Question

Two numbers are such that the ratio between
them is 3:5. If each is increased by 10, the ratio
between the new numbers so formed is 5:7. Find
the original numbers,
​

Answer 1

Given two no.s of the ratio 3:5

Let them be 3x and 5x

when they are increased by 10 ,

3x+10/5x+10 =5/7

Cross multiplying and re arranging , we get

21x+70=25x+50

25x-21x=70-50

4x=20

x=5

Thus the two numbers will be 15 and 25 .

:)

Answer 2

Solution :-

Let the two number be x and y.

According to the first condition :-

$\longrightarrow \sf \dfrac{x}{y} = \dfrac{3}{5} \\\\\longrightarrow x = \dfrac{3}{5} y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

According to the second condition :-

$\longrightarrow \sf \dfrac{x+10}{y+10} = \dfrac{5}{7} \\\\\longrightarrow 7(x+10) = 5(y+10) \\\\\longrightarrow 7x+70 = 5y+50 \\\\\longrightarrow 7x-5y = -20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(2)$

Substitute the value of x in eq (2) :-

$\longrightarrow \sf 7 \times \dfrac{3}{5}y- 5y = -20 \\\\\longrightarrow \dfrac{21y}{5} -5y = -20 \\\\\longrightarrow \dfrac{21y-25y}{5}= -20 \\\\\longrightarrow -4y = -100 \\\\\longrightarrow \bf y = 25$

Substitute y = 25 in eq (1) :-

$\longrightarrow \sf x = \dfrac{3}{5} \times 25 \\\\\longrightarrow x = 3 \times 5 \\\\\longrightarrow \bf x = 15$

The two numbers are 15 and 25 .