# R.D. Sharma Solutions Class 10th: Ch 3 Pair of Linear Equations in Two Variables MCQ's

## Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math MCQ's

**Multiple Choice Questions**

1. The value of k for which the system of equations has a unique solution, is

The given system of equations are

3. The value of k for which the system of equations x + 2y - 3 = 0 and 5x + ky + 7 = 0 has no solution, is

4. The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is

The given system of equations are,

5. If the system of equations has infinitely many solutions, then

For the equations to have infinite number of solutions , a

Let us take a

By cross multilication we get ,

Now take b

8. If the system of equations

The given system of equation is

10. If 2x – 3y = 7 and (a+b) x – (a+b-3)y = 4a + b represent coincident lines, then a and b satisfy the equation

12. The area of the triangle formed by the line x/a + y/b = 1 with the coordinate axes is

14. If the system of equations 2

The given system of equations

The given systems of equations are

Given, 2x + 3y = 12, x – y – 1 = 0 and x = 0

Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.

19. If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4 , then the values of a and b are respectively

The given equation are

The given system of equations is

21. Arun has only 1 and 2 coins with her . If the toatl number of coins that she has is 50 and the amonut of money with her is 75 , then the number of 1 and 2 coins are , respectively

Let the number of ₹1 coins be x and that of ₹2 coins be y.

kx − y = 2

6x − 2y = 3

(a) = 3

(b) ≠ 3

(c) ≠ 0

(d) = 0

**Solution**

The given system of equation are

kx – y = 2

6x – 2y = 3

a

_{1}/a_{2}≠ b_{1}/b_{2}for unique solution
Here a

_{1}= k, a_{2}= 6, b_{1}= -1, b_{2}= -2
k/6 ≠ -1/-2

By cross multiply we get

2k ≠ 6

k ≠ 6/2

k ≠ 3

Hence, the correct choice is b .

2. The value of k for which the system, of equations has infinite number of solutions, is

2x + 3y = 5

4x + ky = 10

(a) 1

(b) 3

(c) 6

(d) 0

**Solution**

The given system of equations are

2x + 3y = 5

4x + ky = 10

For the equation to have infinite number of solutions a

_{1}/a_{2}= b_{1}/b_{2}= c_{1}/c_{2}
Here, a

_{1}= 2, a_{2}= 4, b_{1}= 3, b_{2}= k
Therefore 2/4 = 3/k = 5/10

By cross multiplication of a

_{1},a_{2}= b_{1}/b_{2}we get ,
2/4 = 3/k

2k = 12

k = 12/2

k = 6

And

b

_{1}/b_{2}= c_{1}/c_{2}
3/k = 5/10

30 = 5k

30/5 = k

6 = k

Therefore the value of k is 6

Hence the correct choice is c .

3. The value of k for which the system of equations x + 2y - 3 = 0 and 5x + ky + 7 = 0 has no solution, is

(a) 10

(b) 6

(c) 3

(d) 1

**Solution**

The given system of equations are

x + 2y – 3 = 0

5x + ky + 7 = 0

For the equations to have no solutions,

a

_{1}/a_{2}= b_{1}/b_{2}≠ c_{1}/c_{2}
1/5 = 2/k = -3/7

If we take

a

_{1}a_{2}= b_{1}/b_{2}
1/5 = 2/k

Therefore te value of k is 10

Hence , correct choice is a .

4. The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is

(a) 0

(b) 2

(c) 6

(d) 8

**Solution**

The given system of equations are,

3x + 5y = 0

kx + 10y = 0

Here a

_{1}= 3, a_{2}= k, b_{1}= 5, b_{2}= 10
a

_{1}/a_{2}= b_{1}/b ≠ 0
3/k = 5/10 ≠ 0

By cross multiply we get

30 = 5k

30/5 = k

6 = k

Therefore the value of k is 6,

Hence , the correct choice is c .

5. If the system of equations has infinitely many solutions, then

2x + 3y = 7

(a + b)x + (2a − b)y = 21

(a) a = 1, b = 5

(b) a = 5, b = 1

(c) a = −1, b = 5

(d) a = 5, b = −1

**Solution**

The given systems of equations are

2x + 3y = 7

(a+b) x + (2a-b) y = 21

For the equations to have infinite number of solutions , a

_{1}/a

_{2}= b

_{1}/b

2 = c

_{1}/c_{2}
Here a

_{1}= 2, a_{2}= (a+b), b_{1}= 3, b_{2}= 2a – b, c_{1}= 7, c_{2}= 21,
2/a+b = 3/2a-b = 7/21

Let us take a

_{1}/a

_{2}= b

_{1}/b

_{2}

2/a+b = 3/2a-b

By cross multilication we get ,

2(2a – b) = 3(a+b)

4a – 2b = 3a + 3b

4a – 3a = 3b + 2b

a = 5b …(i)

Now take b

_{1}/b

_{2}= c

_{1}/c

_{2}

3/2a – b = 7/21

3/2a – b = 1/3

By cross multiplication we get,

3 × 3 = 1 × 2a – b

9 = 2a – b …(ii)

Substitute a = 5b in the above equation

9 = 2 × 5b – b

9 = 10b – b

9 = 9b

9/9 = b

1 = b

Substitute b = 1 in equation (i) we get , a = 5b

a = 5 × 1

a = 5

Therefore a = 5 and b = 1

Hence, the correct choice is b .

6. If the system of equations 3x + y = 1 and, (2k – 1) x + (k-1) y = 2k + 1 is inconsistent, then k = 1

(a) 1

(b) 0

(c) -1

(d) 2

**Solution**

The given system of equations is inconsistent,

3x + y = 1

(2k – 1) x + (k-1) y = 2k + 1

If the system of equations is in consistent, we have

a

_{1}b_{2}– a_{2}b_{1}= 0
3×(k-1)-1(2k-1) = 0

3k – 3 – 2k + 1 = 0

1k – 2 = 0

1k = 2

Therefore, the value of k is 2

Hence, the correct choice is d

7. If a

_{m}≠ b_{l}, then the system of equations ax + by = c and, lx + my = n
(a) has a unique solution

(b) has no solution

(c) has infinitely many solutions

(d) may or may not have a solution

**Solution**

Given a

_{m}≠ b_{l}, the system of equation has
ax + by = c

lx + my = n

We know that intersecting lines have unique solution a

_{1}/a_{2}≠ b_{1}/b_{2}
a

_{1}× b_{2}= a_{2}× b_{1}
Here a

_{1}= a, a_{2}= 1, b_{1}= b, b_{2}= m
a/l ≠ b/m

a × m ≠ l×b

Therefore intersecting lines, have unique solution

Hence, the correct choice is a .

8. If the system of equations

2x+ 3y = 7

2ax + (a+b)y = 28

has infinitely many solutions , then

(a) a = 2b

(b) b = 2a

(c) a + 2b = 0

(d) 2a + b = 0

**Solution**

Given the systems of equations are

2x + 3y = 7

2ax + (a+b)y = 28

For the equations to have infinite number of solutions ,

a

_{1}/a_{2}= b_{1}/b_{2}= c_{1}/c_{2}
a

_{1}= 2, a_{2}= 2a , b_{1}= 3, b_{2}= a + b
2/2a = 3/a+b = 7/28

By cross multiplication we have

2/2a = 3/a+b

2(a+b) = 2a(3)

2a + 2b = 6a

2b = 6a – 2a

2b = 4a

Divide both sides by 2. We get b = 21

Hence, the correct choice is b .

9. The value of

*k*for which the system of equations
x + 2y = 5

3x + ky + 15 = 0

has no solution is

(a) 6

(b) -6

(c) 3/2

(d) None of these

**Solution**

The given system of equation is

x + 2y = 5

3x + ky + 15 = 0

If a

_{1}a_{2}= b_{1/}b_{2}≠ c_{1}/c_{2}then the equation have no solution .
1/3 = 2/k = -5/15

By cross multiply we get

k × 1 = 3 × 2

k = 6

Hence, the correct choice is a .

10. If 2x – 3y = 7 and (a+b) x – (a+b-3)y = 4a + b represent coincident lines, then a and b satisfy the equation

(a) a+5b = 0

(b) 5a + b = 0

(c) a – 5b = 0

(d) 5a – b = 0

**Solution**

The given system of equations are

2x – 3y = 7

(a+b) x – (a+b-3)y = 4a + b

For coincident lines, infinite number of solution

a

_{1}a_{2}= b_{1}/b_{2}= c_{1}c_{2}
⇒ 2/(a+b) = -3/-(a+b-3) = 7/4a + b

⇒ 2/(a+b) = 3(a+b-3) = 7/(4a+b)

⇒ 2(a+b-3) = 3(a+b)

⇒ 2a + 2b – 6 = 3a + 3b

⇒ 2a + 2b – 3a – 3b = 6

⇒ - a – b = 6

⇒ a + b = -6 …(i)

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

⇒ 12a + 3b = 7a + 7b – 21

⇒ 5a – 4b = -21 …(ii)

Multiply equation (i) by 5, we get 5a + 5b = -30 …(iii)

Subtract (ii) from (iii),

(5a + 5b) – (5a – 4b) = -30 + 21

⇒ 5a + 5b – 5a + 4b = -9

⇒ 9b = -9

⇒ b = -1

Substitute b = -1 in equation (1)

a + (-1) = -6

⇒ a = -6 + 1 = -5

Option A :

a + 5b = 0

-5 + 5(-1) = -5-5 = -10 ≠ 0

Option B :

5a + b = 0

5(-5) + (-1) = - 25 – 1 = -26 ≠ 0

Option C :

a –b = 0

-5-(-1) = -4 ≠ 0

None of the option satisfies the values .

11. If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) intersecting

(b) parallel

(c) always coincident

(d) intersecting or coincident

**Solution**

If a pair of linear equations in two variables is consistent, then its solution exists.

∴The lines represented by the equations are either intersecting or coincident.

Hence, the correct choice is d .

12. The area of the triangle formed by the line x/a + y/b = 1 with the coordinate axes is

(a) ab

(b) 2ab

(c) 1/2 ab

(d) 1/4 ab

**Solution**

Given the area of the triangle formed by the line x/a + y/b = 1

If in the equation x/a + y/b = 1 either A and B approaches infinity, the line become parallel to either x axis or y axis respectively ,

Therefore

x = a;

y = b

Area of triangle = 1/2 × x × y = 1/2 × a × b

Hence , the correct choice is c .

13. The area of the triangle formed by the lines y = x, x = 6 and y = 0 is

(a) 36 sq. units

(b) 18 sq. units

(c) 9 sq. units

(d) 72 sq. units

**Solution**

Given x = 6 , y = 0 and x = y

We have plotting points as (6,0)(0,0)(6,6) when x = y

Therefore , area of Î”ABC = 1/2 (Base × Height) = 1/2 (CA × AB) = 1/2 (6×6) = 1/2 × 36 = 18

Area of triangle ABC is 18 square units

Hence, the correct choice is b .

14. If the system of equations 2

*x*+ 3

*y*= 5, 4

*x*+

*ky*= 10 has infinitely many solutions, then

*k*=

(a) 1

(b) 1/2

(c) 3

(d) 6

**Solution**

The given system of equations

2x + 3y = 5

4x + ky = 10

a

_{1}a_{2}= 2/4, b_{1}/b_{2}= 3/k , c_{1}c_{2}= 5/10
For the equations to have infinite number of solutions

a

_{1}a_{2}= b_{1}b_{2}= c_{1}c_{2}
2/4 = 3/k = 5/10

If we take

2/4 = 3/4

2k = 12

k = 12/2

k = 6

And

3/k = 5/10

30 = 5k

30/5 = k

6 = k

Therefore, the value of k is 6 .

Hence, the correct choice is d .

15. If the system of equations kx − 5y = 2, 6x + 2y = 7 has no solution, then k =

(a) −10

(b) −5

(c) −6

(d) −15

**Solution**

The given systems of equations are

kx – 5y = 2

6x + 2y = 7

a

_{1}a_{2}= b_{1}b_{2}≠ c_{1}/c_{2}
Here a

_{1}= k , a_{2}= 6, b_{1}= -5 , b_{2}= 2
k/6 = -5/2

2k = -30

k = -30/2

k = -15

Hence , the correct choice is d .

16. The area of the triangle formed by the lines

*x*= 3,*y*= 4 and*x*=*y*is
(a) 1/2 sq. unit

(b) 1 sq. unit

(c) 2 sq . unit

(d) None of these

**Solution**

Given x = 3, y = 4 and x = y

We have plotting points as (3,4)(3,3)(4,4) when x = y

Therefore, area of Î”ABC = 1/2 (Base × Height) = 1/2 (AB × AC) = 1/2 (1×1) = 1/2

Area of triangle ABC is 1/2 square units

Hence, the correct choice is a .

17. The area of the triangle formed by the lines 2

*x*+ 3*y*= 12,*x*−*y*− 1 = 0 and*x*= 0 (as shown in Fig. 3.23), is
(a) 7 sq. units

(b) 7.5 sq. units

(c) 6.5 sq. units

(d) 6 sq. units

**Solution**

Given, 2x + 3y = 12, x – y – 1 = 0 and x = 0

If x = 0 we have plotting points as D(0,-1) B(0,4) P(3,2)

Therefore, area of CBPD = 1/2 (Base × Height) = 1/2 (BP×PM) = 1/2 (5×3) = 1/2(15) = 7.5

Area of triangle ABC is 7.5 square units

Hence, the correct choice is b.

18. The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed . The number is

(a) 25

(b) 72

(c) 63

(d) 36

**Solution**

Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.

Now,

x + y = 9 .....(i)

Also,

10x + y + 27 = 10y + x

⇒ 9x − 9y = −27

⇒ x − y = −3 .....(ii)

Adding (i) and (ii), we get

2x = 6

⇒ x = 3

Putting x = 3 in (i), we get

3 + y = 9

⇒ y = 6

Thus, the required number is 10 × 3 + 6 = 36.

Hence, the correct answer is option (d).

19. If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4 , then the values of a and b are respectively

(a) 3 and 1

(b) 3 and 5

( c) 5 and 3

(d) -1 and -3

**Solution**

The given equation are

x – y = 2 ….(1)

x + y = 4 ….(2)

Adding (1) and (2) , we get

2x = 6

⇒ x = 3

Putting x = 3 in (1) , we get

3 + y = 4

⇒ y = 1

So, x = a = 3 and y = 1.

20. For what the value k, do the equation 3x – y + 8 = 0 and 6x – ky + 16 = represent coincident lines ?

**Solution**

The given system of equations is

3x – y + 8 = 0

6x – ky + 16 = 0

We know that the lines

a

_{1}x + b_{1}y + c_{1}= 0
a

_{2}x + b_{2}y + c_{2}= 0
are coinciden if

a

_{1}/a2 = b_{1}/b_{2}= c_{1/}c_{2}
∴ 3/6 = -1/-k = 8/16

⇒ 1/2 = 1/k = 1/2

⇒ k = 2

Thus, the value of k = 2.

Hence the orrect answer is option c.

21. Arun has only 1 and 2 coins with her . If the toatl number of coins that she has is 50 and the amonut of money with her is 75 , then the number of 1 and 2 coins are , respectively

(a) 35 and 15

(b) 35 and 20

(c) 15 and 35

(d) 25 and 25

**Solution**

Let the number of ₹1 coins be x and that of ₹2 coins be y.

Now,

Total number of coins = 50

So, x + y = 50 .....(i)

Also,

₹1 × x + ₹2 × y = ₹75

∴ x + 2y = 75 .....(ii)

Subtracting (i) from (ii), we get

y = 25

Putting y = 25 in (i), we get

x + 25 = 50

⇒ x = 25

So, the number of ₹1 coins and ₹2 coins are 25 and 25, respectively.

Hence, the correct answer is option (d).

**Note:**The answer given in the book does not match with the one obtained.