Peanut harvest
A group of monkeys eat various fractions of a harvest of peanuts. What fraction is left behind?
Age
11 to 14
Challenge level
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.