Hard GCSE Unit conversion Questions

Read the formula carefully

The formula only works if $m$ is in grams and $L$ is in centimetres.

Check the units given

The mass is in kilograms and the answer is wanted in metres — neither matches the formula.

Convert the mass

There are $1000\,\mathrm{g}$ in a kilogram, so the coil is $2000\,\mathrm{g}$.

Substitute into the formula

Now the units on both sides agree.

Rearrange for L

Divide both sides by $0.8$.

Clear the decimal

Multiplying top and bottom by $10$ makes the division exact.

Work out the length in cm

The wire is $2500\,\mathrm{cm}$ long.

Convert to metres

The coil is $25\,\mathrm{m}$ long.

Check by substituting back

$2000\,\mathrm{g}$ is $2\,\mathrm{kg}$, so the answer satisfies the original formula.

Note the first classic error

Putting the mass in as $2$ (kilograms) gives a $2.5\,\mathrm{cm}$ coil — absurd for $2\,\mathrm{kg}$ of wire.

Note the second classic error

Leaving the answer as $2500$ without converting would claim a coil $2.5\,\mathrm{km}$ long.

Rewrite the formula for metres

If the length is $x$ metres, the mass in grams is $80x$ — the wire weighs $80\,\mathrm{g}$ per metre.

Convert that rule to kilograms

A wire $x$ metres long has mass $0.08x$ kg.

Check the answer with the new rule

A $25\,\mathrm{m}$ coil has mass $2\,\mathrm{kg}$ — the rebuilt formula agrees.

State the answer

The coil of wire is $25$ metres long.

Write down what is given

The rate is in seconds but the running time is in hours and minutes.

Convert the hours

There are $3600$ seconds in an hour.

Convert the minutes

There are $60$ seconds in a minute.

Add the two parts

The machine runs for $13\,500$ seconds.

Check the time another way

Writing the time as a decimal number of hours gives the same $13\,500\,\mathrm{s}$.

Find the number of parts

One part is made every $45$ seconds.

Work out the division

The machine makes $300$ parts.

Check by multiplying back

$300$ parts really do take $13\,500$ seconds.

Find the plastic used

Each of the $300$ parts uses $250\,\mathrm{cm}^3$.

Recall the capacity fact

Now convert cubic centimetres to litres.

Convert to litres

The machine uses $75$ litres of plastic.

Check with a per-part conversion

Route $2$: convert first, then multiply.

Finish route $2$

The second route gives $75$ litres as well.

Express it in cubic metres

Since $1\,\mathrm{m}^3 = 1000$ litres, the plastic occupies $0.075\,\mathrm{m}^3$.

State the answer

The machine uses $75$ litres of plastic in the shift.

Write down what is given

The coverages are in different area units, so they cannot be compared yet.

Choose a common unit

Square metres are the natural unit for wall coverage.

Write down the linear fact

Start every area conversion from the length fact.

Square the linear factor

A square metre holds $10\,000$ square centimetres.

Convert Paint B's coverage

Paint B covers $25\,\mathrm{m}^2$, not the huge area the number $250\,000$ suggests.

Check that conversion

A $25\,\mathrm{m}^2$ square really does measure $250\,000\,\mathrm{cm}^2$, so the factor $10\,000$ is right.

Find the cost per $\mathrm{m}^2$ for A

Paint A costs $\pounds 0.40$ per square metre.

Find the cost per $\mathrm{m}^2$ for B

Paint B costs $\pounds 0.44$ per square metre.

Compare the two

Paint A is cheaper for every square metre painted.

Check the other way round

Square metres per pound: A gives $2.5\,\mathrm{m}^2$, B only $2.27\,\mathrm{m}^2$ — A again wins.

Quantify the difference

Paint A saves $4\mathrm{p}$ on every square metre.

Note the classic error

Using the linear factor would claim Paint B covers $2500\,\mathrm{m}^2$ for $\pounds 11$, making it look $100$ times better than it is.

Sanity-check the coverage

Both paints cover roughly $10 - 12\,\mathrm{m}^2$ per litre, which is realistic.

Answer the question asked

Value for money is decided by the cost of painting the same area.

State the answer

Paint A costs $\pounds 0.40$ per $\mathrm{m}^2$ against $\pounds 0.44$ per $\mathrm{m}^2$ for Paint B.

Write down what is given

The sides are in inches but the area is wanted in square centimetres.

Convert the length

Centimetres are shorter than inches, so the number gets bigger.

Convert the width

Both sides are now in centimetres.

Set up the area

Route $1$: convert the lengths first, then multiply.

Multiply without decimals

Ignoring the decimal points for now keeps the multiplication whole-number.

Replace the decimal point

There are four decimal places altogether, so the answer is $154.8384\,\mathrm{cm}^2$.

Start route $2$: area in square inches

Route $2$: find the area in the original unit first.

Square the conversion factor

For an area the linear factor must be squared.

Apply the squared factor

Route $2$ gives exactly the same area, so the conversion is sound.

Note the classic error

Using the linear factor on an area gives $60.96$ — far too small, and wrong by a factor of $2.54$.

Round to $1$ decimal place

The digit after the tenths place is $3$, so round down.

Sanity-check the size

A rough estimate gives about $150\,\mathrm{cm}^2$, which is close to $154.8\,\mathrm{cm}^2$.

Convert to $\mathrm{mm}^2$ as a check

Since $1\,\mathrm{cm}^2 = 100\,\mathrm{mm}^2$, this is another squared factor at work.

Confirm the units

Multiplying two lengths in centimetres always gives an area in square centimetres.

State the answer

To $1$ decimal place, the photograph has an area of $154.8\,\mathrm{cm}^2$.

Write down what is given

The limit is in kilograms but the cases are weighed in pounds.

Decide the operation

A kilogram is heavier than a pound, so the kilogram number must be smaller — divide.

Convert Jon's case

Since $2.2 \times 20 = 44$, Jon's case is exactly $20\,\mathrm{kg}$.

Check that conversion

Converting back returns $44\,\mathrm{lb}$.

Convert Amy's case

Since $2.2 \times 25 = 55$, Amy's case is exactly $25\,\mathrm{kg}$.

Check that conversion

Converting back returns $55\,\mathrm{lb}$.

Add the two masses

Together the cases weigh $45\,\mathrm{kg}$.

Compare with the allowance

They are inside the $46\,\mathrm{kg}$ limit.

Find the difference

The cases are $1\,\mathrm{kg}$ below the allowance.

Start route $2$: convert the limit

Instead, convert the allowance into pounds.

Add the cases in pounds

The two cases total $99\,\mathrm{lb}$.

Find the difference in pounds

They are $2.2\,\mathrm{lb}$ under the limit.

Convert that difference

The two routes agree: $1\,\mathrm{kg}$ to spare.

Note the classic error

Multiplying instead of dividing would make Jon's case $96.8\,\mathrm{kg}$ — heavier than Jon.

State the answer

The cases total $45\,\mathrm{kg}$ against an allowance of $46\,\mathrm{kg}$.