Question

A number is 56 greater than the average of 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 termsSum of terms=One third of x+Quarter of x+One twelfth of xS=(x/3)+(x/4)+(x/12)S=(4x+3x+x)/12S=8x/12S=2x/3n=3A=(2x/3)×(1/3)A=2x/9As per the problem, the unknown number is 56 more than averagei.e x=56+(2x/9)=> x-(2x/9)=56=> (9x-2x)/9=56=> 7x/9=56=> 7x=504=> x=72Therefore the required number is 72
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