Class 10 Mathematics

# Linear Equations

## Exercise 3.6 (NCERT) Part 3

Question 2: Formulate the following problems as a pair of equations, and hence find their solutions.

(a) 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.

Solution: Let us assume the speed of boat in still water is x and speed of current is y

Hence, downstream speed = x + y and upstream speed = x – y

First condition: Using time = text(distance)/text(speed)

(20)/(x+y)=2

Or, 2x+2y=20
Or, x+y=10

Second Condition: Using time = text(distance)/text(speed)

(4)/(x-y)=2

Or, x-y=2

Adding first and second equation, we get;

x + y + x – y = 10 + 2
Or, 2x = 12
Or, x = 6
Hence, y = 4

Speed of boat = 6 km/h and speed of current = 4 km/h

(b) 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.

Solution: First condition: One day’s work done by 2 women and 5 men;

2w+5m=1/4

Or, 8w+20m=1

Second condition: One day’s work done by 3 women and 6 men;

3w+6m=1/3

Or, 9w+18m=1

From first and second equation, it is clear;

8w + 20m = 9w + 18m
Or, 20m – 18m = 9w – 8w
Or, 2m = w

This means that 1 woman’s work is equal to 2 men’s work. Substituting the value of w in first equation, we get;

8w + 20m = 1
Or, 8w + 10w = 1
Or, 18w = 1

This means that 1 woman will take 18 days to finish the work.

Similarly,

16m + 20m = 1
Or, 36m = 1

This means that 1 man will take 36 days to finish the work.

(c) 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: Let us assume that speed of train is x and speed of bus is y

First condition: She travels 60 km by train and 240 km by bus;

(60)/(x)+(240)/(y)=4

Or, (60y+240x)/(xy)=4

Or, 60y+240x=4xy
Or, 15y+60x=xy

Second condition: She travels 100 km by train and 200 km by bus;

(100)/(x)+(200)/(y)=4(1)/(6)=(25)/(6)

Or, (100y+200x)/(xy)=(25)/(6)

Or, (6)/(25)(100y+200x)=xy
Or, 6(4y+8x)=xy
Or, 24y+48x=xy

Subtracting first equation from second, we get;

24y + 48x – 15y – 60x = xy – xy
Or, 9y – 12x = 0
Or, 9y = 12x
Or, x = (3)/(4)y

Substituting the value of x in first equation, we get;

15y+60x=xy
Or, 15y+60xx(3)/(4)y=(3)/(4)y^2

Or, 15y+45y=(3)/(4)y^2

Or, 60y=(3)/(4)y^2

Or, (4)/(3)xx60y=y^2

Or, 80y=y^2
Or, y=80
Hence, x=(3)/(4)y=(3)/(4)xx80=60`

Speed of train = 60 km/h and speed of bus = 80 km/h

PrevNext

Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 1)

Exercise:3.5 (part 2)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)