Question

Two numbers are respectively $10$% and $40$% more the third number. Find the ratio of the first and the seconds numbers. Also, find the ratio of the third and the first numbers.

Answer 1

Let the third number be 'y'

Let 'a' & 'b' be first & second number

then

$a=y+\frac{104}{100}=\frac{1104}{100}$ ------- (1)

$b=y+\frac{404}{100}=\frac{1404}{100}$ -------- (2)

Now $\frac{a}{b}=\frac{\frac{1104}{100}}{\frac{1404}{100}}=\frac{110}{110}=\frac{11}{14}$

$\therefore$ ratio of first and second number is 11:14

again $\frac{y}{a}=\frac{y}{110y}\times 100=\frac{10}{11}$

$\therefore$ ratio of third number with first number is 10:11.