Percentages · applications

Applications: Percentages

16 question types · Model Method and algebra, side by side

PSLE · O-Level · SAT · GCSE Higher

01

Multi-Tiered Successive Discounts & 9% GST

heuristicSequential Base Reduction / Tax on Discounted Payable

A laptop had an original marked price of $1800. During an IT fair, the store offered a storewide 20% discount. Customers holding a VIP membership card were entitled to an additional 5% discount on the discounted price. A 9% Goods and Services Tax (GST) was then applied strictly to the final payable amount. (a) What was the overall percentage discount given on the original marked price before GST? (b) How much did a VIP member pay in total for the laptop, including GST?

Marked100 units = $1800$1800
The marked price is 100 units, so 1 unit is $18.
Draw 1 whole bar of 100 units representing the $1800 marked price (1 unit = $18).
step 1 of 8

Represent the marked price as 100 units, reduce the first tier, subdivide the remainder into 100 sub-parts, and apply tax to the resulting units.

  1. Draw 1 whole bar of 100 units representing the $1800 marked price (1 unit = $18).
  2. Cut away 20 units (20%): remaining bar is 80 units (80 × 18 = $1440).
  3. Take 5% off the 80 units: 0.05 × 80 = 4 units discount.
  4. Net payable units: 80 − 4 = 76 units of original base.
  5. (a) Total discount: 20 + 4 = 24%.
  6. Net payable value: 76 × $18 = $1368.
  7. Add 9% GST: 1368 × 9100 = $123.12.
  8. (b) Total payable: $1368 + $123.12 = $1491.12.

answer(a) 24%; (b) $1491.12

techniqueTaking a Percent Off · Sales Tax

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the percentage discounts directly (20% + 5% = 25%) instead of calculating the 5% on the already reduced price.
  • Calculating the 9% GST on the original $1800 marked price instead of the net discounted payable price.
  • Subtracting GST from the discounted price rather than adding it.
02

Reverse Percentage with Shifting Markup and Markdown

heuristicWorking Backwards / Multiplier Elimination

A camera retailer marked up the cost price of a camera by 25% to set the display marked price. During a clearance sale, the retailer offered a 16% discount on the marked price. A customer purchased the camera at the sale price for $840. (a) What was the marked price of the camera? (b) How much profit did the retailer make from selling the camera?

Marked84 parts16% off100 parts
After 16% off, 84 parts of the marked price remain.
Represent marked price as 100 parts: after 16% discount, 84 parts = $840.
step 1 of 7

Use unitary division backwards: find 1% of the marked price from the 84% sale price, then relate the 100% marked price to the 125% cost model.

  1. Represent marked price as 100 parts: after 16% discount, 84 parts = $840.
  2. 1 part = 840 ÷ 84 = $10.
  3. 100 parts (Marked Price) = 100 × 10 = $1000.
  4. Represent cost price as 100 units. The marked price is 100 + 25 = 125 units.
  5. 125 units = $1000 ⟹ 1 unit = 1000 ÷ 125 = $8.
  6. Cost price (100 units) = 100 × 8 = $800.
  7. Profit = Sale Price − Cost = 840 − 800 = $40.

answer(a) $1000; (b) $40

techniqueReverse Percentages · Percentage Change

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Assuming the marked price can be found by adding 16% to the $840 sale price (840 × 1.16 = $974.40, which incorrectly uses $840 as the base).
  • Calculating profit as marked price minus cost price (1000 − 800 = $200) instead of sale price minus cost price (840 − 800 = $40).
03

Percentage with Invariant Total (Internal Demographic Shift)

heuristicTotal Invariance / Percentage-to-Unit Conversion

In an auditorium, 40% of the audience were boys and the rest were girls. After 48 boys entered the auditorium and 48 girls left the auditorium, the percentage of boys in the auditorium increased to 60%. (a) How many people were in the auditorium altogether? (b) How many girls were in the auditorium at first?

BeforeBBGGG5 unitsAfterBBBGG5 units
Before: boys 2 units, girls 3, out of 5.
Initial state: Boys : Girls = 40% : 60% = 2 : 3 (Total = 5 units).
step 1 of 6

Because equal numbers entered and left, total units remain constant. Convert percentages to ratios with a common total unit base.

  1. Initial state: Boys : Girls = 40% : 60% = 2 : 3 (Total = 5 units).
  2. Final state: Boys : Girls = 60% : 40% = 3 : 2 (Total = 5 units).
  3. Notice total units are already equal: 5 units before and 5 units after.
  4. Shift in boys: 3u − 2u = 1u = 48 people.
  5. (a) Total people: 5u = 5 × 48 = 240.
  6. (b) Girls at first: 3u = 3 × 48 = 144.

answer(a) 240 people; (b) 144 girls

techniqueFinding a Percent · Ratios with a Constant Total

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Treating 40% and 60% as fractions of changing bases rather than recognizing that the total is constant.
  • Finding the final number of girls (2 × 48 = 96) instead of the initial number of girls (144).
04

Percentage with Single Unchanged Quantity (Moisture Loss)

heuristicInvariant Insoluble Part / Constant Solid Component

A crate of fresh mushrooms weighed 40 kg. Water made up 90% of the total mass of the fresh mushrooms. After being dried under the sun, water made up only 60% of the mass of the dried mushrooms. (a) What was the mass of the mushrooms after drying? (b) What mass of water was evaporated during the drying process?

FreshWWWWWWWWWP40 kg
Fresh: water 9 units, pulp 1, total 40 kg.
Fresh state: Water : Pulp = 90% : 10% = 9 : 1. Total = 10 units = 40 kg.
step 1 of 7

Fix the solid pulp as 1 unit block and scale the total units around the unchanged pulp.

  1. Fresh state: Water : Pulp = 90% : 10% = 9 : 1. Total = 10 units = 40 kg.
  2. Value of 1 pulp unit = 40 ÷ 10 = 4 kg.
  3. Dried state: Water : Pulp = 60% : 40% = 3 : 2.
  4. Since pulp is 4 kg, 2 pulp units = 4 kg ⟹ 1 unit = 2 kg.
  5. Total dried units = 3 (water) + 2 (pulp) = 5 units.
  6. (a) Total dried mass: 5 × 2 = 10 kg.
  7. (b) Evaporated water: 40 − 10 = 30 kg.

answer(a) 10 kg; (b) 30 kg

techniqueRatios with a Constant Part · Finding a Percent

examsSAT

Common pitfalls

  • Assuming the mass drops by 30% because water drops from 90% to 60%, yielding 40 × 0.70 = 28 kg (ignoring base change).
  • Calculating 60% of the initial 40 kg (24 kg) as the dried mass.
05

Equating Residual Percentages of Different Wholes

heuristicEquating Remaining Percentages / Common Unit Scaling

Alan and Dylan had some savings at first. Alan had $60 more savings than Dylan. Alan spent 40% of his savings and Dylan spent 25% of his savings. After spending, they both had the exact same amount of savings left. (a) How much savings did Alan have at first? (b) How much money did they spend altogether?

AlankeptkeptkeptspentspentDylankeptkeptkeptspent
Alan keeps 60%: 3 units of 5.
Alan retains 60% = 35 of his model (3 shaded units out of 5 total units).
step 1 of 8

Equalize the numerators of the remaining fractional parts: 3 units of Alan's remainder equals 3 units of Dylan's remainder.

  1. Alan retains 60% = 35 of his model (3 shaded units out of 5 total units).
  2. Dylan retains 75% = 34 of his model (3 shaded units out of 4 total units).
  3. Since the shaded parts are equal, each unit in Alan's bar is identical in size to each unit in Dylan's bar.
  4. Alan has 5 identical units; Dylan has 4 identical units.
  5. Difference: 5u − 4u = 1u = $60.
  6. (a) Alan's initial savings: 5u = 5 × $60 = $300.
  7. Alan spent 2 units: 2 × 60 = $120. Dylan spent 1 unit: 1 × 60 = $60.
  8. (b) Total spent: 120 + 60 = $180.

answer(a) $300; (b) $180

techniqueFinding a Percent · Ratio, Fraction, Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Setting the ratio of savings as 40 : 25 or 60 : 75 without inverting the proportion to relate the totals.
  • Confusing who had more money and assigning 4 units to Alan and 5 units to Dylan.
06

Profit and Loss on Cost vs Discount on Marked Price

heuristicCompound Markup and Markdown / Margin Balancing

A merchant bought an electric scooter for a certain cost price. He marked up the cost price by 40% to set the display marked price. During a festival sale, he offered a 15% discount on the marked price. If he sold the scooter at this discounted price and made a profit of $114: (a) What was the cost price of the electric scooter? (b) What was the marked price of the scooter?

Cost100 units
The cost price is 100 units.
Cost price = 100 units.
step 1 of 8

Represent cost price as 100 units. A 40% markup gives 140 units. Discounting 15% reduces 140 units by 21 units to 119 units, revealing a 19-unit profit.

  1. Cost price = 100 units.
  2. Marked price = 100 + 40 = 140 units.
  3. Calculate 15% discount on 140 units: 0.15 × 140 = 21 units.
  4. Sale price = 140 − 21 = 119 units.
  5. Profit = 119 units − 100 units = 19 units.
  6. 19 units = $114 ⟹ 1 unit = 114 ÷ 19 = $6.
  7. (a) Cost price: 100 × 6 = $600.
  8. (b) Marked price: 140 × 6 = $840.

answer(a) $600; (b) $840

techniquePercentage Change · Taking a Percent Off

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting the 15% discount directly from the 40% markup (40% − 15% = 25%) and assuming the profit is 25% of cost.
  • Calculating the 15% discount on the cost price rather than on the marked price.
07

Simultaneous Percentage Units and Parts (Dual Subgroup Changes)

heuristicLinear Simultaneous Elimination with Percentage Rates

A bookstore had 360 pens at first, consisting of red pens and blue pens. The owner increased the stock of red pens by 20% and decreased the stock of blue pens by 30%. As a result, the total number of pens in the store became 327. (a) How many red pens were there at first? (b) How many blue pens were in the store in the end?

At first360 pens
360 pens, red and blue.
Total initial pens = 360.
step 1 of 10

Assume all pens decreased by 30%. The difference between the hypothetical total and the actual total reveals the red pen increase.

  1. Total initial pens = 360.
  2. If all 360 pens had decreased by 30%, the total decrease would be 0.30 × 360 = 108 pens.
  3. Hypothetical remaining pens = 360 − 108 = 252.
  4. Actual remaining pens = 327.
  5. Difference between actual and hypothetical = 327 − 252 = 75 pens.
  6. This surplus of 75 arises because red pens increased by 20% instead of decreasing by 30%.
  7. Unit gap per red pen: +20% − (−30%) = 50% = 0.50.
  8. (a) Initial red pens: 75 ÷ 0.50 = 150 red pens.
  9. Initial blue pens: 360 − 150 = 210.
  10. (b) Final blue pens: 210 − (0.30 × 210) = 210 − 63 = 147.

answer(a) 150 red pens; (b) 147 blue pens

techniqueFinding a Percent · Simultaneous by Elimination

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the difference between +20% and −30% as 10% instead of the full 50% algebraic gap.
  • Reporting the initial blue pens (210) instead of the final blue pens (147).
08

Percentage of a Remainder with Fixed Monetary Offsets

heuristicWorking Backwards with Subdivided Percentages

Mrs. Goh had a sum of money. She spent 25% of her money on a vacuum cleaner. She then spent 40% of the remaining money plus another $18 on a rice cooker. She was left with $126. (a) How much money did Mrs. Goh have at first? (b) How much did the rice cooker cost?

Remainder40%40%5 sub-units
The remainder in 5 sub-units: the cooker takes 2 of them and $18 more.
In the remainder bar, 40% is 2 sub-units, leaving 3 sub-units minus $18.
step 1 of 8

Draw a remainder bar partitioned into 5 units (40% = 2 units, 60% = 3 units) shifted by $18.

  1. In the remainder bar, 40% is 2 sub-units, leaving 3 sub-units minus $18.
  2. The leftover is $126, so 3 sub-units = 126 + 18 = $144.
  3. 1 sub-unit = 144 ÷ 3 = $48.
  4. Full remainder (5 sub-units) = 5 × 48 = $240.
  5. Now draw the main bar: vacuum cleaner took 25% = 14, leaving 34.
  6. 3 main units = $240 ⟹ 1 main unit = 240 ÷ 3 = $80.
  7. (a) Total money (4 main units): 4 × 80 = $320.
  8. (b) Rice cooker cost: 2 sub-units + $18 = (2 × 48) + 18 = 96 + 18 = $114.

answer(a) $320; (b) $114

techniqueFinding a Percent · Multiplying and Dividing Undo Each Other

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting $18 from $126 instead of adding it when reconstructing the 60% remainder.
  • Calculating 40% on the initial total money instead of the remaining money.
09

Multi-Subgroup Demographic Percentage Distribution

heuristicHierarchical Percentage Tree / Weighted Component Aggregation

At a science exhibition, 60% of the visitors were adults and the rest were children. 70% of the adults were men. Among the children, 60% were boys. If there were 252 men at the exhibition: (a) How many children attended the exhibition? (b) What percentage of all the visitors at the exhibition were male?

Visitorsmenwomenboysgirls100 unitsadults 60children 40
Everyone is 100 units: adults 60, children 40.
Total visitors = 100 units.
step 1 of 8

Express total visitors as 100 units. Partition into 60 adult units and 40 child units, then determine sub-units for males.

  1. Total visitors = 100 units.
  2. Adults = 60 units; Children = 40 units.
  3. Men = 70% of 60 units = 42 units.
  4. 42 units = 252 men ⟹ 1 unit = 252 ÷ 42 = 6 people.
  5. (a) Children = 40 units = 40 × 6 = 240 children.
  6. Boys = 60% of 40 units = 24 units.
  7. Total male units = 42 (men) + 24 (boys) = 66 units.
  8. (b) Male percentage: 66 units out of 100 units = 66%.

answer(a) 240 children; (b) 66%

techniqueFinding a Percent · Ratio, Fraction, Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Averaging the two percentages directly (70% + 60%2 = 65%) without weighting by the adult and child proportions.
  • Calculating boys as 60% of total visitors rather than 60% of children.
10

Constant Difference in Percentage Context (Age Invariance)

heuristicAge Gap Invariance / Fixed Difference Unit Equalization

In the year 2018, Mr. Tan's age was 400% of his son Ken's age. In the year 2026, Mr. Tan's age will become 250% of Ken's age. (a) How old was Ken in 2018? (b) In what year will Mr. Tan's age be 180% of Ken's age?

2018 Tan4u2018 Ken1u2026 Tan5u2026 Ken2u
2018: father 4 units, Ken 1; difference 3 units.
2018: Father : Ken = 400% : 100% = 4 : 1 (Difference = 3 units).
step 1 of 12

Equalize the difference in units between the two time periods, as age difference never changes.

  1. 2018: Father : Ken = 400% : 100% = 4 : 1 (Difference = 3 units).
  2. 2026: Father : Ken = 250% : 100% = 5 : 2 (Difference = 3 units).
  3. Notice the difference in both ratios is already 3 units.
  4. From 2018 to 2026 is 8 years.
  5. Ken's growth: 2u − 1u = 1u = 8 years.
  6. (a) Ken in 2018 was 1u = 8 years old. (Father was 4u = 32).
  7. Age difference is 32 − 8 = 24 years.
  8. When Father is 180% = 95 of Ken's age, Father has 9 parts and Ken has 5 parts.
  9. Difference = 9p − 5p = 4p = 24 years ⟹ 1p = 6 years.
  10. Ken's age will be: 5p = 5 × 6 = 30 years old.
  11. Years from 2018: 30 − 8 = 22 years.
  12. (b) Target year: 2018 + 22 = 2040.

answer(a) 8 years old; (b) 2040

techniqueRatios with a Constant Difference · Percentage

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Assuming Ken's age in 2026 is simply 8 + 8 = 16 and calculating 250% without verifying the difference consistency.
  • Forgetting that 250% means a ratio of 5:2, not 2.5:1 with unequal unit bases.
11

Tiered Threshold Discounts

heuristicPiecewise Linear Billing / Boundary Offset Evaluation

An online stationery distributor offers a promotion: for any order, the first $60 is charged at full price. Any amount of the purchase exceeding $60 receives a 15% promotional discount. (a) If Marcus places an order with a retail value of $140, how much does he pay? (b) If Chloe paid $102.50 for her order after the discount, what was the original retail value of her purchase?

$140$60 full$80 less 15%−12
Split $140: $60 at full price, $80 that gets the discount.
Part (a): Split $140 into $60 (fixed) and $80 (discountable).
step 1 of 9

Partition the payment bar into a fixed $60 block and an excess block with 85% payable value.

  1. Part (a): Split $140 into $60 (fixed) and $80 (discountable).
  2. 15% discount on $80 = 15100 × 80 = $12.
  3. Discounted excess: 80 − 12 = $68.
  4. (a) Total paid: 60 + 68 = $128.
  5. Part (b): Subtract the un-discounted $60 block: 102.50 − 60 = $42.50.
  6. This $42.50 corresponds to 85 parts out of 100 parts of the excess.
  7. 1 part = 42.50 ÷ 85 = $0.50.
  8. 100 parts (retail excess) = 100 × 0.50 = $50.
  9. (b) Total original retail value: 60 + 50 = $110.

answer(a) $128; (b) $110

techniqueTaking a Percent Off · Finding a Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Applying the 15% discount to the entire $140 order (140 × 0.85 = $119), ignoring the first $60 threshold.
  • Dividing the entire $102.50 by 0.85 to find Chloe's retail value.
12

Annual Simple Interest with Staggered Withdrawals

heuristicMulti-Period Capital Tracking / Interest-Principal Partitioning

Mr. Lim deposited $12000 into a savings account that paid a simple interest rate of 3.5% per annum. At the end of Year 1, the bank credited the first year's interest into his account, after which Mr. Lim withdrew $4420 from the account. The remaining balance stayed in the account for Year 2 at the same interest rate of 3.5% per annum. (a) What was the total interest earned by Mr. Lim across the two years? (b) What was the total amount in Mr. Lim's account at the end of Year 2?

Year 1$12000
Year 1 starts with $12000.
Year 1: Draw a bar of $12000.
step 1 of 7

Track the account balance stage-by-stage as an accumulating bar with credited interest additions and cash deductions.

  1. Year 1: Draw a bar of $12000.
  2. Interest credited (3.5%): 3.5100 × 12000 = $420.
  3. Bar expands to $12420.
  4. Cut off $4420: remaining principal bar is 12420 − 4420 = $8000.
  5. Year 2 interest (3.5% of $8000): 3.5100 × 8000 = $280.
  6. (a) Total interest earned: $420 + $280 = $700.
  7. (b) Final balance: $8000 + $280 = $8280.

answer(a) $700; (b) $8280

techniqueSimple Interest · Finding a Percent

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Calculating Year 2 interest on the original $12000 principal rather than on the reduced $8000 balance.
  • Deducting the $4420 withdrawal from the initial $12000 before adding the Year 1 interest.
13

Cycling Shares Given as a Fraction, a Decimal and a Percentage

heuristicConvert to Percentages / Percentage of Each School's Total

Three schools recorded how many of their students cycle to school. At Northfield, which has 600 students, 720 of the students cycle. At Eastbrook, which has 750 students, the share of students who cycle is 0.32. At Westgate, which has 400 students, 36% of the students cycle. (a) Rank the three schools by the share of their students who cycle, largest share first. (b) Which school has the greatest number of students who cycle, and how many more cyclists does it have than the school with the largest share?

Northfield35 of 100do not cycle35%
Northfield: 720 = 35100, so 35 parts of 100 are shaded.
Northfield: 720 = 7 × 520 × 5 = 35100, so 35 parts out of 100 are shaded.
step 1 of 8

Draw each school as a bar of 100 equal parts and shade the parts that cycle, so that the three shares are out of the same whole. Then find the size of one part at each school to count the cyclists.

  1. Northfield: 720 = 7 × 520 × 5 = 35100, so 35 parts out of 100 are shaded.
  2. Eastbrook: 0.32 means 32 hundredths, so 32 parts out of 100 are shaded.
  3. Westgate: 36% means 36 out of 100, so 36 parts out of 100 are shaded.
  4. (a) Every bar has 100 parts, so the shaded parts can be compared directly. The order is Westgate (36%), Northfield (35%), Eastbrook (32%).
  5. Now count the cyclists. One part is 1% of a school. At Northfield, 1% of 600 is 6 students, so 35% is 35 × 6 = 210 students.
  6. At Eastbrook, 1% of 750 is 7.5 students, so 32% is 32 × 7.5 = 240 students.
  7. At Westgate, 1% of 400 is 4 students, so 36% is 36 × 4 = 144 students.
  8. (b) Eastbrook has the most cyclists: 240 − 144 = 96 more than Westgate. Eastbrook's share is the smallest, but it is a share of the most students. Check: 144400 = 0.36 and 240750 = 0.32.

answer(a) Westgate 36%, Northfield 35%, Eastbrook 32%; (b) Eastbrook, 96 more

techniqueComparing with Percent · Comparing Percents, Fractions and Decimals · Finding a Percent

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Comparing the numbers as they are written, so that 720 looks smaller than 0.32 or 36%. The three shares must be written in the same form, such as out of 100, before they can be compared.
  • Taking the school with the largest share to have the most cyclists. Each share is out of that school's own number of students, so a smaller share of a larger school can be more students.
14

A School's Enrollment Rise, Worked Back and Carried a Year On

heuristicPercentage Increase as a Multiplier / A New Base for the Next Rise

After a new wing opened, a school's enrollment rose by 25% from last year to this year, and the school now has 1000 students. The school expects its enrollment to rise next year by the same number of students as it rose this year. (a) How many students did the school have last year? (b) By what percentage of this year's enrollment is the school expected to grow next year?

Last year25%25%25%25%100%This year25%25%25%25%+25%125%
25% is a quarter: last year is 4 units and the rise is 1 unit.
25% = 14, so draw last year's enrollment as 4 units and this year's rise as 1 more unit.
step 1 of 6

Since 25% = 14, draw last year's enrollment as 4 units and the rise as 1 more unit. Next year's rise is one more unit, now compared with this year's 5 units.

  1. 25% = 14, so draw last year's enrollment as 4 units and this year's rise as 1 more unit.
  2. This year's enrollment is 4 + 1 = 5 units, and 5 units are 1000 students.
  3. 1 unit = 1000 ÷ 5 = 200 students.
  4. (a) Last year: 4 units = 4 × 200 = 800 students.
  5. Next year the school expects another rise of 1 unit, or 200 students, on top of this year's 5 units: 1000 + 200 = 1200 students.
  6. (b) The rise is 1 unit out of this year's 5 units: 15 = 20100 = 20%. Check: 2001000 × 100% = 20%.

answer(a) 800 students; (b) 20%

techniquePercentage Increase · Reverse Percentages · Percentage Change

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking 25% off 1000 to get 750 students for last year. The 25% was a percentage of last year's enrollment, not of this year's, so this year's 1000 students are 125% of last year's.
  • Answering 25% for next year because the number of new students is the same. The same 200 students are now compared with 1000 students instead of 800, so they make a smaller percentage.
15

A Bakery Oven Losing a Fixed Percentage of Its Value Each Year

heuristicRepeated Percentage Decrease / One Year at a Time

A bakery bought an oven for $25000. Each year, the oven loses 20% of the value it had at the start of that year. (a) What is the oven worth at the end of the third year? (b) By what percentage of the price the bakery paid has the oven's value fallen over the three years?

Year 15000500050005000lost$20000
Year 1: $25000 is 5 units of $5000. One is lost, so $20000 remains.
Year 1: $25000 is 5 units of $5000. One unit is lost, so 4 units remain: 4 × 5000 = $20000.
step 1 of 6

Since 20% = 15, draw each year's starting value as 5 equal units. One unit is lost, and the other 4 units carry on to the next year, where they are split into 5 new, smaller units.

  1. Year 1: $25000 is 5 units of $5000. One unit is lost, so 4 units remain: 4 × 5000 = $20000.
  2. Year 2: $20000 is 5 units of $4000. One unit is lost, so 4 × 4000 = $16000 remains.
  3. Year 3: $16000 is 5 units of $3200. One unit is lost, so 4 × 3200 = $12800 remains.
  4. (a) At the end of the third year, the oven is worth $12800.
  5. The value lost over the three years is 5000 + 4000 + 3200 = $12200, which agrees with 25000 − 12800 = 12200.
  6. (b) As a percentage of the price paid: 1220025000 × 100% = 48.8%. This is less than 3 × 20% = 60%, because each year's 20% is taken from a smaller value.

answer(a) $12800; (b) 48.8%

techniqueDepreciation · Percentage Change · Finding a Percent

examsSAT · GCSE Higher

Common pitfalls

  • Taking $5000, which is 20% of the price paid, off every year to reach $10000. Each year's 20% is of the value at the start of that year, and that value gets smaller every year.
  • Adding the three percentages to say that the value fell by 60%. The three 20% drops are of three different values, so they cannot be added as percentages of the price paid.
16

Charity Fair Money Shared Out as a Fraction, a Decimal and a Percentage

heuristicConvert to One Form / A Part of the Amount Given Away

A club raised $4800 at a charity fair. It gave 38 of the money to a children's hospital, 0.3 of the money to an animal shelter and 20% of the money to the town library, and kept the rest for next year's fair. (a) How much money did the club keep? (b) What fraction of the money that the club gave away went to the children's hospital?

Raised40 units = $4800Hospital15 units3/8
The $4800 is 40 units. Hospital: 38 = 1540, so 15 units.
Hospital: 38 = 3 × 58 × 5 = 1540, so the hospital gets 15 units.
step 1 of 6

The shares are in eighths, tenths and hundredths, and 40 is the smallest number of equal units that shows 38, 0.3 and 20% exactly. Draw the $4800 as 40 units and give each share its number of units.

  1. Hospital: 38 = 3 × 58 × 5 = 1540, so the hospital gets 15 units.
  2. Shelter: 0.3 = 310 = 1240, so the shelter gets 12 units.
  3. Library: 20% = 20100 = 15 = 840, so the library gets 8 units.
  4. Given away: 15 + 12 + 8 = 35 units, so the club kept 40 − 35 = 5 units.
  5. 1 unit = 4800 ÷ 40 = $120. (a) The club kept 5 × 120 = $600.
  6. (b) The club gave away 35 units, and 15 of them went to the hospital: 1535 = 37. Check: $1800 + $1440 + $960 + $600 = $4800.

answer(a) $600; (b) 37

techniqueFractions, Decimals and Percents · Comparing Percents, Fractions and Decimals · Finding a Percent

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Adding the numbers as they are written, 38 + 0.3 + 20, which mixes three forms. Each share must be written in the same form, such as fortieths or percentages, before the shares can be added.
  • Answering (b) with 38. That is the hospital's share of all the money raised, but (b) asks for its share of the $4200 given away, which is a smaller whole.
Mr. Chalk Drill percentages