# Peanut Harvest

A group of monkeys eat various fractions of a harvest of peanuts. What fraction is left behind?

## Problem

My four pet monkeys and I harvested a large pile of peanuts.

Monkey A woke in the night and ate half of them;

then Monkey B woke and ate one third of what remained;

then Monkey C woke and ate one quarter of the rest;

finally Monkey D ate one fifth of the much diminished remaining pile.

What fraction of the original harvest was left in the morning?

This problem is taken from the UKMT Mathematical Challenges.

## Student Solutions

**Answer**: $\frac15$

**Working out how much is left after each stage**

After A has eaten the fraction left is $\frac{1}{2}$.

B eats $\frac{1}{3}$ of this which leaves $\frac{1}{2} - (\frac{1}{2} \times \frac{1}{3}) = \frac{1}{2} - \frac{1}{6}= \frac{1}{3}$.

C eats $\frac{1}{4}$ which leaves $\frac{1}{3} - (\frac{1}{3} \times \frac{1}{4}) = \frac{1}{3} - \frac{1}{12} = \frac{1}{4}$.

D eats $\frac{1}{5}$ which similarly leaves $\frac{1}{5}$.

**Multiplying the fractions not eaten**

A eats $\frac12$ and leaves $\frac12$

B eats $\frac13$ so leaves $\frac23$

And so on, so the fraction left at the end will be $\frac12\times\frac23\times\frac34\times\frac45=\frac15$ since most numerators cancel with the previous denominator.