Tag: RD Sharma Maths Solutions Class 10

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

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

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.

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 5.6

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

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

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) a_n = 3 + 4n
(ii) b_n = 5 + 2n
(iii) x_n = 6 – n
(iv) y_n = 9 – 5n
Solution:

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

Question 7.
Find the sum of the first 25 terms of an A.P. whose nth term is given by a_n = 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 a_n = 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 a_n = 50.
(ii) n and a, if an = 4, d = 2 and S_n = -14.
(iii) d, if a = 3, n = 8 and S_n = 192.
(iv) a, if an = 28, S_n = 144 and n = 9.
(v) n and d, if a = 8, a_n = 62 and S_n = 210.
(vi) n and a_n, if a = 2, d = 8 and S_n = 90.
(vii) k, if S_n = 3n² + 5n and a_k = 164.
Solution:
In an A.P. a is the first term, d, the common difference a is the nth term and S_n is the sum of first n terms,

Question 57.
If S_n 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.
S_n = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
S_n = \(\frac { 12 }{ 2 }\) [2 x 450 + (12 – 1) x 20]
=> S_n = 6[900 + 240]
=> S_n = 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
S_n = \(\frac { n }{ 2 }\) [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 S_n 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., S_n

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

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

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.

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 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

• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

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.

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 4.7

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

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

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 – y² = 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)² + (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

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

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

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

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.

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 5.4

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

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

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
a_n = 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.
a_n = 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.
a_n = 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 (a_n) = 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 (a_n) = 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
a_n = 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 a_n – a_k 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.
a_n = 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
a_n = 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
a_n = 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
a_n = 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
a_n = 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
a_n = 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 a_n = a + (n – 1) d
Now a₃₆ = a + (36 – 1) d = 9 + 35 x 3 = 9 + 105 = 114
Let the an be the required term
a_n = a + (n – 1) d
= 9 + (n – 1) x 3 = 9 + 3n – 3 = 6 + 3n
But their difference is 39
a_n – 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 a_n = 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 a_n = 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
T_n = 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
a_n = 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 T_n = 0
T_n = 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,
T_n = 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 (a_m) = 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, a_n = 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 a_n = l – (n – 1 )d,
where l is the last term and d is the common difference. 12th term from the end,
a_n = -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., a_n = 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 a_n and b_n are the nth terms of first and second A.P.
Then b_n – a_n = [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.

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