The 10% comes off the reduced price, not off the original price, so the true saving has to be worked out by multiplying, not by adding the percentages.

Take a price of £100. After 10% off it is \(£90\), and after a further 10% off that is \(90 \times 0.9 = £81\).

The saving is \(£100 - £81 = £19\), which is 19% of £100.

Check with multipliers: \(0.9 \times 0.9 = 0.81\), and \(1 - 0.81 = 0.19\), so the true saving is 19%.

The sign says 23%, which is not the same as the true saving, so the sign is wrong. The answer is No.

Take the two changes one at a time. A rise of 25% multiplies by \(1.25\), and a fall of 5% multiplies by \(0.95\).

After the first year: \(160,000 \times 1.25 = 200,000\)

After the second year: \(200,000 \times 0.95 = 190,000\)

So the flat is worth £190,000.

Notice that the 5% is taken off £200,000, not off the £160,000 Callum paid — which is why the overall change is not simply a 20% rise. The value has ended up £30,000 more than he paid.

An increase of 8.2% each year means the amounts are multiplied by \(1 + \frac{ 8.2 }{100} = 1.082\) each year.

Over 14 years this is done 14 times, so the total multiplier is a single power:

\(1.082^{ 14 } \approx 3.014\)

They need to reach 3 times their present level, and 3.014 is greater than 3.

So after 14 years the ticket prices at the park will be at least treble their present level.

Adding 8.2% once for each of the 14 years would give only \(14 \times 8.2 = 114.8\%\), short of the \(200\%\) rise needed. But each year's rise is taken on a larger amount, so the rises must be compounded as above.

Each change applies to the number left by the one before, so use a multiplier for each year.

An 8% increase multiplies by \(1.08\). A 7% fall multiplies by \(0.93\).

This year: \(50000 \times 1.08 = 54000\)

Next year: \(54000 \times 0.93 = 50220\)

The predicted number of passengers next year is 50,220, which is fewer than 50,300. So the answer is Yes.

Adding the percentages, \(8 - 7 = 1\%\), would treat the 7% as a change on the original 50,000 rather than on 54,000. That would give 50,500, which is more than 50,300 and points to the wrong answer.

Each month the amount owed falls by 15%, so it is multiplied by the same multiplier each month:

\(1 - \frac{ 15 }{100} = 0.85\)

After 1 month: \(£1000 \times 0.85 = £850.00\)

After 3 months, the multiplier is used 3 times:

\(£1000 \times 0.85^{ 3 } \approx £614.13\)

£614.13 is less than £630.00, so Erin's claim is correct. The answer is Yes.

A rise of 6.1% leaves 106.1% of the amount, so each year the number is multiplied by \(1.061\).

Doing this 8 years running multiplies by \(1.061\) 8 times, which is \(1.061^8\).

\(900 \times 1.061^8 \approx 1445.33\)

To the nearest whole number there will be about \(1445\) members.

A change of 15% multiplies by \(1.15\), and a change of 40% multiplies by \(0.6\).

Doing one and then the other multiplies by both:

\(1.15 \times 0.6 = 0.69\)

One number does the work of both changes, whichever order they happen in.

A multiplier of exactly \(1\) would mean no change at all, so compare the single multiplier with 1:

\(0.69 - 1 = -0.31\)

\(-0.31 \times 100 = -31\)

So the overall change is a 31% decrease.

It is not a 25% decrease. Percentages of two different amounts cannot be added: the 40% in 2019 was worked out from the population of the village at the end of 2018, not from the population of the village at the start of 2018.

Take the two changes one at a time.

An increase of 15% multiplies by \(\frac{ 115 }{100} = 1.15\):

\(104 \times 1.15 = 119.60\)

A reduction of 15% multiplies by \(\frac{ 85 }{100} = 0.85\):

\(119.60 \times 0.85 = 101.66\)

The sale price is £101.66.

The two percentages are percentages of different amounts.

The increase added 15% of £104, which is £15.60.

The reduction took away 15% of £119.60, which is £17.94.

£17.94 is more than £15.60, so the price finishes £2.34 below where it started — a 2.25% fall overall. The same thing happens whatever the starting price: an increase and a decrease of 15% always leave you at \(0.9775\) times the original, never at 1.

Each year has its own multiplier: \(1.09\), then \(1.06\), then \(1.03\).

Apply all three to the starting amount in one calculation, and keep every figure in the calculator until the end:

\(4,900 \times 1.09 \times 1.06 \times 1.03 \approx 5831.30\)

The investment is worth £5831.30 to the nearest penny.

Rounding part-way — to the nearest penny after each year — can shift the final answer by a penny or two, so round once, at the end.

Multiply the three multipliers together to get the single multiplier for the whole three years:

\(1.09 \times 1.06 \times 1.03 = 1.190062\)

A multiplier of \(1.190062\) is an increase of

\((1.190062 - 1) \times 100 = 19.0062\%\)

which is \(19.0\%\) to 1 decimal place.

It is not 18%, the three rates added. Each year's percentage is worked out from a different amount, so they cannot be added — and because every year's amount is larger than the last, the true figure is always a little more than the sum.

Work through the two years one step at a time — interest first, then the top-up.

A rate of 4.5% multiplies the balance by \(1.045\).

In the first year:

Interest: \(£7100.00 \times 1.045 = £7419.50\)

Top-up: \(£7419.50 + £1900.00 = £9319.50\)

In the second year:

Interest: \(£9319.50 \times 1.045 = £9738.8775\)

Top-up: \(£9738.8775 + £1900.00 = £11638.88\)

So the account holds £11638.88 at the end of the second year, to the nearest penny.

The £1,900 paid in after year 1 then sits in the account for the whole of year 2, so it earns interest too. Treating the two years as "two lots of interest on £7,100, plus two top-ups" gives only £11553.38 — £85.50 too little, because it misses the interest earned on that first top-up.