Question

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

Answer 1

Let the unknown number be 'x'
Let the sum of terms be 'S'
Let the no.of terms be 'n'
Let the average be 'A'
Average=Sum of terms/No.of terms
Sum of terms=One third of x+Quarter of x+One twelfth of x
S=(x/3)+(x/4)+(x/12)
S=(4x+3x+x)/12
S=8x/12
S=2x/3
n=3
A=(2x/3)×(1/3)
A=2x/9
As per the problem, the unknown number is 56 more than average
i.e x=56+(2x/9)
=> x-(2x/9)=56
=> (9x-2x)/9=56
=> 7x/9=56
=> 7x=504
=> x=72
Therefore the required number is 72

Answer 2

Answer:

72

Step-by-step explanation:

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