Challenge Level

Jack from Stoke by Nayland Middle School sent in a solution to the problem that makes, in his words, "the World's most bland cereal bar":

Use $500$ grams of seeds and $500$ grams of nuts this will give you a mixture that weighs $1000$ grams and costs $ £5$

Anja from Stoke by Nayland Middle School, John and Andrew from Lazonby C of E School and Esther (school not given) managed to combine all 3 ingredients as follows:

$60\%$ of seeds $= £2.40 = 600 \, \text{g}$$20\%$ of nuts $= £1.20 = 200 \, \text{g}$

$20\%$ of apricots $= £1.40 = 200 \,\text{g} $

$100%$ of mix $ £5.00 = 1000\,\text{g}$ or $1\,\text{kg}$

Anja says:

I started out by experimenting with different mixtures until I found this one which adds up to exactly $ £5.00$.Esther observes that:

The ratio of apricots : nuts : seeds is $1 : 1 : 3$

Taylor & Nia from Llandaff City Church in Wales Primary School also combined the 3 ingredients, but in a different way:

$1/2 \, \text{kg}$ seeds $ £2.00$$1/8 \, \text{kg}$ seeds $ £0.50 $

$1/8 \, \text{kg}$ nuts $ £0.75$

$1/4 \, \text{kg}$ dried apricots $ £1.75$

We solved this problem through the process of trial and error

The ratio that Taylor and Nia have worked out is:

Apricots : nuts : seeds = $2 : 1 :
5$

The two solutions above make sense
since you need seeds to match the nuts

(they cost $ £6$ and $ £4$, so equal amounts of each will average out to £5),

plus twice as many seeds as apricots

(they cost $ £7$ and $ £4$ so two $4$s and one $7$ will average out to £5).

So altogether the seeds must amount to the nuts plus double the apricots.

(they cost $ £6$ and $ £4$, so equal amounts of each will average out to £5),

plus twice as many seeds as apricots

(they cost $ £7$ and $ £4$ so two $4$s and one $7$ will average out to £5).

So altogether the seeds must amount to the nuts plus double the apricots.

The problem can also be defined in one
equation

(where $a =$ weight of apricots, $n =$ weight of nuts, $s =$ weight
of seeds):

$$7a + 6n + 4s = 5(a + n + s)$$

$$2a+ n = s$$

Any combination that fits the relationship $s = 2a + n$ will satisfy the criteria (see the examples above). Notice that this is a more general solution than the ratios offered above.

You can now use the equation to find many more combinations.