 44.
The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is
(a) 3
(b) 5
(c) 4
(d) 6 [CBSE 2014]
Solution:
Diameter of solid sphere = 6 cm
∴ Radius = \((\frac { 6 }{ 2 } )\) = 3 cm
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 48

Question 45.
Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 49
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 50

Question 46.
A right circular cylinder of radius r and height h (h > 2r) just encloses a sphere of diameter
(a) r
(b) 2r
(c) h
(d) 2h
Solution:
Because the sphere enclose in the cylinder, therefore the diameter of sphere is equal to diameter of cylinder which is 2r. (b)

Question 47.
In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) circle
(b) frustum of a cone
(c) sphere
(d) hemisphere
Solution:
We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone. (b)

Question 48.
If two solid-hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is
(a) 4πr²
(b) 6πr²
(c) 3πr²
(d) 8πr²
Solution:
Because curved surface area of a hemisphere is 2πr² and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.
Hence, the curved surface area of new solid = 2πr² + 2 πr² = 4πr². (a)

Question 49.
The diameters of two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is
(a) 32.7 litres
(b) 33.7 litres
(c) 34.7 litres
(d) 31.7 litres
Solution:
Given, diameter of one end of the bucket, 2R = 44 ⇒ R = 22 cm    [∵ diameter, r = 2 x radius]
and diameter of the other end,
2r = 24 ⇒ r = 12 cm   [∵ diameter, r = 2 x radius]
Height of the bucket, h = 35 cm
Since, the shape of bucket is look like as frustum of a cone.
∴ Capacity of the bucket = Volume of the frustum of the cone
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 51

Question 50.
A spherical ball of radius r is melted to make 8 new identical balls each of radius r,. Then r:rl =
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
Solution:
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS 52

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS are helpful to complete your math homework.

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Online Education for RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3

Online Education for RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3

These Solutions are part of Online Education RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3

Other Exercises

Question 1.
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.
Solution:
Radius of the circle (r) = 4 cm
Length of the chord AB = 4 cm
∴  In ΔOAB
OA = OB = AB    (each = 4 cm)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 1
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 2

Question 2.
A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
Solution:
Length of chord PQ = 12 cm
Angle at the centre (θ) = 120°
∵  Draw OD ⊥ DQ
which bisects PQ at D and also bisects ∠POQ
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 3
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 4
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 5

Question 3.
A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.
Solution:
Radius of the circle (r) = 14 cm
Angle at the centre (θ) = 90°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 6
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 7

Question 4.
A ehord 10 cm long is drawn in a circle whose radius is 5\(\sqrt { 2 } \)
cm. Find area of both the segments. (Take π = 3.14).

Solution:
Radius of the circle (r) = 5\(\sqrt { 2 } \)  cm
And length of chord AB = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 8
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 9
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 10

Question 5.
A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)
Solution:
Radius of the circle (r) – 14 cm
Angle at the centre subtended in the fnui
AB = 60°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 11
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 12

Question 6.
Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°. [NCERT Exemplar]
Solution:
Given that, radius of circle (r) = 14 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 13
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 14

Question 7.
A chord of a circle of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. (Use π = 3.14) [NCERT Exemplar]
Solution:
Let AB be the chord of a circle of radius 10 cm,
with O as the centre of the circle.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 15
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 16
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 17

Question 8.
The radius of a circle with centre O is 5 cm (see figure). Two radii OA and OB are drawn at right angles to each other. Find the areas of the segments made by the chord AB (Take π = 3.14).
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 18
Solution:
Radius of the circle (r) = 5 cm
∵   OA and OB are at right angle
∴ ∠AOB = 90°
∵  Chord AB makes two segments which are minor segment and major segment Now area of minor segment
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 19

Question 9.
AB is the diameter of a circle, centre O. C is a point on the circumference such that ∠COB = 0. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 20
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 21
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 22
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 23

Question 10.
A chord of a circle subtends an angle of 0 at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 24
Solution:
Let chord AB subtends angle 0 at the centre
of a circle with radius r
Now area of the circle = nr1
and area of the minor segment ACB
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 25
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 26

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

A Letter to God Extra Questions and Answers Class 10 English First Flight

Online Education for A Letter to God Extra Questions and Answers Class 10 English First Flight

Here we are providing Online Education for A Letter to God Extra Questions and Answers Class 10 English First Flight, Extra Questions for Class 10 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-english/

Online Education for A Letter to God Extra Questions and Answers Class 10 English First Flight

A Letter to God Extra Questions and Answers Very Short Answer Type

A Letter To God Extra Questions Question 1.
What objects did Lencho call “New coins”?
Answer:
Lencho called the “rain-drops’ as new coins.

Letter To God Extra Questions Question 2.
Where was Lencho’s house located?
Answer:
Lencho’s house was located at the crest of a low hill.

A Letter To God Extra Questions And Answers Pdf Question 3.
What could be seen from Lencho’s house?
Answer:
The fields of the ripe com could be seen from his house.

A Letter To God Class 10 Extra Questions Question 4.
What did the earth need?
Answer:
The earth needed some rain.

A Letter To God Short Questions And Answers Question 5.
Why did Lencho keep gazing at the sky?
Answer:
Lencho expected rain and so he kept on gazing at the sky.

Extra Questions Of A Letter To God Question 6.
When did it start raining?
Answer:
It started raining when the family was having dinner.

A Letter To God Long Questions And Answers Question 7.
How did Lencho feel when it started raining?
Answer:
Lencho was very happy and felt that the raindrops were coins, promising money.

A Letter To God Extra Questions And Answers Question 8.
Was the rain really a blessing?
Answer:
No, the rain proved to be a curse.

A Letter To God Short Question Answer Question 9.
What destroyed the crop?
Answer:
The hailstorm which lasted for an hour destroyed the crops.

Letter To God Class 10 Extra Questions Question 10.
How did Lencho and his family react to the calamity?
Answer:
They were filled with sorrow but they had faith in God.

A Letter To God Questions And Answers Question 11.
What was the only ‘One hope’ Lencho bad?
Answer:
His only one hope was help from God.

A Letter To God Very Short Question Answer Question 12.
How did Lencho appeal to God?
Answer:
Lencho wrote a letter to God, asking for one hundred pesos.

A Letter To God Additional Question Answer Question 13.
How did postmaster react seeing Lencho’s letter?
Answer:
The postmaster laughed heartily on seeing Lencho’s letter.

Question 14.
How did the postmaster react on reading Lencho’s letter.
Answer:
He felt that such faith in God should be sustained.

Question 15.
What was the reaction of Lencho after receiving lesser amount of money?
Answer:
Lencho thought that the post office employees had taken, away 30 pesos.

A Letter to God Extra Questions and Answers Short Answer Type

Question 1.
What are the raindrops compared to and why?
Answer:
Raindrops are compared to new coins – ten-cent and five cent pieces. It is because they promised a good harvest and as such good money.

Question 2.
How was Lencho sure that it was going to rain?
Answer:
Lencho knew his fields intimately and he could predict the weather by looking at the sky. His experienced eyes saw clouds and he predicted rain.

Question 3.
What promised a good harvest?
Answer:
Lencho felt that the field of ripe corn dotted with flowers always promised a good harvest. That is why he was anticipating a good harvest.

Question 4.
How did Lencho’s prediction about rain come true?
Answer:
Lencho had looked towards the north-east and remarked that they would get some water. His prediction came true when it started raining in the evening.

Question 5.
Why did Lencho go out?
Answer:
Lencho was eagerly awaiting for the rain which could be good for his crops. So he went out to have the pleasure of feeling the rain on his body.

Question 6.
How did Lencho regard his field? Why?
Answer:
When it started raining Lencho regarded his field with satisfaction. He was happy to see his crop draped in a curtain of rain. He hoped to reap a good harvest.

Question 7.
“It’s really getting bad now.” What was getting bad and how?
Answer:
Lencho was happy when it started raining. But the rain and the weather were getting bad. A strong wind was blowing and hailstones were falling.

Question 8.
Why was Lencho’s soul filled with sorrow?
Answer:
The hail and hailstones rained on the valley for an hour. It had left the field totally covered with snow. The crop was completely destroyed. It made Lencho’s soul filled with sorrow.

Question 9.
Why was the family not really upset?
Answer:
Though Lencho’s family was facing ruin, yet they were not really upset. It was because having immense faith in God, they were confident that God would help them.

Question 10.
Why did the postmaster send a reply to Lencho’s first letter addressed to God?
Answer:
The postmaster “was a generous and amiable man. He was impressed by Lencho’s firm faith in God. To sustain that faith, the postmaster sent a reply to Lencho.

Question 11.
Why was Lencho angry when he received the letter?
Answer:
The postmaster could raise only 70 pesos which he sent to Lencho, supposedly from God. Lencho was angry to find 30 pesos less than the amount he asked for. He was angry because he thought the post office employees had kept 30 pesos sent to him by God.

Question 12.
How was Lencho helped?
Answer:
The postmaster, a kind-hearted and generous man, read Lencho’s letter addressed to God. To sustain his faith, the postmaster raised money from friends and colleagues and sent it to Lencho, in the name of God.

Question 13.
Do you think the post-office employees were ‘a bunch of crooks’?
Answer:
No, they were kind, generous and helpful persons. The postmaster contributed a part of his salary and motivated his friends and colleagues for a charitable cause. So they were not a bunch of crooks.

Question 14.
How would you describe Lencho?
Answer:
Lencho was a simple, naive and a hard-working farmer. He was not only energetic, he had full faith in God also. He became a victim of natural calamity, befit God helped him indirectly.

Question 15.
What did Lencho hope for?
Answer:
Lencho hoped for getting some rain from the sky.

Question 16.
Why did Lencho say the raindrops were like ‘new coins’?
Answer:
Lencho said that the raindrops were like new coins. It was because the raindrops could cause a good harvest that his field needed most. Thus he would get a rich crop. Further the raindrops had a brightness like new coins.

Question 17.
How did the rain change? What happened to Lencho’s fields?
Answer:
The rain changed into heavy hailstone. In no time the whole valley was covered with hailstones. His fields became white as if covered with salt. His com was totally destroyed.

Question 18.
What were Lencho’s feelings when the hail stopped?
Answer:
When the hail stopped, Lencho’s soul was filled with sadness. He was very much troubled. His com was totally destroyed. He said, “the hail has left nothing. This year we will have no com. We will all go hungry”.

Question 19.
Who or what did Lencho have faith in? What did he do?
Answer:
Lencho had deep faith in God. He believed that God would help them all. No one would die of hunger as. He sees everything. Lebcho decided to write a letter to God.

Question 20.
Who read the letter?
Answer:
The postman and the postmaster read the letter.

Question 21.
What did the postmaster do then?
Answer:
The postmaster burst into laughter on seeing the deep faith of Lencho in God. He collected money from his employees. He even gave part of his salary. He put all the money in an envelope addressed to Lencho and wrote a letter containing a single word “God”.

Question 22.
Was Lencho surprised to find a letter for him with money in it?
Answer:
Lencho was not at all surprised to find a letter with money in it.

Question 23.
What made him angry?
Answer:
Lencho was angry on counting the money since it was less than the amount he had asked for. He had deep faith that God could neither make a mistake nor could deny what was requested.

Question 24.
Why does the postmaster send money to Lencho? Why does he sign the letter “God”?
Answer:
The postmaster was a very generous fellow. In order not to shake Lencho’s faith in God, he collected money and sent it to Lencho. The postmaster signed the letter “God”, lest Lencho should think that the money had not been sent by God.

Question 25.
Did Lencho try to find out who had sent the money to him? Why/why not?
Answer:
Lencho was a simple-minded person. He had firm faith in God. So he did not try to find out who had sent the money. He was of the view that the money was sent by God and none else could do such an act.

A Letter to God Extra Questions and Answers Long Answer Type

Question 1.
What was Leneho’s pride and possession?
Answer:
Lencho lived in the solitary house on the crest of a low hill in the valley. From here, he could survey and see the river and his pride, the field of ripe corn dotted with flowers. These always promised a good harvest.

Their only wish was a downpour and Lencho, who knew his fields intimately, predicted rain. His prediction turned true when it started raining during dinner. He went out to feel the rain and surveyed with satisfaction, his crop draped in a curtain of rain. He regarded rain drops as a new silver coins.

Question 2.
Bring out Lencho’s immense faith in God.
Answer:
Lencho wanted some rain for his crop. The ripe corn stood proud in the field and promised a good harvest. Rain came, as he predicted but turned to hail and ruined his crop. Lencho’s heart was filled with sorrow. But he did not lose heart. He had firm belief in God.

He wrote a letter to God, asking him to send 100 pesos so that he could sow his field again. He dropped ‘ the letter in the mail. The postmaster read the letter addressed to ‘God’ and to preserve theman’s faith in God, he raised 70 pesos and sent them to Lencho.

Lencho could not believe that God had made a mistake. He wrote again asking God to send the rest of the money, but not through mail. He believed that the post office employees were ‘a bunch of crooks’.

Question 3.
How was Lencho’s crop destroyed? How did he ask God for help?
OR
How did the hailstones affect Lencho’s field? What was Lencho’s only hope?
Answer:
Being a farmer, Lencho was completely dependent upon the crops of his field. Once a heavy downpour occurred. Along with rain large hailstones also started falling. As a result the field turned white as if it had been covered with salt all over. His annual crop was completely destroyed.

Even there was not a single flower left on the plants. In the entire village there was no one to help him in the lurch. Being a firm believer in God, he turned to the Almighty for the help. He wrote a letter requesting him to send 100 pesos so that he might sow his fields again till the next crop comes. He had a belief that God would certainly help him with the money.

Question 4.
“I wish I had the faith of the man who wrote this letter.” In the light of this statement describe Lencho’s character.
Answer:
As soon as the postmaster received and read the letter written by Lencho to God, he expressed his feelings in the words by referring Lencho’s faith in God. Lencho, the writer of the letter was a simple farmer. He had a firm belief in God. Once, when his crop was destroyed by hailstones, he turned to God for help.

He wrote a letter believing that God would not leave them to die of hunger and starvation. When he received a packet full of money, he was not the least surprised. Being simple in mind and generous by soul, he never knew that some generous soul had sent him the money in the name of God.

Question 5.
As the postmaster, write how you felt when Lencho accused you and your men of stealing money. Comment.
Answer:
I was shocked and discouraged at the first sight when I grabbed the letter of complaint from Lencho in which he accused me and my employees of stealing money. I could not expect such kind of a treatment. Although I had involved every man in my office in that generous work, yet Lencho regarded us as a bunch of crooks.

This man proved very ungrateful. I was so downcast that I decided not to send the rest of the money. If I did so he would be convinced that we had already cheated him. In the name of God we should let him suffer and face the situation boldly. I pray to God to have pity on him.

Question 6.
Why did Lencho write a letter to God?
Answer:
Lencho was a hardworking, simple farmer. He had sown a field of com and was waiting for rains. His joy knew no bounds when it started raining. He expected^ good harvest. But his joy turned to sorrow when rain gave way to hail. After an hour, the field of golden corn was covered with snow. The crop was totally destroyed. Lencho faced ruin. The year seemed bad without any food. This simple, god-fearing man had immense faith in the Almighty. He wrote a letter to God asking for 100 pesos to sow his field again.

Question 7.
What did the postmaster need to answer the letter? How did he collect it? How did Lencho react to the help?
OR
How did post office employees help Lencho? How did Lencho react to their help?
Answer:
The postmaster needed something more than ink and paper to answer the letter. He needed 100 pesos. He gave up a part of his salary and asked all other employees to help. Even friends were made to contribute for a cause of charity. Thus, he managed to collect 70 pesos. Lencho wasn’t surprised to receive the letter. But he was angry to receive 70 pesos instead of 100. He didn’t doubt God, such was his confidence. So he wrote another letter to God asking him to send the rest of the money. He warned him not to send it through the post office because those people were a ‘bunch of crooks’.

Question 8.
How do you think the postmaster felt when he received Lencho’s second letter? What do you think he did?
Answer:
The postmaster must have felt shocked and let-down. The contentment, which he had felt when Lencho had taken the letter, would have been replaced by consternation. All his generosity and desire to maintain Lencho’s faith had been wasted.

Yet, I believe, he must also have been amused at Lencho’s unshakeable faith in God. He must have made greater effort to collect the rest of the money. He would have sent a letter along with the money, saying that the postal employees were not a ‘bunch of crooks’.

Question 9.
If you had been Lencho, and this incident would have happened with you, what would have you done? Describe your feelings in simple words.
OR
We are faced with difficult situations at some points of time in life. God’s help comes to our rescue then. But God helps those who help themselves. Comment.
Answer:
If I had been Lencho and this incident would have happened with me, the way of solving my problem would have been different. I am quite familiar with the realities of the world. I know that God cannot receive any letter by the post and neither can God help me directly. “God only helps those who help themselves.”
I would have tried to search another work for some time so that I could survive and feed my family. Then I would have tried to save some amount to sow my field again.

Question 10.
Think about the statement:
“Faith can move mountains.” Do you think that this feeling had been in Lencho’s mind and so he could have been able to write a letter to God? Throw light on his feelings.
Answer:
Without any doubt, I can say that this statement has a great importance. This statement can give a great strength to anyone who is about to fall deep down in earth, reason may be different. According to me, this is true that Lencho has deep faith in God. Because of his faith in God, he wrote a letter to God. When he got seventy pesos, once again he wrote a letter to God to get the remaining amount. So we can say that his feelings for God were very powerful.

Question 11.
The reader may also be impressed with Lencho’s faith as the postmaster was. Can we see such an example in present time? If you were in place of postmaster what would have you done?
Answer:
Yes, it is quite obvious that the reader may get impressed to see Lencho’s faith in God because it is rare. It is also possible that the thoughts might be different. Now the time is very fast and no one has so simple and pure feelings. It is very difficult to find out such an example at present. The example of the postmaster is also very rare. Now even our close relatives do not help us in our need.

If I were in place of postmaster, I too would have helped Lencho. But my pattern would have been different. I would have called Lencho and handed over the amount to him. I would try to make him realise that God does not help us directly and “God helps only those who help themselves.

Question 12.
Who does Lencho have complete faith in? Which sentences in the story tell you this?
Answer:
Lencho has complete faith in God. The following sentences tell us this.
(a) Lencho thought only of his one hope – the help of God, whose eyes, as he had been instructed, see everything even what is deep in one’s ‘conscience’.
(b) He wrote “God, if you don’t help me, my family and I will go hungry this year”.
(c) “God, the money that I asked for, only seventy pesos reached me, send me the rest”.

Question 15.
Who does Lencho think has taken the rest of the money? What is the irony in the situation? (Remember that the irony of a situation is an unexpected aspect of it. An ironic situation is strange or amusing because it is the opposite of what is expected?)
Answer:
Lencho has thought that the rest of the money has been taken by the post office employees. He believed that God could not send him less money than what he had asked for. Here one can find the irony that the money has been collected and paid by the post-office employees. He does not believe in them. He calls them “a bunch of crooks”. He thinks that thirty pesos have been kept by the post office employees. He has got no idea that even the amount of seventy pesos has been sent by them.

An irony is an amusing or a strange situation because we find it quite opposite in nature from what we duly expect. Here the post office employees collect and send money to Lencho. He calls them a group of cheats on not receiving the full amount.

Question 16.
Are there people like Lencho in the real world? What kind of a person would you say he is? You may select appropriate words from the box to answer the question.
greedy, naive, stupid, ungrateful, selfish, comical, Unquestioning.
Answer:
One can find many people like Lencho in the real world. In true sense, Lencho is a simple minded fellow. He is quite ignorant of worldly events. Like an innocent person, he has firm faith in God. He thinks “God sees everything, even what is deep in one’s conscience”. He believes in his work and is like an ox of a man who goes on working and minding his own business. In reality he is an unquestioned worshipper and follower of God. He is a hardworking person. He lives-with his family on a hill. He is a naive, comical and unquestioning person.

Question 17.
There are two kinds of conflicts in the story: between humans and nature, and between humans themselves. How are these conflicts illustrated?
Answer:
We can find two kinds of conflicts in the story namely:

  • between humans and nature
  • between humans themselves Let us see about them:

Between humans and nature: Lencho is a hardworking farmer. He needs a shower of rain for his crop. He waits for the rain to come. By chance the rain starts pouring with hailstones. It hailed heavily. The crop was totally destroyed. He was totally upset. This is one conflict.

Between humans themselves: After complete destruction, Lencho writes a letter to God for money. The postmaster opens the letter addressed to God. In order not to shake the writer’s faith in God, he collects money from his employees. He too contributes from his salary. He sends more than half to Lencho signed as God. On receiving the money, Lencho gets angry. He believes that the post office employees have taken some of his money. This is a conflict between humans themselves.

A Letter to God Extra Questions and Answers Reference to Context

Read the following passages and answer the questions that follow:

Question 1.
The house—the only one in the entire valley—sat on the crest of a low hill. From this height one could see the river and the field of ripe com dotted with the flowers that always promised a good harvest. The only thing the earth needed was, a downpour or at least a shower. Throughout the morning Lencho— who knew his fields intimately—had done nothing else but see the sky towards the north-east. “Now we’re really going to get some water, woman.” The woman who was preparing supper, replied, “Yes, God willing”.
(i) Where was the house located?
(ii) What did the field of com dotted with flowers promise?
(iii) What did the earth need?
(iv) Pick out the words from the passage which mean the same as
(a) Peak (b) Rain (c) Food
Answer:
(i) The house was located on the crest of a low hill.
(ii) The field of com dotted with flowers promise a good harvest.
(iii) The earth needed a downpour or at least a shower.
(iv) (a) crest (b) downpour (c) supper

Question 2.
The older boys were working in the field, while the smaller ones were playing near the house until the woman called to them all, “Come for dinner”. It was during the meal that, just as Lencho had predicted, big drops of rain began to fall. In the north-east huge mountains of clouds could be seen approaching. The air was fresh and sweet. The man went out for no other reason than to have the pleasure of feeling the rain on his body, and when he returned he exclaimed, “These aren’t raindrops falling from the sky, they are new coins. The big drops are ten cent pieces and the little ones are fives.”
(i) What were the boys doing?
(ii) What had Lencho predicted?
(iii) Why did Lencho go out?
(iv) What did he compare to new coins?
Answer:
(i) The boys were working in the field and the younger ones were playing near the house.
(ii) Lencho had predicted that it would rain.
(iii) Lencho went out to enjoy the rain.
(iv) He compared the raindrops to new coins.

Question 3.
With a satisfied expression he regarded the field of ripe corn with its flowers, draped in a curtain of rain. But suddenly a strong wind began to blow and along with the rain very large hailstones began to fall. These truly did resemble new silver coins. The boys, exposing themselves to the rain, ran out to collect the frozen pearls.
(i) Who is ‘he’ in the passage?
(ii) What did it suddenly change?
(iii) What resembled new silver coins?
(iv) Why did the children run out?
Answer:
(i) Lencho is ‘he’ in the passage.
(ii) Suddenly, a strong wind began to blow and along with the rain very large hailstones began to fall.
(iii) Hailstones resembled the new silver coins.
(iv) The children ran out to collect the frozen pearls i.e., the hailstones.

Question 4.
“It’s really getting bad now,” exclaimed the man. “I hope it passes quickly. “It did not pass quickly. For an hour the hail rained on the house, the garden, the hillside, the cornfield, on the whole valley. The field was white, as if covered with salt. Not a leaf remained on the trees. The com was totally destroyed. The flowers were gone from the plants. Lencho’s soul was filled with sadness. When the storm had passed, he stood in the middle of the field and said to his sons, “A plague of locusts would have left more than this. The hail has left nothing. This year we will have no com.”
(i) What was really getting bad?
(ii) Why did he wish for it to pass quickly?
(iii) Why will they have no com that year?
(iv) Why was Lencho’s soul filled with sadness?
Answer:
(i) The continuous falling of hailstones was getting really bad.
(ii) He wished it to pass quickly because it was not good for his crops.
(iii) They would have no com that year because the com in the field was totally destroyed by the hailstones.
(iv) Lencho’s soul was filled with sadness because his total com was destroyed.

Question 5.
All through the night, Lencho thought only of his one hope: the help of God, whose eyes, as he had been instructed, see everything, even what is deep in one’s conscience. Lencho was an ox of a man, working like an animal in the fields, but still he knew how to write. The following Sunday, at daybreak, he began to write a letter which he himself would carry to town and place in the mail. It was nothing less than a letter to God.

“God,” he wrote, “if you don’t help me, my family and I will go hungry this year. I need a hundred pesos in order to sow my fields again and to live until the crop comes, because the hailstorm…”
(i) What did Lencho think of all through the night?
(ii) Which sentence shows that Lencho was a hardworking farmer?
(iii) What did Lencho decide to do?
(iv) What did he ask God to do for him?
Answer:
(i) All through the night, Lencho thought only of his one hope—the help of God.
(ii) “Lencho was an ox of a man,’ working like an animal in the field.” This sentence shows that he was a hardworking farmer.
(iii) Lencho decided to write a letter to God.
(iv) He asked God to help him by sending one hundred pesos.

Question 6.
He wrote “To God’ on the envelope, put the letter inside and, still troubled, went to town. At the post office, he placed a stamp on the letter and dropped it into the mailbox. One of the employees, who was a postman and also helped at the post office, went to his boss laughing heartily and showed him the letter to God. Never in his career as a postman had he known that address. The postmaster—a fat, amiable fellow—also broke out laughing, but almost immediately he turned serious and, tapping the letter on his desk, commented, “What faith! I wish I had the faith of the man who wrote this letter. Starting up a correspondence with God!”
(i) Who is he in the passage?
(ii) Why did he write a letter to God?
(iii) Who received the letter? Why did he laugh?
(iv) Why did the postmaster become serious?
Answer:
(i) Lencho is ‘he’ in the passage.
(ii) He wrote a letter to God asking for help.
(iii) The postmaster received the letter. He laughed heartily to see a letter addressed to God. Nobody ever wrote to God.
(iv) The postmaster became serious when he realised the deep faith of the writer in God.

Question 7.
So, in order not to shake the writer’s faith in God, the postmaster came up with am idea: answer the letter. But when he opened it, it was evident that to answer it he needed something more than goodwill, ink and paper. But he stuck to his resolution: he asked for money from his employees, he himself gave part of his salary, and several friends of his were obliged to give something ‘for an act of charity’. It was impossible for him to gather together the hundred pesos, so he was able to send the farmer only a little more than half. He put the money in an envelope addressed to Lencho and with it a letter containing only a single word as a signature: God.
(i) Why did the postmaster decide to answer the letter?
(ii) What was required to answer the letter?
(iii) What did the postmaster do?
(iv) How much did he collect for Lencho?
Answer:
(i) The postmaster did not want to shake the writer’s faith in God. So, he decided to answer the letter.
(ii) One hundred pesos were required to answer the letter.
(iii) The postmaster decided to help Lencho. He asked his employees and his friends to contribute and he also gave part of his salary.
(iv) The postmaster could collect only seventy pesos for Lencho.

Question 8.
The following Sunday Lencho came a bit earlier than usual to ask if there was a letter for him. It was the postman himself who handed the letter to him while the postmaster, experiencing the contentment of a man who has performed a good deed, looked on from his office. Lencho showed not the slightest surprise on seeing the money; such was his confidence—but he became angry when he counted the money. God could not have, made a mistake, nor could he have denied Lencho what he had requested.
(i) Why did Lencho come to the post office?
(ii) Why was the postmaster happy and content?
(iii) Why did Lencho show no surprise on seeing the money?
(iv) Why did Lencho get angry?
Answer:
(i) Lencho came to the post office to see if there was any letter from him.
(ii) The postmaster was happy and contented because he had done an act of charity.
(iii) Lencho did not show any surprise because he had deep faith in God.
(iv) Lencho got angry when he counted the money, he found that it was less than the amount he had requested. He was sure that God had not made the mistake.

Question 9.
Immediately, Lencho went up to the window to ask for paper and ink. On the public writing-table, he started to write, with much wrinkling of his brow, caused by the effort he had to make to express his ideas. When he finished, he went to the window to buy a stamp which he licked and then affixed to the envelope with a blow of his fist. The moment the letter fell into the mailbox the postmaster went to open it. It said: “God: Of the money that I asked for, only seventy pesos reached me. Send me the rest, since I need it very much. But don’t send it to me through the mail because the post office employees are a bunch of crooks. Lencho.”
(i) Why did Lencho go up to the window?
(ii) What made Lencho angry? What did he think?
(iii) Why did Lencho ask God not to send money by post?
(iv) What did Lencho call the employees of the post office?
Answer:
(i) Lencho went to the window to write another letter to God.
(ii) Lencho was angry because he found less money in the envelope. He thought that thirty pesos had been taken out by the employees of the post office.
(iii) Lencho requested God not to send money by post because he thought that the employees of the ‘ post office were dishonest.
(iv) Lencho called them ‘a bunch of crooks’.

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Check the below Online Education for NCERT MCQ Questions Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development Pdf free download. https://ncertmcq.com/extra-questions-for-class-10-social-science/

Online Education for Resource and Development Class 10 Extra Questions Geography Chapter 1

Class 10 Geography Chapter 1 One Mark Questions Question 1.
What do you mean by soil?
Answer:
The uppermost layer of the earth’s crust, which is loose, fragmented and useful for plants, is known as soil.

Class 10 Geography Chapter 1 Extra Questions Question 2.
Point out the factors which contribute to the formation of soil.
Answer:
The main factors which contribute in the formation of soil are the following-

  • Rocks
  • Climate
  • Plants
  • Animals
  • Local Topography
  • Long period of time.

Class 10 Geography Chapter 1 Very Short Questions And Answers Question 3.
In which tyes the soils were classified in ancient India?
Answer:
In the ancient India, soils were classified on the fertility basis-

  • Fertile (Urvara)
  • Sterile (Utara).

Class 10 Geography Chapter 1 Extra Questions And Answers Question 4.
How are soil classified on the basis of texture?
Answer:
Soils are classified into the following types or the basis of their texture

  • Sandy
  • Clayey
  • Loamy.

Geography Class 10 Chapter 1 Extra Questions Question 5.
What do you mean by Khadar?
Answer:
Khadar is meant by new alluvial soil which is sandy and light in colour and is found close to rivers.

Class 10 Geography Chapter 1 Extra Questions Pdf Question 6.
What is Bhangar?
Answer:
Bhangar is the old alluvial soil which is clayey and dark. It is found away from rivers.

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Class 10 Geography Chapter 1 Extra Questions Pdf Download Question 7.
Write down the other name of black soil.
Answer:
Regur.

Extra Questions Of Resources And Development Class 10 Question 8.
What is meant by Ravines?
Answer:
Ravines are the badlands in the Chambal basin.

Class 10 Geography Chapter 1 Important Questions Question 9.
What gives red colour to the red soil?
Answer:
The red soil gets its colour from the presence of iron.

Class 10 Geography Chapter 1 Extra Question Answer Question 10.
To which state is black soil predominantly confined?
Answer:
Black soils are confined to certain parts of Maharashtra, Madhya Pradesh, Gujarat and Tamil Nadu.

Class 10 Geo Ch 1 Extra Questions Question 11.
How can land degradation in industrial and suburban areas can be reduced?
Answer:
Land degradation in industrial and suburban areas may be reduced by proper discharge and disposal of industrial effluents and wastes after treatment.

Ncert Class 10 Geography Chapter 1 Extra Questions Question 12.
Why is soil considered a prime resource?
Answer:
Soil is considered a prime resource because it supports life on the earth. All of the living organisms like plants, animals as well as humans are dependent on soil for their food.

Extra Questions For Class 10 Geography Chapter 1 Question 13.
Though India has a very little pasture land yet it has the highest number of cattle. Do you agree?
Answer:
India had only 4% of its land under pastures. Yet it manages the largest number of the cattle. The main reason for this is that the cattle in India are reared mainly on husk, gain chaff and farm waste and a few fodder crops. This is the most economical way to have a large number of drought animals.

Geography Chapter 1 Class 10 Extra Questions Question 14.
Point out the important factors determining the utilisation of land.
Answer:
There are two sets of factors that determine the use of land in India :

  1. Physical factors including topography, climate, and nature of the soil,
  2. Human factors mainly density of population and the level of economic exploitation of natural resources.

Extra Questions Of Geography Class 10 Chapter 1 Question 15.
Point out the main causes of increase in wasteland.
Answer:
Wasteland is that land which is agriculturally unproductive. The land which is either sandy, rocky or arid is a wasteland. Some lands become wasteland because of human negligence.

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Cbse Class 10 Geography Chapter 1 Extra Questions Question 16.
Suggest appropriate steps for resource planning.
Answer:
Resources are very valuable forces. If we do not plan them properly we wouldn’t have them anymore in future. Hence the development in future would stop. Because of this fact we must plan our resources very wisely. The following steps may be taken for achieving this goal-

  • Wet and marshy lands should be reclaimed and in the same time available lands should be put to optimum use.
  • Rivers should be linked to each other and in desired direction.
  • The changing of the directions of the river would prevent the water going waste into the sea.
  • The wastewater from the factories should be recycled to be used over and over again.
  • More special attention should be given to the non-renewable resources like minerals etc.
  • For the above purpose wastage during mining and processing should be reduced to the minimum.
  • Whenever and wherever possible, alternative products like wood and plastic should be used.
  • To prolong the life of our limited metallic mineral resources, scrap should be recycled. Processing iron scrap into steel is a good example.
  • New methods should be devised for utilising ores of lower grade.
  • To make ourselves sure that the coming generations are not deprived of the benefits of the minerals resources, we should use the available stock of mineral wealth most judiciously.
    If we take above steps wisely, we would be able to plan and save our resources in the wisest way.

Class 10 Geography Chapter 1 One Mark Questions Pdf Question 17.
Briefly introduce the soils found in India.
Or
Which are the main types of soils that are found in India?
Answer:
India is very rich in matter of soils. Indian soils have been classified in various ways. In ancient India, soils were classified on the basis of the fertility as urvara and usara. In the present time soils are classified on the basis of colour.

A brief introduction of the Indian soils may be presented in the following words-
1. Alluvial soil-This type of soils is found over large parts of the country. These are confined mainly to the northern plains, coastal strips and Chhattisgrah basin. This type of soil is generally of two types-

  • Khadar-the newer alluvium.
  • Bhangar-the older alluvium.

The Khadar is sandy and light in colour. On the other hand the Bhangar is clayey and dark,

2. Black soils-This is also known as Regur soils. These are known for their cultivation value of cotton. These soils have been derived from the Deccan traps and these occur mainly in Maharashtra, Western Madhya Pradesh and Gujarat. In these areas, the black soils are generally deep. Black soils are also found in parts of Karnataka, Andhra Pradesh and Tamil Nadu.

3. Red soils-These are formed in the areas of igneous and metamorphic rocks. These have in fact developed as the result of weathering. Its red colour is due to the presence of iron in it. These are highly porous and fertile when they are fine-grained and deep. Red soil mainly occur in parts of Tamil Nadu, Karnataka, Andhra Pradesh, Orissa and Jharkhand.

4. Laterite soil-These are intensively leached soils of the monsoon climate. They lack in elements of fertility and they are normally of low value of crop production. They are red in colour and composed of little clay and much gravel of red sandstones. These soils are well developed in hills of the Deccan, Karnataka, Kerala, Orissa and even in the parts of Assam and Meghalaya.

5. Mountain Soils-These soils are found in mountainous regions of the country, particularly in Meghalaya, Arunachal Pradesh, eastern ranges, Uttaranchal, Himachal Pradesh and Jammu and Kashmir.

This type of soils is characterised mainly by deposition of organic materials derived from vegetative cover. These soils are heterogeneous in nature and vary from place to place. For example in the Shiwalik hills, the soils are shallow and immature containing large proportion of decomposed mineral grains. These are also sandy with gravel, porous and devoid of humus.

6. Desert soils-These soils are found in the arid and the semiarid conditions. These are mainly found in the arid areas of Rajasthan, Punjab and Haryana. The sands are partly of local origin and partly have been blown in from the Indus valley. In some areas, the soils also contain high percentage of soluble salt lack in organic matter. Scarcity of water is the main limiting factor for agriculture.

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Multiple Choice Questions

Question 1.
India has the following percentage of plateau land.
(a) 27%
(b) 43%
(c) 30%
(d) 25%
Answer:
(d) 25%

Question 2.
The plateau land area does not have the following.
(a) minerals
(b) forests
(c) perennial rivers
(d) fossil fuels.
Answer:
(c) perennial rivers

Question 3.
Black soil is found in one of the following states of India.
(a) Punjab
(b) Maharashtra
(c) Haryana
(d) Himachal Pradesh.
Answer:
(b) Maharashtra.

Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development

Question 4.
Yellow and red soils are found in.
(a) Chhattisgarh and Orissa
(b) Orissa and Punjab
(c) Punjab and Chattisgarh
(d) Chattisgarh and Haryana
Answer:
(a) Chhattisgarh and Orissa.

Question 5.
Gully erosion is quite pronounced in the following state.
(a) Uttar Pradesh and Madhya Pradesh
(b) Madhya Pradesh and Punjab
(c) Punjab and Rajasthan
(d) Rajasthan and Jammu-Kashmir.
Answer:
(a) Uttar Pradesh and Madhya Pradesh

Map Skills

Question 1.
On an outline map of India, show the distribution of soils.
Answer:
Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development 1

Question 2.
On an outline map of India show the following:
(i) Areas of black soil
(ii) Chambal ravines
(iii) Thar desert,
Answer:
Class 10 Geography Chapter 1 Extra Questions and Answers Resource and Development 2

Extra Questions for Class 10 Social Science

Online Education NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

In Online Education NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 10
Chapter Name Circles
Exercise Ex 10.2
Number of Questions Solved 13
Category NCERT Solutions

Online Education NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm Sol.
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 1

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 2
∠OPT = 90°
∠OQT = 90°
∠POQ = 110°
TPOQ is a quadrilateral,
∴ ∠PTQ + ∠POQ = 180° ⇒ ∠PTQ + 110° = 180°
⇒∠PTQ = 180°- 110° = 70°
Hence, correct option is (b).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
In AOAP and AOBP
OA = OB [Radii]
PA = PB
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 3
[Lengths of tangents from an external point are equal]
OP = OP [Common]
∴ ∆OAP ≅ ∆OBP [SSS congruence rule]
∠AOB + ∠APB = 180° ⇒ ∠AOB + 80° = 180°
⇒∠AOB = 180° – 80° = 100°
From eqn. (i), we get
⇒∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°
Hence, correct option is (a)

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
AB is a diameter of the circle, p and q are two tangents.
OA ⊥ p and OB ⊥ q
∠1 = ∠2 = 90°
⇒ p || q ∠1 and ∠2 are alternate angles]
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 4

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
XY tangent to the circle C(0, r) at B and AB ⊥ XY. Join OB.
∠ABY = 90° [Given]
∠OBY = 90°
[Radius through point of contact is perpendicular to the tangent]
∴ ∠ABY + ∠OBY = 180° ⇒ ABOiscollinear
∴ AB passes through centre of the circle.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 5

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
OA = 5 cm, AP = 4 cm OP = Radius of the circle
∠OPA = 90° [Radius and tangent are perpendicular]
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 6

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Radius of larger circle = 5 cm Radius of smaller circle = 3 cm
OP ⊥ AB
[Radius of circle is perpendicular to the tangent]
AB is a chord of the larger circle
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 7

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.
Solution:
AP = AS … (i)
[Lengths of tangents from an external point are equal]
BP = BQ … (ii)
CR = CQ … (iii)
DR = DS … (iv)
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 8
Adding equations (i), (ii), (iii) and (iv), we get
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 9
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence proved.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 10
Solution:
Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X’Y’ at B To Prove: ∠AOB = 90° with point of contact C intersects XY at A and X’Y’ at B
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 11

To Prove: ∠AOB = 90°
Construction: Join OA, OB and OC
Proof: In ∆AOP and ∆AOC
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 12

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
PA and PB are two tangents, A and B are the points of contact of the tangents.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 13

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Parallelogram ABCD circumscribing a circle with centre O.
OP ⊥ AB and OS ⊥ AD
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 14

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 15
Solution:
BD = 8 cm and DC = 6 cm
BE = BD = 8 cm
CD = CF = 6 cm
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 16
Let AE = AF = x cm
In ∆ABC, a = 6 + 8 = 14 cm
b = (x + 6) cm
c = (x + 8) cm
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 17

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
AB touches at P.
BC, CD and DA touch the circle at Q, R and S.
Construction: Join OA, OB, OC, OD and OP, OQ, OR, OS.
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 18
Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.

We hope the NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 10 Circles E 10.2, drop a comment below and we will get back to you at the earliest.

Online Education for RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

Online Education for RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

These Solutions are part of Online Education  RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

Other Exercises

Question 1.
Find two consecutive numbers whose squares have the sum 85. (C.B.S.E. 2000)
Solution:
Let first number = x
Then second number = x + 1
According to the condition
x² + (x + 1)2 = 85
=> x² + x² + 2x + 1 = 85
=> 2x² + 2x + 1 – 85 = 0
=> 2x² + 2x – 84 = 0
=> x² + x – 42 = 0
=> x² + 7x – 6x – 42 = 0
=> x (x + 7) – 6 (x + 7) = 0
=> (x + 7) (x – 6) = 0
Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6
(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6
(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7
Hence numbers are -7, -6 or 6, 7

Question 2.
Divide 29 into two parts so that the sum of the squares of the parts is 425.
Solution:
Total = 29
Let first part = x
Then second part = 29 – x
According to the condition
x² + (29 – x)2 = 425
=> x² + 841 + x² – 58x = 425
=> 2x² – 58x + 841 – 425 = 0
=> 2x² – 58x + 416 = 0
=> x² – 29x + 208 = 0 (Dividing by 2)
=> x² – 13x – 16x + 208 = 0
=> x(x – 13) – 16 (x – 13) = 0
=> (x – 13) (x – 16) = 0
Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16
(i) If x = 13, then First part =13 and second part = 29 – 13 = 16
(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13
Parts are 13, 16

Question 3.
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2. Find the sides of the squares. (C.B.S.E. 1997)
Solution:
Side of the first square = x cm
Its area = (side)2 = x² cm2
Side of the second square = (x + 4) cm
Its area = (x + 4)2 cm2
According to the condition,
x² + (x + 4)2 = 656
=> x² + x² + 8x + 16 = 656
=> 2x² + 8x + 16 – 656 = 0
=> 2x² + 8x – 640 = 0
=> x² + 4x – 320 = 0 (Dividing by 2)
=> x² + 20x – 16x – 320 = 0
=> x (x + 20) – 16 (x + 20) = 0
=> (x + 20) (x – 16) = 0
Either x + 20 = 0, then x = -20 Which is not possible being negative
or x – 16 = 0, then x = 16
Side of the first square = 16 cm
and side of the second square = 16 + 4 = 20 cm

Question 4.
The sum of two numbers is 48 and their product is 432. Find the numbers.
Solution:
Sum of two numbers = 48
Let first number = x
The second number = 48 – x
According to the condition,
x (48 – x) = 432
=> 48x – x² = 432
=> – x² + 48x – 432 = 0
=> x² – 48x + 432 = 0
=> x² – 12x – 36x + 432 = 0
=> x (x – 12) – 36 (x – 12) = 0
=> (x – 12) (x – 36) = 0
Either x – 12 = 0, then x = 12 or x – 36 = 0, then x = 36
(i) If x = 12, then First number = 12 and second number = 48 – 12 = 36
(ii) If x = 36, then First number = 36 and second number = 48 – 36 = 12
Numbers are 12, 36

Question 5.
If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.
Solution:
Let the given integer be = x
According to the condition
x² + x = 90
=> x² + x – 90 = 0
=> x² + 10x – 9x – 90 = 0
=> x (x + 10) – 9 (x + 10) = 0
=> (x + 10) (x – 9) = 0
Either x + 10 = 0, then x = -10 or x – 9 = 0, then x = 9.
The integer will be -10 or 9

Question 6.
Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.
Solution:
Let the given whole number = x
Then its reciprocal = \(\frac { 1 }{ x }\)
According to the condition,
x – 20 = 69 x \(\frac { 1 }{ x }\)
=> x – 20 = \(\frac { 69 }{ x }\)
=> x² – 20x = 69
=> x² – 20x – 69 = 0
=> x² – 23x + 3x – 69 = 0
=> x (x – 23) + 3 (x – 23) = 0
=> (x – 23) (x + 3) = 0
Either x – 23 = 0, then x = 23
or x + 3 = 0, then x = -3, but it is not a whole number
Required whole number = 23

Question 7.
Find two consecutive natural numbers whose product is 20.
Solution:
Let first natural number = x
Then second number = x + 1
According to the condition,
x (x + 1) = 20
=> x² + x – 20 = 0
=> x² + 5x – 4x – 20 = 0
=> x (x + 5) – 4 (x + 5) = 0
=> (x + 5) (x – 4) = 0
Either x + 5 = 0, then x = -5 which is not a natural number
or x – 4 = 0, then x = 4
First natural number = 4 and second number = 4 + 1=5

Question 8.
The sum of the squares of two consecutive odd positive integers is 394. Find them.
Solution:
Let first odd number = 2x + 1
Then second odd number = 2x + 3
According to the condition
(2x + 1)2 + (2x + 3)2 = 394
=> 4x² + 4x + 1 + 4x² + 12x + 9 = 394
=> 8x² + 16x + 10 = 394
=> 8x² + 16x + 10 – 394 = 0
=> 8x² + 16x – 384 = 0
=> x² + 2x – 48 = 0 (Dividingby8)
=> x² + 8x – 6x – 48 = 0
=> x(x + 8) – 6(x + 8) = 0
=> (x + 8) (x – 6) = 0
Either x + 8 = 0, then x = 8 but it is not possible as it is negative
or x – 6 = 0, then x = 6
First odd number = 2x + 1 = 2 x 6 + 1 = 13
and second odd number = 13 + 2 = 15

Question 9.
The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Solution:
Sum of two numbers = 8
Let first number = x
Then second number = 8 – x
According to the condition,
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 1
(ii) If x = 5, then First number = 5 and second number = 8 – 5 = 3
Numbers are 3, 5

Question 10.
The sum of a number and its positive square root is \(\frac { 6 }{ 25 }\). Find the number.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 2
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 3

Question 11.
The sum of a number and its square is \(\frac { 63 }{ 4 }\) , find the numbers.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 4

Question 12.
There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers ?
Solution:
Let first integer = x
Then second integer = x + 1
and third integer = x + 2
According to the condition,
x² + (x + 1) (x + 2) = 154
=> x² + x² + 3x + 2 = 154
=> 2x² + 3x + 2 – 154 = 0
=> 2x² – 16x + 19x – 152 = 0
=> 2x(x – 8) + 19 (x – 8) = 0
=> (x – 8) (2x + 19) = 0
Either x – 8 = 0, then x = 8
or 2x + 19 = 0, then 2x = -19 => x = \(\frac { -19 }{ 2 }\) But it is not an integer
First number = 8
Second number = 8 + 1=9
and third number = 8 + 2 = 10

Question 13.
The product of two successive integral multiple of 5 is 300. Determine the multiplies.
Solution:
Let first multiplie of 5 = 5x
Then second multiple = 5x + 5
According to the condition,
5x (5x + 5) = 300
=> 25 x² + 25x – 300 = 0
=> x² + x – 12 = 0 (Dividing by 25)
=> x² + 4x – 3x – 12 = 0
=> x (x + 4) – 3 (x + 4) = 0
=> (x – 4) (x – 3) = 0
Either x + 4 = 0, then x = -4
or x – 3 = 0, then x = 3
(i) When x = -4, then
Required multiples of 5 will be
5 (-4) = -20, -20 + 5 = -15
or when x = 3, then
Required multiples will be
5 x 3 = 15, 15 + 5 = 20
Required number are -20, -15 or 15, 20

Question 14.
The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the nqmbers.
Solution:
Let first number = x
Then second number = 2x – 3
According to the condition,
x² + (2x – 3)2 = 233
=> x² + 4x² – 12x + 9 = 233
=> 5x² – 12x + 9 – 233 = 0
=> 5x² – 12x – 224 = 0
=> 5x² – 40x + 28x – 224 = 0
=> 5x (x – 8) + 28 (x – 8) = 0
=> (x – 8) (5x + 28) = 0
Either x – 8 = 0, then x = 8
or 5x + 28 = 0, then 5x = -28 => x = \(\frac { -28 }{ 5 }\) But it is not possible
x = 8
First number = 8
Second number = 2x – 3 = 2 x 8 – 3 = 16 – 3 = 13
Number are 8, 13

Question 15.
Find two consecutive even integers whose squares have the sum 340.
Solution:
Let first even integer = x
The second even integer = x + 2
According to the condition,
x² + (x + 2)2 = 340
x² + x² + 4x + 4 = 340
=> 2x² + 4x + 4 – 340 = 0
=> 2x² + 4x – 336 = 0
=> x² + 2x – 168 = 0
=> x² + 14x – 12x – 168 = 0
=> x (x + 14) – 12 (x + 14) = 0
=> (x + 14) (x – 12) = 0
Either x + 14 = 0, then x = -14
or x – 12 = 0, the x = 12
(i) If x = -14, then
First number = -14
and second number = -14 + 2 = -12
(ii) If x = 12, then
First number =12
and second number =12 + 2 = 14
Hence even numbers are 12, 14 or -14, -12

Question 16.
The difference of two numbers is 4. If the difference of their reciprocals is \(\frac { 4 }{ 21 }\), find the numbers. (C.B.S.E. 2008)
Solution:
Let first number = x
Then second number = x – 4
According to the condition,
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 5
Either x – 7 = 0, then x = 7
or x + 3 = 0, then x = -3
(i) If x = 7, then
First number = 7
and second number = 7 – 4 = 3
(ii) If x = -3, then
First number = -3
and second number = -3 – 4 = -7
Number are 7, 3 or -3, -7

Question 17.
Find two natural numbers which differ by 3 and whose squared have the sum 117.
Solution:
Let first number = x
Then second number = x – 3
According to the condition,
x² + (x – 3)2 = 117
=> x² + x² – 6x + 9 = 117
=> 2x² – 6x + 9 – 117 = 0
=> 2x² – 6x – 108 = 0
=> x² – 3x – 54 = 0 (Dividing by 2)
=> x² – 9x + 6x – 54 = 0
=> x (x – 9) + 6 (x – 9) = 0
=> (x- 9) (x + 6) = 0
Either x – 9 = 0, then x = 9
or x + 6 = 0, then x = -6 which is not a natural number
First natural number = 9
and second number = 9 – 3 = 6

Question 18.
The sum of squares of three consecutive natural numbers is 149. Find the numbers.
Solution:
Let first number = x
Then second number = x + 1
and third number = x + 2
According to the condition,
x² + (x + 1)2 + (x + 2)2 = 149
=> x² + x² + 2x + 1 + x2 + 4x + 4 = 149
=> 3x² + 6x + 5 – 149 = 0
=> 3x² + 6x – 144 = 0
=> x² + 2x – 48 = 0 (Dividing by 3)
=> x² + 8x – 6x – 48 = 0
=> x (x + 8) – 6 (x + 8) = 0
=> (x + 8) (x – 6) = 0 .
Either x + 8 = 0, then x = -8, But it is not a natural number
or x – 6 = 0, then x = 6
Numbers are 6, 6 + 1 = 7, 6 + 2 = 8 or 6, 7, 8

Question 19.
The sum of two numbers is 16. The sum of their reciprocals is \(\frac { 1 }{ 3 }\). Find the numbers. (C.B.S.E. 2005)
Solution:
Sum of two numbers = 16
Let first number = x
Then second number = 16 – x
According to the condition,
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 6
Either x – 12 = 0, then x = 12
or x – 4 = 0, then x = 4
(i) If x = 12, then
First number = 12
and second number = 16 – 12 = 4
(ii) If x = 4, then First number = 4
and second number = 16 – 4 = 12
Hence numbers are 4, 12

Question 20.
Determine two consecutive multiples of 3 whose product is 270.
Solution:
Let first multiple of 3 = 3x
Then second multiple of 3 = 3x + 3
According to the condition,
3x (3x + 3) = 270
=> 9x² + 9x – 270 = 0
=> x² + x – 30 = 0 (Dividing by 9)
=> x² + 6x – 5x – 30 = 0
=> x (x + 6) – 5 (x + 6) = 0
=> (x + 6) (x – 5) = 0
Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
(i) When x = -6, then
First number = 3x = 3 x (-6) = -18 and second number = -18 + 3 = -15
(ii) If x = 5, then
First number = 3x = 3 x 5 = 15 and second number =15 + 3 = 18
Hence numbers are 15, 18 or -18, -15

Question 21.
The sum of a number and its reciprocal is \(\frac { 17 }{ 4 }\) , Find the number.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 7

Question 22.
A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Product of two digits = 8
Let units digit = x
Then tens digit = \(\frac { 8 }{ x }\)
Number = x + 10 x \(\frac { 8 }{ x }\) = x + \(\frac { 80 }{ x }\)
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 8

Question 23.
A two-digit number is such that the product of the digits is 12. When 36 is added to the number the digits interchange their places. Determine the number.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 9
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 10

Question 24.
A two-digit number is such that the product of the digits is 16. When 54 is subtracted from the number, the digits are interchanged. Find the number.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 11
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 12

Question 25.
Two numbers differ by 3 and their product is 504. Find the numbers. (C.B.S.E. 2002C)
Solution:
Difference of two numbers = 3
Let first number = x
Then second number = x – 3
According to the condition,
x (x – 3) = 504
=> x² – 3x – 504 = 0
=> x² – 24x + 21x – 504 = 0
=> x (x – 24) + 21 (x – 24) = 0
=> (x – 24) (x + 21) = 0
Either x – 24 = 0, then x = 24
or x + 21 = 0, then x =-21
(i) If x = 24, then
First number = 24
and second number = 24 – 3 = 21
(ii) If x =-21, then
First number = -21
and second number = -21 – 3 = -24
Hence numbers are 24, 21 or -21, -24

Question 26.
Two numbers differ by 4 and their product is 192. Find the numbers. (C.B.S.E. 2000C)
Solution:
Let first number = x
Then second number = x – 4
According to the condition,
x (x – 4) = 192
=> x² – 4x – 192 = 0
=> x² – 16x + 12x – 192 = 0
=> x (x – 16) + 12 (x – 16) = 0
=> (x – 16) (x + 12) = 0
Either x – 16 = 0, then x = 16
or x + 12 = 0, then x = -12
(i) If x = 16, then
First number = 16
and second number = 16 – 4 = 12
(ii) If x = -12, then
First number = -12
and second number = -12 – 4 = -16
Hence numbers are 16, 12 or -12, -16

Question 27.
A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number. (C.B.S.E. 1999C)
Solution:
Let units digit of the number = x
and tens digit = y
Number = x + 10y
According to the given conditions,
x + 10y = 4 (x + y)
=> x + 10y = 4x + 4y
=> 10y – 4y = 4x – x
=> 3x = 6y
=>x = 2y …(i)
and x + 10y = 2xy ….(ii)
Substituting the value of x in (i)
2y + 10y = 2 x 2y x y
=> 12y = 4y2
=> 4y2 – 12y = 0
=> y2 – 3y = 0
=> y (y – 3) = o
Either y = 0, but it is not possible because y is tens digit number
or y – 3 = 0, then y = 3
x = 2y = 2 x 3 = 6
and number = x + 10y = 6 + 10 x 3 = 6 + 30 = 36

Question 28.
The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers. [CBSE 2014]
Solution:
Let first large number = x
and smaller number = y
According to the condition,
x2 – y= 180 …(i)
and y2 = 8x
From (i) and (ii),
x2 – 8x – 180 = 0
=> x2 – 18x + 10x – 180 = 0
=> x (x – 18)+ 10 (x – 18) = 0
=> (x – 18) (x + 10) = 0
Either x – 18 = 0, then x = 18
or x + 10 = 0, then x = -10 But it is not possible being negative
x = 18
First number =18
Then second number y2 = 8x
y2 = 8 x 18 = 144 = (12)2
=> y = 12
Numbers are 18, 12

Question 29.
The sum of two numbers is 18. The sum of their reciprocals is \(\frac { 1 }{ 4 }\). Find the numbers. (C.B.S.E. 2005)
Solution:
Sum of two numbers = 18
Let one number = x
Then second number = 18 – x
According to the condition,
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 13
=> x2 – 12x – 6x + 72 = 0
=> x (x – 12) – 6 (x – 12) = 0
=> (x – 12) (x – 6) = 0
Either x – 12 = 0, then x = 12
or x – 6 = 0, then x = 6
(i) If x = 12, then
First number = 12
Second number =18 – 12 = 6
(ii) If x = 6, then
First number = 6
Then second number = 18 – 6 = 12
Numbers are 6, 12

Question 30.
The sum of two numbers a and b is 15, and the sum of their reciprocals \(\frac { 1 }{ a }\) and \(\frac { 1 }{ b }\) is \(\frac { 3 }{ 10 }\). Find the numbers a and b. (C.B.S.E. 2005)
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 14
or b – 5 = 0, then b = 5
(i) a = 15 – 10 = 5
(ii) or a = 15 – 5 = 10
Numbers are 5, 10 or 10, 5

Question 31.
The sum of two numbers is 9. The sum of their reciprocals is \(\frac { 1 }{ 2 }\). Find the numbers. [CBSE 2012]
Solution:
Sum of two numbers = 9
Let first number = x
Then second number = 9 – x
According to the condition,
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 15
By cross multiplication
18 = 9x – x2
=> x2 – 9x + 18 = 0
=> x2 – 6x – 3x + 18 = 0
=> x (x – 6) – 3 (x – 6) = 0
=> (x – 6) (x – 3) = 0
Either x – 6 = 0, then x = 6
or x – 3 = 0, then x = 3
Numbers are 6 and (9 – 6) = 3, or 3 and (9 – 3) = 6
Numbers are 3, 6

Question 32.
Three consecutive positive integers are such that the sum of the square of the’ first and the product of other two is 46, find the integers. [CBSE 2010]
Solution:
Let first number = x
Then second number = x + 1
and third number = x + 2
According co the condition,
(x)+ (x+ 1) (x + 2) = 46
x2 + x2 + 3x + 2 = 46
=> 2x2 + 3x + 2 – 46 = 0
=> 2x2 + 3x – 44 = 0
=> 2x2 + 11x – 8x – 44 = 0
=> x (2x + 11) – 4 (2x + 11) = 0
=> (2x + 11) (x – 4) = 0
Either 2x + 11 = 0, then x = \(\frac { -11 }{ 2 }\) which is not possible being fraction
or x – 4 = 0, then x = 4
Numbers are 4, 5, 6

Question 33.
The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers. [CBSE 2010]
Solution:
Let smaller number = x
Then larger number = 2x – 5
According to the condition,
(2x – 5)2 – x2 = 88
=> 4x2 – 20x + 25 – x2 – 88 = 0
=> 3x2 – 20x – 63 = 0
=> 3x2 – 27x + 7x – 63 = 0
=> 3x (x – 9) + 7 (x – 9) = 0
=> (x – 9) (3x + 7) = 0
Either x – 9 = 0, then x = 9
or 3x + 7 = 0, then x = \(\frac { -7 }{ 3 }\) which is not possible
Smaller number = 9
and greater number = 2x – 5 = 2 x 9 – 5 = 18 – 5 = 13
Hence numbers are 13, 9

Question 34.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find two numbers. [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 16
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 17
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 18

Question 35.
Find two consecutive odd positive integers, sum of whose squares is 970.
Solution:
Let two consecutive positive integers be x and x + 2
A.T.Q.,
(x)2 + (x + 2)2 = 970
=> x2 + x2 + 4x + 4 – 970 = 0
=> 2x2 + 4x – 966 = 0
=> x2 + 2x – 483 = 0 (Dividing by 2)
=> x2 + 23x – 21x – 483 = 0
=> x (x + 23) – 21 (x + 23) = 0
=> (x – 21) (x + 23) = 0
Either x – 21 = 0 or x + 23 = 0
x = 21 or x = – 23 (rejected being -ve)
As integers should be +Ve
x = 21 and x + 2 = 21 + 2 = 23
Hence integers are 21, 23

Question 36.
The difference of two natural numbers is 3 and the difference of their reciprocals is \(\frac { 3 }{ 28 }\). Find the numbers.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 19
y(y + 7) – 4(y + 7) = 0
(y – 4) (y + 7) = 0
y – 4 = 0 or y + 7 = 0
y = 4 or y = -7 (rejected being natural no.)
When y = 4, x = 3 + 4 = 7 [From (ii)]
Number are 7, 4

Question 37.
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution:
Let two consecutive positive integers be x and x + 2
A.T.Q.,
(x)2 + (x + 2)2 = 394
x2 + x2 + 4x + 4 – 394 = 0
2x2 + 4x – 390 = 0
x2 + 2x – 195 = 0 (Dividing by 2)
x2 + 15x – 13x – 195 = 0
x (x + 15) – 13 (x + 15) = 0
(x – 13) (x + 15) = 0
Either x – 13 = 0 or x + 15 = 0
x = 13 or x = -15 (rejected)
Number should be x = 13 and x = 13 + 2 = 15
or x = -15 and x = -15 + 2 = -13
Hence odd numbers are 13, 15 or -15, -13

Question 38.
The sum of the squares of two consecutive multiple of 7 is 637. Find the multiples. [ICSE 2014]
Solution:
Let first multiple of 7 = 7x
Then second = 7x + 7
(7x)2 + (7x + 7) = 637
49x2 + 49x2 + 98x + 49 = 637
98x2 + 98x + 49 – 637 = 0
98x2 + 98x – 588 = 0
x2 + x – 6 = 0 (dividing by 98)
x2 + 3x – 2x – 6 = 0
x (x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
Either x + 3 = 0, then x = -3, but not possible being negative
or x – 2 = 0, then x = 2
Numbers will be 14, 21

Question 39.
The sum of the squares of two consecutive even numbers is 340. Find the numbers. [CBSE 2014]
Solution:
Let first even number = 2x
Then second number = 2x + 2
(2x)2 + (2x + 2)2 = 340
4x2 + 4x2 + 8x + 4 – 340 = 0
8x2 + 8x – 336 = 0
x2 + x – 42 = 0 (Dividing by 8)
x2 + 7x – 6x – 42 = 0
x (x + 7) – 6 (x + 7) = 0
=> (x + 7) (x – 6) = 0
Either x + 7 = 0, then x = -7 but not possible being negative
or x – 6 = 0, then x = 6
Numbers are 12, 14

Question 40.
The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \(\frac { 29 }{ 20 }\). Find the original fraction.
Solution:
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 20
RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 21

Question 41.
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. [NCERT Exemplar]
Solution:
Let n be a required natural number.
Square of a natural number diminished by 84 = n2 – 84
and thrice of 8 more than the natural number = 3 (n + 8)
Now, by given condition,
n2 – 84 = 3 (n + 8)
=> n2 – 84 = 3n + 24
=> n2 – 3n – 108 = 0
=> n2 – 12n + 9n – 108 = 0 [by splitting the middle term]
=> n (n – 12) + 9 (n – 12) = 0
=> (n – 12) (n + 9) = 0
=> n = 12 [n ≠ – 9 because n is a natural number]
Hence, the required natural number is 12.

Question 42.
A natural number when increased by 12 equals 160 times its reciprocal. Find the number. [NCERT Exemplar]
Solution:
Let the natural number be x.
According to the question,
x + 12 = \(\frac { 160 }{ x }\)
On multiplying by x on both sides, we get
=> x2 + 12x – 160 = 0
=> x2 + (20x – 8x) – 160 = 0
=> x2 + 20x – 8x – 160 = 0 [by factorisation method]
=> x (x + 20) – 8 (x + 20) = 0
=> (x + 20) (x – 8) = 0
Now, x + 20 = 0 => x = -20 which is not possible because natural number is always greater than zero
and x – 8 = 0 => x = 8.
Hence, the required natural number is 8.

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