These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 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 3.6

Question 1.

Solve the following pairs of equations by reducing them to a pair of linear equations:

Solution:

**
**By cross multiplication method

**By cross multiplication method**

**By cross multiplication method**

**By cross multiplication method**

**solving for u and v by cross multiplication method:**

**By cross multiplication method:**

**By cross multiplication method:**

(viii) We have

(i) and (iii), we get

P + q = – \(\frac { 3 }{ 4 }\) … (iii)

\(\frac { 1 }{ 2 }\)P – \(\frac { 1 }{ 2q }\) = \(\frac { -1 }{ 8 }\) … (iv)

By eliminating method Adding equations (iii) and (iv) we get

q = \(\frac { 1 }{ 2 }\)

But P = \(\frac { 1 }{ 3x+y }\)

and q = \(\frac { 1 }{ 3x – y }\)

\(\frac { 1 }{ 3x+y }\) = \(\frac { 1 }{ 4 }\)

and

3x + y = 4 … (v)

3x – y = 2 … (vi)

By eliminating method

Adding equations (v) and (vi) we get

6x = 6

x = 1

Question 2.

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row’ downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

(i) Let Ritu’s speed in still water = x km/h

Speed of current = y km/h

During downstream, speed = (x + y) km/h

During upstream, speed = (x – y) km/h

In the first case

The time taken in hour be f, then

By eliminating method

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

2x = 12

x = 6

Putting this value in equation (ii), we get

We get

– y = 2 – 6

y = 4

Then x = 6 and y = 4 where x and y are respectively speed (in km/h) of rowing and current.

(ii) Let the number of days taken by one woman by n and the number of days taken by one man by m.

According to question

\(\frac { 2 }{ n }\) + \(\frac { 5 }{ m }\) = \(\frac { 1 }{ 4 }\) … (i)

\(\frac { 3 }{ n }\) + \(\frac { 6 }{ m }\) = \(\frac { 1 }{ 3 }\) … (i)

Putting \(\frac { 1 }{ n }\) = p and \(\frac { 1 }{ m }\) = q in equations (i) and (ii) we get

2p + 5q = \(\frac { 1 }{ 4 }\) … (iii)

3p + 6q = \(\frac { 1 }{ 3 }\) … (iv)

or 8p + 20q – 1 = 0 … (iii)

9p + 18q -1 = 0 … (iv)

By cross multiplication method

(iii) Let the speed of the train be u km/h

and the speed of the bus be v km/h

In the first case

\(\frac { 60 }{ u }\) + \(\frac { 240 }{ v }\) = 4 … (i)

In the second case

\(\frac { 100 }{ u }\) + \(\frac { 200 }{ v }\) = \(\frac { 25 }{ 6 }\) … (ii)

Putting \(\frac { 1 }{ u }\) = p and \(\frac { 1 }{ v }\) = q in equations (i) and (ii)

we get

60p + 240q = 4 … (iii)

100p + 200p = \(\frac { 25 }{ 6 }\) … (iv)

or 60p + 240p – 4 = 0 … (v)

600p + 120q – 25 = 0 … (vi)

By using elemination method multiply equation (ii) by 10 we get

600p + 2400 – 40 = 0 … (vii)

Substract equation (iv) from equations (v)

1200q = 15

1200q = 15

q = \(\frac { 15 }{ 12000 }\)

q = \(\frac { 1 }{ 80 }\)

Putting this value in equations (iv)

P = \(\frac { 1 }{ 60 }\)

But p = \(\frac { 1 }{ u }\) and q = \(\frac { 1 }{ v }\)

\(\frac { 1 }{ u }\) = \(\frac { 1 }{ 60 }\) and \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 80 }\)

u = 60 and v = 80