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Stolen Pension

Age 14 to 16
ShortChallenge Level Yellow starYellow starYellow star
Secondary curriculum
  • Problem
  • Solutions

Answer: £2057


Algebra
Started with $A$

                        $\frac6{11}A$ stolen,          $\frac{5}{11}A$ left
                  $\swarrow$      $\searrow$
$\frac6{11}$ of $\frac6{11}A$ spent, $\frac5{11}$ of $\frac6{11}A$ recovered
Difference between money left and money recovered is $ £425$ $$\begin{align}
\tfrac5{11}A-\tfrac5{11}\times \tfrac6{11}A&=425\\
\Rightarrow \tfrac5{11}\left(1-\tfrac6{11}\right)A&=425\\
\Rightarrow \tfrac5{11}\times\tfrac5{11}\times A&=425\\
\Rightarrow A &= \tfrac{425}{25}\times121\\
\Rightarrow A&=2057
\end{align}$$


Using a diagram to represent the story
This diagram represents the story.


  

To compare the recovered and kept money, it would be useful if the pieces were the same size.

The 'stolen' money is split into 6 parts by the pensioner, but into 11 parts by the thief. This means that we can split it into 66 pieces.

  

55$-$30 = 25, so £425 is equivalent to 25 pieces.

If 25 pieces are worth £425, then each piece is worth £425$\div$25 = £17.

So 121 pieces are worth 121$\times$ £17 = £2057.

You can find more short problems, arranged by curriculum topic, in our short problems collection.

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