Question

a number is 56 greater than the average of its one third, one fourth, one twelth. Find the number ​

Answer 1

Answer:

Let the unknown number be 'x'

Let the sum of terms be 'S'

Let the no.of terms be 'n'

Let the average be 'A'

Average=Sum of terms/No.of terms

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

S=(x/3)+(x/4)+(x/12)

S=(4x+3x+x)/12

S=8x/12

S=2x/3

n=3

A=(2x/3)×(1/3)

A=2x/9

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

i.e x=56+(2x/9)

=> x-(2x/9)=56

=> (9x-2x)/9=56

=> 7x/9=56

=> 7x=504

=> x=72

Therefore the required number is 72

Answer 2

Answer :-

$\large{\boxed{\underline{\sf The \: number \: is \: 72}}}$

Explanation :-

Given :

• A number is 56 greater than the average of it's one third, one fourth and one twelfth

To find :

The number

Solution :

Let the number be x

According to question,

$\sf\longrightarrow x = \dfrac{ \frac{x}{3} + \frac{x}{4} + \frac{x}{12} }{3} + 56$

$\sf\longrightarrow x = \dfrac{ \frac{4x + 3x + x}{12} }{3} + 56$

$\sf\longrightarrow x = \dfrac{ \frac{8x}{12} }{3} + 56$

$\sf\longrightarrow x = 56 + \dfrac{8x}{12} \times \dfrac{1}{3}$

$\sf\longrightarrow x = 56 + \dfrac{8x}{36}$

$\sf\longrightarrow x = \dfrac{2016 + 8x}{36}$

$\sf\longrightarrow 36x = 2016 + 8x$

$\sf\longrightarrow 36x - 8x = 2016$

$\sf\longrightarrow 28x = 2016$

$\sf\longrightarrow x = \dfrac{2016}{28}$

$\sf\longrightarrow x = 72$

Hence, the number is 72 respectively