# Perkin's age problem [closed]

Perkin and Poise's combined age is 22 which is 3 years more than the combined age of Perkin and Pootle which is 2 years more than Poise's and Pootle's age combined.

The question is -

In how many years will the sum of all their ages be 56?

## closed as off-topic by JMP, Glorfindel, w l, Bass, QuantumTwinkieSep 28 '18 at 23:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – JMP, Glorfindel, w l, Bass, QuantumTwinkie
If this question can be reworded to fit the rules in the help center, please edit the question.

• I'm not seeing a reason for these close votes. It might not be a particularly inspired puzzle, but it is a puzzle. – Chris Cudmore Sep 27 '18 at 16:52
• Why the downvotes? – Quark-epoch Sep 28 '18 at 9:43

its-

9 years

Explanation -

Perkin $$+$$ Poise $$= 22$$
Perkin $$+$$ Pootle $$= 19$$ [$$22-3$$]
Poise $$+$$ Pootle $$= 17$$ [$$19-2$$]
Solving them we get,
Perkin = 12, Poise = 10, Pootle = 7, sum= 29, required sum 56,
difference = 27, sum will increase by 3 each year so required time = 9 years

9 years

a=Perkin
b=Poise
c=Pootle

a+b=22
a+c=19
b+c=17

2(a+b+c)=58

a+b+c=29

29 + 3*years=56

years = 9

Twice the current combined age of Perkin, Poise and Pootle is $$3\times22-3-(3+2)=66-3-5=58$$. Their combined age twice will be $$2\times56=112$$, in $$112-58=54$$ years. There are $$3$$ of them, we are counting each person twice, hence the event will occur in $$\frac{54}{2\times3}=9$$ years time.