a number is 56 greater than the average of its one third quarter and one 12 find the number
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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
THANKYOU
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
THANKYOU
aryandhankar:
thanks for helping
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