Question

what is bumber of which the third part exceeds the fifth part by 6

Answer 1

$\huge\mathcal{Heya}$
$<b><i>$
Let the number be x.

Third part of the number = 1 / 3 of x = x / 3

Fifth part of the number = 1 / 5 of x = x / 5

A/q

x / 3 = x / 5 + 6

On further solving, we get

$\frac{x}{3} = \frac{x + 6(5)}{5} \\ \\ \frac{x}{3} = \frac{x + 30}{5}$

On cross multiplication, we get

$5(x) = 3(x + 30) \\ \\ 5x = 3x + 90 \\ \\ 5x - 3x = 90 \\ \\ 2x = 90 \\ \\ x = 45$

Hence, the required number is 45.

$\huge\mathbb{Hope \ this \ helps.}$

Answer 2

$\frac{x}{3} + 6 = \frac{x}{5} \\ = >x = 45$
45/3=15. & 45/5=9.
15 exceeds 9by6