Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Objective Questions

 

Algebraic Solution

1. 1. Half the perimeter of a rectangular room is 46 m, and its length is 6 m more than its breadth. What is the length and breadth of the room?
      1. 2m, 20m
      2. 2m, 3m
      3. 56m, 40m
      4. 26m, 20m

Answer: (D) 26m, 20m

Solution: Let l and b be the length and breadth of the room. Then, the perimeter of the room = 2(l+b) metres

From question, l=6+b… (1)

×2(l+b) =46⟹l+b=46… (2)

Using the Substitution method:

Substituting the value of l from (1) in (2), we get

6+b+b = 46

⇒ 6+2b=46⟹2b = 40

⟹b =20 m.

Thus, l = 26 m

2. 2. Solve the following pair of equations: 2x+y=73x+2y = 12. Choose the correct answer from the given options.
      1. (-3,2)
      2. (1,0)
      3. (3,2)
      4. (2,3)

Answer: (D) (2, 3)

Solution: We have,

2x+y=7    … (1)

3x+2y=12… (2)

Multiply equation (1) by 2, and we get:

2(2x+y) =2(7)

⇒4x+2y=14… (3)

Subtracting (2) from (3), we get,

x=2

Substituting the value of x in (1), we get,

2(2) +y=7⟹y = 3

Thus, the solution for the given pair of linear equations is (2, 3).

3. 3. Solve
      1. Y = 51/19
      2. X = 51/19
      3. Y = 94/57
      4. X = 117/54

Answer: (A) Y = 51/19

Solution: (3x/2) – (5y/3) =2

LCM of 2 and 3 is 6. Multiply by 6 on both sides

9x−10y=−12 ————- (1)

(X/2) + (y/2) = 13/6

LCM is 6, so multiply by 6 on both sides

3x+3y=13 ————–(2)

Multiply equation (2) by 3 to eliminate x, so we get,

9x+9y=39……………(3)

Subtract (3) from (1) we have

−19y=−51⇒y = 51/19

Substitute this in one of the equations, and we get

4. 4. Given: 3x–5y=4; 9x=2y+7

Solve the above equations by the elimination method and find the value of x.

4. 4. 1. X = 9/13
      2. Y = 5/13
      3. X = (-5) / 13
      4. Y = 9/13

Answer: (A) X = 9/13

Solution: Given:

3x–5y=4……(1)

9x=2y+7

9x−2y=7…..(2)

Multiply equation (1) by 3

⇒9x–15y=12…….(3)

Subtracting (2) from (3), we get,

−13y=5

Y= – ( (-5) / 13)

Substituting the value of y in (2)

All about Lines

5. 5. Choose the pair of equations which satisfy the point (1,-1)
      1. 4x–y = 3, 4x+y = 3
      2. 4x+y = 3, 3x+2y = 1
      3. 2x+3y = 5, 2x+3y = −1
      4. 2x+y = 3, 2x–y = 1

Answer: (B) 4x+y = 3, 3x+2y = 1

Solution: For a pair of equations to satisfy a point, the point should be their unique solution of them.

Solve the pair equations 4x+y=3,3x+2y = 1

let 4x+y=3…..(1)

and 3x+2y=1 …..(2)

y=3−4x   [ From (1)]

Substituting value if y in (2)

3x+2y = 1

3x+2(3−4x) = 1

3x+6−8x = 1

−5x=−5⇒x = 1

Substituting x = 1 in (1),

4(1)+y=3⇒y = −1

⇒ (1,-1) is the solution of pair of equations.

∴ The pair of equations which satisfy the point is (1,-1).

Note: We can also substitute the value (1,-1) in the given equations and check if it satisfies the pair of equations or not. In this case, it only satisfies the pair of equation 4x+y=3, 3x+2y=1, and hence (1, -1) is the unique solution of the equation.

6. 6. 54 is divided into two parts such that the sum of 10 times the first part and 22 times the second part is 780. What is the bigger part?
      1. 34
      2. 32
      3. 30
      4. 24

Answer: (A) 34

Solution: Let the 2 parts of 54 be x and y

x+y = 54…. (i)

And 10x + 22y = 780 ——————– (ii)

Multiply (i) by 10, and we get

Substituting y = 20 in x + y = 54, we have x + 20 = 54; x = 34

Hence, x = 34 and y = 20.

7. 7. What are the values of a, b and c for the equation y = 0.5x+√7 when written in the standard form: ax+by+c=0?
      1. 0.5, 1, √7
      2. 0.5, 1, – √7
      3. 0.5, -1, √7
      4. -0.5, 1, √7

Answer: (C) 0.5, -1, √7

Solution: Y= 0.5X + √7

⇒0.5x−y+ √7= 0

The general form of an equation is ax+by+c = 0.

Here, on comparing, we get

a = 0.5, b = −1 and c = √7

8. 8. Which of the following pairs of linear equations has infinite solutions?

Answer: (D) x+2y = 7; 3x+6y = 21

Solution: If two equations are consistent and overlapping, they will have infinite solutions. Option A consists of two equations where the second equation can be reduced to an equation which is the same as the first equation.

x+2y=7…. (i)

3x+6y=21….. (ii)

Dividing equation (ii) by 3, we get

x+2y=7 which is the same as equation (i).

The equations coincide and will have an infinite solution.

Alternate Method:

Let the two equations be

Basics Revisited

9. 9. Which of the following points lie on the line 7x+8y=61?
      1. (3, 4)
      2. (2, 5)
      3. (-3, 7)
      4. (3, 5)

Answer: (D) (3, 5)

Solution: Substituting the value of x = 3 and y = 4,

7x+8y = 61 = 7(3) +8(4) = 53.

Substituting the value of x = 2 and y = 5,

7x+8y=61 = 7(2) +8(5) = 54.

Substituting the value of x = -3 and y = 7,

7x+8y=61 = 7(−3) +8(7) =35.

Substituting the value of x = 3 and y = 5,

7x+8y=61 = 7(3) +8(5) =61

Hence, (3, 5) lies on the given line

10. 10. Which of these following equations have x = −3, y = 2 as solutions?
        1. 3x−2y = 0
        2. 3x+2y = 0
        3. 2x+3y = 0
        4. 2x−3y = 0

Answer: (C) 2x+3y = 0

Solution: Substituting the values in LHS,

L.H.S = 2x+3y

L.H.S = 2(−3) +3(2)

L.H.S = 0 = R.H.S

Hence, x = −3, y = 2 is the solution of the equation 2x+3y = 0

11. 11. If y = 1/2 (3x+7) is rewritten in the form ax+by+c = 0, what are the values of a, b and c?
        1. ½, 7/2, 3/2
        2. 7, 2, 3
        3. -2, 3, -7
        4. -3, 2,-7

Answer: (D) -3, 2,-7

Solution: The given equation is:

y = 1/2 (3x+7)

Simplifying the equation, we get the following:

2y−3x−7=0

⇒−3x+2y−7=0   (1)

Thus, the value of a, b and c is -3, 2 and -7, respectively.

The equation can also be written as,

3x−2y+7=0

Thus, the value of a, b and c is +3, -2 and +7, respectively.

Option D, -3, 2 and -7, is correct [Since +3, -2 and +7 is not an option]

12. 12. x−y=0 is a line:
        1. Passing through origin
        2. Passing through (1,-1)
        3. || to the y-axis
        4. || to the x-axis

Answer: (A) Passing through origin

Solution: x−y = 0, is a line passing through the origin as point (0, 0) satisfies the given equation

Graphical Solution

13. 13. 

Statement 1: This is the condition for inconsistent equations

Statement 2: There exist infinitely many solutions

Statement 3: The equations satisfying the condition are parallel

Which of the above statements is true?

13. 13. 1. S1 only
        2. S1 and S2
        3. S1 and S3
        4. S2 only

Answer: (C) S1 and S3

The condition is for inconsistent pair of equations which are parallel and have no solution.

∴ Statements 1 and 3 are correct.

14. 14. 

Choose the correct statement.

14. 14. 1. For k ≠ 3(m/5), a unique solution exists.
        2. For k = 3(m/5), infinitely many solutions exist
        3. For k =  5( m/3), a unique solution exists
        4. Fork = 5(m/3), infinitely many solutions exist

Answer: (D) For k = 5(m/3), infinitely many solutions exist

15. 15. The figure shows the graphical representation of a pair of linear equations. On the basis of the graph, the pair of linear equations gives ____________ solutions.

15. 15. 1. Four
        2. Only one
        3. Infinite
        4. Zero

Answer: (B) Only one

Solution: If the graph of linear equations represented by the lines intersects at a point, this point gives the unique solution. Here, the lines meet at the point (1,-1), which is the unique solution of the given pair of linear equations.

16. 16. For what value of k do the pair of linear equations 3x+ky = 9 and 6x+4y = 18 have infinitely many solutions?
        1. -5
        2. 6
        3. 1
        4. 2

Answer: (D) 2

Solution: Given equations gives infinitely many solutions if,

The given linear equations are:

3x+ky=9; 6x+4y=18.

⇒ 3/6 = k/4 = (-9)/ (-18)

⇒ ½ = k/4

⇒ k = 2

Solving Linear Equations

17. 17. Solve the following pairs of linear equation

17. 17. 1. 9, 8
        2. 4, 9
        3. 3, 2
        4. ½, 1/3

Answer: (B) 4, 9

Solution: The pair of equations is not linear. We will substitute 1/x as u2 and 1/y as v2 then we will get the equation as

2u+3v=2

4u−9v=−1

We will use the method of elimination to solve the equation.

Multiply the first equation by 3, and we get

6u+9v=6

4u−9v=−1

Adding the above two equations

10u=5

u= ½

Substituting u in equation 4u−9v=−1 we get v= 1/3

18. 18. Solve the following pairs of the equation

(7x-2y) / xy=5

(8x+6y) /xy =15

18. 18. 1. None of these
        2. 2, not defined
        3. 5/2, not defined
        4. (-2) /5, not defined

Answer: (A) None of these

Solution:

19. 19. 
        1. x = -2,  y = 2
        2. x = 7, y = -8
        3. x = 0, y = 8
        4. x = -1/5, y = 19/7

Answer: (D) x = -1/5, y = 19/7

20. Solve the following pairs of equations:

(1/x) + (3/y) =1

(6/x)- 12/y = 2

20. 20. 1. x = 5/3 , y = 15/2
        2. x = 4, y = 9
        3. x = 3, y = 11
        4. x = ¾ , y = 7/3

Answer: (A) x = 5/3, y = 15/2

Solution: Let 1/x = a and 1/y = b

(As x ≠ 0, y ≠ 0)

Then, the given equations become

a+3b = 1 … (1)

6a−12b = 2 … (2)

Multiplying equation (1) by 4, we get 4a+12b = 4 … (3)

On adding equation (2) and equation (3), we get 10a = 6

⇒a = 3/5

Putting a = 3/5 in equation (1), we get

(3/5) + 3b =1

⇒b= 2/15

Hence, x = 5/3 and y = 15/2