SOLUTION: Allen is 3 times as old as Betty. Catherine is 5 yrs younger than Allen. 3 yrs ago sum of all ages is 56. What is each of their present age?

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Question 38940: Allen is 3 times as old as Betty. Catherine is 5 yrs younger than Allen. 3 yrs ago sum of all ages is 56. What is each of their present age?
Answer by Earlsdon(6294) (Show Source):
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Start by letting:
A = Allen's present age.
B = Betty's present age.
C = Catherine's present age.
From the problem desciption:
1) A = 3B (Allen is 3 times as old as Betty)
2) C = A-5 (Catherine is 5 years younger than Allen).
3) (A-3)+(B-3)+(C-3) = 56 (Three years ago, the sum of their ages was 56).
Simplify equation 3)
A+B+C-9 = 56 Add 9 to both sides.
A+B+C = 65 Substitute equation 1) for A.
3B+B+C = 65 Rewrite equation 2 as: C = 3B-5 and substitute for C.
3B+B+3B-5 = 65 Simplify.
7B-5 = 65 Add 5 to both sides
7B = 70 Divide both sides by 7.
B = 10 This is Betty's age.
A = 3B = 3(10) = 30 This is Allen's age.
C = A-5 = 30-5 = 25 This is Catherine's age.