Henry Ernest Dudeney/Puzzles and Curious Problems/43 - Mike's Age/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $43$

Mike's Age
Pat O'Connor is $1 \tfrac 1 3$ times as old as when he built the pigsty.
Little Mike, who was $40$ months old when Pat built the sty, is now two years more than half as old as Pat's wife, Biddy, was when Pat built the sty,
so that when Little Mike is as old as Pat was when he built the sty,
their three ages combined will amount to just one hundred years.
How old is Little Mike?


Solution

Mike is currently $10 \tfrac {16} {21}$ years old.

Pat was $22 \tfrac 2 7$ when he built the pigsty, which was $7 \tfrac 3 7$ years ago, and is now $29 \tfrac 5 7$ years old.

Biddy was $17 \tfrac {11} {21}$ when Pat built the pigsty, and is now $24 \tfrac {20} {21}$ years old


Proof

Let $P$ years be Pat's age when he built the pigsty.

Let $M$ years be Mike's age now.

Let $B$ years be Biddy's age when Pat built the pigsty.


First we note that from:

Pat O'Connor is $1 \tfrac 1 3$ times as old as when he built the pigsty

$P$ built the pigsty $\dfrac P 3$ years ago.


We have:

\(\text {(1)}: \quad\) \(\ds M\) \(=\) \(\ds \dfrac {40} {12} + \dfrac P 3\) Little Mike, who was $40$ months old when Pat built the sty, ...
\(\text {(2)}: \quad\) \(\ds M\) \(=\) \(\ds \dfrac B 2 + 2\) ... is now two years more than half as old as Pat's wife, Biddy, was when Pat built the sty,
\(\text {(3)}: \quad\) \(\ds 100\) \(=\) \(\ds M + \paren {P - M} + B + \dfrac P 3 + \paren {P - M} + \dfrac {4 P} 3 + \paren {P - M}\) so that when Little Mike is as old as Pat was when he built the sty, their three ages combined will amount to just one hundred years.
\(\ds \) \(=\) \(\ds 4 \tfrac 2 3 P + B - 2 M\) simplifying
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 300\) \(=\) \(\ds 14 P + 3 B - 6 M\) simplifying
\(\ds \leadsto \ \ \) \(\ds B\) \(=\) \(\ds 2 M - 4\) rearranging $(2)$ with a view towards eliminating $B$
\(\ds \leadsto \ \ \) \(\ds 300\) \(=\) \(\ds 14 P + 3 \paren {2 M - 4} - 6 M\) Eliminating $B$ from $(4)$
\(\ds \leadsto \ \ \) \(\ds 300\) \(=\) \(\ds 14 P - 12\) simplifying
\(\ds \leadsto \ \ \) \(\ds P\) \(=\) \(\ds \dfrac {156} 7\) simplifying
\(\ds \) \(=\) \(\ds 22 \tfrac 2 7\)
\(\ds \leadsto \ \ \) \(\ds M\) \(=\) \(\ds \dfrac {156} {3 \times 7} + \dfrac {40} {12}\) substituting into $(1)$
\(\ds \) \(=\) \(\ds \dfrac {52} 7 + \dfrac {10} 3\) cancelling down
\(\ds \) \(=\) \(\ds \dfrac {52 \times 3 + 10 \times 7} {21}\) common denominator
\(\ds \) \(=\) \(\ds \dfrac {226} {21}\) simplifying
\(\ds \) \(=\) \(\ds 10 \tfrac {16} {21}\) converting to mixed fraction


The remaining ages are calculated hence.

$\blacksquare$


Sources

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