Advanced Ratio, Proportion, and Scaling Problems

Hello again. In the previous lesson, fractions, decimals and percentages described parts of a whole. This lesson develops a closely related idea: ratio compares quantities, while proportion tells us that a relationship stays in the same balance as quantities change.

These are questions 21–26 of your hard Year 6 SATs challenge. You will learn how to split totals in a ratio, find missing values, spot wording traps, scale recipes and rates, convert units, and enlarge similar shapes. These methods also appear regularly in Year 7 maths.

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Keep the relationship balanced

A ratio compares quantities in a particular order. For example:

$$\text{red : blue}=4:7$$

means that every group has $4$ red items for every $7$ blue items. The order matters:

$$4:7\ne 7:4$$

When a ratio is scaled, every part is multiplied or divided by the same number. So these are equivalent ratios:

$$2:3=6:9=20:30$$

The key SATs question is usually: What number has this part been multiplied or divided by?

Model 1: one part of a ratio is known

A craft club uses red and black card in the ratio:

$$4:7$$

It uses $63$ black sheets. Since black corresponds to $7$ parts, find the scale factor:

$$63÷7=9$$

So each ratio part represents $9$ sheets. The number of red sheets is:

$$4\times 9=36$$

The complete ratio is:

$$36:63$$

A quick check is to divide both numbers by $9$:

$$36:63=4:7$$

The check works.

Model 2: a total must be shared in a ratio

A school receives $288$ books. Fiction books and non-fiction books are shared in the ratio:

$$5:3$$

First find the total number of ratio parts:

$$5+3=8$$

Each part is:

$$288÷8=36$$

Now multiply by the correct number of parts:

$$\text{fiction}=5\times 36=180$$ $$\text{non-fiction}=3\times 36=108$$

The answer is $180$ fiction books and $108$ non-fiction books.

A reliable way to set this out is:

Quantity	Ratio parts	Actual amount
Fiction	$5$	$180$
Non-fiction	$3$	$108$
Total	$8$	$288$

Do not divide $288$ by $5$ or $3$ first. The $288$ represents all eight parts together.

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Read the words before writing the ratio

Hard SATs questions often contain a sentence that looks like a ratio but is actually describing a part and a whole.

Model 3: the “out of every” trap

“Out of every $5$ seeds planted, $3$ grow into plants. If $42$ seeds grow, how many seeds were planted?”

The statement does not mean planted : grown is $5:3$, because the $3$ seeds that grow are included within the $5$ seeds planted.

Instead, each group of $5$ contains:

• $3$ seeds that grow;
• $2$ seeds that do not grow.

So the ratio of growing to not growing seeds is:

$$3:2$$

We know that $3$ parts represent $42$ growing seeds:

$$42÷3=14$$

One group is scaled by $14$, so the number planted is:

$$5\times 14=70$$

Therefore, $70$ seeds were planted.

The useful habit is to ask: What does each number count? In this example, “planted” is the total of both outcomes.

Another version of this trap occurs in queues. If there are $37$ children in a queue including Maya, and there are twice as many children ahead of Maya as behind her, do not use $37$ as the total for the ratio. Maya is neither ahead nor behind herself.

You would first remove Maya:

$$37-1=36$$

Then split $36$ in the ratio:

$$2:1$$

There are $3$ parts, so each part is:

$$36÷3=12$$

There are $2$ parts ahead:

$$2\times 12=24$$

So $24$ children are ahead of Maya.

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Proportion: scaling a complete situation

A proportional situation keeps the same rate or recipe. If you make three times as much paint, you need three times as much of every ingredient. If a machine runs for fifteen times as long, it produces fifteen times as much output at the same rate.

Model 4: scale a rate, then convert the unit

A juice machine pours $250\text{ ml}$ every $4$ seconds. How many litres does it pour in $1$ minute?

One minute is:

$$60\text{ seconds}$$

Find how many lots of $4$ seconds fit into $60$ seconds:

$$60÷4=15$$

The machine pours $15$ equal batches, so the total volume is:

$$250\times 15=3750\text{ ml}$$

The question asks for litres. Since:

$$1000\text{ ml}=1\text{ L}$$

we calculate:

$$3750÷1000=3.75$$

The machine pours:

$$\boxed{3.75\text{ L}}$$

A frequent exam mistake is stopping at $3750\text{ ml}$. The calculation is correct, but the unit is not what the question requested.

Model 5: scale a price in pence, then write pounds

Cheese costs $84\text{ p}$ per $125\text{ g}$. What does $750\text{ g}$ cost?

First find the scale factor:

$$750÷125=6$$

So $750\text{ g}$ costs six lots of $84\text{ p}$:

$$84\times 6=504\text{ p}$$

Now change pence to pounds:

$$504\text{ p}=\pounds 5.04$$

So the cost is:

$$\boxed{\pounds 5.04}$$

This works because the price is proportional to the mass: twice as much cheese costs twice as much, provided the rate stays the same.

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Scale factors for similar shapes

A scale factor tells you how much every corresponding length changes.

To find a scale factor from an original shape to a new shape, use corresponding lengths:

$$\text{scale factor}=\frac{\text{new corresponding length}}{\text{original corresponding length}}$$

A scale factor:

• greater than $1$ makes a shape larger;
• between $0$ and $1$ makes a shape smaller;
• equal to $1$ leaves the shape unchanged.

When shapes are similar, their corresponding angles are equal and their corresponding side lengths have the same scale factor.

Model 6: find a scale factor and use the perimeter

A small triangular sign has side lengths:

$$4.8\text{ cm},6.2\text{ cm},7.6\text{ cm}$$

A similar larger sign has a side corresponding to the $4.8\text{ cm}$ side that measures $12\text{ cm}$.

First find the scale factor:

$$12÷4.8=2.5$$

Every side length is multiplied by $2.5$:

$$6.2\times 2.5=15.5$$ $$7.6\times 2.5=19$$

So the larger triangle has side lengths:

$$12\text{ cm},15.5\text{ cm},19\text{ cm}$$

Its perimeter is:

$$12+15.5+19=46.5$$

Therefore, the perimeter is:

$$\boxed{46.5\text{ cm}}$$

You could also scale the original perimeter. The small perimeter is:

$$4.8+6.2+7.6=18.6\text{ cm}$$

Then:

$$18.6\times 2.5=46.5\text{ cm}$$

Both methods agree.

Be precise with the phrase “scale factor $4$.” It means every length is multiplied by $4$. It does not mean the area is multiplied by $4$. If both length and width are multiplied by $4$, the area is multiplied by:

$$4\times 4=16$$

For a scale factor $k$:

$$\text{new perimeter}=k\times \text{original perimeter}$$ $$\text{new area}=k^2\times \text{original area}$$

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Your SATs method for unfamiliar ratio questions

When a question has lots of information, use this order:

1. Underline what is being asked. Is it one missing amount, a total, a cost, a time, a perimeter, or a measurement in a different unit?

2. Label the quantities. Write, for example, “red : blue,” not just $4:7$. This prevents reversing the answer.

3. Decide which structure you have.

   • One ratio part is known: divide to find the scale factor.
   • A total is known: add the parts, divide the total, then multiply.
   • A rate or recipe changes: find how many equal batches are needed.
   • Similar shapes are involved: find the scale factor from matching lengths.

4. Deal with units at the end. Common conversions include:

   $$1000\text{ g}=1\text{ kg}$$ $$1000\text{ ml}=1\text{ L}$$ $$100\text{ p}=\pounds 1$$

5. Check the relationship. For a ratio, simplify your final amounts. For a scaled quantity, check that both quantities changed by the same factor.

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Key takeaways

Ratio and proportion questions become manageable when you identify what each number represents.

• A ratio compares quantities in a fixed order.
• Equivalent ratios are made by multiplying or dividing every part by the same factor.
• To split a total, add the ratio parts before dividing.
• Read “out of every” wording carefully: one number may be part of the other total.
• In proportional situations, all linked quantities scale by the same factor.
• Convert into the requested unit only after completing the calculation.
• For similar shapes, all side lengths and perimeters scale by the scale factor; areas scale by the square of it.

Next, you will move into early algebra: using sequences, rules, symbols, and missing values to describe patterns and solve problems efficiently.