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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?

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closed as off-topic by JMP, Glorfindel, w l, Bass, QuantumTwinkie Sep 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.

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

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5
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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

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1
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The answer is:

9 years

My reasoning is:

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.

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