Question

a number is 56 greater than the average of his third quarter and one twelfth find it

Answer 1

Answer:

Step-by-step explanation:

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:

Mark it as brainliest

Step-by-step explanation:

let the number be x

x=(3/4x+1/12x)/2+56

2x=3x/4+x/12+56

2x=(9x+x)/12+56

2x-5x/6=56

7x/6=56

x=48