SOLUTION: Betty is 5 years older than twice the age of Bambam. Fred is 3 years older than Betty. Their ages equal 63. How old is each family member?
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Question 502856: Betty is 5 years older than twice the age of Bambam. Fred is 3 years older than Betty. Their ages equal 63. How old is each family member? Answer by Maths68(1474) (Show Source):
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Present age of Betty = b
Present age of Bambam = m
Present age of Fred = f
Given
Betty is 5 years older than twice the age of Bambam
b=2m + 5 .............(1)
Fred is 3 years older than Betty
f = b + 3 ............(2)
Their ages equal 63
b+m+f=63 .............(3)
Put the value of b from (1) to (2)
b=2m + 5 .............(1)
f = b + 3 ............(2)
f=2m + 5 + 3
f=2m+8 ...............(4)
Put the value f from (4) and value of b from (1) to (3)
b+m+f=63 .............(3)
(2m+5)+m+(2m+8)=63
2m+5+m+2m+8=63
5m+13=63
5m=63-13
5m=50
m=50/5
m=10
Put the value of m in (1)
b=2m + 5
b=2*10 + 5
b=20+5
b=25
Put the value of b in (2)
f = b + 3 ............(2)
f = 25 +3
f = 28
Present age of Betty = 25 years
Present age of Bambam = 10 years
Present age of Fred = 28 years
