## 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, isÂ

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 â€“ 3y = 7 and (a+b) x â€“ (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.Â If the system of equations 2

The given system of equations

The given systems of equations areÂ

Given, 2x + 3y = 12, x â€“ y â€“ 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 â€“ y = 2 and x + y = 4 , then the values of a and b are respectively

The given equation are

The given system of equationsÂ Â 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 â‚¹1 coins be x and that of â‚¹2 coins be y.

kx âˆ’ y = 2

6x âˆ’ 2y = 3

(a) = 3

(b) â‰ 3

(c) â‰ 0

(d) = 0

**Solution**

The given system of equation areÂ

kx â€“ y = 2Â

6x â€“ 2y = 3Â

a

_{1}/a_{2}Â â‰ b_{1}/b_{2}Â for unique solutionÂ
Here a

_{1}Â = k, a_{2}Â = 6, b_{1}Â = -1, b_{2}Â = -2Â
k/6 â‰ -1/-2Â

By cross multiply we getÂ

2k â‰ 6Â

k â‰ 6/2Â

k â‰ 3Â

Hence, the correct choice is b .Â

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}Â = b_{1}/b_{2}Â = c_{1}/c_{2}
Here, a

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

By cross multiplication of a

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

2k = 12

k = 12/2

k = 6

And

b

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

30 = 5k

30/5 = k

6 = k

Therefore the value of k is 6

Hence the correct choice is c .Â

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

(a) 10

(b) 6

(c) 3

(d) 1

**Solution**

The given system of equations are

x + 2y â€“ 3 = 0Â

5x + ky + 7 = 0Â

For the equations to have no solutions,Â

a

_{1}/a_{2}Â = b_{1}/b_{2}Â â‰ c_{1}/c_{2}Â
1/5 = 2/k = -3/7Â

If we takeÂ

a

_{1}a_{2}Â = b_{1}/b_{2}Â
1/5 = 2/kÂ

Therefore te value of k is 10Â

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) 8Â

**Solution**

The given system of equations are,

3x + 5y = 0

kx + 10y = 0

Here a

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

_{1}/a_{2}Â = b_{1}/b â‰ 0
3/k = 5/10 â‰ 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 .Â

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

2x + 3y = 7

(a + b)x + (2a âˆ’ b)y = 21

(a) a = 1, b = 5

(b) a = 5, b = 1

(c) a = âˆ’1, b = 5

(d) a = 5, b = âˆ’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}Â = b

_{1}/b

2 = c

_{1}/c_{2}
Here a

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

Let us take a

_{1}/a

_{2}Â = b

_{1}/b

_{2}

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

By cross multilication we get ,

2(2a â€“ b) = 3(a+b)

4a â€“ 2b = 3a + 3b

4a â€“ 3a = 3b + 2b

a = 5b â€¦(i)

Now take b

_{1}/b

_{2}Â = c

_{1}/c

_{2}

3/2a â€“ b = 7/21

3/2a â€“ b = 1/3

By cross multiplication we get,

3 Ã— 3 = 1 Ã— 2a â€“ b

9 = 2a â€“ b â€¦(ii)

Substitute a = 5b in the above equation

9 = 2 Ã— 5b â€“ b

9 = 10b â€“ b

9 = 9b

9/9 = b

1 = b

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

a = 5 Ã— 1

a = 5

Therefore a = 5 and b = 1

Hence, the correct choice is b .

6. If the system of equations 3x + yÂ Â = 1 and, (2k â€“ 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 â€“ 1) x + (k-1) y = 2k + 1

If the system of equations is in consistent, we have

a

_{1}b_{2}Â â€“ a_{2}b_{1}Â = 0
3Ã—(k-1)-1(2k-1) = 0

3k â€“ 3 â€“ 2k + 1 = 0

1k â€“ 2 = 0

1k = 2

Therefore, the value of k is 2

Hence, the correct choice is dÂ Â Â

7. If a

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

lx + my = n

We know that intersecting lines have unique solution a

_{1}/a_{2}Â â‰ b_{1}/b_{2}
a

_{1}Â Ã— b_{2}Â = a_{2}Â Ã— b_{1}
Here a

_{1}Â = a, a_{2}Â = 1, b_{1}Â = b, b_{2}Â = m
a/l â‰ b/m

a Ã— m â‰ lÃ—b

Therefore intersectingÂ Â 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}Â = b_{1}/b_{2}Â = c_{1}/c_{2}
a

_{1}Â = 2, a_{2}Â = 2a , b_{1}Â = 3, b_{2}Â = 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 â€“ 2a

2b = 4a

Divide both sides by 2. We get b = 21

Hence, the correct choice is b .

9. The value ofÂ

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

3x + ky + 15 = 0

has no solution is

(a) 6

(b) -6

(c) 3/2

(d) None of theseÂ

**Solution**

The given system of equation is

x + 2y = 5

3x + ky + 15 = 0

If a

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

By cross multiply we get

k Ã— 1 = 3 Ã— 2

k = 6

Hence, the correct choiceÂ Â is a .Â

10. If 2x â€“ 3y = 7 and (a+b) x â€“ (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)Â Â a â€“ 5b = 0

(d) 5a â€“ b = 0

**Solution**

The given system of equations are

2x â€“ 3y = 7

(a+b) x â€“ (a+b-3)y = 4a + b

For coincident lines, infinite number of solution

a

_{1}a_{2}Â = b_{1}/b_{2}Â = c_{1}c_{2}
â‡’Â 2/(a+b) = -3/-(a+b-3) = 7/4a + b

â‡’ 2/(a+b) = 3(a+b-3) = 7/(4a+b)

â‡’ 2(a+b-3) = 3(a+b)

â‡’ 2a + 2b â€“ 6 = 3a + 3b

â‡’ 2a + 2b â€“ 3a â€“ 3b = 6

â‡’ - a â€“ b = 6

â‡’ a + b = -6 â€¦(i)

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

â‡’ 12a + 3b = 7a + 7b â€“ 21

â‡’ 5a â€“ 4b = -21 â€¦(ii)

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

Subtract (ii) from (iii),

(5a + 5b) â€“ (5a â€“ 4b) = -30 + 21

â‡’ 5a + 5b â€“ 5a + 4b = -9

â‡’ 9b = -9

â‡’ b = -1

Substitute b = -1 in equation (1)

a + (-1) = -6

â‡’ a = -6 + 1 = -5

Option A :

a + 5b = 0

-5 + 5(-1) = -5-5 = -10 â‰ 0

Â Option B :

5a + b = 0

5(-5) + (-1) = - 25 â€“ 1 = -26 â‰ 0

Option C :

a â€“b = 0

-5-(-1) = -4 â‰ 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.

âˆ´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**Â

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 Ã— x Ã— yÂ = 1/2 Ã— a Ã— b

Hence , the correct choice is c .Â Â Â Â Â

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 Ã— Height) = 1/2 (CA Ã— AB) = 1/2 (6Ã—6) = 1/2 Ã— 36 = 18

Area of triangle ABC is 18 square units

Hence, the correct choice is b .

14.Â If the system of equations 2

*x*Â + 3

*y*Â = 5, 4

*x*Â +Â

*ky*Â = 10 has infinitely many solutions, thenÂ

*k*Â =

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

a

_{1}a_{2}Â = b_{1}b_{2}Â = 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,Â Â theÂ Â correct choice is d .

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

(a) âˆ’10

(b) âˆ’5

(c) âˆ’6

(d) âˆ’15

**Solution**

The given systems of equations areÂ

kx â€“ 5y = 2Â

6x + 2y = 7Â

a

_{1}a_{2}Â = b_{1}b_{2}Â â‰ c_{1}/c_{2}Â
Here a

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

2k = -30Â

k = -30/2Â

k = -15Â

Hence , the correct choice is d .

16. The area of the triangle formed by the linesÂ

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

(b) 1 sq. unit

(c) 2 sq . unit

(d) None of these

**Solution**

GivenÂ Â x = 3, y = 4 and x = y

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

Therefore,Â Â area ofÂ Â Î”ABC = 1/2 (Base Ã— Height) = 1/2 (AB Ã— AC) = 1/2 (1Ã—1) = 1/2

Area of triangle ABC is 1/2 square unitsÂ

Hence, the correct choice is a .

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

*x*Â + 3*y*Â = 12,Â*x*Â âˆ’Â*y*Â âˆ’ 1 = 0 andÂ*x*Â = 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. unitsÂ

**Solution**

Given, 2x + 3y = 12, x â€“ y â€“ 1 = 0 and x = 0

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

Therefore, area of CBPD = 1/2 (Base Ã— Height) = 1/2 (BPÃ—PM) = 1/2 (5Ã—3) = 1/2(15) = 7.5

Area of triangle ABC is 7.5 square units

Hence, the correct choice is b.Â

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 Â Â Â Â .....(i)

Also,

10x + y + 27 = 10y + x

â‡’Â 9xÂ âˆ’ 9y = âˆ’27

â‡’Â x âˆ’ y =Â âˆ’3 Â Â Â Â Â Â Â .....(ii)

Adding (i) and (ii), we get

2x = 6

â‡’Â x = 3

Putting x = 3 in (i), we get

3 + y = 9

â‡’Â y = 6

Thus, the required number is 10 Ã— 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 â€“ 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 -3Â

**Solution**

The given equation are

x â€“ y = 2 â€¦.(1)

x + y = 4 â€¦.(2)

Adding (1) and (2) , we get

2x = 6

â‡’ x = 3

Putting x = 3 in (1) , we get

3 + y = 4

â‡’ y = 1

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

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

**Solution**

The given system of equationsÂ Â is

3x â€“ y + 8 = 0

6x â€“ ky + 16 = 0

We know that the lines

a

_{1}x + b_{1}y + c_{1}Â = 0
a

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

a

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

â‡’ 1/2 = 1/k = 1/2

â‡’ k = 2

Thus, the value of k = 2.

Hence the orrect answer is option c.Â

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 â‚¹1 coins be x and that of â‚¹2 coins be y.

Now,

Total number of coins = 50

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

Also,

â‚¹1 Ã— x + â‚¹2 Ã— y = â‚¹75

âˆ´ x + 2y = 75 .....(ii)

Subtracting (i) from (ii), we get

y = 25

Putting y = 25 in (i), we get

x + 25 = 50

â‡’ x = 25

So, the number of â‚¹1 coins and â‚¹2 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.