5.01d - Using ratios in problems

5.01d - Using ratios in problems

A recipe for four people must feed six: should you add the same amount to every ingredient or multiply them all? Learn to scale recipes, connect ratios through a shared quantity and turn a fixed comparison into a linear input-output rule.

Scaling a recipe

To preserve a recipe, multiply every ingredient by the same scale factor; adding the same amount changes its composition.

Suppose four portions need 300 g300\,\mathrm{g} flour, 180 ml180\,\mathrm{ml} milk and 2 eggs. Doubling the portions doubles all three quantities. A non-integer scale factor works in exactly the same way.

Recipe scale factor

For equal-sized portions made to the same recipe,

scale factor=required number of portionsoriginal number of portions.\text{scale factor}=\frac{\text{required number of portions}}{\text{original number of portions}}.

Multiply every original ingredient amount by this factor. The original portion count must be nonzero. Not on the 2027 formula sheet: reason with equal portions.

This assumes the portion size and recipe stay fixed, with no extra wastage. Convert stock and required amounts to matching units before comparing them. Where a recipe calls for indivisible items or whole batches, respect those restrictions rather than rounding each ingredient separately.

Worked example 1: increasing by a fractional factor

Question. Four portions need 300 g300\,\mathrm{g} flour, 180 ml180\,\mathrm{ml} milk and 2 eggs. Find the amounts for six portions.

Plan. Six is 64=1.5\frac64=1.5 times four, so multiply every amount by 1.5.

flour=300×1.5=450 gmilk=180×1.5=270 mleggs=2×1.5=3\begin{aligned} \text{flour}&=300\times1.5=450\,\mathrm{g} \\ \text{milk}&=180\times1.5=270\,\mathrm{ml} \\ \text{eggs}&=2\times1.5=3 \end{aligned}

Answer. 450 g450\,\mathrm{g} flour, 270 ml270\,\mathrm{ml} milk and 3 eggs.

Check. Per portion, the flour is 450÷6=75 g450\div6=75\,\mathrm{g}, the same as 300÷4300\div4. Milk is 45 ml45\,\mathrm{ml} per portion in both recipes, and 3 eggs for 6 gives the same half-egg per portion.

Keep the recipe but reverse what is known: available ingredients now limit the number of portions. Every ingredient must be sufficient.

Worked example 2: finding the limiting ingredient

Question. The same four-portion recipe needs 300 g300\,\mathrm{g} flour, 180 ml180\,\mathrm{ml} milk and 2 eggs. There are 750 g750\,\mathrm{g} flour, 600 ml600\,\mathrm{ml} milk and 6 eggs available. What is the greatest number of whole equal-sized portions possible if ingredients may be measured proportionally?

Plan. Find how many portions each ingredient could support; the smallest limit decides.

flour limit=4×750300=10milk limit=4×600180=1313egg limit=4×62=12\begin{aligned} \text{flour limit}&=4\times\frac{750}{300}=10 \\ \text{milk limit}&=4\times\frac{600}{180}=13\frac13 \\ \text{egg limit}&=4\times\frac62=12 \end{aligned}

Answer. 10 portions; flour is the limiting ingredient.

Check. Ten portions need scale factor 10÷4=2.510\div4=2.5: 750 g750\,\mathrm{g} flour, 450 ml450\,\mathrm{ml} milk and 5 eggs, all available. Eleven portions need 825 g825\,\mathrm{g} flour, which is too much.

If only complete four-portion batches were allowed, these stocks would permit two batches, or eight portions. The restriction matters. In general, choose the smallest ingredient limit and round down to the permitted whole portion or batch count only at the end.

Solving linked ratio problems

Two ratios can describe one situation, but the entries for their shared quantity must represent the same number of equal parts before you join them.

If red to blue is 2:32:3 and blue to green is 3:53:5, the blue entries already agree. You can write red to blue to green as 2:3:52:3:5. If the blue entries differ, multiply each whole ratio until they agree; never change just the shared entry.

Problem-solving

Name the quantities in order, match the shared quantity, then use the actual total, share or difference to find one part. Check the finished amounts in both original ratios.

This is more than joining numbers: the given total must include exactly the groups being combined. A ratio involving only some of a collection cannot use the whole collection's total until the remaining groups are accounted for.

Worked example 3: shared entries already match

Question. A box contains red, blue and green beads only. Red to blue is 2:32:3 and blue to green is 3:53:5. There are 200 beads altogether. Find the number of green beads.

Plan. The blue entries match, so join the ratios and share the total.

red : blue : green=2:3:5total parts=2+3+5=10one part=200÷10=20green=5×20=100\begin{aligned} \text{red : blue : green}&=2:3:5 \\ \text{total parts}&=2+3+5=10 \\ \text{one part}&=200\div10=20 \\ \text{green}&=5\times20=100 \end{aligned}

Answer. 100 green beads.

Check. The counts are red 40, blue 60 and green 100. Their sum is 200; 40:60=2:340:60=2:3 and 60:100=3:560:100=3:5.

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