Pair Of Linear Equations In Two Variables| Worksheet Solutions| Class 10

Worksheet solutions for chapter: Pair Of Linear Equations In Two Variables for class 10 are provided below.

 

(1) The cost of the 2kg apple and 1kg orange is Rs 200. After a month, the cost of 2 kg apple and 4 kg oranges is Rs 400. Represent the situation algebraically and geometrically.

Ans-

Let cost of 1kg apple = Rs x

Let cost of 1kg orange = Rs y

Condition –

The cost of the 2kg apple and 1kg orange is Rs 200.

2x + y = RS 200

y = 200 – 2x…(1)

The cost of 2 kg apple and 4 kg oranges is Rs 400.

2x + 4y= 400…(2)

Put value of y in equation (2)

2x + 4(200 – 2x) = 400

2x + 800 – 8x = 400

-6x = 400 – 800

-6x = -400

Hence, cost of 1 kg apple and 1 kg orange is Rs 66.66

 

(2) Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:

x + y = 8

2x + 2y = 12

Ans-

From equations

a1 = 1, b1 = 1 and c1 = -8

a2 = 2, b2 = 2 and c2 = -12

Those pairs which do not have a common solution are known as inconsistent pair of linear equations.

Graphically-

Solutions for equation (1), ie, x + y – 8 = 0

x 0 8
y 8 0

 

Solutions for equation (2), ie, 2x + 2y -12 = 0

x 0 6
y 6 0

 

Both lines are parallel.

 

(3) The difference between two numbers is 36 and one number is four times the other. Find them.

Ans-

Let x and y are two numbers.

x – y = 36 (From conditions)…(1)

x = 4y (from conditions)…(2)

Put value of x from equation 2 to equation (1)

x – y = 36

4y – y = 36

3y = 36

y = 12

Put value of y in equation (2)

x = 4(12)

= 48

Hence, value of x is 48 and value of y is 12.

 

(4) Solve equation by substitution method –

x + 2y = 8

5x + 6y = 60

Ans-

x + 2y = 8…(1)

5x + 6y = 60…(2)

From equation (1)

x = 8 – 2y

Put value of x in equation (2)

5(8 – 2y) + 6y = 60

40 – 10y + 6y = 60

-4y = 20

y = -5

Put value of y in equation (1)

x + 2(-5) = 8

x – 10 = 8

x = 18

Hence, value of x = 18 and y = -5.

 

(5) Ten years ago, Neena was twice as old as Rina. Five years later, Neena will be half as old as Rina. How old are Neena and Rina?

Ans-

Let age of Neena = x years

Let age of Rina = y years

Ten years ago, age of Neena = x – 10 years

Ten years ago, age of Rina = y – 10 years

Condition-

Ten years ago, Neena was twice as old as Rina.

x – 10 = 2(y – 10)

x – 10 = 2y – 20

x – 2y = – 10 …(1)

Five years later, age of Neena = x + 5

Five years later, age of Rina = y + 5

Five years later, Neena will be half as old as Rina

x + 5 =

2x + 10 = y + 5

2x – y = 5 – 10

2x – y = -5 …(2)

From equation (1)

x = -10 + 2y

Put value of x in equation (2)

2(-10 + 2y) – y = -5

-20 + 4y –  y = -5

3y = – 5 + 20

3y = 15

y = 5

Put value of y in equation (1)

x – 2(5) = -10

x – 10 = -10

x = 0

Hence, value of x = o and y = 5.

 

(6) Solve the equation by cross – multiplication method –

4x + 5y = 9

2x + 3y = 10

Ans-

4x + 5y – 9 =0

2x + 3y -10 = 0

a1 = 4, b1 = 5, c1 = -9

a2 = 2, b2 =3, c2 = – 10

(7) Ritu can row downstream 10 km in 2 hours, and upstream 2 km in 2 hours. Find her speed or rowing in still water and the speed of the current.

Ans-

Let, speed of the boat in still water = x km/h

Let, speed of the stream = y km/h

Therefore,

Speed of the boat downstream = x + y km/h

Speed of the boat upstream = x – y km/h

x – y = 1 ….(2)

Subtract equation (2) from equation (1)

(x + y) – (x – y) = 5 – 1

x + y – x + y = 4

2y = 4

y = 2

Put value of y in equation (1)

x + y = 5

x + 2 = 5

x = 3

Hence, speed of the boat in still water = 3 km/h

Speed of the current  = 2km /h

 

(8) 4 women and 5 men can together finish an embroidery work in 6 days, while 5 women and 3 men can finish work in 1 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Ans-

Let time taken by 1 woman = x days

Let time taken by 1 man = y days

Condition-

4 women and 5 men can together finish an embroidery work in 2 days

4x + 5y = 6..(1)

5 women and 3 men can finish work in 5 days.

5x + 3y = 1…(2)

Multiply equation (1) by 5

5(4x) + 5(5y) = 5(6)

20x + 25y = 30…(3)

Multiply equation (2) by 4

4(5x) + 4(3y) = 4(1)

20x + 12y = 4…(4)

Subtract equation 4 from equation 3

20x + 25y – (20x + 12y )= 30 – 4

20x + 25y – 20x – 12y = -10

13y = 26

y = 2 days

Put value of y in equation (1)

4x + 5y = 2

4x + 5(2) =2

4x = 2 + 10

4x = 12

x = 3

Hence, time taken by 1 woman is  3 days and I man is 2 days.

 

5p + 2q = 40…(3)

Equation (2) can be written as

p + q = 10..(4)

q = 10 – p  {From equation 4)

Put value of q in equation (3)

5p + 2(10 – p) = 40

5p + 20 – 2p = 40

3p = 20

(10) Solve the equation by elimination method-

8x + 10 y = 12

x + y = 10

Ans-

8x + 10 y = 12..(1)

x + y = 10..(2)

Multiply equation (2) by 8

8x + 8y = 80…(3)

Subtract equation (3) from equation (1)

8x + 10y –(8x + 8 y) = 80 – 12

8x + 10 y – 8x – 8y = 68

2y = 68

y = 34

Put value of y in equation(2)

x + 34= 10

x = -24

Hence, value of x is – 24 and y is 10.

 

Helping Topics

Pair of linear equations in two variables

NCERT Solutions Class 10

Worksheet Class 10