Question

A 3.0 x 10^7 kg ship collides elastically with a 2.5 x 10^7 kg ship moving north at 4.0 km/h. After the collision, the first ship moves north at 3.1 km/h and the second ship moves south at 6.9 km/h. Find the unknown velocity.

Answer 1

Given :-

• The numerator of fraction is 10 less than the denominator.
• If 5 added to both numerator and denominator the fraction will be 4/9.

To FinD:-

• The fraction.

SolutioN:-

Let the denominator of fraction be x.

Then, the numerator of fraction will be x - 10.

$\small\sf \: So \: , \: fraction = \frac{x - 10}{x}$

It is given that the fraction will be 4/9 when 5 is added to both numerator and denominator.

$\small\sf \implies \: \frac{x - 10 + 5}{x + 5} = \frac{4}{9}$

$\small\sf \implies \: \frac{x - 5}{x + 5} = \frac{4}{9}$

By cross multiplication , we have

$\small \sf \implies \:9(x - 5) = 4(x + 5)$

$\small\sf \implies \:9x - 45= 4x + 20$

By transportung 4x to left side , we have

$\small\sf \implies \: 9x - 4x - 45 = 20$

$\small\sf \implies \:5x - 45 = 20$

By transporting 45 to right side , we have

$\small \sf \implies \: 5x = 20 + 45$

$\small \sf \implies \:5x = 65$

By transporting 5 to the right side , we have

$\small \sf \implies \: x = \frac{ 65}{5}$

$\small \sf \implies \: x = 13$

By putting the value of x , the fraction

$\small \sf \: = \frac{x - 10}{x}$

$\small \sf \: = \frac{13 - 10}{13}$

$\small \sf \: = \frac{3}{13}$

Therefore , the required fraction is 3/13