This problem is in three parts. If you are feeling confident about working with fractions, you might want to skip straight to part (ii) or part (iii).

**(i)** A little monkey had 60 peaches.

On the **first** day he decided to keep ${\bf \frac{3}{4}}$ of his peaches.

He gave the rest away. Then he ate one.

On the **second** day he decided to keep ${\bf \frac{7}{11}}$ of his peaches.

He gave the rest away. Then he ate one.

On the **third** day he decided to keep ${\bf \frac{5}{9}}$ of his peaches.

He gave the rest away. Then he ate one.

On the **fourth** day he decided to keep ${\bf \frac{2}{7}}$ of his peaches.

He gave the rest away. Then he ate one.

He gave the rest away. Then he ate one.

On the **fifth** day he decided to keep ${\bf \frac{2}{3}}$ of his peaches.

He gave the rest away. Then he ate one.

**How many did he have left at the end?**

**(ii)** A little monkey had 75 peaches.

Each day, he kept a fraction of his peaches, gave the rest away, and then ate one.

These are the fractions he decided to **keep: **$$ \frac{1}{2} \qquad \frac{1}{4} \qquad \frac{3}{4} \qquad \frac{3}{5} \qquad \frac{5}{6} \qquad \frac{11}{15}$$

**In which order did he use the fractions so that he was left with just one peach at the end?**

I wonder how long he could make his peaches last for...

Here are his rules:

- Each fraction must be in its simplest form and must be less than 1.
- The denominator can never be the same as the number of peaches left.

For example, if there were 45 peaches left, he could not choose to keep $\frac{44}{45}$ of them.

Click here for a poster of part (ii) of this problem.