Solving Complex Measurement Problems: Conversions, Geometry, Time, and Money

Hello again. Last lesson focused on algebra: using rules, formulas, and unknown values rather than guessing. Measurement questions use the same careful thinking, but there is an extra trap: the numbers can be correct while the units are wrong.

These are Questions 33–38 of your hard Year 6 SATs and early Year 7 challenge. You will solve problems involving conversion, perimeter, area, volume, time, and money. The goal is not merely to remember formulas. It is to decide what each question is really asking, organise the information, and show enough working to earn method marks.

---

The first rule: make units agree

In a measurement problem, do not begin calculating until you have checked the units. You cannot sensibly add $3.75\text{ kg}$ and $240\text{ g}$ until they are both written in the same unit.

Here are the facts worth knowing securely:

Measure	Key conversion
Mass	$1\text{ kg}=1000\text{ g}$
Capacity	$1\text{ l}=1000\text{ ml}$
Length	$1\text{ km}=1000\text{ m}$
Length	$1\text{ m}=100\text{ cm}$
Length	$1\text{ cm}=10\text{ mm}$
Volume and capacity	$1000{\text{ cm}}^3=1\text{ l}$
Time	$1\text{ hour}=60\text{ minutes}$

A strong SATs routine is:

1. Underline the question being asked.
2. Circle every unit.
3. Convert to one sensible unit.
4. Write each calculation on a separate line.
5. Check that the final unit matches the answer required.

Question 33: Mass conversion before division

A cleaning company has $3.75\text{ kg}$ of washing powder. Before any cleaning begins, $240\text{ g}$ is spilled. Each room needs exactly $125\text{ g}$ of powder.

How many complete rooms can the company clean?

The mass is given in two units, so convert kilograms to grams first:

$$3.75\text{ kg}=3750\text{ g}$$

Subtract the spilled powder:

$$3750-240=3510$$

So $3510\text{ g}$ remains. Now divide by the amount needed for one room:

$$3510÷125$$

Check nearby multiples rather than trying to guess:

$$125\times 28=3500$$ $$125\times 29=3625$$

There is enough for $28$ rooms, but not $29$. There would be $10\text{ g}$ left over.

$$\boxed{28\text{ complete rooms}}$$

The word complete matters. Do not give a decimal answer when the question is asking how many whole rooms can be cleaned.

---

Perimeter: the distance around the outside

Perimeter is the total distance around the outer boundary of a shape. For a rectangle:

$$\text{perimeter}=2\times (\text{length}+\text{width})$$

Do not confuse this with area. Area measures the space inside a shape and uses square units such as ${\text{m}}^2$. Perimeter uses ordinary length units such as $\text{m}$ or $\text{cm}$.

Question 34: Fencing a garden

A rectangular garden is $18\text{ m}$ long. Its width is $3\text{ m}$ shorter than its length. There is a gate $1.2\text{ m}$ wide, so no fencing is needed across the gate. Fencing is sold only in rolls of $5\text{ m}$.

What is the smallest number of rolls needed?

First find the width:

$$18-3=15$$

The garden measures $18\text{ m}$ by $15\text{ m}$. Its full perimeter is:

$$2\times (18+15)$$ $$2\times 33=66$$

But the gate does not need fencing:

$$66-1.2=64.8$$

So $64.8\text{ m}$ of fence is needed. Now decide how many $5\text{ m}$ rolls are enough:

$$64.8÷5=12.96$$

You cannot buy $0.96$ of a roll. Twelve rolls would give only $60\text{ m}$, which is not enough. Therefore, round up to the next whole roll:

$$\boxed{13\text{ rolls}}$$

This is a common real-life maths idea: when buying whole items, round up whenever a partial item is needed.

---

Area: the space inside a shape

For a rectangle:

$$\text{area}=\text{length}\times \text{width}$$

Area is measured in square units because it counts how many unit squares cover a surface. For example, $12\text{ m}\times 5\text{ m}$ gives $60{\text{ m}}^2$, not $60\text{ m}$.

Questions about paths, borders, and frames are often solved by finding:

1. the area of the whole shape;
2. the area of the inner shape;
3. the difference between them.

Question 35: A paved path

A rectangular playground is $24\text{ m}$ long and $18\text{ m}$ wide. A path $1.5\text{ m}$ wide is built all the way around the inside edge of the playground.

The paving slabs each cover $0.25{\text{ m}}^2$. How many slabs are needed to cover the path?

Start with the area of the full playground:

$$24\times 18=432$$ $$\text{Whole area}=432{\text{ m}}^2$$

The path takes $1.5\text{ m}$ from both opposite sides. The remaining grass length is:

$$24-1.5-1.5=21$$

The remaining grass width is:

$$18-1.5-1.5=15$$

So the grass area is:

$$21\times 15=315$$ $$\text{Grass area}=315{\text{ m}}^2$$

Now find the path area:

$$432-315=117$$ $$\text{Path area}=117{\text{ m}}^2$$

Each slab covers $0.25{\text{ m}}^2$, so divide:

$$117÷0.25=468$$ $$\boxed{468\text{ slabs}}$$

A useful check: dividing by $0.25$ should make the answer larger, because each slab covers less than $1{\text{ m}}^2$.

---

Volume: filling three-dimensional space

A cuboid has three dimensions: length, width, and height. Its volume is:

$$\text{volume}=\text{length}\times \text{width}\times \text{height}$$

If the measurements are in centimetres, the answer is in cubic centimetres, written ${\text{cm}}^3$. For water containers, convert cubic centimetres to litres using:

$$1000{\text{ cm}}^3=1\text{ l}$$

Question 36: How much space remains in a tank?

A fish tank measures $80\text{ cm}$ by $45\text{ cm}$ by $50\text{ cm}$. It is initially $\frac{2}{3}$ full of water. Then $17.5\text{ l}$ of water is added.

How much more water can the tank hold?

First calculate the tank’s total volume:

$$80\times 45\times 50=180000$$ $$\text{Total volume}=180000{\text{ cm}}^3$$

Convert to litres:

$$180000÷1000=180$$ $$\text{Tank capacity}=180\text{ l}$$

It starts $\frac{2}{3}$ full:

$$180÷3=60$$ $$60\times 2=120$$

So it contains $120\text{ l}$ at first. Add the extra water:

$$120+17.5=137.5$$

Now subtract from the full capacity:

$$180-137.5=42.5$$ $$\boxed{42.5\text{ l}}$$

Notice the order: find the full capacity, find the starting amount, add the new amount, then find the empty space. Writing those stages prevents a very common error: subtracting from the wrong amount.

---

Time: use the clock carefully

Time questions become difficult when they cross midnight. The safest method is to split a journey into hours and minutes rather than treating time like ordinary decimals.

For example, $2$ hours $30$ minutes is not $2.30$ hours in decimal notation. An hour has $60$ minutes, not $100$.

Question 37: A journey after midnight

A coach leaves at $22:46$. Its driving time is $2$ hours $38$ minutes, and it makes a rest stop lasting $27$ minutes.

What time does the coach arrive, using the $24$-hour clock?

Add the $2$ hours first:

$$22:46+2\text{ hours}=00:46$$

The date has changed because the time has passed midnight. Add the remaining $38$ minutes:

$$00:46+38\text{ minutes}=01:24$$

Now add the rest stop:

$$01:24+27\text{ minutes}=01:51$$ $$\boxed{01:51}$$

A quick check: the total time is a little over $3$ hours, and $22:46$ plus just over $3$ hours should be a little before $02:00$. So $01:51$ is sensible.

---

Money: compare what you actually need

Money problems may combine price, mass, and rounding up. Be precise with decimal points, and remember that you cannot buy part of a packet unless the question says the shop sells items by weight.

Question 38: Which shop is cheaper?

A café needs at least $4.5\text{ kg}$ of pasta.

• Shop A sells $750\text{ g}$ packs for $\pounds 1.35$ each.
• Shop B sells $1.2\text{ kg}$ packs for $\pounds 2.04$ each.

The café can buy from either shop or use a mixture of both. What is the cheapest possible cost? How much change will it receive from $\pounds 10$?

First convert Shop A’s pack mass:

$$750\text{ g}=0.75\text{ kg}$$

At Shop A, the number of packs needed is:

$$4.5÷0.75=6$$

The cost is:

$$6\times \pounds 1.35=\pounds 8.10$$

At Shop B, three packs would weigh:

$$3\times 1.2=3.6\text{ kg}$$

That is not enough, so four packs are needed:

$$4\times 1.2=4.8\text{ kg}$$ $$4\times \pounds 2.04=\pounds 8.16$$

A mixture comes close: two packs from Shop B and three packs from Shop A give enough pasta:

$$(2\times 1.2)+(3\times 0.75)=4.65\text{ kg}$$

Its cost would be:

$$(2\times \pounds 2.04)+(3\times \pounds 1.35)$$ $$\pounds 4.08+\pounds 4.05=\pounds 8.13$$

So Shop A alone is still the cheapest choice:

$$\boxed{\pounds 8.10}$$

Finally, find the change:

$$\pounds 10.00-\pounds 8.10=\pounds 1.90$$ $$\boxed{\pounds 1.90\text{ change}}$$

The important reasoning point is that Shop B’s packet is bigger, but that does not automatically make it better value for the amount required.

---

Key takeaways

Hard measurement questions are usually several small maths tasks joined together.

• Convert quantities into matching units before adding, subtracting, multiplying, or dividing.
• Perimeter is the distance around a shape; area is the space inside it.
• For a border or path, subtract the inner area from the outer area.
• For a cuboid, multiply three dimensions, then convert ${\text{cm}}^3$ to litres if needed.
• When a time calculation passes midnight, write the new time clearly and use the $24$-hour clock.
• For money problems involving packets, check that you have bought enough, then compare the total costs.
• In practical problems involving rolls, slabs, or packs, you often need to round up to a whole item.

Next, you will move to Questions 39–44: geometry problems involving angles, coordinates, transformations, and properties of shapes.