Bookshop
If Clara spends £23 on books and magazines, how many of each does she buy?
Age
11 to 14
Challenge level
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.
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:
$340b+160m=2300$
Both sides of this equation can be divided by $20$, to give:
$17b+8m=115$
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.