Question

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

Answer 1

Hi,

Let the number = x

Third part of the number = x / 3

Quarter of the number = x / 4

Twelfth part of the number = x / 12

Average of third , quarter and

twelfth numbers = ( x/3 + x/4 + x/12)/3

= [ (4x+3x+x)/12 ]/3

= 8x/36

= 2x/9 -----( 1 )

According to the problem given,

x is 56 greater than ( 1 )

Therefore ,

x = 56 + ( 1 )

x = 56 + 2x/9

x - 2x/9 = 56

( 9x - 2 x ) /9 = 56

7x /9 = 56

x = ( 56 × 9 ) / 7

x = 8 × 9

x = 72

Required number = x = 72

I hope this helps you.

:)