## Online Education for RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4

Other Exercises

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
Question 1.
x² – 4 √2x + 6 = 0
Solution:

Question 2.
2x² – 7x + 3 = 0
Solution:

Question 3.
3x² + 11x + 10 = 0
Solution:

Question 4.
2x² + x – 4 = 0
Solution:

Question 5.
2x² + x + 4 = 0
Solution:

Question 6.
4x² + 4√3x + 3 = 0
Solution:

Question 7.
√2 x² – 3x – 2√2 = 0
Solution:

Question 8.
√3 x² + 10x + 7√3 = 0
Solution:

Question 9.
x² – (√2 + 1)x + √2 = 0
Solution:

Question 10.
x² – 4ax + 4a² – b² = 0
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4 are helpful to complete your math homework.

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## 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 :

Solution:

Question 2.
Find the sum to n term of the A.P. 5, 2, – 1, 4, -7, …,
Solution:

Question 3.
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 – 6n.
Solution:

Question 4.
Find the sum of last ten terms of the A.P.: 8, 10, 12, 14,…, 126. [NCERT Exemplar]
Solution:

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:

Question 6.
Find the sum of first 20 terms of the sequence whose nth term is an = An + B.
Solution:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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

Question 18.
Find the sum of n terms of the series

Solution:

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:

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

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

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

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:

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.

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

= 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:

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:

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:

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

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:

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:

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:

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:

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:

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:

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:

Question 49.
Find the sum of first n odd natural numbers.
Solution:

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

Question 51.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Solution:

Hence proved.

Question 52.
Find the sum of all integers between 84 and 719, which are multiples of 5.
Solution:

Question 53.
Find the sum of all integers between 50 and 500, which are divisible by 7.
Solution:

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

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:

= 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,

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:

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

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:

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:

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

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:

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

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

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:

=> (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

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

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:

Question 71.
Solve the question: (-4) + (-1) + 2 + 5 + … + x = 437. [NCERT Exemplar]
Solution:
Given equation is,

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:

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

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)

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 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:

Question 2.
Write the value of k for which the system of equations 2x – y = 5 6x + ky = 15 has infinitely many solutions.
Solution:

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

Question 4.
Write the value of k for which the system of equations x + ky = 0, 2x – y = 0 has unique solution.
Solution:

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:

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:

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)

Question 8.
Write the number of solutions of the following pair of linear equations :
x + 3y – 4 = 0
2x + 6y = 7
Solution:

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

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## 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:

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

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

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

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

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

surface area of the tent = curved surface area of conical part + curved surface area of cylindrical part =

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:

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

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

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

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
Then volume of the larger cone = $$(\frac { 1 }{ 3 } )$$ πr1²h1

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

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

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

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:

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

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

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

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

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:

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

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

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

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

Solution:

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

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

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)

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

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
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

Solution:
Let r be the radius of the solid = height of the conical part = h

Question 31.
The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is

Solution:
and height = r

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

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
Height of frustum = h2

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:

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

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

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

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

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:

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

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

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

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:

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

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:

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

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)

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

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°

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

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°

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

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.

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).

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

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

Solution:

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

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

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

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## 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,

(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:

Question 11.
The sum of a number and its square is $$\frac { 63 }{ 4 }$$ , find the numbers.
Solution:

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,

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,

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:

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 + $$\frac { 80 }{ x }$$

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:

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:

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,

=> 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:

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,

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:

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:

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:

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.

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 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 for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5

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 5.5

Other Exercises

Question 1.
Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P.
Solution:
(8x + 4), (6x – 2) and (2x + 7) are in A.P.
(6x – 2) – (8x + 4) = (2x + 7) – (6x – 2)
=> 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2
=> -2x – 6 = -4x + 9
=> -2x + 4x = 9 + 6
=> 2x = 15
Hence x = $$\frac { 15 }{ 2 }$$

Question 2.
If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Solution:
x + 1, 3x and 4x + 2 are in A.P.
3x – x – 1 = 4x + 2 – 3x
=> 2x – 1 = x + 2
=> 2x – x = 2 + 1
=> x = 3
Hence x = 3

Question 3.
Show that (a – b)², (a² + b²) and (a + b)² are in A.P.
Solution:
(a – b)², (a² + b²) and (a + b)² are in A.P.
If 2 (a² + b²) = (a – b)² + (a + b)²
If 2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab
If 2 (a² + b²) = 2a² + 2b² = 2 (a² + b²)
Which is true
Hence proved.

Question 4.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Solution:
Let the three terms of an A.P. be a – d, a, a + d
Sum of three terms = 21
=> a – d + a + a + d = 21
=> 3a = 21
=> a = 7
and product of the first and 3rd = 2nd term + 6
=> (a – d) (a + d) = a + 6
a² – d² = a + 6
=> (7 )² – d² = 7 + 6
=> 49 – d² = 13
=> d² = 49 – 13 = 36
=> d² = (6)²
=> d = 6
Terms are 7 – 6, 7, 7 + 6 => 1, 7, 13

Question 5.
Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
Solution:
Let the three numbers of an A.P. be a – d, a, a + d
According to the conditions,
Sum of these numbers = 27
a – d + a + a + d = 27
=> 3a = 27

Question 6.
Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Solution:
Let the four terms of an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d)
Now according to the condition,
Sum of these terms = 50
=> (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50
=> a – 3d + a – d + a + d + a – 3d= 50
=> 4a = 50
=> a = $$\frac { 25 }{ 2 }$$
and greatest number = 4 x least number
=> a + 3d = 4 (a – 3d)
=> a + 3d = 4a – 12d
=> 4a – a = 3d + 12d

Question 7.
The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Solution:

Question 8.
Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6. [CBSE 2016]
Solution:

Question 9.
The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.
Solution:
Let the four angles of a quadrilateral which are in A.P., be
a – 3d, a – d, a + d, a + 3d
Common difference = 10°
Now sum of angles of a quadrilateral = 360°
a – 3d + a – d + a + d + a + 3d = 360°
=> 4a = 360°
=> a = 90°
and common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d
2d = 10°
=> d = 5°
Angles will be
a – 3d = 90° – 3 x 5° = 90° – 15° = 75°
a – d= 90° – 5° = 85°
a + d = 90° + 5° = 95°
and a + 3d = 90° + 3 x 5° = 90° + 15°= 105°
Hence the angles of the quadrilateral will be
75°, 85°, 95° and 105°

Question 10.
Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623. [NCERT Exemplar]
Solution:
Let the three parts of the number 207 are (a – d), a and (a + d), which are in A.P.
Now, by given condition,
=> Sum of these parts = 207
=> a – d + a + a + d = 207
=> 3a = 207
a = 69
Given that, product of the two smaller parts = 4623
=> a (a – d) = 4623
=> 69 (69 – d) = 4623
=> 69 – d = 67
=> d = 69 – 67 = 2
So, first part = a – d = 69 – 2 = 67,
Second part = a = 69
and third part = a + d = 69 + 2 = 71
Hence, required three parts are 67, 69, 71.

Question 11.
The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles. [NCERT Exemplar]
Solution:
Given that, the angles of a triangle are in A.P.

Question 12.
The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number. [NCERT Exemplar]
Solution:

or, d = ± 2
So, when a = 8, d = 2,
the numbers are 2, 6, 10, 14.

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5 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 for RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6

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 4.6. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Other Exercises

Question 1.
Determine the nature of the roots of following quadratic equations :
(i) 2x² – 3x + 5 = 0 [NCERT]
(ii) 2x² – 6x + 3 = 0 [NCERT]
(iii) $$\frac { 3 }{ 5 }$$ x² – $$\frac { 2 }{ 3 }$$ x + 1 = 0
(iv) 3x² – 4√3 x + 4 = 0 [NCERT]
(v) 3x² – 2√6 x + 2 = 0
Solution:

Question 2.
Find the values of k for which the roots are real and equal in each of the following equations :
(i) kx² + 4x + 1 = 0
(ii) kx² – 2√5 x + 4 = 0
(iii) 3x² – 5x + 2k = 0
(iv) 4x²+ kx + 9 = 0
(v) 2kx² – 40x + 25 = 0
(vi) 9x² – 24x + k = 0
(vii) 4x² – 3kx +1 = 0
(viii) x² – 2 (5 + 2k) x + 3 (7 + 10k) = 0
(ix) (3k + 1) x² + 2(k + 1) x + k = 0
(x) kx² + kx + 1 = – 4x² – x
(xi) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0
(xii) x² – 2kx + 7k – 12 = 0
(xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0
(xiv) 5x² – 4x + 2 + k (4x² – 2x – 1) = 0
(xv) (4 – k) x² + (2k + 4) x (8k + 1) = 0
(xvi) (2k + 1) x² + 2 (k + 3) x (k + 5) = 0
(xvii) 4x² – 2 (k + 1) x + (k + 4) = 0
(xviii) 4x² (k + 1) x + (k + 1) = 0
Solution:

Question 3.
In the following, determine the set of values of k for which the given quadratic equation has real roots :
(i) 2x² + 3x + k = 0
(ii) 2x² + x + k = 0
(iii) 2x² – 5x – k = 0
(iv) kx² + 6x + 1 = 0
(v) 3x² + 2x + k = 0
Solution:

Question 4.
Find the values of k for which the following equations have real and equal roots :
(i) x²- 2(k + 1) x + k² = 0 [CBSE 2001C, 2013]
(ii) k²x² – 2 (2k – 1) x + 4 = 0 [CBSE 2001C]
(iii) (k + 1) x² – 2(k – 1) x + 1 = 0 [CBSE 2002C]
(iv) x² + k(2x + k – 1) + 2 = 0 [CBSE 2017]
Solution:

Question 5.
Find the values of k for which the following equations have real roots
(i) 2x² + kx + 3 = 0 [NCERT]
(ii) kx (x – 2) + 6 = 0 [NCERT]
(iii) x² – 4kx + k = 0 [CBSE 2012]
(iv) kx(x – 2√5 ) + 10 = 0 [CBSE 2013]
(v) kx (x – 3) + 9 = 0 [CBSE 2014]
(vi) 4x² + kx + 3 = 0 [CBSE 2014]
Solution:
(i) 2x² + kx + 3 = 0
Here a = 2, b = k, c = 3

Question 6.
Find the values of k for which the given quadratic equation has real and distinct roots :
(i) kx² + 2x + 1 = 0
(ii) kx² + 6x + 1 = 0
Solution:

Question 7.
For what value of k, (4 – k) x² + (2k + 4) x + (8k + 1) = 0, is a perfect square.
Solution:
(4 – k) x² + (2k + 4) x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1

Question 8.
Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:

Question 9.
Find the value of k for which the quadratic equation (3k + 1) x² + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.
[CBSE 2014]
Solution:

Question 10.
Find the values of p for which the quadratic equation (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has equal roots. Also, find these roots.
Solution:

Question 11.
If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. [CBSE 2014]
Solution:

Question 12.
If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k. [CBSE 2014]
Solution:

=> 16k = 16
k = 16

Question 13.
If 1 is a root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b=0 has equal roots, find the value of b.
Solution:

Question 14.
Find the value of p for which the quadratic equation (p + 1) x² – 6 (p + 1) x + 3 (p + q) = 0, p ≠ -1 has equal roots. Hence, find the roots of the equation. [CBSE 2015]
Solution:

Question 15.
Determine the nature of the roots of following quadratic equations :
(i) (x – 2a) (x – 2b) = 4ab
(ii) 9a²b²x² – 24abcdx + 16c²d² = 0, a ≠ 0, b ≠ 0
(iii) 2 (a² + b²) x² + 2 (a + b) x + 1 = 0
(iv) (b + c) x² – (a + b + c) x + a = 0
Solution:

Question 16.
Determine the set of values of k for which the given following quadratic equation has real roots :
(i) x² – kx + 9 = 0
(ii) 2x² + kx + 2 = 0
(iii) 4x² – 3kx +1=0
(iv) 2x² + kx – 4 = 0
Solution:

Question 17.
If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c. [CBSE 2002C]
Solution:

=> a + c = 2b
=> 2b = a + c
Hence proved.

Question 18.
If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + (c² + d²) = 0 are equal. prove that $$\frac { a }{ b }$$ = $$\frac { c }{ d }$$
Solution:

Question 19.
If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then prove that b² = ac
Solution:

Question 20.
If p, q are real and p ≠ q, then show that the roots of the equation (p – q) x² + 5(p + q) x – 2(p – q) = 0 are real and unequal.
Solution:

Question 21.
If the roots of the equation (c² – ab) x² – 2 (a² – bc) x + b² – ac = 0 are equal, prove that either a = 0 or a3 + b3 + c3 = 3abc.
Solution:

Question 22.
Show that the equation 2 (a² + b²) x² + 2 (a + b) x + 1 = 0 has no real roots, when a ≠ b.
Solution:

Question 23.
Prove that both the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are real but they are equal only when a = b = c.
Solution:

Question 24.
If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax² + bx + c = 0 and – ax² + bx + c = 0 has real roots.
Solution:

Question 25.
If the equation (1 + m²) x² + 2mcx + (c² – a²) = 0 has equal roots, prove that c² = a² (1 + m²). (C.B.S.E. 1999)
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6 are helpful to complete your math homework.

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## Online Education for RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4

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 5.4

Other Exercises

Question 1.
(i) 10th term of the A.P. 1, 4, 7, 10, ………
(ii) 18th term of the A.P. √2 , 3√2 , 5√2 , ……….
(iii) nth term of the A.P. 13, 8, 3, -2, ……..
(iv) 10th term of the A.P. -40, -15, 10, 35, ……..
(v) 8th term of the A.P. 117, 104, 91, 78, ………..
(vi) 11th term of the A.P. 10.0 , 10.5, 11.0, 11.5, ……….
(vii) 9th term of the A.P. $$\frac { 3 }{ 4 }$$ , $$\frac { 5 }{ 4 }$$ , $$\frac { 7 }{ 4 }$$ , $$\frac { 9 }{ 4 }$$ , ………
Solution:

Question 2.
(i) Which term of the A.P. 3, 8, 13, …… is 248 ?
(ii) Which term of the A.P. 84, 80, 76, ….. is 0 ?
(iii) Which term of the A.P. 4, 9, 14, ….. is 254 ?
(iv) Which term of the A.P. 21, 42, 63, 84, ….. is 420 ?
(v) Which term of the A.P. 121, 117, 113, ….. is its first negative term ?
Solution:
(i) A.P. is 3, 8, 13, …, 248
Here first term (a) = 3
and common difference (d) = 8 – 3 = 5

Question 3.
(i) Is 68 a term of the A.P. 7, 10, 13, …… ?
(ii) Is 302 a term of the A.P. 3, 8, 13, ….. ?
(ii) Is -150 a term of the A.P. 11, 8, 5, 2, …… ?
Solution:

Question 4.
How many terms are there in the A.P. ?
(i) 7, 10, 13, … 43
(ii) -1, – $$\frac { 5 }{ 6 }$$ , – $$\frac { 2 }{ 3 }$$ , – $$\frac { 1 }{ 2 }$$ , …….., $$\frac { 10 }{ 3 }$$
(iii) 7, 13, 19, …, 205
(iv) 18, 15$$\frac { 1 }{ 2 }$$ , 13, …, -47
Solution:

Question 5.
The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.
Solution:
The first term of an A.P. (a) = 5
and common difference (d) = 3
Last term = 80
Let the last term be nth
an = a + (n – 1) d
=> 80 = 5 + (n – 1) x 3
=> 80= 5 + 3n – 3
=> 3n = 80 – 5 + 3 = 78
=> n = 26
Number of terms = 26

Question 6.
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Solution:
6th term of A.P. = 19
and 17th term = 41
Let a be the first term, and d be the common difference
We know that

Question 7.
If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
Solution:

Question 8.
If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.
Solution:
Let a, a + d, a + 2d, a + 3d, ……… be an A.P.
an = a + (n – 1) d
Now a10 = a + (10 – 1) d = a + 9d
and a15 = a + (15 – 1) d = a + 14d

Question 9.
The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
Solution:

Question 10.
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Solution:
Let a, a + d, a + 2d, a + 3d, …….. be an A.P.
an = a + (n – 1) d
10th (a10) = a + (10 – 1) d = a + 9d
and 24th term (a24) = a + (24 – 1) d = a + 23d
24th term = 2 x 10th term
a + 23d = 2 (a + 9d)
=> a + 23d = 2a + 18d
=> 2a – a = 23d – 18d
=> a = 5d ….(i)
Now 72nd term = a + (72 – 1)d = a + 71d
and 34th term = a + (34 – 1) d = a + 33d
Now a + 71d – 5d + 71d = 76d
and a + 33d = 5d+ 33d = 38d
76d = 2 x 38d
72th term = 2 (34th term) = twice of the 34th term
Hence proved.

Question 11.
The 26th, 11th and last term of an A.P. are 0, 3 and – $$\frac { 1 }{ 5 }$$ , respectively. Find the common difference and the number of terms. [NCERT Exemplar]
Solution:
Let the first term, common difference and number of terms of an A.P. are a, d and n, respectively.
We know that, if last term of an A.P. is known, then
l = a + (n – 1) d ……(i)
and nth term of an A.P is

Question 12.
If the nth term of the A.P. 9, 7, 5, … is same as the nth term of the A.P. 15, 12, 9, … find n.
Solution:
In A.P 9, 7, 5, ………
Here first term (a) = 9 and d = 7 – 9 = -2 {or 5 – 7 = -2}
nth term (an) = a + (n – 1) d = 9 + (n – 1) (-2) = 9 – 2n + 2 = 11 – 2n
Now in A.P. 15, 12, 9, …..
Here first term (a) = 15 and (d) = 12 – 15 = -3
nth term (an) = a + (n – 1) d = 15 + (n – 1) x (-3)
The nth term of first A.P. = nth term of second A.P.
11 – 2n = 18 – 3n
=> -2n + 3n = 18 – 11
=> n = 7
Hence n = 7

Question 13.
Find the 12th term from the end of the following arithmetic progressions :
(i) 3, 5, 7, 9, … 201
(ii) 3, 8, 13,…, 253
(iii) 1, 4, 7, 10, …, 88
Solution:
(i) In the A.P. 3, 5, 7, 9, … 201
First term (a) = 3, last term (l) = 201
and common difference (d) = 5 – 3 = 2
We know that nth term from the last = l – (n – 1 ) d
12th term from the last = 201 – (12 – 1) x 2 = 201 – 11 x 2 = 201 – 22 = 179
(ii) In the A.P. 3, 8, 13, …, 253
First term (a) = 3
Common difference (d) = 8 – 3 = 5
and last term = 253
The nth term from the last = l – (n – 1) d
12th term from the last = 253 – (12 – 1) x 5 = 253 – 11 x 5 = 253 – 55 = 198
(iii) In the A.P. 1, 4, 7, 10, …, 88
First term (a) = 1
Common difference (d) = 4 – 1 = 3
and last term = 88
The nth term from the last = l – (n – 1) d
12th term from the last = 88 – (12 – 1) x 3 = 88 – 11 x 3 = 88 – 33 = 55

Question 14.
The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
Solution:

Question 15.
Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.
Solution:
In an A.P.
6th term (a6) = 12
and 8th term (a8) = 22
Let a be the first term and d be the common difference, then

Question 16.
How many numbers of two digit are divisible by 3 ?
Solution:
Let n be the number of terms which are divisible by 3 and d are of two digit numbers
Let a be the first term and d be the common difference, then
a = 12, d = 3, last term = 99
an = a + (n – 1) d
99 = 12 + (n – 1) x 3
=> 99 = 12 + 3n – 3
=> 3n = 99 – 9
=> n = 30
Number of terms = 30

Question 17.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.
Solution:
In an A.P.
n = 60
First term (a) = 7 and last term (l) = 125
Let d be the common difference, then
a60 = a + (60 – 1) d
=> 125 = 7 + 59d
=> 59d = 125 – 7 = 118
Common difference = 2
Now 32nd term (a32) = a + (32 – 1) d = 7 + 31 x 2 = 7+ 62 = 69

Question 18.
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
Solution:

Question 19.
The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P.
Solution:
First term of an A.P. = 5
and 100th term = -292

Question 20.
Find a30 – a20 for the A.P.
(i) -9, -14, -19, -24, …
(ii) a, a + d, a + 2d, a + 3d, …
Solution:

Question 21.
Write the expression an – ak for the A.P. a, a + d, a + 2d, ……
Hence, find the common difference of the A.P. for which
(i) 11th term is 5 and 13th term is 79.
(ii) a10 – a5 = 200
(iii) 20th term is 10 more than the 18th term.
Solution:
In the A.P. a, a + d, a + 2d, …..

Question 22.
Find n if the given value of x is the nth term of the given A.P.

Solution:

Question 23.
The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
Solution:

Question 24.
Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.
Solution:
Let a, a + d, a + 2d, a + 3d, ………. be the A.P.
an = a + (n – 1) d
But a3 = 16

Question 25.
The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P. [CBSE 2004]
Solution:
Let a, a + d, a + 2d, a + 3d, be the A.P.
Here a is the first term and d is the common difference
an = a + (n – 1) d
Now a7 = a + (7 – 1) d = a + 6d = 32 ….(i)
and a13 = a + (13 – 1) d = a + 12d = 62 ….(ii)
Subtracting (i) from (ii)
6d = 30
=> d = 5
a + 6 x 5 = 32
=> a + 30 = 32
=> a = 32 – 30 = 2
A.P. will be 2, 7, 12, 17, ………..

Question 26.
Which term of the A.P. 3, 10, 17, … will be 84 more than its 13th term ? [CBSE 2004]
Solution:

Question 27.
Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Solution:

Question 28.
For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,… and 3, 10, 17, … are equal ? (C.B.S.E. 2008)
Solution:
In the A.P. 63, 65, 67, …
a = 63 and d = 65 – 63 = 2
an = a1 + (n – 1) d = 63 + (n – 1) x 2 = 63 + 2n – 2 = 61 + 2n
and in the A.P. 3, 10, 17, …
a = 3 and d = 10 – 3 = 7
an = a + (n – 1) d = 3 + (n – 1) x 7 = 3 + 7n – 7 = 7n – 4
But both nth terms are equal
61 + 2n = 7n – 4
=> 61 + 4 = 7n – 2n
=> 65 = 5n
=> n = 13
n = 13

Question 29.
How many multiples of 4 lie between 10 and 250 ?
Solution:
All the terms between 10 and 250 are multiple of 4
First multiple (a) = 12
and last multiple (l) = 248
and d = 4
Let n be the number of multiples, then
an = a + (n – 1) d
=> 248 = 12 + (n – 1) x 4 = 12 + 4n – 4
=> 248 = 8 + 4n
=> 4n = 248 – 8 = 240
n = 60
Number of terms are = 60

Question 30.
How many three digit numbers are divisible by 7 ?
Solution:
First three digit number is 100 and last three digit number is 999
In the sequence of the required three digit numbers which are divisible by 7, will be between
a = 105 and last number l = 994 and d = 7
Let n be the number of terms, then
an = a + (n – 1) d
994 = 105 + (n – 1) x 7
994 = 105 + 7n – 7
=> 7n = 994 – 105 + 7
=> 7n = 896
=> n = 128
Number of terms =128

Question 31.
Which term of the arithmetic progression 8, 14, 20, 26, … will be 72 more than its 41st term ? (C.B.S.E. 2006C)
Solution:
In the given A.P. 8, 14, 20, 26, …

Question 32.
Find the term of the arithmetic progression 9, 12, 15, 18, … which is 39 more than its 36th term (C.B.S.E. 2006C)
Solution:
In the given A.R 9, 12, 15, 18, …
First term (a) = 9
and common difference (d) = 12 – 9 = 3
and an = a + (n – 1) d
Now a36 = a + (36 – 1) d = 9 + 35 x 3 = 9 + 105 = 114
Let the an be the required term
an = a + (n – 1) d
= 9 + (n – 1) x 3 = 9 + 3n – 3 = 6 + 3n
But their difference is 39
an – a36 = 39
=> 6 + 3n – 114 = 39
=> 114 – 6 + 39 = 3n
=> 3n = 147
=> n = 49
Required term is 49th

Question 33.
Find the 8th term from the end of the A.P. 7, 10, 13, …, 184. (C.B.S.E. 2005)
Solution:
The given A.P. is 7, 10, 13,…, 184
Here first term (a) = 7
and common difference (d) = 10 – 7 = 3
and last tenn (l) = 184
Let nth term from the last is an = l – (n – 1) d
a8= 184 – (8 – 1) x 3 = 184 – 7 x 3 = 184 – 21 = 163

Question 34.
Find the 10th term from the end of the A.P. 8, 10, 12, …, 126. (C.B.S.E. 2006)
Solution:
The given A.P. is 8, 10, 12, …, 126
Here first term (a) = 8
Common difference (d) = 10 – 8 = 2
and last tenn (l) = 126
Now nth term from the last is an = l – (n – 1) d
a10 = 126 – (10 – 1) x 2 = 126 – 9 x 2 = 126 – 18 = 108

Question 35.
The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P. (C.B.S.E. 2009)
Solution:

Question 36.
Which term of the A.P. 3, 15, 27, 39, …. will be 120 more than its 21st term ? (C.B.S.E. 2009)
Solution:
A.P. is given : 3, 15, 27, 39, …….
Here first term (a) = 3
and c.d. (d) = 15 – 3 = 12
Let nth term be the required term
Now 21st term = a + (n – 1) d = 3 + 20 x 12 = 3 + 240 = 243
According to the given condition,
nth term – 21 st term = 120
=> a + (n – 1) d – 243 = 120
=> 3 + (n – 1) x 12 = 120 + 243 = 363
=> (n – 1) 12 = 363 – 3 = 360
=> n – 1 = 30
=> n = 30 + 1 = 31
31 st term is the required term

Question 37.
The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.[CBSE 2012]
Solution:

Question 38.
Find the number of ail three digit natural numbers which are divisible by 9. [CBSE 2013]
Solution:
First 3-digit number which is divisible by 9 = 108
and last 3-digit number = 999
d= 9
a + (n – 1) d = 999
=> 108 + (n – 1) x 9 = 999
=> (n – 1) d = 999 – 108
=> (n – 1) x 9 = 891
=> n – 1 = 99
=> n = 99 + 1 = 100
Number of terms = 100

Question 39.
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P. [CBSE 2013]
Solution:

Question 40.
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P. [CBSE 2013]
Solution:
Let a be the first term and d be the common difference and
Tn = a + (n – 1) d

Question 41.
The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2013]
Solution:

Hence 72nd term = 4 times of 15th term

Question 42.
Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. [CBSE 2014]
Solution:
Numbers divisible by both 2 and 5 are 110, 120, 130, ………. , 990
Here a = 110, x = 120 – 110 = 10
an = 990
As a + (n – 1) d = 990
110 + (n – 1) (10) = 990
(n – 1) (10) = 990 – 110 = 880
n – 1 = 88
n = 88 + 1 = 89

Question 43.
If the seventh term of an AP is $$\frac { 1 }{ 9 }$$ and its ninth term is $$\frac { 1 }{ 7 }$$ , find its (63) rd term. [CBSE 2014]
Solution:

Question 44.
The sum of 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP. [CBSE 2014]
Solution:

Question 45.
Find where 0 (zero) is a term of the AP 40, 37, 34, 31, …… [CBSE 2014]
Solution:
AP 40, 37, 34, 31, …..
Here a = 40, d = -3
Let Tn = 0
Tn = a + (n – 1) d
=> 0 = 40 + (n – 1) (-3)
=> 0 = 40 – 3n + 3
=> 3n = 43
=> n = $$\frac { 43 }{ 3 }$$ which is in fraction
There is no term which is 0

Question 46.
Find the middle term of the A.P. 213, 205, 197, …, 37. [CBSE2015]
Solution:

Question 47.
If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P. [BTE2015]
Solution:
We know that,
Tn = a + (n – 1 )d
T5 = a + 4d => a + 4d = 31 ……(i)
and T25 = a + 24d
=>a + 24d = 140 + T5
=> a + 24d = 140 + 31 = 171 …..(ii)
Subtracting (i) from (ii),
20d= 140
and a + 4d = 31
=> a + 4 x 7 = 31
=> a + 28 = 31
=> a = 31 – 28 = 3
a = 3 and d = 7
AP will be 3, 10, 17, 24, 31, ……..

Question 48.
Find the sum of two middle terms of the

Solution:

Question 49.
If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.
Solution:

Question 50.
If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).
Solution:
In an A.P.
Number of terms = n
First term = a
and nth term = l
mth term (am) = a + (m – 1) d
and mth term from the end = l – (m – 1)d
Their sum = a + (m – 1) d + l – (m – 1) d = a + l
Hence proved.

Question 51.
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? [NCERT Exemplar]
Solution:
Here, the first number is 11, which divided by 4 leave remainder 3 between 10 and 300.
Last term before 300 is 299, which divided by 4 leave remainder 3.
11, 15, 19, 23, …, 299
Here, first term (a) = 11,
common difference (d) = 15 – 11 = 4
nth term, an = a + (n – 1 ) d = l [last term]
=> 299 = 11 + (n – 1) 4
=> 299 – 11 = (n – 1) 4
=> 4(n – 1) = 288
=> (n – 1) = 72
n = 73

Question 52.
Find the 12th term from the end of the A.P. -2, -4, -6, …, -100. [NCERT Exemplar]
Solution:
Given, A.P., -2, -4, -6, …, -100
Here, first term (a) = -2,
common difference (d) = -4 – (-2)
and the last term (l) = -100.
We know that, the nth term an of an A.P. from the end is an = l – (n – 1 )d,
where l is the last term and d is the common difference. 12th term from the end,
an = -100 – (12 – 1) (-2)
= -100 + (11) (2) = -100 + 22 = -78
Hence, the 12th term from the end is -78

Question 53.
For the A.P.: -3, -7, -11,…, can we find a30 – a20 without actually finding a30 and a20 ? Give reasons for your answer. [NCERT Exemplar]
Solution:
True.
nth term of an A.P., an = a + (n – 1)d
a30 = a + (30 – 1 )d = a + 29d
and a20 = a + (20 – 1 )d = a + 19d …(i)
Now, a30 – a20 = (a + 29d) – (a + 19d) = 10d
and from given A.P.
common difference, d = -7 – (-3) = -7 + 3 = -4
a30 – a20 = 10(-4) = -40 [from Eq- (i)]

Question 54.
Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why? [NCERT Exemplar]
Solution:
Let the same common difference of two A.P.’s is d.
Given that, the first term of first A.P. and second A.P. are 2 and 7 respectively,
then the A.P.’s are 2, 2 + d, 2 + 2d, 2 + 3d, … and 7, 7 + d, 7 + 2d, 7 + 3d, …
Now, 10th terms of first and second A.P.’s are 2 + 9d and 7 + 9d, respectively.
So, their difference is 7 + 9d – (2 + 9d) = 5
Also, 21st terms of first and second A.P.’s are 2 + 20d and 7 + 20d, respectively.
So, their difference is 7 + 20d – (2 + 9d) = 5
Also, if the an and bn are the nth terms of first and second A.P.
Then bn – an = [7 + (n – 1 ) d] – [2 + (n – 1) d = 5
Hence, the difference between any two corresponding terms of such A.P.’s is the same as the difference between their first terms.

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4 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.

## RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm
Solution:
(d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent

OP ⊥ PQ
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (12)² = (5)2 + PQ²
=> 144 = 25 + PQ²
PQ² = 144 – 25 = 119
PQ = √119

Question 2.
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
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
(a) Let PQ be the tangent from Q to the circle with O as centre
PQ = 24 cm
OQ = 25 cm

OQ ⊥ PQ
Now in right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = r² + (24)²
=> 625 = r² + 576
=> r² = 625 – 576 = 49 = (7)²
r = 7
Radius of the circle = 7 cm

Question 3.
The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
(a) √7 cm
(b) 7 cm
(c) 5 cm
(d) 25 cm
Solution:
(c) Let AB be the tangent from A to the circle of centre O, then
OB = 3 cm
BA = 4 cm

OB ⊥ BA
In right ∆OBA,
OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²
OA = 5
Distance of A from the centre O = 5 cm

Question 4.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(a) PA and PB are the tangents to the circle from P and ∠APB = 80°

∠AOB = 180° – ∠APB = 180°- 80° = 100°
But OP is the bisector of ∠AOB
∠POA = ∠POB = $$\frac { 1 }{ 2 }$$ ∠AOB
=> ∠POA = $$\frac { 1 }{ 2 }$$ x 100° = 50°

Question 5.
If TP and TQ are 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:
(b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°

But ∠POQ + ∠PTQ = 180°
=> 110° + ∠PTQ = 180°
=> ∠PTQ = 180° – 110° = 70°

Question 6.
PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(b) In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ

OP is radius of the circle
OP ⊥ PQ
OP = PQ
∠POQ = ∠OQP
But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)
∠OQP = ∠POQ = 45°

Question 7.
Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(d) Two circles with centres O and O’ touch each other at C externally
A common tangent is drawn which touches the circles at A and B respectively.
Join OA, O’B and O’O which passes through C

AO = BO’ (radii of the equal circle)
AB || OO’
=> AOO’B is a rectangle
Draw another common tangent through C which intersects AB at D, then DA = DC = DB
AC and BC are the diagonals of equal square
AC = BC
∠DAC = ∠DBC = 45°
∠ACB = 90°

Question 8.
ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ∆ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution:
(b) In a right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm

AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
An incircle is drawn with centre 0 which touches the sides of the triangle ABC at P, Q and R
OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle
OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA
OPBQ is a square
Let r be the radius of the incircle
PB = BQ = r
AR = AP = 8 – r,
CQ = CR = 6 – r
AC = AR + CR
=> 10 = 8 – r + 6 – r
10 = 14 – 2r
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 9.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(c) PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°

OQ is radius and PQ is tangent to the circle
OQ ⊥ QP or ∠OQP = 90°
But ∠QOP + ∠POR = 180° (Linear pair)
=> ∠QOP + 120° = 180°
∠QOP = 180° – 120° = 60°
Now in ∆POQ
∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)
=> 60° + 90° + ∠OPQ = 180°
=> 150° + ∠OPQ = 180°
=> ∠OPQ = 180° – 150° = 30°

Question 10.
If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Solution:
(b) A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides
AB + CD = BC + AD

Question 11.
The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(a) √7 cm
(b) 2√7cm
(c) 10 cm
(d) 5 cm
Solution:
(b) Radius of the circle = 6 cm
and distance of the external point from the centre = 8 cm
Length of tangent = √{(8)² – (6)²}
= √(64 – 36) = √28
= √(4 x 7) = 2√7 cm

Question 12.
AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
(c) AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D

CD = 4 cm
DA and DC are tangents to the first circle from D
CD = AD = 4 cm
Similarly DC and DB are tangents to the second circle from D
CD = DB = 4 cm
AB = AD + DB = 4 + 4 = 8 cm

Question 13.
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 3AD = AB + BC + CA
(d) 4AD = AB + BC + CA
Solution:
(b) AD, AE and BC are the tangents to the circle at D, E and F respectively
D and AE are tangents to the circle from A
Similarly, CD = CF and BE = BF ….(ii)
Now AB + AC + BC = AE – BE + AD – CD + CF + BF
= AD – BE + AD – CD + BE + BE
= 2AD [From (i) and (ii)]
or 2 AD = AB + BC + CA

Question 14.
In the figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =

(a) 8 cm
(b) 3 cm
(c) 2.5 cm
(d) 5 cm
Solution:
(d) In the figure, 0 is the centre of the circle
QR is tangent to the circle and QOS is a diameter SQ = 6 cm, QR = 4 cm

OQ = $$\frac { 1 }{ 2 }$$ QS = $$\frac { 1 }{ 2 }$$ x 6 = 3 cm
OQ ⊥ QR
Now in right ∆OQR
OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OR = 5 cm

Question 15.
In the figure, the perimeter of ∆ABC is
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 15 cm

Solution:
(a) ∆ABC is circumscribed of circle with centre O
AQ = 4 cm, CP = 5 cm and BR = 6 cm
AQ and AR the tangents to the circle AQ = AR = 4 cm
Similarly BP and BR are tangents,
BP = BR = 6 cm
and CP and CQ are the tangents
CQ = CP = 5 cm

AB = AR + BR = 4 + 6 = 10 cm
BC = BP + CP = 6 + 5 = 11 cm
AC = AQ + CQ = 4 + 5 = 9 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

Question 16.
In the figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =
(a) 2√2 cm
(b) 2 cm
(c) 2√3 cm
(d) 3√2 cm

Solution:
(c) In the figure, AP is the tangent to the circle with centre O such that
OP = 4 cm, ∠OPA = 30°
Join OA, let AP = x

Question 17.
AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =
(a) 12 cm
(b) 18 cm
(c) 24 cm
(d) 36 cm
Solution:
(c) OP is radius, PA is the tangent
OP ⊥ AP

Now in right ∆OAP,
OA² = OP² + AP²
(15)² = (9)² + AP²
225 = 81 + AP²
=> AP² = 225 – 81 = 144 = (12)²
AP = 12 cm
But AP = AQ = 12 cm (tangents from A to the circle)
AP + AQ = 12+ 12 = 24 cm

Question 18.
At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:
(d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm
From P, at a distance of 8 cm AB is a chord drawn parallel to XY
To find the length of AB
Join OA

XY is tangent and OP is the radius
OP ⊥ XY or PQ ⊥ XY
AB || XY
OQ is ⊥ AB which meets AB at R
Now in right ∆OAR,
OA² = OR² + AR²
(5)² = (3)² + AR²
25 = 9 + AR²
=> AR² = 25 – 9 = 16 = (4)²
AR = 4 cm
But R is mid-point of AB
AB = 2 AR = 2 x 4 = 8 cm

Question 19.
If PT is tahgent drawn froth a point P to a circle touching it at T and O is the centre of the circle, then ∠OPT + ∠POT =
(a) 30°
(b) 60°
(c) 90°
(d) 180°
Solution:
(c) In the figure, PT is the tangent to the circle with centre O.
OP and OT are joined

PT is tangent and OT is the radius
OT ⊥ PT
Now in right ∆OPT
∠OTP = 90°
∠OPT + ∠POT = 180° – 90° = 90°

Question 20.
In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =
(a) 5 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm

Solution:
(d) In the figure, ∆ABC is the circumscribed a circle
AB = 12 cm, BC = 8 cm and AC = 10 cm
Let AD = a, DB = b and EC = c, then
AF = a, BE = b and FC = c

But AB + BC + AC = 12 + 8 + 10 = 30
a + b + b + c + c + a = 30
=> 2 (a + b + c) = 30
a + b + c = 15
Subtracting BC or b + c from this a = 15 – 8 = 7

Question 21.
In the figure, if AP = PB, then
(a) AC = AB
(b) AC = BC
(c) AQ = QC
(d) AB = BC

Solution:
(b) In the figure, AP = PB
But AP and AQ are the tangent from A to the circle

AP = AQ
Similarly PB = BR
But AP = PB (given)
AQ = BR ….(i)
But CQ and CR the tangents drawn from C to the circle
CQ = CR
AQ + CQ = BR + CR
AC = BC

Question 22.
In the figure, if AP = 10 cm, then BP =
(a) √91 cm
(b) √127 cm
(c) √119 cm
(d) √109 cm

Solution:
(b) In the figure,
OA = 6 cm, OB = 3 cm and AP = 10 cm

OA is radius and AP is the tangent
OA ⊥ AP
Now in right ∆OAP
OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136
Similarly BP is tangent and OB is radius
OP² = OB² + BP²
136 = (3)² + BP2
=> 136 = 9 + BP²
=> BP² = 136 – 9 = 127
BP = √127 cm

Question 23.
In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
(a) 110°
(b) 100°
(c) 120°
(d) 90°

Solution:
(c) In the figure, PR is the tangent to the circle at P.
O is the centre of the circle ∠QPR = 60°

OP is the radius and PR is the tangent OPR = 90°
=> ∠OPQ + ∠QPR = 90°
=> ∠OPQ + 60° = 90°
=> ∠OPQ = 90° – 60° = 30°
OP = OQ (radii of the circle)
∠OQP = 30°
In ∆OPQ,
∠OPQ + ∠OQP + ∠POQ = 180°
=> 30° + 30° + ∠PQR = 180°
=> 60° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°

Question 24.
In the figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =
(a) PQ
(b) QR
(c) PR
(d) PS

Solution:
(a) In the figure, quadrilateral PQRS is circumscribed a circle

PD = PA (tangents from P to the circle)
Similarly QA = QB
PD + QB = PA + QA = PQ

Question 25.
In the figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =
(a) 9 cm
(b) 18 cm
(c) 15 cm
(d) 13.5 cm

Solution:
(a) In the figure, two equal circles touch, each other externally at T
QR is the common tangent
QP = 4.5 cm

PQ = PT (tangents from P to the circle)
Similarly PT = PR
PQ = PT = PR
Now QR = PQ + PR = 4.5 + 4.5 = 9 cm

Question 26.
In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is
(a) 100°
(b) 120°
(c) 140°
(d) 150°

Solution:
(a) In the figure, APB is a tangent to the circle with centre O

∠QPB = 50°
OP is radius and APB is a tangent
OP ⊥ AB
=> ∠OPB = 90°
=> ∠OPQ + ∠QPB = 90°
∠OPQ + 50° = 90°
=> ∠OPQ = 90° – 50° = 40°
But OP = OQ
∠OPQ = OQP = 40°
∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

Question 27.
In the figure, if tangents PA and PB are drawn to a circle such that ∠APB = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =
(a) 60°
(b) 90°
(c) 30°
(d) None of these

Solution:
(c) In the figure, PA and PB are the tangents to the circle with centre O

∠APB = 30°
Chord AC || BP,
AB is joined
PA = PB
∠PAB = ∠PBA
But ∠PAB + ∠PBA = 180° – 30° = 150°
=> ∠BPA + ∠PBA = 150°
=> 2 ∠PBA = 150°
=> ∠PBA = 75°
AC || BC
∠BAC = ∠PBA = 75°
But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)
∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

Question 28.
In the figure, PR =
(a) 20 cm
(b) 26 cm
(c) 24 cm
(d) 28 cm

Solution:
(b) In the figure, two circles with centre O and O’ touch each other externally
PQ and RS are the tangents drawn to the circles

OQ and O’S are the radii of these circles and
OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm
Now in right ∆OQP
OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OP = 5 cm
Similarly in right ∆RSO’
(O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)²
O’R = 13 cm
Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm

Question 29.
Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. Then AO : AO’ =
(a) $$\frac { 3 }{ 2 }$$
(b) 2
(c) 3
(d) $$\frac { 1 }{ 4 }$$

Solution:
(c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A

From A, AT is the tangent to the circle (O, r)
O’Q ⊥ AT
Now AO : AO’ = 3r : r
= 3 : 1 = $$\frac { 3 }{ 1 }$$

Question 30.
Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm

Solution:
(c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O
Chord BC touches the inner circle at P
Draw a tangent AB to the inner circle
Join OQ and OA

OQ is radius and AQB is the tangent
OQ ⊥ AB and OQ bisects AB
AQ = QB
Similarly, BP = PC or P is mid-point of BC
But BQ and BP are tangents from B
QB = BP = AQ
In right ∆OAQ,
OA² = AQ² + OQ²
(5)² = AQ² + (3)²
=> AQ² = (5)² – (3)²
=> AQ² = 25 – 9 = 16 = (4)²
AQ = 4 cm
BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

Question 31.
In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to
(a) 10 cm
(b) 12 cm
(c) 15 cm
(d) 18 cm

Solution:
(c) In the figure, two concentric circles with centre O
From a point P on the outer circle,
PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively
PR = 7.5 cm
Join OR and OQ

PT is chord and OR is radius
R is mid-point of PT
Similarly Q is mid-point of PS
But PR = PQ (tangents from P)
PT = 2 PR and PS = 2 PQ
PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

Question 32.
In the figure, if AB = 8 cm and PE = 3 cm, then AE =
(a) 11 cm
(b) 7 cm
(c) 5 cm
(d) 3 cm

Solution:
(c) In the figure, AB and AC are the tangents to the circle from A
DE is another tangent drawn from P
AB = 8 cm, PE = 3 cm

AB = AC (tangents drawn from A to the circle)
Similarly PE = EC and DP = DB
Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

Question 33.
In the figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then
(a) a = 30°, b = 60°
(b) a = 35°, b = 55°
(c) a = 40°, b = 50°
(d) a = 45°, b = 45°

Solution:
(b) In the figure, PQ and PR are the tangents drawn from P to the circle with centre O
∠OPQ = 35°
PO is joined

PQ = PR (tangents from P to the circle)
∠OPQ = ∠OPR
=> 35° = a
=> a = 35°
OQ is radius and PQ is tangent
OQ ⊥ PQ
=> ∠OQP = 90°
In ∆OQP,
∠POQ + ∠QPO = 90°
=> b + 35° = 90°
=> b = 90° – 35° = 55°
a = 35°, b = 55°

Question 34.
In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then
(a) 25°
(b) 30°
(c) 40°
(d) 60°

Solution:
(b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O
OP, OQ and PQ are joined
∠TQP = 60°
TP = TQ (Tangents from T to the circle)
∠TQP = ∠TPQ = 60°
∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°
and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°
But OP = OQ (radii of the same circle)
∠OPQ = ∠OQP
But ∠OPQ + ∠OQP = 180° – 120° = 60°
But ∠OPQ = 30°

Question 35.
In the figure, the sides AB, BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012]

(a) 11 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Solution:
(b) In the figure,
PA = 4 cm, BP = 3 cm, AC = 11 cm
AP and AR are the tangents from A to the circle
AP = AR
=> AR = 4 cm
Similarly BP and BQ are tangents
BQ = BP = 3 cm
AC =11 cm
AR + CR = 11 cm
4 + CR =11 cm
CR = 11 – 4 = 7 cm
CQ and CR are tangents to the circle
CQ = CR = 7 cm
Now, BC = BQ + CQ = 3 + 7 = 10 cm

Question 36.
In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012]
(a) 18 cm
(b) 13.5 cm
(c) 12 cm
(d) 9 cm

Solution:
(a) In ∆DEF
DF touches the circle at H
and circle touches ED and EF Produced at K and M respectively
EK = 9 cm
EK and EM are the tangents to the circle
EM = EK = 9 cm
Similarly DH and DK are the tangent
DH = DK and FH and FM are tangents
FH = FM
Now, perimeter of ∆DEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18 cm

Question 37.
In the figure DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is [CBSE 2013]

(a) 3 cm
(b) 5 cm
(c) 4 cm
(d) 6 cm
Solution:
(b) If figure, DE and DF are tangents to the circle drawn from D.
A is the centre of the circle.
DE = 5 cm and DE ⊥ DF
Join AE, AF

DE is the tangent and AE is radius
AE ⊥ DE
Similarly, AF ⊥ DF
But ∠D = 90° (given)
AFDE is a square
AE = DE (side of square)
But DE = 5 cm
AE = 5 cm
Radius of circle is 5 cm

Question 38.
In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013]

(a) 11
(b) 18
(c) 6
(d) 15
Solution:
(a) In the figure, a circle touches the sides of a quadrilateral ABCD
∠B = 90°, OP = OQ = r
AB = 29 cm, AD = 23 cm, DS = 5 cm
∠B = 90°
BA is tangent and OQ is radius
∠OQB = 90°
Similarly OP is radius and BC is tangents
∠OPB = 90°
But ∠B = 90° (given)
PBQO is a square
DS = 5 cm
But DS and DR are tangents to the circles
DR = 5 cm
AR = 23 – 5= 18 cm
AR = AQ (tangents to the circle from A)
AQ = 18 cm
But AB = 29 cm
BQ = 29 – 18 = 11 cm
OPBQ is a square
OQ = BQ = 11 cm
Radius of the circle = 11 cm

Question 39.
In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c)

Question 40.
Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of ∠APB is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)
∠TAP = ∠TPA = x (say)
and ∠TBP = ∠TPB = y (say)
Also, in triangle APB,
x + x + x + y + y = 180°
=> 2x + 2y = 180°
=> x + y = 90°
=> ∠APB = 90°

Question 41.
In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals
(a) 67°
(b) 134°
(c) 44°
(d) 46°

Solution:
(b) ∠OQP = 90°
[Tangent is ⊥ to the radius through the point of contact]
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 42.
In the figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is [CBSE2014]
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9

Solution:
(b) In the figure, QR is common tangent to the two circles touching each other externally at T
Tangent at T meets QR at P
PT = 3.8 cm
PT and PQ are tangents from P
PT = PQ = 3.8 cm
Similarly PT and PR are tangents
PT = PR = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm

Question 43.
In the figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x =
(a) 10
(b) 9
(c) 8
(d) 7 (CBSE 2014)

Solution:
(b) In the given figure,
ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively
AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm
CR and CQ are tangents to the circle from C
CR = CQ = 3 cm
BQ = BC – CQ = 7 – 3 = 4 cm
BQ = and BP are tangents from B
BP = BQ = 4 cm
AS and AP are tangents from A
AP = AS = 5 cm
AB = AP + BP = 5 + 4 = 9 cm
x = 9 cm

Question 44.
If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar)
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
(b) O is the centre of the circle.
Given, ∠POQ = 130°
PT and QT are tangents drawn from external point T to the circle.

∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]
∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°
=> ∠PTQ + 90° + 90° + 130° = 360°
=> ∠PTQ = 360° – 310° = 50°
Thus, the angle between the tangents is 50°.

Question 45.
If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar]
(a) $$\frac { 3\surd 3 }{ 2 }$$ cm
(b) 6 cm
(c) 3 cm
(d) 3√3 cm
Solution:
(d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.
Join OA and OP.

Also, OP is a bisector of line ∠APC
∠APO = ∠CPO = 30°
Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm

Question 46.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is [NCERT Exemplar]
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
(b) Let O be the centre of two concentric circles C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm.
Now, we draw a chord AC of circle C2, which touches the circle C1 at B.
Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

Now, in right angled ∆OBC, by using Pythagoras theorem,
OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]
=> 5² = BC² + 4²
=> BC² = 25 – 16 = 9
=> BC = 3 cm
Length of chord AC = 2 BC = 2 x 3 = 6 cm

Question 47.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is [NCERT Exemplar]
(a) 4 cm
(b) 5 cm
(b) 6 cm
(d) 8 cm
Solution:
(d) First, draw a circle of radius 5 cm having centre O.
A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
Now, ∠OAY = 90°
[Tangent and any point of a circle is perpendicular to the radius through the point of contact]
∠OAY + ∠OED = 180°
[sum of cointerior is 180°]
=> ∆OED = 180°
Also, AE = 8 cm, Join OC
Now, in right angled ∆OBC
OC² = OE² + EC²
=> EC² = OC² – OE² [by Pythagoras theorem]
EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]
EC² = 25 – 9 = 16
=> EC = 4 cm
Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm
[Since, perpendicular from centre to the chord bisects the chord]

Question 48.
From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangent PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is [NCERT Exemplar]
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
(a) Firstly, draw a circle of radius 5 cm having centre O.
P is a point at a distance of 13 cm from O.
A pair of tangents PQ and PR are drawn.
OQ ⊥ QP [since, AP is a tangent line]
In right angled ∆PQO,
OP² = OQ² + QP²
=> 13² = 5² + QP²
=> QP² = 169 – 25 = 144 = 12²
=> QP = 12 cm

Now, area of ∆OQP = $$\frac { 1 }{ 2 }$$ x QP x QO = $$\frac { 1 }{ 2 }$$ x 12 x 5 = 30 cm²
Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

Question 49.
If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]

In ∆PAB,
∠P + ∠A + ∠B = 180°
[since, sum of angles of a triangle = 180°
50°+ θ + θ = 180°
2θ = 180° – 50° = 130°
θ = 65°
Also, OA ⊥ PA
[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠PAO = 90°
=> ∠PAB + ∠BAO = 90°
=> 65° + ∠BAO = 90°
=> ∠BAO = 90° – 65° = 25°

Question 50.
The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is [NCERT Exemplar]
(a) 10 cm
(b) 7.5 cm
(c) 5 cm
(d) 2.5 cm
Solution:
(c)

Question 51.
In the figure, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]

(a) 45°
(b) 35°
(c) 55°
(d) 62$$\frac { 1 }{ 2 }$$°
Solution:
(c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°
=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

Question 52.
In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar]

(a) 20°
(b) 40°
(c) 35°
(d) 45°
Solution:
(b) Given, AB || PR
∠ABQ = ∠BQR = 70° [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB
∠QDA = ∠QDB [each 90°]
QD = QD [common side]
∆ADQ = ∆BDQ [by SAS similarity criterion]
Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]
Also, ∠ABQ = ∠BQR [alternate interior angle]
∠ABQ = 70° [∠BQR = 70°]
Hence, ∠QAB = 70° [from Eq. (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠Q = 180°
=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 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.

## RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If 6 is an acute angle such that tan2 6 = 8 $$\frac { 8 }{ 7 }$$, then the value of

Solution:

Question 9.
If 3 cos θ = 5 sin 6, then the value of

Solution:

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =

Solution:

Question 11.
The value of cos217° – sin2 73° is

Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.

Solution:

Question 13.

Solution:

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1

(c) -1
(d) 2,

Solution:

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to

Solution:

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =

Solution:

Question 18.
If 6 is an acute angle such that sec2 θ =

Solution:

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1

(c) 0
(d) None of these

Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of

Solution:

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to

Solution:

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:

Question 25.
If A + B = 90°, then
Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°

(c) 45°
(d) 60°

Solution:

Question 29.

Solution:

Question 30.
If A, B and C are interior angles of a

Solution:

Question 31.
If cos θ = $$\frac { 2 }{ 3 }$$, then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to

Solution:

Question 33.

(a)-2
(b) 2
(c) 1
(d) 0
Solution:

Question 34.
In the figure, the value of cos Φ is

Solution:

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 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.