Question 1 of 24
Find 18% of 350.
Show a hint
Convert 18% to 0.18, then multiply by 350.
Cambridge O Level · Mathematics 4024
Use percentages to compare and change quantities, solve profit, loss, interest and reverse-percentage problems, and model repeated growth or decay with multipliers. This original practice pack covers sections 1.13 and 1.17; it is not a complete course and has not been academically reviewed.
2025 / 2026 / 2027 · Academic review not recorded · Published 2026-09-08
AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.
Guided tutorial
21 lesson sections
Explanation
A percentage compares a quantity with a chosen whole, called the base or reference quantity. The word percent means per hundred, so p% is p/100. Before calculating, decide which value represents 100%; changing the base can change the percentage completely.
To find p% of Q, calculate (p/100) × Q. To express A as a percentage of B, calculate (A/B) × 100%, after converting matching measurements to the same units. The quantity after the word 'of' is often the base, but the context must decide.
Worked example
Find 17.5% of 640. Convert the percentage to a decimal: 17.5% = 0.175. Then 0.175 × 640 = 112, so the required amount is 112.
A mental check supports the calculation: 10% of 640 is 64, 5% is 32 and 2.5% is 16. Their sum is 64 + 32 + 16 = 112. The check also confirms that the answer is sensibly below 20% of 640.
Explanation
In the calculation (compared quantity/reference quantity) × 100%, the denominator is the value being treated as 100%. For example, an increase from 50 to 65 adds 15, and that increase is measured against the original 50, not the new 65.
A percentage can exceed 100%. If a new quantity is 1.4 times the reference quantity, it is 140% of that reference. This does not mean a 140% increase: the increase itself is only the extra 40% above the original 100%.
Worked example
Express 54 kg as a percentage of 45 kg. The units match, and 45 kg is the stated reference, so calculate (54/45) × 100% = 1.2 × 100% = 120%.
The answer exceeds 100% because 54 kg is greater than 45 kg. It says that 54 kg is 120% of 45 kg, which is equivalent to being 20% greater than 45 kg. Those two statements use different percentage descriptions of the same comparison.
Explanation
An increase of r% keeps the original 100% and adds r%, giving multiplier 1 + r/100. A decrease of r% keeps 100 − r percent, giving multiplier 1 − r/100. Multiplying once gives the new quantity directly.
For a 12% increase, use 1.12; for an 18% decrease, use 0.82. A decrease multiplier must lie between 0 and 1 when the rate is between 0% and 100%. Writing 1.18 for an 18% decrease would move the value in the wrong direction.
Worked example
A fee of MVR 800 rises by 12%. The multiplier is 1.12, so the new fee is 800 × 1.12 = MVR 896. The increase itself is MVR 96, which is 0.12 × 800.
A price of MVR 750 is reduced by 18%. The multiplier is 0.82, so the sale price is 750 × 0.82 = MVR 615. Checking by subtraction gives an MVR 135 reduction and 750 − 135 = 615.
Explanation
Percentage change is (change/original) × 100%. Use the magnitude of the change, then describe it as an increase or decrease. The original value belongs in the denominator because it is the reference value before the change occurred.
For a sale, profit = selling price − cost price and loss = cost price − selling price. Unless a question states another basis, profit percentage and loss percentage are measured against cost price. A margin measured against selling price is a different calculation.
Worked example
An item costs MVR 160 and sells for MVR 196. The profit is 196 − 160 = MVR 36. Profit percentage = (36/160) × 100% = 22.5%.
Dividing by 196 would give about 18.4%, but that uses the selling price as the base and does not answer the usual cost-based profit question. A check is 22.5% of 160 = 36, and 160 + 36 = 196.
Explanation
A reverse-percentage problem gives the quantity after a change and asks for the original. If final = original × multiplier, then original = final ÷ multiplier. The final value is not normally 100%, so taking the stated percentage of it cannot undo the change.
After a 30% discount, 70% of the original remains, so divide the sale price by 0.70. After a 15% increase, the final value is 115% of the original, so divide by 1.15. Substitute the recovered original into the forward calculation to check it.
Worked example
A jacket costs MVR 504 after a 30% discount. The sale price is 70% of the original, so original price = 504 ÷ 0.70 = MVR 720.
Subtracting 30% of 504 would give MVR 352.80 and is incorrect because 504 is the reduced price, not the original base. The forward check is 30% of 720 = 216 and 720 − 216 = 504.
Worked example
A trader sells an item for MVR 690 after making a 15% profit on cost. The selling price is 115% of cost, so cost price = 690 ÷ 1.15 = MVR 600.
The profit is MVR 90. Checking against the correct base gives 90/600 × 100% = 15%. Calculating 15% of the selling price would use the wrong reference quantity and would not recover the cost.
Explanation
With simple interest, each equal time period earns interest on the original principal only. If principal is P, annual rate is r% and time is t years, the interest is P × (r/100) × t. The accumulated amount is principal plus interest.
Keep the rate and time units consistent. An annual rate requires time in years unless the problem gives a convention for part-years. Distinguish the interest earned from the final balance: they differ by the original principal.
Worked example
MVR 3600 is invested at 5.5% simple interest per year for 4 years. Interest = 3600 × 0.055 × 4 = MVR 792.
The final amount is 3600 + 792 = MVR 4392. A yearly check gives MVR 198 interest each year; four identical interest payments total MVR 792 because simple interest does not add earlier interest to the principal.
Explanation
Repeated percentage change applies the multiplier to the latest value each time. For initial value P, rate r% per period and n periods, growth gives P(1 + r/100)^n and decay gives P(1 − r/100)^n. The exponent counts completed changes.
Compound interest is repeated growth because interest from one period becomes part of the next period's balance. Keep the unrounded calculator value through the calculation and round only the final money amount as required. Form the multiplier from the stated rate rather than trying to recall an unexplained formula.
Worked example
MVR 2400 is invested at 6% compound interest per year for 3 years. The annual multiplier is 1.06, so the amount is 2400 × 1.06^3 = MVR 2858.4384.
Rounded to the nearest laari, the amount is MVR 2858.44. The three year-end balances before final rounding are 2544, 2696.64 and 2858.4384. Using 2400 × (1 + 3 × 0.06) would calculate simple interest instead.
Explanation
Exponential growth occurs when the same positive percentage acts on the current value during each equal interval. The absolute increase therefore becomes larger as the quantity grows, even though the percentage rate stays constant.
Write the initial value, multiplier and number of completed intervals explicitly. A population growing by 2.8% per year uses multiplier 1.028. After n years, its model is initial population × 1.028^n, provided the stated model assumptions continue.
Worked example
A model starts with a population of 18 500 and assumes growth of 2.8% each year. After 4 years, the model gives 18 500 × 1.028^4 = 20 660.659819136.
If a whole-number population is required, this rounds to 20 661. The exponent is 4 because four annual changes have occurred. This is a mathematical projection based on a fixed rate, not a claim that a real population must follow the model exactly.
Explanation
Exponential decay keeps a fixed percentage of the current value each interval. A 18% annual depreciation leaves 82%, so its multiplier is 0.82. Repeating this multiplier produces decreasing absolute losses because it acts on a smaller balance each time.
Depreciation value is not the same as total depreciation. The value after n periods is initial value × decay multiplier^n; subtract that result from the initial value only if the problem asks how much value has been lost altogether.
Worked example
A machine is valued at MVR 45 000 and depreciates by 18% each year. After 3 years, its modelled value is 45 000 × 0.82^3 = MVR 24 811.56.
The total modelled loss is 45 000 − 24 811.56 = MVR 20 188.44. Subtracting 3 × 18% = 54% of the original would incorrectly assume equal absolute losses instead of applying depreciation to each new value.
Explanation
For successive percentage changes, multiply their factors in time order. Do not usually add or subtract the rates because each later rate acts on a changed base. Equal percentage increases and decreases do not cancel: (1 + r)(1 − r) is less than 1 for any non-zero decimal rate r.
The combined multiplier gives the net percentage change. If the product is 1.04, the final value is 104% of the original, a net 4% increase. If the product is 0.96, the final value is 96% of the original, a net 4% decrease.
Worked example
A quantity of 1200 increases by 30% and then decreases by 20%. Apply both multipliers: 1200 × 1.30 × 0.80 = 1248.
The combined multiplier is 1.30 × 0.80 = 1.04, so the net change is a 4% increase, not 10%. The second change is 20% of 1560, the value after the first change, rather than 20% of the original 1200.
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Question 1 of 24
Convert 18% to 0.18, then multiply by 350.
Quick recall
Card 1 of 16
Front
Review the essentials
First identify the base: use p/100 × quantity for a percentage of, compared/reference × 100% for a percentage comparison, and change/original × 100% for percentage change. For profit or loss, check whether the question defines the percentage against cost price.
Use 1 + r/100 for an increase and 1 − r/100 for a decrease. Divide the final value by its multiplier for reverse percentages. Use a single repeated multiplier with an exponent for compound interest, population change and depreciation, but use the original principal each period for simple interest.
Check direction, size, units and periods. An increase should not produce a smaller answer; a reverse discount should produce an original price above the sale price; the exponent should equal the number of completed changes; and intermediate rounding should be avoided.
Authorship: original ai assisted.
Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.