# 10th cbse maths solution for exercise 3.5 part 4

This page 10th cbse maths solution for exercise 3.5 part 4 is going to provide you solution for every problems that you find in the exercise no 3.5

## 10th CBSE maths solution for Exercise 3.5 part 4

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the car travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other ,they meet in 1 hour. What are the speeds of the two cars?

Solution:

Let “x km/hr” be the speed of 1st car

Let “y km/hr” be the speed of the 2nd car

Time = Distance/Speed

Speed of both cars while they are traveling in the same direction = (x – y)

Speed of both cars while they are traveling in the opposite direction = (x + y)

5 (x – y) = 100

x – y = 100/5

x -  y = 20 ------(1)

1(x + y) = 100

x + y = 100 ------(2)  x = 120/2

x = 60

Substitute x = 60 in the first or second equation to get the value of y

60 – y = 20

-y = 20 – 60

-y = -40

Y = 40

Therefore the speed of first car = 60 km/hr

Speed of second car = 40 km/hr

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Area of any rectangle = Length x breadth

Let “x” be the length of rectangle

Let “y” be the breadth of rectangle

(x – 5) (y + 3) = xy – 9

x y + 3 x -  5 y – 15 = x y – 9

xy – xy  + 3 x – 5 y – 15 + 9 = 0

3 x – 5 y – 6 = 0 -------(1)

(x + 3) (y + 2) = xy  + 67

xy  + 2 x + 3 y + 6 – xy – 67 = 0

2 x + 3 y –61 = 0 -------(2)  x/(305+18) = y/(-12 + 183) = 1/(9 + 10)

x/323 = y/171 = 1/19

x/323 = 1/19                y/171 = 1/19

x = 323/19                     y = 171/19

x = 17                             y = 9

Therefore the length of rectangle = 17 units

Breadth of rectangle = 9 units