Peaches Today, Peaches Tomorrow...

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 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?
 

• 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.