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If Clara spends £23 on books and magazines, how many of each does she buy?
Problem
A book costs £3.40 and a magazine costs £1.60. If Clara spends exactly £23 on books and magazines, how many of each does she buy?
If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
Answer: 3 books and 8 magazines
Making whole numbers of pounds
Look for the smallest multiples of books/magazines etc required to make a whole number of pounds:
| Number of Books ( £3.40 each) | Number of Magazines ( £1.60 each) | Total Cost |
| 5 | 0 | £17 |
| 0 | 5 | £8 |
| 1 | 1 | £5 |
Observe that £23 = £15 + £8 = 3$\times$ £5 + £8. This is 3 lots of the bottom row plus the middle row.
So she buys $3$ books, and $3+5=8$ magazines.
Trying to fill £23 with magazines
Magazine: £1.60 Book: £3.40 = 2 magazines + 20p
Imagine buying only magazines:
£23$\div$ £1.60 = 230$\div16 = 14 remainder 6 so 16 magazines remainder 60p
= 14 magazines + 3$\times$20p
= 14 magazines + 3 books - 6 magazines
= 8 magazines and 3 books
Algebra
Suppose Clara buys $b$ books and $m$ magazines. Then we know that, working in pence:
Both sides of this equation can be divided by $20$, to give:
At this point, it is worth remembering that $m$ and $b$ must both be non-negative integers.
Since $8m$ is even and $115$ is odd, $17b$ must be odd, which in turn means that $b$ must be odd. Also, as $17 \times 7 = 119 > 115$, $b$ must be smaller than $7$. This means the possible values for be are $1$, $3$ and $5$.
If $b=1$, $17+8m=115$, so $8m = 98$, but $98$ is not divisible by $8$. So, $b \neq 1$.
If $b=3$, $51+8m=115$, so $8m = 64$ and $m=8$. This gives one solution.
If $b=5$, $85+8m=115$, so $8m = 30$, but $30$ is not divisible by $8$. So, $b \neq 5$.
Therefore the only solution is that Clara bought $3$ books and $8$ magazines.