Question

A number is $56$ greater than the average of its one-third, one-quarter, and one-twelfth. Find the number.

Answer 1

Let the unknown number be ${}^{\prime }{x}^{\prime }$
Let the sum of terms be ${}^{\prime }{S}^{\prime }$
Let the no.of terms be ${}^{\prime }{n}^{\prime }$

Let the average be ${}^{\prime }{A}^{\prime }$

Average $=\frac{\text{Sum of terms}}{\text{No. of terms}}$

Sum of terms $=$ One third of $x+$ Quarter of $x+$ One twelfth of $x$

S $=\frac{x}{3}+\frac{x}{4}+\frac{x}{12}$

$=\frac{(4x+3x+x)}{12}$

$=\frac{8x}{12}$

$=\frac{2x}{3}$

$n=3$

A $=\frac{2x}{3}÷3=\frac{2x}{3}\times \frac{1}{3}=\frac{2x}{9}$

As per the problem, the unknown number is $56$ more than average.

i.e $x=56+\frac{2x}{9}$

$\Rightarrow x-\frac{2x}{9}=56$

$\Rightarrow \frac{7x}{9}=56$

$\Rightarrow 7x=504$

$\Rightarrow x=72$

Therefore the required number is $72$