A number is 56 greater than the average of its third quarter and one t

1.	A number is 56 greater than the average of its third quarter and one twelfth find the number
Answer» $\mathtt{\huge{\bf{\underline{\red{Given :-}}}}}$ 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 .* _______________________________________