Linear Equations in two variables Class 10 MCQ with Solutions Maths

Linear Equations in two variables Class 10 MCQ with Solutions

1. If x^y= y^x, Then ${\left( {\frac{x}{y}} \right)^{x/y}}$ is equal to :

(A) x^{x/y + 1}

(B) x^1/y

(D) x^x/y–1

Ans. (D)

Sol. Given x^y = y^x

Now ${\left( {\frac{x}{y}} \right)^{x/y}}$ = $\frac{{{x^{x/y}}}}{{{{({y^x})}^{1/y}}}} = \frac{{{x^{x/y}}}}{{{{({x^y})}^{1/y}}}}$ (As x^y = y^x)

= $\frac{{{x^{x/y}}}}{{{x^1}}} = {x^{\frac{x}{y}\, - \,1}}$

2. If a + b = 5 and ab = 2, then a⁴ + b⁴ = :

(A) 433

(B) 437

(D) None of these

Ans. (A)

Sol. Given a + b = 5 & ab = 2

Now a⁴ + b⁴ = (a² + b²)² – 2a²b²

= [(a + b)² – 2ab]² – 2a²b²

= [5² – 2 × 2]² – 2(2)²

= (25 – 4)² – 8

= 21² – 8 = 441 – 8 = 433

3. If p + q = 4 and p² – q²= 1, then p – q = :

(A) 4

(B) $\frac{1}{4}$

(D) None of these

Ans. (B)

Sol. Given p + q = 4, p²– q² = 1

Now (p – q) (p + q) = p² – q² = 1

⇒ p –q = $\frac{1}{{p + q}} = \frac{1}{4}$

4. If x + y = 5 and x² + y² = 13, then x – y = :

(A) 2

(B) 1

(D) 1 or – 1

Ans. (D)

Sol. Given x + y = 5, x²+ y² = 13 & x – y = ?

Now x² + y² = (x + y)² – 2xy

⇒ 13 = 25 – 2xy

⇒ 2xy = 12

xy = 6

∴ (x – y)² = (x + y)² – 4xy

= 25 – 4 × 6 = 1

5. Which of the following system of equations has infinitely many solutions :

(A) 5x – 4y = 20, 7.5x – 6y = 30

(B) 2x – 3y = 5, 3x – 4.5y = 7.5

(D) All of the above

Ans. (B)

Sol. As 2x – 3y = 5

3x – 4.5y = 7.5

Now $\frac{2}{3}$ = $\frac{{ - 3}}{{ - 4.5}} = \frac{5}{{7.5}} = \frac{1}{{1.5}} = k$ (constant)

Hence above two equs. are conicidents lines.

6. For what value of k, the system of equations kx – y = 2, 6x – 2y = 3 has infinitely many solutions :

(A) k = 3

(B) k $\ne$ 4

(D) Does not exist

Ans. (D)

Sol. For infinitely in any solutions

$\frac{k}{6}$ = $\frac{{ - 1}}{{ - 2}} \ne \frac{2}{3}$

⇒ k = 3, k = 4 two different values coming.

and $\frac{{ - 1}}{{ - 2}} \ne \frac{2}{3}$ Hence not possible.

7. The condition to represent two non parallel lines by a system of linear equations given by a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 will be :

(A) $\frac{{{a_1}}}{{{a_2}}}$ = $\frac{{{b_1}}}{{{b_2}}}$ = $\frac{{{c_1}}}{{{c_2}}}$

(B) $\frac{{{a_1}}}{{{a_2}}}$ $\ne$ $\frac{{{b_1}}}{{{b_2}}}$

(D) None of these

Ans. (B)

Sol. For non-parallel lines we know

$\frac{{{a_1}}}{{{a_2}}}$ $\ne$ $\frac{{{b_1}}}{{{b_2}}}$ i.e. unique solution.

8. For what value of k will the equations 2x + ky = 1 and 5x – 7y = 5 represent two non parallel lines :

(A) $-\frac{{14}}{5}$

(B) $\frac{{14}}{5}$

(D) All real numbers except $-\frac{{14}}{5}$

Ans. (D)

Sol. For non parallel lines

$\frac{2}{5}$ ≠ $\frac{k}{{ - 7}}$

⇒ k ≠ – Hence all value except $-\frac{{14}}{5}$

9. The lines 3x – 2y = 1, 4x + 3y = 7 and x = 1 intersect in :

(A) One point

(B) Two points

(D) Not any point

Ans. (A)

Sol. 3x – 2y = 1 …(1)

4x + 3y = 7 …(2)

⇒

Hence solution is same as (x = 1)

10. Linear equation 2x + 3 = 7 is :

(A) Parallel to x – axis

(B) Parallel to y – axis

(D) None of these

Ans. (B)

Sol. Here 2 x = 7 – 3

⇒ 2 x = 4

⇒ (x = 2)

i.e. parallel to y-axis.

11. The solution of the system of equations $\frac{4}{x} + 5y$ = 7 and $\frac{3}{x} + 4y$ = 5 is :

(A) x = – $\frac{1}{3}$ , y = – 1

(B) x = $\frac{1}{3}$ , y = – 1

(D) x = $\frac{1}{3}$ , y = 1

Ans. (B)

Sol. $\frac{4}{x} + 5y$ = 7 …… (1)

$\frac{3}{x} + 4y$ = 5 …… (2)

on slving (1) & (2)

x = $\frac{1}{3}$ , y = – 1

12. The equations $\frac{{x + y}}{{xy}}$ = 2 and $\frac{{x - y}}{{xy}}$ = 6 are satisfied if :

(A) x = $\frac{1}{4}$ – , y = $\frac{1}{2}$

(B) x = – $\frac{1}{2}$ , y = $\frac{1}{4}$

(D) x = $\frac{1}{4}$ , y = $\frac{1}{2}$

Ans. (B)

Sol. $\frac{{x + y}}{{xy}}$ = 2 …… (1)

$\frac{{x - y}}{{xy}}$ = 6 …… (2)

on solving (1) & (2)

x = – $\frac{1}{2}$ , y = $\frac{1}{4}$

13. The equations 3x – 4y = 5 and 12x – 16y = 20 have :

(A) More than two solutions

(B) Exactly two solutions

(D) No solution

Ans. (A)

Sol. 3x – 4y = 5

12x – 16y = 20

$\frac{3}{{12}}$ = $\frac{{ - 4}}{{ - 16}} = \frac{5}{{20}}$

So, they have infinite solution.

14. The condition for which the system of equations kx – y = 2 and 6x – 2y = 3 has a unique solution :

(A) k = 0

(B) k $\ne$ 0

(D) k $\ne$ 3

Ans. (D)

Sol. kx – y = 2

6x – 2y = 3

condition for unique solution

$\frac{k}{6}$ $\ne$ $\frac{{ - 1}}{{ - 2}}$

$\Rightarrow$ k $\ne$ 3

15. The solution of the system of equations 2x + 3y + 5 = 0 and 3x – 2y – 12 = 0 is :

(A) x = –3, y = 2

(B) x = 2, y = –3

(D) x = –2, y = 3

Ans. (B)

Sol. 2x + 3y + 5 = 0 …… (1)

3x – 2y – 12 = 0 …… (2)

on solving (1) & (2)

x = 2, y = – 3

16. The solution of the system of equations 2x – 3y + 4xy = 0 and 6x + 5y – 2xy = 0 is :

(A) x = 0, y = 0

(B) x = 1, y = –2

(D) None of these

Ans. (C)

Sol. 2x – 3y + 4xy = 0 …… (1)

6x + 5y – 2xy = 0 …… (2)

on solving

x = 0, y = 0

or x = 1, y = – 2

17. Determine the value of K for which the following pair of equations has no solutions

(3k + 1) x + 11y – 22 = 0

x + (2k – 1) y – 4 = 0

(A) $\frac{2}{3}$

(B) $- \frac{4}{3}$

(D) None of these

Ans. (B)

Sol. (3k + 1) x + 11y – 22 = 0

x + (2k – 1) y – 4 = 0

for no solution

$\frac{{3k + 1}}{1}$ = $\frac{{11}}{{2k - 1}} \ne \frac{{ - 22}}{{ - 4}}$

so, (3k + 1) (2k – 1) = 11

$\Rightarrow$ 6k² – k – 1 = 11

6k² – k – 12 = 0

6k² – 9k + 8k – 12 = 0

3k (2k – 3) + 4 (2k – 3) = 0

(3k + 4) (2k – 3) = 0

k = $- \frac{4}{3}$

k = $\frac{3}{2}$

but k $\ne$ $\frac{3}{2}$ because if k = $\frac{3}{2}$ , eq. have infinite solution.

So, k = $- \frac{4}{3}$

18. Find the value of x if y = 7 in given equation 4x + 5y = 7 :

(A) –3

(B) –7

(D) +7

Ans. (B)

Sol. 7x + 5y = 7

4x + 5.7 = 7

x = – 7

19. Find the value of x in terms of y in given equation $3x\, - \,y\, + \,7\, = \,0$ :

(A) $\frac{{y\, - \,7}}{3}$

(B) $\frac{{y\, + \,7}}{3}$

(D) $\frac{{ - y\, - \,7}}{3}$

Ans. (A)

Sol. 3x – y + 7 = 0

x = $\frac{{y\, - \,7}}{3}$

20. Find the value of y in terms of x in given equation 8x – 7y = 12 :

(A) $\frac{{8x\, + \,12}}{7}$

(B) $\frac{{8x\, - \,12}}{7}$

(D) $\frac{{ - 8x\, - \,12}}{7}$

Ans. (B)

Sol. 8x – 7y = 12

y = $\frac{{8x\, - \,12}}{7}$

21. Find the value of x in given equation $\frac{{3x\, + \,y}}{3}\,\, = \,\,\frac{{ - y\, + \,7\, + \,7x}}{8}$ :

(A) $\frac{{21\,y\, + \,11\,y}}{3}$

(B) $\frac{{21\,y\, - \,11\,}}{3}$

(D) $\frac{{21\,y\, + \,11\,}}{3}$

Ans. (C)

Sol. $\frac{{3x + y}}{3}$ = $\frac{{ - y + 7 + 7x}}{8}$

x = $\frac{{ - 11y + 21}}{3}$

22. Which of the following system of equations has no solution :

(A) 3x – y = 2, 9x – 3y = 6

(B) 4x – 7y + 28 = 0, 5x – 7y + 9 = 0

(D) None of these

Ans. (C)

Sol. 3x – 5y – 11 = 0

6x – 10y – 7 = 0

$\frac{3}{6}$ = $\frac{{ - 5}}{{ - 10}} \ne \frac{{ - 11}}{{ - 7}}$

they are having no solution.

23. Find the value of k for which, the following system of equations has infinitely many solutions :

2x + 3y – 5 = 0

6x + ky – 15 = 0

(A) 6

(B) 7

(D) – 9

Ans. (C)

Sol. 2x + 3y – 5 = 0

6x + ky – 15 = 0

for infinite solution.

$\frac{2}{6}$ = $\frac{3}{k}$ = $\frac{{ - 5}}{{ - 15}}$

so, k = 9

24. Find the value of k for which the following system of equations has infinitely many solutions :

4x + 5y = 3

kx + 15y = 9

(A) 12

(B) – 12

(D) – 11

Ans. (A)

Sol. 4x + 5y = 3

kx + 15y = 9

for infinite solution.

$\frac{4}{k}$ = $\frac{5}{{15}} = \frac{3}{9}$

so, k = 12

25. Find the value of y :

x + 2y = 7

–2x + y = 9

(A) $\frac{{23}}{5}$

(B) – $\frac{{23}}{5}$

(D) $\frac{9}{5}$

Ans. (A)

Sol. x + 2y = 7 …… (1)

–2x + y = 9 …… (2)

on solving (1) & (2)

y = $\frac{{23}}{5}$

26. The coordinates of the point where the lines 3x + 2y = 4, 6x – y = 10 meet the y-axis :

(A) (0, 2) (0, –10)

(B) (–5, 0) (–10, 0)

(D) (0, –5) (0, –10)

Ans. (A)

Sol. 3x + 2y = 4

putting x = 0

0 + 2y = 4

y = 2

so, (0, 2)

and 6x – y = 10

putting x = 0

y = – 10

point (0, –10)

27. The coordinates of the point where the line 2(x – 3) = y – 8, meet the x-axis :

(A) (3, 0)

(B) (2, 0)

(D) None of these

Ans. (C)

Sol. 2(x – 3) = y – 8

putting y = 0

2(x – 3) = – 8

x = – 1

so, point = (–1, 0)

28. The coordinates of the point where the line 2x + 3y = 4y + 2, meet the y-axis :

(A) (0, –2)

(B) (0, –1)

(D) None of these

Ans. (A)

Sol. 2x + 3y = 4y + 2

putting x = 0

0 + 3y = 4y + 2

y = – 2

so point (0, –2)

29. If the lines represented by the system of equations 3x + y = 1, (2k – 1)x + (k – 1)y = 2k + 1, are parallel then k = :

(A) 2

(B) – 2

(D) – 1

Ans. (A)

Sol. 3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

$\frac{3}{{2k - 1}}$ = $\frac{1}{{k - 1}} \ne \frac{1}{{2k + 1}}$

so, 3(k – 1) = 2k – 1

k = 2

30. For what value of k will the equations x + 2y + 7 = 0, 2x + ky + 14 = 0 represent coincident lines :

(A) 1

(B) 2

(D) 4

Ans. (D)

Sol. x + 2y + 7 = 0

2x + ky + 14 = 0

for co-incident lines

$\frac{1}{2}$ = $\frac{2}{k}$ = $\frac{7}{{14}}$

k = 4

31. The sum of two numbers is 58. The greater number exceeds twice the smaller number by 1, then smaller number is :

(A) 39

(B) 19

(D) 9

Ans. (B)

Sol. x + y = 58 …… (1)

x – 2y = 1 …… (2)

subtracting (2) from (1)

3y = 57

y = 19

32. The sum of the present ages of Ram and his mother is 89 years. After 11 years, mother’s age will be twice Ram’s age, then present age of Ram is (in years) :

(A) 23

(B) 24

(D) 26

Ans. (D)

Sol. Let x = Ram’s age,

y = his mother’s age

x + y = 89 …… (1)

2[x + 11] = y + 11 …… (2)

on solving (1) & (2)

x = 26

33. The perimeter of a rectangle is 44 cm. Its length exceeds twice its breadth by 4 cm, then area of the rectangle is :

(A) 80 cm²

(B) 96 cm²

(D) 102 cm²

Ans. (B)

Sol. 2(x + y) = 44 …… (1)

x = 2y + 4 …… (2)

solving (1) & (2) x = 16, y = 6

so, area = x×y = 16 × 6

= 96 cm²

34. Divide 62 into two parts such that fourth part of the first and two fifth part of the second are in the ratio 2 : 3

(A) 24, 38

(B) 32, 30

(D) 40, 22

Ans. (B)

Sol. x + y = 62 …… (1)

$\frac{{\frac{x}{4}}}{{\frac{2}{5}y}}$ = $\frac{2}{3}$ …… (2)

on solving (1) & (2)

x = 32, y = 30

35. If (5, k) is solution of 2x + y – 6 = 0, then the value of k is equal to :

(A) 6

(B) 4

(D) None of these

Ans. (D)

Sol. 2x + y – 6 = 0

2 × 5 + k – 6 = 0

k = – 4

36. If a + b = 5 and 3a + 2b = 20, then 3a + b will be :

(A) 25

(B) 20

(D) 10

Ans. (A)

Sol. a + b = 5 …… (1)

3a + 2b = 20 …… (2)

on solving (1) & (2)

a = 10, b = – 5

so, 3a + b = 3 × 10 + (–5)

= 25

37. Which of the following respective values of x and y satisfy the following equations I and II?

3x + y = 19

x – y = 9

(A) 7, 2

(B) 7, – 2

(D) – 7, – 2

Ans. (B)

Sol. 3x + y = 19

x – y = 9

on solving x = 7, y = – 2

38. If 3x – 5 y = 5 and $\frac{x}{{x + y}}$ = $\frac{5}{7}$ , then what is the value of x – y?

(A) 9

(B) 6

(D) 3

Ans. (D)

Sol. 3x – 5y = 5 …… (1)

$\frac{x}{{x + y}}$ = $\frac{5}{7}$

7x = 5x + 5y

2x = 5y

2x – 5y = 0 …… (2)

3x – 5y = 5

subtracting (2) from (1)

x = 5

so from (2) 2 × 5 – 5y = 0

y = 2

so, x – y = 5 – 2

= 3

39. If 3y + 2x = 47 and 11x = 7y, then what is the value of y – x ?

(A) 4

(B) 5

(D) 7

Ans. (A)

Sol. 3y + 2x = 47 …… (1)

11x = 7y …… (2)

on solving (1) & (2)

x = 7, y = 11

so, y – x = 11 – 7

= 4

40. The solution of the simultaneous equations $\frac{1}{2}x + \frac{1}{3}y$ = 2 and x + y = 1 is :

(A) x = 1, y = 0

(B) x = 0, y = 1

(D) x = $\frac{2}{3}$ , y = $\frac{3}{2}$

Ans. (C)

Sol. $\frac{1}{2}x + \frac{1}{3}y$ = 2 …… (1)

x + y = 1 …… (2)

on solving (1) & (2)

x = 10, y = – 9

41. If 4x + 6y = 32 and 4x – 2y = 4, then the value of 8y is :

(A) 24

(B) 28

(D) 42

Ans. (B)

Sol. 4x + 6y = 32 …… (1)

4x – 2y = 4 …… (2)

subtracting (2) from (1)

8y = 28

42. If 2x + 3y = 29 and y = x + 3, what is the value of x ?

(A) 4

(B) 5

(D) 7

Ans. (A)

Sol. 2x + 3y = 29 …… (1)

y = x + 3 …… (2)

on solving (1) & (2)

x = 4

43. If $\frac{x}{4} + \frac{y}{3}$ = $\frac{5}{{12}}$ , $\frac{x}{2} + y$ = 1, then the value of x + y is :

(A) $\frac{1}{2}$

(B) 1

(D) 2

Ans. (C)

Sol. $\frac{x}{4} + \frac{y}{3}$ = $\frac{5}{{12}}$ …… (1)

$\frac{x}{2} + y$ = 1 …… (2)

on solving (1) & (2)

x = 1, y = $\frac{1}{2}$

so, x + y = $1 + \frac{1}{2}$

= $\frac{3}{2}$

44. The solution of the two simultaneous equation 2x + y = 8 and 3y = 4 + 4x is :

(A) x = 3, y = – 4

(B) x = 1, y = 4

(D) x = 4, y = 1

Ans. (C)

Sol. 2x + y = 8 …… (1)

3y = 4 + 4x …… (2)

from (1) and (2)

x = 2, y = 4

45. The solution of the simultaneous equations $\frac{x}{2} + \frac{y}{3}$ = 4 and x + y = 10 is given by :

(A) (– 6, 4)

(B) (6, – 4)

(D) (6, 4)

Ans. (C)

Sol. $\frac{x}{2} + \frac{y}{3}$ = 4 …… (1)

x + y = 10 …… (2)

from (1) & (2)

x = 4, y = 6

46. If x + y = 6 and 3x – y = 4, then x – y is equal to :

(A) – 1

(B) 0

(D) 4

Ans. (A)

Sol. x + y = 6 …… (1)

3x – y = 4 …… (2)

from (1) and (2)

x = $\frac{5}{2}$ , y = $\frac{7}{2}$

so, x – y = $\frac{5}{2} - \frac{7}{2}$ = – 1

47. The solution of 2x + 3y = 2 and 3x + 2y = 2 can be represented by a point in the co-ordinate planes in :

(A) First quadrant

(B) Second quadrant

(D) Fourth quadrant

Ans. (A)

Sol. 2x + 3y = 2 …… (1)

3x + 2y = 2 …… (2)

from (1) & (2)

x = $\frac{2}{5}$ , y = $\frac{2}{5}$

so, point will be in first quadrant.

48. Values of x and y satisfying the equations 2x – $\frac{3}{y}$ = 9 and 3x + $\frac{7}{y}$ = 2 are :

(A) x = 3, y = – 1

(B) x = 0, y = $\frac{1}{3}$

(D) x = 0, y = 3 $\frac{1}{2}$

Ans. (A)

Sol. 2x – $\frac{3}{y}$ = 9 …… (1)

3x + $\frac{7}{y}$ = 2 …… (2)

from (1) & (2)

x = 3, y = – 1

49. The solution of the system of equations given below is

\frac{x}{2} + \frac{y}{9}

= 11,

\frac{x}{3} + \frac{y}{6}

= 9

(A) 9, 9

(B) 18, 9

(D) 36, 9

Ans. (C)

Sol. $\frac{x}{2} + \frac{y}{9}$ = 11 …… (1)

$\frac{x}{3} + \frac{y}{6}$ = 9 …… (2)

on solving x = 18, y = 18

50. The system of equations x + 2y = 3 and 2x + 4y = 3 has :

(A) Exactly two solutions

(B) No solution

(D) A unique solution

Ans. (B)

Sol. x + 2y = 3 …… (1)

2x + 4y = 3 …… (2)

$\frac{1}{2}$ = $\frac{2}{4} \ne \frac{3}{3}$

so, lines are parallel, so No solution.

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