These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

Question 1.

The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

Let the age of Ani by x then the age of Biju is x – 3 and if the age of Ani’s father Dharam be y.

The age of Cathy = \(\frac { 1 }{ 2 }\) the age of Biju then the age of Cathy is \(\frac { x-3 }{ 2 }\).

According to question

The age of Ani is 19 years and the age of Biju = x – 3

the age of Biju = 19 – 3 = 16

Hence the age of Ani is 19 years and the age of Biju is 16 years.

Question 2.

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?

Solution:

Let the two friends have ₹ x and ₹ y.

According to the first condition:

One friend has an amount = ₹(x + 100)

Other has an amount = ₹ (y – 100

∴ (x + 100) =2 (y – 100)

⇒ x + 100 = 2y – 200

⇒ x – 2y = – 300 …(i)

According to the second condition:

One friend has an amount = ₹(x – 10)

Other friend has an amount =₹ (y + 10)

∴ 6(x – 10) = y + 10

⇒ 6x – 60 = y + 10

⇒ 6x-y = 70 …(ii)

Multiplying (ii) equation by 2 and subtracting the result from equation (i), we get:

x – 12x = – 300 – 140

⇒ -11x = -440

⇒ x = 40

Substituting x = 40 in equation (ii), we get

6 x 40 – y = 70

⇒ -y = 70- 24

⇒ y = 170

Thus, the two friends have ₹ 40 and ₹ 170.

Question 3.

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the original speed of the train be x km/h

and the time taken to complete the journey be y hours. ‘

Then the distance covered = xy km

Case I : When speed = (x + 10) km/h and time taken = (y – 2) h

Distance = (x + 10) (y – 2) km

⇒ xy = (x + 10) (y – 2)

⇒ 10y – 2x = 20

⇒ 5y – x = 10

⇒ -x + 5y = 10 …(i)

Case II : When speed = (x – 10) km/h and time taken = (y + 3) h

Distance = (x – 10) (y + 3) km

⇒ xy = (x – 10) (y + 3)

⇒ 3x- 10y = 30 …(ii)

Multiplying equation (i) by 3 and adding the result to equation (ii), we get

15y – 10y = 30

⇒ 5y = 60

⇒ y = 12

Putting y = 12 in equation (ii), we get

3x- 10 x 12= 30

⇒ 3x = 150

⇒ x = 50

∴ x = 50 and y = 12

Thus, original speed of train is 50 km/h and time taken by it is 12 h.

Distance covered by train = Speed x Time

= 50 x 12 = 600 km.

Question 4.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of rows be x and the number of students in each row be y.

Then the total number of students = xy

Case I : When there are 3 more students in each row

Then the number of students in a row = (y + 3)

and the number of rows = (x – 1)

Total number of students = (x – 1) (y + 3)

∴ (x – 1) (y + 3) = xy

⇒ 3x – y = 3 … (i)

Case II : When 3 students are removed from each row

Then the number of students in each row = (y-3)

and the number of rows = (x + 2)

Total number of students = (x + 2) (y – 3)

∴ (x + 2) (y – 3) = xy

⇒ – 3x + 2y = 6 … (ii)

Adding the equations (i) and (ii), we get

– y + 2y = 3 + 6

⇒ y = 9

Putting y = 9 in the equation (ii), we get

– 3x + 18 = 6

– 3x + 18 = 6

⇒ x = 4

∴ x = 4 and y = 9

Hence, the total number of students in the class is 9 x 4 = 36.

Question 5.

In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.

Solution:

Let ∠A = x° and ∠B = y°.

Then ∠C = 3∠B = (3y)°.

Now ∠A + ∠B + ∠C = 180°

⇒ x + y + 3y = 180°

⇒ x + 4y = 180° …(i)

Also, ∠C = 2(∠A + ∠B)

⇒ 3y – 2(x + y)

⇒ 2x – y = 0° …(ii)

Multiplying (ii) by 4 and adding the result to equation (i), we get:

9x = 180°

⇒ x = 20°

Putting x = 20 in equation (i), we get:

20 + 4y = 180°

⇒ 4y = 160°

⇒ y = \(\frac { 160 }{ 40 }\) = 40°

∴ ∠A = 20°, ∠B = 40° and ∠C = 3 x 40° = 120°.

Question 6.

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Solution:

We have

Hence the vertices of the triangle are (1,0), (0, – 3), (0, – 5).

Question 7.

Solve the following pairs of linear equations:

Solution:

(i) The given equations are

px + qy = p – q …(1)

qx – py = p + q …(2)

Multiplying equation (1) by p and equation (2) by q and then adding the results, we get:

x(p^{2} + q^{2})

x = 1

Putting this value in equation (1),

we get p + qy = p – q

qy = – q

y = – 1

(ii) We have

ax + by =c …(i)

bx + ay = 1 + c …(ii)

Multiply equation (i) by b and equation (ii) by a, we get

abx + b²y = bc … (iii)

abc + a²y = a + bc … (iv)

Subtract equation (ii) by equation (iii) we get

(b² – a²)y = bc – a – ac

(b² – a²)y = c(b – a) – a

(iii) We have

\(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 0 … (i)

ax + by = a² + b²

Multiply equation (i) by ab we get

bx – ay = 0 … (iii)

Multiply equation (iii) by b and equation (ii) by (a), we get

b²x – aby = 0 … (iv)

a²x + aby = a² + ab² … (v)

Adding equation (iv) and (v), we get

(a² + b²)x = a³ + ab²

(a² + b²)x = a(a² + b²)

x = a

Putting this value in equation (v), we get y = b

(iv) We have

(a – b) x + (a + b) y = a² – 2ab – b²

(a + b)(x + y) = a² + b²

The above system of equations can be written as

(a – b) x + (a + b) y – (a² – 2ab – b)² = 0

ax + ay + bx + by – (a² + b²) = 0

(a – b) x + (a + b) y – (a² – 2ab – b²) = 0

(a + b)x + (a + b)y – (a² + b²) = 0

Applying the cross multiplication method, we get

(v) The given equations may be written as:

76x – 189y = – 37 … (1)

– 189x + 76y = – 302 … (2)

Multiplying equation (1) by 76 and equation (2) by 189, we get:

5776x – 14364y = – 2812 … (3)

– 35721x + 14364y = -57078 … (4)

Adding equations (3) and (4), we get:

5776x – 35721x = – 2812 – 57078

⇒ – 29945x = – 59890

⇒ x = 2

Putting x = 2 in equation (1), we get:

76 x 2 – 189y = – 37

⇒ 152 – 189y = – 37

⇒ – 189y = – 189

⇒ y = 1

Thus, x = 2 and y = 1 is the required solution.

Question 8.

ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:

In a cyclic quadrilateral sum of the opposite angle is 180°,

So, ∠A + ∠C = 180°

4y + 20 – 4x = 180°

4y – 4x = 160° … (i)

∠B + ∠D= 180°

3y – 5 + 5 – 7x = 180° … (ii)

3y – 7x = 180°

Multiply equation (i) by 3 and equation (ii) by 4 then subtract equation (ii) from equation (i) we get,

16x = – 240

x = – 15

Putting this value ie equation we get

y = 25

∠A = 4y + 20

∠A = 120

∠B = 3y – 5

∠B = 70°

∠C = – 4x = 60°

∠D = 110°