Question

A number is 56 greater than the average of its third quarter and one twelfth find the number
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Answer 1

$\mathtt{\huge{\bf{\underline{\red{Given :-}}}}}$

• A number is 56 greater than the average of its third quarter and one twelfth .

$\mathtt{\huge{\bf{\underline{\green{To\:Find :-}}}}}$

• The number in the question.

$\mathtt{\huge{\bf{\underline{\blue{Solution :-}}}}}$

Let the number be 'x'

The one- third of the number will be = $\dfrac{x}{3}$

The quarter of the number will be its $\dfrac{1}{4}$ . So, it is = $\dfrac{x}{4}$

The one- twelth of the number will be = $\dfrac{x}{12}$

According to the question ,

• The average of its third quarter and one twelfth . (Given )

The average of the number will becomes,

➝ $\dfrac{ \dfrac{x}{3} + \dfrac{x}{4} + \dfrac{x}{12} }{3}$

➝ $\dfrac{1}{3}$ × ( ${ \dfrac{x}{3} + \dfrac{x}{4} + \dfrac{x}{12}$ )

• A number is 56 greater than its average. (Given)

➝ x = $\dfrac{1}{3}$ × ( ${ \dfrac{x}{3} + \dfrac{x}{4} + \dfrac{x}{12}$ ) + 56

➝ x = ${ \dfrac{x}{9} + \dfrac{x}{12} + \dfrac{x}{36}$ + 56

➝ x - ${ \dfrac{x}{9} + \dfrac{x}{12} + \dfrac{x}{36}$ = 56

➝ x - ${ \dfrac{x}{9} - \dfrac{x}{12} - \dfrac{x}{36}$ = 56

➝ 36x - 4x - 3x - x = 36 × 56

➝ 36x - 4x - 3x - x = 2016

➝ 36x - 8 x = 2016

➝ 36x - 8 x = 2016

➝ 28 x = 2016

➝ x = $\dfrac{2016}{28}$

➝ x = 72

The required number is 72 .

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