# 10th cbse maths solution for exercise 4.3 part 3

This page 10th cbse maths solution for exercise 4.3 part 3 is going to provide you solution for every problems that you find in the exercise no 4.3 part 3

## 10th CBSE maths solution for Exercise 4.3 part 3

(6) The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side,find the sides of the field.

Solution:

Let 'x' be the length of shorter side of the rectangle

length of diagonal = x + 60

length of longer side = x + 30 (x + 60)²  = x² + (x + 30)²

x² + 60² + 2 x (60) = x² + x² + 2 x (30) + 30²

x² + 3600 + 120 x = 2 x² + 60 x + 900

2 x² - x² + 60 x - 120 x + 900 - 3600 = 0

x² - 60 x - 2700 = 0

x² - 90 x + 30 x - 2700 = 0

x (x - 90) + 30 (x - 90) = 0

(x + 30) (x - 90) = 0

x + 30 = 0           x - 90 = 0

x = -30                x = 90

breadth of rectangle = 90 m

length of rectangle = 90 + 30 = 120 m

(7) The  difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let 'x' be the larger number

let 'y' be the smaller number

y² = 8 x

The difference of squares of two numbers = 180

x² - y² = 180

x² - 8 x = 180

x² - 8 x - 180 = 0

x² - 18 x + 10 x - 180 = 0

x (x - 18) + 10 (x - 18) = 0

(x - 18) (x + 10) = 0

x - 18 = 0         x + 10 = 0

x = 18             x = -10

y² = 8 (18)

y = 8 (18)

y = 2 x 2 x 2 x 3 x 3 x 2

y = 2 x 2 x 3

y = 12

Therefore the larger number = 18

and smaller number = 12

(8) A train travels 360 km at a uniform speed. If the speed had been 5 km/h more,it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let 'x' be the speed of the train

If the speed had been 5 km/h more,it would have taken 1 hour less for the same journey

Time = distance/speed

Distance to be covered = 360 km

T₁ = 360/x

T₂ = 360/(x + 5)

T₁ - T₂ = 1 hour

[360/x] -  [360/(x + 5)] = 1

360 [ (1/x) - (1/(x + 5))] = 1

360 [ (x + 5 - x)/x(x + 5) ] = 1

360[5/(x² + 5 x)] = 1

1800/(x² + 5 x) = 1

1800 = (x² + 5 x)

x² + 5 x - 1800 = 0

x² - 40 x + 45 x - 1800 = 0

x (x - 40) + 45 (x - 40) = 0

(x - 40) (x + 45) = 0

x - 40 = 0     x + 45 = 0

x = 40   and x = -45

Therefore speed of the train = 40 km/hr