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Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6 | Set 1
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Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 3

Last Updated : 07 Apr, 2021
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Question 27. For what value of a, the following system of the equation has no solution:

ax + 3y = a − 3

12x + ay = a

Solution: 

Given that,

ax + 3y = a − 3  ...(1)

12x + ay = a ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0 ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = a, b1 = 3, c1 = - (a − 3) 

a2 = 12, b2 = a, c2 = − a

For unique solution, we have

a1/a2 = b1/b2 ≠ c1/c2

a/12 = 3/a ≠ -(a - 3)/-a

 a-3 ≠ 3

 a ≠ 6

And,

a/12 = 3/a

a2 = 36

a = + 6 or – 6?

a  ≠ 6, a = – 6

Hence, when a = -6 the given set of equations will have no solution.

Question 28. Find the value of a, for which the following system of equation have

(i) Unique solution

(ii) No solution

kx + 2y = 5

3x + y = 1

Solution: 

Given that,

kx + 2y − 5 = 0 ...(1)

3x + y − 1 = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = k, b1 = 2, c1 = −5 

a2 = 3, b2 = 1, c2 = −1

(i) For unique solution, we have

a1/a2 ≠ b1/b2

k/3 ≠ 2

k ≠ 6

Hence, when k ≠ 6 the given set of equations will have unique solution.

(ii) For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

k/3 = 2/1 ≠ -5/-1

k/3 = 2/1 

k = 6

Hence, when k = 6 the given set of equations will have no solution.

Question 29. For what value of c, the following system of equation have infinitely many solutions (where c ≠ 0 c ≠ 0):

6x + 3y = c − 3

12x + cy = c

Solution: 

Given that,

6x + 3y − (c − 3) = 0  ...(1)

12x + cy − c = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 6, b1 = 3, c1 = −(c − 3) 

a2 = 12, b2 = c, c2 = - c

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

6/12 = 3/c = -(c + 3)/-c

6/12 = 3/c and 3/c = -(c + 3)/-c 

c = 6 and c – 3 = 3

c = 6 and c = 6

Hence, when c = 6 the given set of equations will have infinitely many solution.

Question 30. Find the value of k, for which the following system of equation have

(i) Unique solution

(ii) No solution

(iii) Infinitely many solution

2x + ky = 1

3x − 5y = 7

Solution: 

Given that,

2x + ky = 1 ...(1)

3x − 5y = 7 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = k, c1 = −1 

a2 = 3, b2 = −5, c2 = −7

(i) For unique solution, we have

a1/a2 ≠ b1/b2

2/3 ≠ -k/-5 ≠ -10/3

Hence, when k ≠ -10/3 the given set of equations will have unique solution.

(ii) For no solution, we have

a1/a2 = b1/b2 ≠ c1/c2

2/3 = k/-5 ≠ -1/-7

2/3 = k/-5 and k/-5 ≠ -1/-7

k = -10/3 and k ≠ -5/7

k = -10/3

Hence, when k = -10/3 the given set of equations will have no solution.

(iii) For the given system to have infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/3 = k/-5 = -1/-7

Clearly a1/a2 ≠ c1/c2

Hence. there is no value of k for which the given set of equation has infinitely many solution.

Question 31. For what value of k, the following system of equation will represent the coincident lines:

x + 2y + 7 = 0

2x + ky + 14 = 0

Solution: 

Given that,

x + 2y + 7 = 0  ...(1)

2x + ky + 14 = 0  ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0   ...(3)

a2x + b2y − c2 = 0  ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 1, b1 = 2, c1 = 7 

a2 = 2, b2 = k, c2 = 14

The given system of equation will represent the coincident

lines if they have infinitely many solutions

So, 

a1/a2 = b1/b2 = c1/c2

1/2 = 2/k = 7/14

1/2 = 2/k = 7/14

k = 4

Hence, when k = 4 the given set of equations will have infinitely many solution.

Question 32. Find the value of k, for which the following system of equation have unique solution:

ax + by = c,

lx + my = n

Solution: 

Given that,

ax + by − c = 0 ...(1)

lx + my − n = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0 ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = a, b1 = b, c1 = − c

a2 = l, b2 = m, c2 = − n

For unique solution, we have

a1/a2 ≠ b1/b2

a/l ≠ b/m

am ≠ bl

Hence, when am ≠ bl the given set of equations will have unique solution.

Question 33. Find the value of a and b such that the following system of linear equation have infinitely many solutions:

(2a − 1)x + 3y − 5 = 0,

3x + (b − 1)y − 2 = 0

Solution: 

Given that, 

(2a − 1)x + 3y − 5 = 0 

3x + (b − 1)y − 2 = 0

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a2 = 3, b2 = b − 1, c2 = −2

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

(2a - 1)/3 = 3/(b - 1) = -5/-2

(2a - 1)/3 = 3/(b - 1) and 3/(b - 1) = -5/-2 

2(2a − 1) = 15 and 6 = 5(b − 1)

4a − 2 = 15 and 6 = 5b − 5 

4a = 17 and 5b = 11

So, a = 17/4 and b = 11/5

Question 34. Find the value of a and b such that the following system of linear equation have infinitely many solution:

2x − 3y = 7,

(a + b)x − (a + b − 3)y = 4a + b

Solution: 

Given that, 

2x − 3y − 7 = 0 ...(1)

(a + b)x − (a + b − 3)y − (4a + b) = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = −3, c1 = −7 

a2 = (a + b), b2 = −(a + b − 3), c2 = −(4a + b)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a + b) = -3/-(a + b - 3) = -7/-(4a + b)

2/(a + b) = -3/-(a + b - 3) and -3/-(a + b - 3) = -7/-(4a + b)

2(a + b − 3) = 3(a + b) and 3(4a + b) = 7(a + b − 3)

2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21

a + b = −6 and 5a − 4b = −21

a = − 6 − b

Substituting the value of a in 5a − 4b = −21, and we will get,

5( -b - 6) - 4b = -21

− 5b − 30 − 4b = − 21

9b = − 9 ⇒ b = −1

As a = – 6 – b

a = − 6 + 1 = − 5

So, a = – 5 and b = –1.

Question 35. Find the value of p and q such that the following system of linear equation have infinitely many solution:

2x − 3y = 9,

(p + q)x + (2p − q)y = 3(p + q + 1)

Solution: 

Given that, 

2x − 3y − 9 = 0 ...(1)

(p + q)x + (2p − q)y − 3(p + q + 1) = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = −9 

a2 = (p + q), b2 = (2p − q), c2 = -3(p + q + 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(p + q) = 3/(2p - q) = -9/-3(p + q + 1)

2/(p + q) = 3/(2p - q) and 3/(2p - q) = -9/-3(p + q + 1)

2(2p - q) = 3(p + q) and (p + q + 1) = 2p - q

4p - 2q = 3p + 3q and -p + 2q = -1

p = 5q and p - 2q = 1

Substituting the value of p in p - 2q = 1, we have

3q = 1

q = 1/3

Substituting the value of p in p = 5q we have

p = 5/3

So, p = 5/3 and q = 1/3.

Question 36. Find the values of a and b for which the following system of equation has infinitely many solution:

(i)  (2a − 1)x + 3y = 5,

3x + (b − 2)y = 3

(ii)  2x − (2a + 5)y = 5,

(2b + 1)x − 9y = 15

(iii) (a − 1)x + 3y = 2,

6x + (1 − 2b)y = 6

(iv) 3x + 4y = 12,

(a + b)x + 2(a − b)y = 5a – 1

(v) 2x + 3y = 7,

(a − 1)x + (a + 1)y = 3a − 1

(vi) 2x + 3y = 7,

(a − 1)x + (a + 2)y = 3a

(vii) 2x + 3y = 7,

(a - b)x + (a + b)y = 3a + b + 2

(viii) x + 2y = 1,

(a − b)x + (a + b)y = a + b - 2

(ix) 2x + 3y = 7,

2ax + a y = 28 - by

Solution: 

(i) Given that, 

(2a − 1)x + 3y − 5 = 0 ...(1)

3x + (b − 2)y − 3 = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2a − 1, b1 = 3, c1 = −5 

a2 = 3, b2 = b − 2, c2 = -3(p + q + 1)

For infinitely many solution, we have

a1/a2 = b1/b2 = c1/c2

(2a - 1)/3 = -3/(b - 2) = -5/-3

(2a - 1)/3 = -3/(b - 2) and -3/(b - 2) = -5/-3 

2a - 1 = 5 and - 9 = 5(b - 2)

a = 3 and -9 = 5b - 10  

a = 3 and b = 1/5

So, a = 3 and b = 1/5.

(ii) Given that, 

2x − (2a + 5)y = 5 ...(1)

(2b + 1)x − 9y = 15 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = - (2a + 5), c1 = −5 

a2 = (2b + 1), b2 = −9, c2 = −15

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(2b + 1) = (-2a + 5)/-9 = -5/-15

2/(2b + 1) = (-2a + 5)/-9 and (-2a + 5)/-9 = -5/-15

6 = 2b + 1 and 2a + 5

3 = b = 5/2 and a = -1 

So, a = - 1 and b = 5/2.

(iii) Given that, 

(a − 1)x + 3y = 2 ...(1)

6x + (1 − 2b)y = 6 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = a - 1, b1 = 3, c1 = −2 

a2 = 6, b2 = 1 − 2b, c2 = −6

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

(a - 1)/6 = 3/(1 - 2b) = 2/6

(a - 1)/6 = 3/(1 - 2b) and 3/(1 - 2b) = 2/6

a - 1 = 2 and 1 - 2b = 9

a - 1 = 2 and 1 - 2b = 9  

a = 3 and b = -4

a = 3 and b = -4

So, a = 3 and b = −4.

(iv) Given that, 

3x + 4y − 12 = 0 ...(1)

(a + b)x + 2(a − b)y − (5a − 1) = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 3, b1 = 4, c1 = −12 

a2 = (a + b), b2 = 2(a − b), c2 = – (5a − 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

3/(a + b) = 4/2(a - b) = 12/(5a - 1)

3/(a + b) = 4/2(a - b) and 4/2(a - b) = 12/(5a - 1)

3(a - b) = 2a + 2b and 2(5a - 1) = 12(a - b)

a = 5b and -2a = -12b + 2

On substituting a = 5b in -2a = -12b + 2, we have

-2(5b) = -12b + 2

−10b = −12b + 2 ⇒ b = 1

Thus a = 5

So, a = 5 and b = 1.

(v) Given that, 

2x + 3y − 7 = 0 ...(1)

(a − 1)x + (a + 1)y − (3a − 1) = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = −7 

a2 = (a − 1), b2 = (a + 1), c2 = - (3a − 1)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a - b) = 3/(a + 1) = -7/(3a - 1)

2/(a - b) = 3/(a + 1) and 3/(a + 1) = -7/(3a - 1)  

2(a + 1) = 3(a - 1) and 3(3a - 1) = 7(a + 1)

2a - 3a = -3 - 2 and 9a - 3 = 7a + 7

a = 5 and a = 5

So, a = 5 and b = 1.

(vi) Given that, 

2x + 3y − 7 = 0 ...(1)

(a − 1)x + (a + 2)y − 3a = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = −7 

a2 = (a − 1), b2 = (a + 2), c2 = −3a

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a - b) = 3/(a + 2) = -7/-3a

2/(a - b) = 3/(a + 2) and 3/(a + 2) = -7/-3a

2(a + 2) = 3(a - 1) and 3(3a) = 7(a + 2)

2a + 4 = 3a - 3 and 9a = 7a + 14

a = 7 and a = 7

So, a = 7 and b = 1.

(vii) 2x + 3y - 7 = 0,  ...(1)

(a - b)x + (a + b)y - 3a - b + 2 = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = −7 

a2 = (a − b), b2 = (a + b), c2 = −(3a + b - 2)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/(a - b) = 3/(a + b) = -7/−(3a + b - 2)

2/(a - b) = 3/(a + b) and 3/(a + b) = -7/−(3a + b - 2)

2(a + b) = 3(a - b) and 3(3a + b - 2) = 7(a + b)

2a + 2b = 3a - 3b and 9a + 3b - 6 = 7a + 7b

So, a = 5 and b = 1.

(viii) x + 2y - 1 = 0  ...(1)

(a − b)x + (a + b)y - a - b + 2 = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 1, b1 = 2, c1 = −1 

a2 = (a − b), b2 = (a + b), c2 = −(a + b - 2)

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

1/(a - b) = 2/(a + b) = -1/−(a + b - 2)

1/(a - b) = 2/(a + b) and 2/(a + b) = -1/−(a + b - 2)

(a + b) = 2(a - b) and 2(a + b - 2) = (a + b)

a + b = 2a - 2b and 2a + 2b - 4 = a + b

So, a = 3 and b = 1.

(ix) 2x + 3y - 7 = 0 ...(1)

2ax + ay - 28 + by = 0 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = 2, b1 = 3, c1 = −7 

a2 = 2a, b2 = (a + b), c2 = −28

For infinitely many solutions, we have

a1/a2 = b1/b2 = c1/c2

2/2a = 3/(a + b) = -7/−28

2/2a = 3/(a + b) and 3/(a + b) = 7/28

2(a + b) = 6a and 84 = 7(a + b)

So, a = 4 and b = 8.

Question 37. For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have

(i) no solution?

(ii) infinitely many solutions?

(iii) a unique solution? 

Solution: 

Given that, 

λx + y = λ2 ...(1)

x + λy = 1 ...(2)

So, the given equations are in the form of:

a1x + b1y − c1 = 0  ...(3)

a2x + b2y − c2 = 0 ...(4)

On comparing eq (1) with eq(3) and eq(2) with eq (4), we get

a1 = λ, b1 = 1, c1 = -λ2 

a2 = 1, b2 = λ, c2 = -λ2 

(i) For no solution,

a1/a2 = b1/b2 ≠ c1/c2

λ = 1/λ ≠ -λ2/-1

λ2 - 1 = 0

λ = 1, -1

Here we will take only λ = -1 because at λ = 1 the linear 

equation will have infinite many solutions.

(ii) For infinite many solutions,

 a1/a2 = b1/b2 = c1/c2

λ = 1/λ = λ2 /1

λ(λ - 1) = 0

When λ ≠ 0 then λ = 1

(iii) For Unique solution,

a1/a2 ≠ b1/b2

λ ≠ 1/λ


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    The Polynomials are fundamental algebraic expressions consisting of variables raised to the whole-number powers and combined using addition, subtraction, and multiplication. They form the basis of many concepts in algebra including solving equations, graphing functions, and mathematical modeling. In
    15+ min read
    Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.1 | Set 2
    In this section, we delve into Chapter 2 of the Class 10 RD Sharma textbook, focusing on Polynomials. Exercise 2.1 is designed to help students understand the fundamental concepts of polynomials, including their types, properties, and various operations performed on them.Class 10 RD Sharma Solutions
    12 min read
    Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.2
    In this article, we will explore the solutions to Exercise 2.2 from Chapter 2 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Polynomials". This chapter is fundamental for understanding the basics of polynomials their properties and how to manipulate them. By solving Exercise 2.2
    7 min read
    Class 10 RD Sharma Solutions - Chapter 2 Polynomials - Exercise 2.3
    Chapter 2 of RD Sharma's Class 10 Mathematics book delves into the polynomials a fundamental algebraic concept. This chapter introduces polynomials, their types and their properties laying the groundwork for the solving polynomial equations. Exercise 2.3 focuses on the solving polynomial equations u
    11 min read

    Chapter 3: Pair of Linear Equations in Two Variables

    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1
    Question 1. Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she h
    7 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 1
    Question 1. Solve the following of equation graphicallyx + y = 32x + 5y = 12 Solution: Given that, 2x + 5y = 12 and x + y = 3 We have, x + y = 3, When y = 0, we get x = 3 When x = 0, we get y = 3 So, the following table giving points on the line x + y = 3: x03y30 Now, 2 + 5y = 12 y = (12 - 2x)/5 Whe
    11 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 2
    In Chapter 3 of RD Sharma's Class 10 Mathematics, students explore the concept of solving pairs of linear equations in two variables. Exercise 3.2 | Set 2 presents various problems designed to reinforce these concepts through practical application. This article comprehensively addresses each problem
    14 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2 | Set 3
    Question 23. Solve the following system of linear equations graphically and shade the region between the two lines and the x-axis.(i) 2x + 3y = 12 and x - y = 1 Solution: Given that, 2x + 3y = 12 and x - y = 1 Now, 2x + 3y = 12 x = (12-3y)/2 When y = 2, we get x = 3 When y = 4, we get x = 0 So, the
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3 | Set 1
    Solve the following system of equations:Question 1. 11x + 15y + 23 = 0 and 7x – 2y – 20 = 0 Solution: 11x +15y + 23 = 0 …………………………. (i) 7x – 2y – 20 = 0 …………………………….. (ii) From (ii) 2y = 7x – 20 ⇒ y = (7x −20)/2 ……………………………… (iii) Putting y in (i) we get, ⇒ 11x + 15((7x−20) / 2) + 23 = 0 ⇒ 11x + (10
    9 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3 | Set 2
    Solve the following system of equations:Question 14. 0.5x + 0.7y = 0.74 and 0.3x + 0.5y = 0.5 Solution: 0.5x + 0.7y = 0.74……………………… (i) 0.3x – 0.5y = 0.5 ………………………….. (ii) Multiply LHS and RHS by 100 in (i) and (ii) 50x +70y = 74 ……………………….. (iii) 30x + 50y = 50 …………………………… (iv) From (iii) 50x = 74
    9 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4 | Set 1
    Solve each of the following systems of equations by the method of cross multiplication.Question 1. x + 2y + 1 = 0 and 2x – 3y – 12 = 0 Solution: Given that, x + 2y + 1 = 0 2x - 3y - 12 = 0 On comparing both the equation with the general form we get a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = −3, c2 = -12 N
    11 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4 | Set 2
    Chapter 3 of RD Sharma’s Class 10 Mathematics book delves into the topic of Pair of Linear Equations in Two Variables. This chapter is fundamental in understanding how to solve systems of linear equations and their applications. Exercise 3.4 | Set 2 provides the practice problems to strengthen stude
    14 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 1
    Question 1. In each of the following systems of equations determine whether the system has a unique solution, no solution, or infinite solutions. In case there is a unique solution, find it from 1 to 4:x − 3y − 3 = 0, 3x − 9y − 2 = 0.Solution: Given that,x − 3y − 3 = 0 ...(1) 3x − 9y − 2 = 0 ...(2)
    9 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 2
    Chapter 3 of RD Sharma's Class 10 Mathematics textbook deals with the "Pair of Linear Equations in the Two Variables." This chapter explores methods for solving systems of linear equations involving the two variables. Understanding these methods is crucial for solving the problems in algebra and geo
    11 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5 | Set 3
    Question 27. For what value of a, the following system of the equation has no solution:ax + 3y = a − 312x + ay = a Solution: Given that, ax + 3y = a − 3 ...(1) 12x + ay = a ...(2) So, the given equations are in the form of: a1x + b1y − c1 = 0 ...(3) a2x + b2y − c2 = 0 ...(4) On comparing eq (1) with
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6 | Set 1
    Question 1. 5 pens and 6 pencils together cost ₹ 9 and 3 pens and 2 pencils cost ₹ 5. Find the cost of 1 pen and 1 pencil. Solution: Let the cost of 1 pen be Rs. x and the cost of 1 pencil be Rs. y Now, 5 pens and 6 pencils cost Rs. 9 => 5x + 6y = 9 (1) And, 3 pens and 2 pencils cost Rs. 5 =>
    9 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6 | Set 2
    Linear equations in two variables form a critical part of algebraic problems in mathematics. These equations which represent lines on a graph provide the foundation for solving various real-world problems involving relationships between the two variables. Chapter 3 of RD Sharma's Class 10 mathematic
    7 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7
    In Class 10 mathematics, solving linear equations is a fundamental concept that underpins many advanced topics. Chapter 3 of RD Sharma's textbook titled "Pair of Linear Equations in Two Variables" explores methods to solve systems of equations involving the two variables. Understanding these concept
    14 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.8
    In this section, we explore Chapter 3 of the Class 10 RD Sharma textbook, which focuses on Pairs of Linear Equations in Two Variables. Exercise 3.8 is designed to help students understand the methods for solving pairs of linear equations, enhancing their problem-solving skills in algebra.Class 10 RD
    9 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.9
    In Class 10 Mathematics, Chapter 3 focuses on Pair of Linear Equations in Two Variables. This chapter deals with the solving systems of the equations where each equation is linear and has two variables. The importance of this topic lies in its applications in various fields such as economics, engine
    11 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.10
    Question 1. Points A and B are 70km. apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars. Solution: Let’s assume that th
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.11 | Set 1
    Question 1. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle. Solution: Let’s
    15 min read
    Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.11 | Set 2
    In this article, we will discuss the solutions to Exercise 3.11 Set 2 from Chapter 3 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Pair of Linear Equations in the Two Variables". This chapter is vital as it introduces students to methods for solving systems of linear equations a
    11 min read

    Chapter 4: Triangles

    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.1
    Question 1. (Fill in the blanks using the correct word given in brackets: (i) All circles are _____________ (congruent, similar).(ii) All squares are _____________ (similar, congruent).(iii) All _____________ triangles are similar (isosceles, equilaterals) :(iv) Two triangles are similar, if their c
    2 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.2
    Question 1: In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC. Solution: Given: Δ ABC where Length of side AD = 6 cm, DB = 9 cm, AE = 8 cm Also, DE ∥ BC, To find : Length of side AC By using Thales Theorem we get,
    15 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.3
    Problem 1: In a ∆ABC, AD is the bisector of ∠A, meeting side BC at D.(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC Solution: Given: Length of side BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm. To find: Length of side DC In Δ ABC, AD is the bisector of ∠A, meeting side BC at D. Since, AD is ∠A bi
    11 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.4
    Question 1. (i) In fig., if AB || CD, find the value of x. Solution: Given, AB∥ CD. To find the value of x. Now, AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other] ⇒ 4/ (4x – 2) = (x +1)/ (2x + 4) 4(2x + 4) = (4x – 2)(x +1) 8x + 16 = x(4x – 2) + 1(4x – 2) 8x + 16 = 4x2 – 2x + 4x – 2 -4
    3 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 1
    The Triangles are fundamental geometric shapes characterized by three sides, three angles and three vertices. They are categorized based on their side lengths and angles leading to the various types such as the equilateral, isosceles, and scalene triangles. The study of the triangles includes unders
    6 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 2
    Chapter 4 of the Class 10 RD Sharma Mathematics textbook, "Triangles," covers the properties and theorems related to triangles, including similarity, congruence, and the Pythagorean theorem. Exercise 4.5 focuses on applying these properties to solve problems related to triangles.RD Sharma Solutions
    10 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 1
    This exercise contains questions regarding the similarity of triangles. Questions in this exercise are based on the concepts of properties of two similar triangles. The RD Sharma Solutions Class 10 provides all solutions required for a quick preparation for exams. The following questions and their s
    7 min read
    Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 2
    Chapter 4 of RD Sharma's Class 10 mathematics book focuses on "Triangles" which is a fundamental concept in the geometry. This chapter deals with the various properties and theorems related to triangles such as the congruence similarity and Pythagorean theorem. Understanding these concepts is essent
    9 min read
    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 1
    Question 1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.Solution: According to the question The sides of triangle are: AB = 3 cm BC = 4 cm AC = 6 cm According to Pythagoras Theorem:  AB2 = 32 = 9 BC2 = 42 = 16 AC2 = 62 = 36 Sinc
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    Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 2
    Question 15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal. Solution: Given, In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles Each side = 10 cm and one diagonal AC = 16 cm ∴ AO = OC = 16/2 = 8 cm Now in ∆AOB, By us
    11 min read

    Chapter 5: Trigonometric Ratios

    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 1
    Chapter 5 of the Class 10 RD Sharma Mathematics textbook, "Trigonometric Ratios," introduces students to the basic trigonometric functions and their applications. Exercise 5.1 focuses on solving problems involving the fundamental trigonometric ratios of angles in a right-angled triangle.RD Sharma So
    9 min read
    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2 | Set 2
    Evaluate each of the following(14-19)Question 14. \frac{sin30°-sin90°+2cos0°}{tan30°tan60°} Solution: Given:\frac{sin30°-sin90°+2cos0°}{tan30°tan60°} -(1) Putting the values of sin 30° = 1/2, tan 30° = 1/√3, tan 60° = √3, sin 90° = cos 0° = 1 in eq(1) = \frac{(\frac{1}{2})-(1)+2(1)}{(\sqrt3)(\frac{1
    4 min read
    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.1 | Set 3
    Trigonometry is a fundamental branch of mathematics that deals with the relationships between the angles and sides of triangles. Chapter 5 of RD Sharma's Class 10 textbook focuses on the Trigonometric Ratios which are the foundation of trigonometry. In this chapter, students will learn to apply thes
    5 min read
    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.2 | Set 1
    Evaluate each of the following(1-13)Question 1. sin 45° sin 30° + cos 45° cos 30° Solution: Given: sin 45° sin 30° + cos 45° cos 30° -(1) Putting the values of sin 45° = cos 45°= 1/√2, sin 30° = 1/2, cos 30° = √3/2 in eq(1) = (1/√2)(1/2) + (1/√2})(√3/2) = (1/2√2) + (√3/2√2) = (1 + √3)/2√2 Question 2
    4 min read
    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.3 | Set 1
    Question 1. Evaluate the following:(i) sin 20°/cos 70° Solution: Given: sin 20°/cos 70° = sin(90° − 70°)/cos 70° = cos 70°/cos 70° -(∵ sin (90° - θ) = cos θ) = 1 Hence, sin 20°/cos 70° = 1 (ii) cos 19°/sin 71° Solution: Given: cos 19°/sin 71° = cos(90° − 71°)/sin 71° = sin 71°/sin 71° -(∵ cos (90° -
    8 min read
    Class 10 RD Sharma Solutions - Chapter 5 Trigonometric Ratios - Exercise 5.3 | Set 2
    Question 8. Prove the following:(i) sin θ sin (90° - θ) - cosθ cos (90° - θ) = 0 Solution: We have to prove that sin θ sin (90° - θ) - cosθ cos (90° - θ) = 0 Taking LHS = sin θ sin (90° - θ) - cosθ cos (90° - θ) -(∵ sin (90° - θ) = cos θ) = sin θ cosθ - cosθ sinθ = 0 LHS = RHS Hence proved (ii) \fra
    9 min read

    Chapter 6: Trigonometric Identities

    Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 1
    Prove the following trigonometric identities:Question 1. (1 – cos2 A) cosec2 A = 1 Solution: We have, L.H.S. = (1 – cos2 A) cosec2 A By using the identity, sin2 A + cos2 A = 1, we get, = (sin2 A) (cosec2 A) = sin2 A × (1/sin2 A) = 1 = R.H.S. Hence proved. Question 2. (1 + cot2 A) sin2 A = 1 Solution
    7 min read
    Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 2
    Prove the following trigonometric identities:Question 29. \frac{1+secθ}{secθ}=\frac{sin^2θ}{1-cosθ} Solution: We have, L.H.S. = \frac{1+secθ}{secθ} = \frac{1+\frac{1}{cosθ}}{\frac{1}{cosθ}} = \frac{cosθ+1}{cosθ}×{\frac{cosθ}{1}} = 1 + cos θ = \frac{(1+cosθ)(1-cosθ)}{1-cosθ} = \frac{1-cos^2θ}{1-cosθ}
    7 min read
    Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.1 | Set 3
    The Trigonometric identities are fundamental formulas in trigonometry that relate the angles and sides of the triangles. These identities simplify trigonometric expressions and solve the equations involving trigonometric functions. In Class 10 RD Sharma's Chapter 6 on the Trigonometric Identities Ex
    13 min read
    Class 10 RD Sharma Solutions - Chapter 6 Trigonometric Identities - Exercise 6.2
    In Chapter 6 of RD Sharma's Class 10 Mathematics textbook, we explore Trigonometric Identities a fundamental topic in trigonometry. Exercise 6.2 focuses on applying various trigonometric identities to solve problems and simplify expressions. Understanding these identities is crucial for solving comp
    8 min read

    Chapter 7: Statistics

    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 1
    Problem 1: Calculate the mean for the following distribution:x:56789f:4 814113 Solution: x ffx542068487 1498811889327 N = 40 ∑ fx = 281 We know that, Mean = ∑fx/ N = 281/40 = 7.025 Problem 2: Find the mean of the following data :x:19212325272931f:13151618161513 Solution: x ffx19132472115315231636825
    4 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 2
    Problem 11: Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.No. of heads per tossNo. of tosses 03811442
    3 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.2
    Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:Number of calls (x) 0 1 2 3 4 5 6 Number of intervals (f) 15 24 29 46 54 43 39 Compute the mean number of calls per interval. Solution: Let
    6 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1
    Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application
    8 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 2
    Question 14. Find the mean of the following frequency distribution:Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59Frequency:1422166534 Solution: Let’s consider the assumed mean (A) = 42 Class intervalMid – value xidi = xi – 42ui = (xi – 42)/5fifiui25 – 2927-15-314-4230 – 3432-10-222
    13 min read
    Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1
    In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various field
    8 min read
    Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2
    Question 11. An incomplete distribution is given below :Variable10-2020-3030-4040-5050-6060-7070-80Frequency1230-65-2518You are given that the median value is 46 and the total number of items is 230. (i) Using the median formula fill up missing frequencies. (ii) Calculate the AM of the completed dis
    9 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1
    Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10
    9 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2
    Question 11. Find the mean, median, and mode of the following data:Classes0-5050-100100-150150-200200-250250-300300-350Frequency2356531 Solution: Let mean (A) = 175 Classes Class Marks (x) Frequency (f) c.f. di = x - A A = 175 fi * di0-502522-150-30050-1007535-100-300100-150125510-50-250150-200175-A
    8 min read
    Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.6
    Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to
    8 min read

    Chapter 8: Quadratic Equations

    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.1
    Question 1. Which of the following are quadratic equations?(i) x2 + 6x - 4 = 0 Solution: As we know that, quadratic equation is in form of: ax2 + bx + c = 0 Here, given equation is x2 + 6x - 4 =0 in quadratic equation. So, It is a quadratic equations. (ii) √3x2 − 2x + ½= 0 Solution: As we know that,
    12 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.2
    Quadratic equations are a fundamental concept in algebra that involves a polynomial of degree two. They play a crucial role in various fields of mathematics, physics, engineering, and many other disciplines. Understanding quadratic equations is essential for solving problems related to motion, area,
    5 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1
    Solve the following quadratic equations by factorizationQuestion 1. (x - 4) (x + 2) = 0 Solution: We have equation, (x - 4) (x + 2) = 0 Implies that either x - 4 = 0 or x + 2 = 0 Therefore, roots of the equation are 4 or -2. Question 2. (2x + 3) (3x - 7) = 0 Solution: We have equation, (2x + 3) (3x
    8 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 2
    The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharma’s Class 10 Mathematics bo
    10 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.4
    Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 . Solution: Given: x^2-4\sqrt{2x}+6=0 We have to make the equation a perfect square. => x^2-4\sqrt{2x} = -6 => x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt
    5 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.5
    Question 1. Find the discriminant of the following quadratic equations:(i) 2x2 - 5x + 3 = 0 Solution: Given quadratic equation: 2x2 - 5x + 3 = 0 ....(1) As we know that the general form of quadratic equation is ax2 + bx + c = 0 ....(2) On comparing eq(1) and (2), we get Here, a = 2, b = -5 and c = 3
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 1
    Question 1. Determine the nature of the roots of following quadratic equations : (i) 2x² – 3x + 5 = 0 (ii) 2x² – 6x + 3 = 0 (iii) 3/5 x² – 2/3 x + 1 = 0 (iv) 3x² – 4√3 x + 4 = 0 (v) 3x² – 2√6 x + 2 = 0 Solution: (i) 2x² – 3x + 5 = 0 Here a=2, b=-3, c=5 D=b2-4ac=(-3)2-4*2*5 =-9-40=-31 D<0 Roots ar
    9 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 2
    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. Solution: 2x² + px – 15 = 0 -5 is its one root It will satisfy it 2(-5)2+p(-5)-15=0 2*25-5p-15=0 50-15-5p=0 -5p+35=0 -5p=-35 p=-35/-5=7 Now in
    6 min read
    Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations - Exercise 8.7 | Set 1
    Question 1: Find two consecutive numbers whose squares have the sum 85. Solution: Let first number is = x ⇒ Second number = (x+1) Now according to given condition— ⇒ Sum of squares of the numbers = 85 ⇒ x2 + (x+1)2 = 85 ⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2] ⇒ 2x2 + 2x + 1 - 85 =
    8 min read
    Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2
    Question 11: The sum of a number and its square is 63/4 , find the numbers. Solution: Let the number is = x so it's square is = x2 Now according to condition- ⇒(number)+(number)2=63/4 ⇒ x+x2 = 63/4 ⇒ x2+x-63/4=0 Multiplying by 4- ⇒ 4x2+4x-63=0 ⇒ 4x2 +(18-14)x - 63 =0 [because 63*4=252 so 18*14=252
    5 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.8
    Chapter 8 of RD Sharma's Class 10 textbook focuses on the Quadratic Equations an essential topic in mathematics that lays the foundation for understanding more advanced algebraic concepts. In this chapter, students will learn how to solve quadratic equations using various methods such as factorizati
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9
    The Quadratic equations are a fundamental topic in algebra often introduced at the Class 10 level. They are equations of the form ax2 +bx+c=0 where a, b, and c are constants and x is the variable. Understanding quadratic equations is crucial as they form the basis for many mathematical concepts and
    9 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.10
    Question 1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of other two sides of the triangle be x and y. Therefore, according to the question, x - y = 5 or, x =
    5 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.11
    Question 1: The perimeter of the rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle? Solution: Given: Perimeter = 82 m and its area = 400 m2 Let the breadth of the rectangle be 'b' m. As we know, Perimeter of a rectangle = 2×(length + breadth) 82 = 2×(length + b) 41
    6 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.12
    Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. Solution: Let us assume that B takes 'x' days to complete a piece of work. So, B’s 1 day work = 1/x Now, A takes
    3 min read
    Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.13
    Question 1. A piece of cloth costs ₹ 35. If the piece were 4 m longer and each meter costs ₹ 1 less, the cost would remain unchanged. How long is the piece? Solution: Let us considered the length of piece of cloth = x m Given: The total cost = ₹ 35 So, the cost of 1 m cloth = ₹ 35/x According to the
    11 min read

    Chapter 9: Arithmetic Progressions

    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progression Exercise 9.1
    Problem 1: Write the first terms of each of the following sequences whose nth term are:(i) an = 3n + 2 Solution: Given: an = 3n + 2 By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4)
    7 min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2
    Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (
    7 min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3
    Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solut
    10 min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 1
    In arithmetic progressions (AP), the terms follow a consistent pattern where each term is derived from the previous one by adding a fixed value, known as the common difference. To analyze various aspects of APs, such as finding specific terms or determining the number of terms, certain formulas and
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2
    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: Let’s assume that the first term and the common difference of the A.P to be a and d respectively. Given that, 4th term of an A.P.
    10 min read
    Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5
    Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4
    5 min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 1
    Class 10 RD Sharma Solutions are comprehensive guides to the exercises and problems presented in the RD Sharma Mathematics textbook, which is a popular reference for students studying mathematics at the 10th class level. These solutions provide step-by-step explanations and answers to the questions
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 2
    Question 25. In an A.P. the first term is 22, nth term is –11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 =>
    15+ min read
    Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3
    Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 =
    15+ min read
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