# Peaches today, Peaches tomorrow...

*Peaches Today, Peaches Tomorrow printable worksheet*

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.

**fourth**day he decided to keep ${\bf \frac{2}{7}}$ 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?**

**(iii)**Whenever the monkey has some peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one.

- 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.

**Can you start with fewer than 100 peaches and choose fractions so that there is at least one peach left after a week?**

**Starting with fewer than 100, what is the longest you can make the peaches last?**

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

The monkey will always keep a whole number of peaches.

For part (ii), you could use a tree diagram: at each stage, branch off the fractions it would be possible to try next so that all possibilities are checked.

For Part (iii), why not try a starting value and see what happens? Then decide whether you need to start with a bigger or smaller starting value next time.

**(i) 60 peaches**

Well done to everyone who worked out how many peaches the monkey was left with. This is Angela's work, from Westridge School in the USA:

1st day: $\frac34 \times 60=45$ $45-1=44$

2nd day: $\frac7{11} \times 44=28$ $28-1=27$

3rd day: $\frac59 \times 27=15$ $15-1=14$

4th day: $\frac27 \times 14=4$ $4-1=3$

5th day: $\frac23 \times 3=2$ $2-1=1$

The monkey had $1$ peach at the end of 5 days.**(ii) 75 peaches**

Well done to Anh Minh from British Vietnamese International School Hanoi, Siddhant from Singapore International School in India, Iqra from Uphall Primary School, Robyn and Connie from St Stephen's CE Primary School, EDSR from Priestlands School, Tommy and Sam from Widey Court Primary School and Jessica from Edmund de Moundeford (all in the UK) who put the fractions in the correct order. Saya and
Addison from Westridge described their method:

First, we started with 75 peaches (like said in the problem). Then, we looked at all the denominators and wondered which could divide 75 without a remainder. The options were $\frac{11}{15}$ or $\frac35$ because 75 could be divided by their denominator evenly. We decided with $\frac{11}{15}$ because that would give us the largest amount of peaches to keep out of the two. Then, we kept going with
this idea to look at the denominators to see which would be the best choice for the next day. The order that we ended up with was $\frac{11}{15},\frac56,\frac34, \frac12, \frac35,$ and then $\frac14.$

X

Making a tree diagram helped me a lot. With a tree diagram, one for each fraction, you can rule out numbers that take you to a dead end or numbers that might work if you use them a different way.

Keane and Michael from Bangkok Patana School in Thailand considered which fractions could go at the end:

On the last day, the fraction he wants to keep needs to be one of the unit fractions, or in other words, $\frac12$ or $\frac14$, or a fraction with the numerator as 2. This is because if you have any other fraction, like $\frac35$, and you have a number like 5, the number would be 3, and if you subtract 1, it would still be 2 peaches. But if you had $\frac25$, and the number was 5, the result would be 2, and minus 1 would be 1. Same applies for the unit fractions, but the denominator has to be half the amount starting, so for example, you would have to have 8 for $\frac14$ to work, or 4 for the $\frac12.$

**(iii) Making the peaches last as long as possible**

Click on the links below to see the answers we received:

7 days: D from Humanitree, EDSR (with more peaches left over to last more days)

8 days: Mateo and Ian from Humanitree

11 days: Shaufie and Xaviera from Humanitree

12 days: Burapa and Jesse from Bagkok Patana School

13 days: Dana and A from Humanitree

14 days: Lorea from Humanitree, and Siddhant

15 days: Marcelo from Humanitree

16 days: Hiren from Crestwood Community School in the UK, and Keane and Michael

Burapa and Jesse chose their starting number of peaches carefully:

We used the trial and error technique but started off with the numbers under 100 with the most factors which were 60, 72, 84, 96

Since 96 was the biggest number, we thought that it would be able to be the number that could be divided the most.

Most other people used a combination of these ideas decribed by Keane and Michael:

Our method was to find the biggest factor of the number of peaches we had, [which would] be the denominator of the amount of peaches we keep, and use the biggest possible numerator. For instance, 32/33 of 99. Then, we subtract 1. If the fraction makes a prime number after subtracting 1, then we make the numerator 1 less and see if it works then. If is still doesn't work, we keep going with the process. We kept doing that until we could no longer do it.

### Why do this problem?

This problem is in three parts, with each part becoming more open-ended and requiring more reasoning, giving students a chance to develop their skills at solving problems with fractions and then applying those skills in a more challenging context. By listening to others' approaches, students will be encouraged to persevere and continue to improve on their solution to the final part of the
problem.

This task has a playful context which we hope will draw curious learners in, and then offers lots of opportunities to practise routine fraction calculations while making progress towards solving the problem.

### Possible approach

Introduce the first part of the problem to the class, and give them a little bit of time to work in pairs to solve it. All three parts of this problem can be displayed on these slides (available as PowerPoint or PDF): *Peaches Peaches*

You may also find this printable worksheet useful: Peaches Worksheet

Once students have had a chance to work on the first challenge, share strategies and discuss any difficulties that arose.

Next, introduce the second task. This is more challenging, as students are given the fractions but not told which order to use them in. While they are working, circulate and look out for good examples of strategic thinking to share with the rest of the class. For example:*The denominator of the fraction must be a factor of the number of peachesWe started with 75 peaches so there are only two options for the first fractionWe could write each fraction on a piece of paper and then move them around to find an order that works!*

Once each group of students is convinced they have found the correct order, they can make a start on the third challenge. They could start by trying to find a number of peaches less than 100 and a set of fractions so that after a week there is still at least one peach remaining. This would work very well as a 'simmering activity' set for students to think about beyond a single lesson -
perhaps this could be set as a homework. The best solution could be displayed on the classroom wall, and students could be challenged to improve on it.

### Key questions

### Possible support

Students could be presented with a shorter version of part (ii) of the problem:*A little monkey had 44 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:**$$ \frac{1}{2} \qquad \frac{1}{4} \qquad \frac{3}{4} \qquad \frac{3}{5}$$

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

### Possible extension

Proving that they have found the longest possible chain requires some detailed mathematical reasoning.

For part (ii), if students are struggling to record their thinking effectively, suggest that they could use a tree diagram: at each stage, branch off the fractions it would be possible to try next so that all possibilities are checked.