Question

A number is 56 greater than the average of its one-third, one-fourth and one-twelfth. Find the number. Please please do this sum in a paper and also so its steps clearly thank you

Answer 1

Let,

• Unknown number = x
• Sum of terms = S
• Number of terms = n
• Average = A

Average = Sum of terms/Number of terms

Sum of terms = 1/3rd of x + 1/4th of x + 1/12th of x

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

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

$\implies\bold{S=\frac{8x}{12}}$

$\implies\bold{S=\frac{2x}{3}}$

$\implies\bold{n=3}$

---------------------------------------------------------------------------------------

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

$\implies\bold{A = \frac{2x}{9}}$

As per the problem,

The unknown number is 56 more than average.

i.e.,

$\bold{x = 56 + \frac{2x}{9}}$

$\implies\bold{x-\frac{2x}{9}=56}}$

$\implies\bold{\frac{9x-2x}{9}=56}$

$\implies\bold{\frac{7x}{9}=56}$

$\implies\bold{7x=504}$

$\implies\bold{x=72}$

∴ The required number is 72

Answer 2

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REFER TO ATTACHMENT

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