Question

Seperate the number 53 into 3 parts so that the second number is 3 less than the first number and the third number is twice the first number. Give the value of the second number.

Answer 1

Hello,

Let a the first, b the second and c the third.
a+b+c=53
b=a/3
c=2a
==>a+a/3+2a=53
==>10a/3=53
==>a=15.9
c=2*a=2*15.9=31.8
b=a/3=15.9/3=5.3

Proof: a+b+c=15.9+5.3+31.8=53

The second number is 5.3