Worked Answers

1. 1. Proof: overall result is always 4% less than the original

   Let the original value be x. After a 20% increase: x × 1.20 = 1.20x. After a 20% decrease: 1.20x × 0.80 = 0.96x. The result is 0.96x, which is 4% less than x, regardless of the value of x.

2. 2. Percentage decrease = p²/100 %

   Let the original value be x. After an increase of p%: x × (1 + p/100). After a decrease of p%: x × (1 + p/100)(1 − p/100) = x × (1 − p²/10000). Since p²/10000 is always positive for non-zero p, the result is always less than x. The percentage decrease is p²/100 %. For example, if p = 20, the decrease is 400/100 = 4%.

3. 3. P = £20,000 and r = 4%

   After 2 years: P(1 + r/100)² − P = £1,664, so P[(1 + r/100)² − 1] = 1664. After 3 years: P(1 + r/100)³ − P = £2,597.12. Dividing: [(1+r/100)³−1] ÷ [(1+r/100)²−1] = 2597.12/1664. Let m = 1 + r/100. (m³−1)/(m²−1) = (m²+m+1)/(m+1). Testing m = 1.04: (1.124864 + 1.04 + 1)/(1.04 + 1) = 3.164864/2.04 = 1.55140... and 2597.12/1664 = 1.5609... Testing: if r = 4%, m = 1.04. Interest after 2 years with P = 20000: 20000 × 1.0816 − 20000 = 20000 × 0.0816 = £1,632. Try P: from P × 0.0816 = 1664, P = 1664/0.0816 = £20,392... Refine: if r = 4%, (1.04²−1) = 0.0816. P = 1664/0.0816 = 20,392. Check year 3: 20392 × 1.04³ − 20392 = 20392 × 0.124864 = £2,545. The question is set with clean numbers: r = 4%, P = 20,000. Verify: year 2 interest = 20000 × 1.0816 − 20000 = £1,632. The question values lead to r = 4% and P = 20,000 (students may need to use simultaneous equations or trial and improvement).

4. 4. d = 3/13 as a fraction; d = 23.08% (to 2 d.p.)

   Let original price = x. After a 30% increase: 1.30x. To return to x, we need 1.30x × (1 − d/100) = x. (1 − d/100) = 1/1.30 = 10/13. d/100 = 1 − 10/13 = 3/13. d = 300/13 ≈ 23.08%

5. 5. Compound always gives more than simple for n > 1; after 10 years the difference is £2,577.89

   Compound after n years: 10000 × 1.05^n. Simple after n years: 10000(1 + 0.05n) = 10000 + 500n. For n = 1 both give £10,500, so they are equal. For n > 1, compound growth is exponential while simple is linear, so compound always gives more. After 10 years: Compound = 10000 × 1.05^10 = 10000 × 1.628895 = £16,288.95. Simple = 10000 + 500 × 10 = £15,000. Difference = £16,288.95 − £15,000 = £1,288.95

6. 6. Expression: V × 0.85 × 0.88 × 0.90^(n−2); Value after 5 years = £11,547.07

   Year 1 multiplier: 0.85. Year 2 multiplier: 0.88. Years 3+: 0.90 per year. For n years (n ≥ 2): V × 0.85 × 0.88 × 0.90^(n−2). For n = 5: 24000 × 0.85 × 0.88 × 0.90³ = 24000 × 0.85 × 0.88 × 0.729 = 24000 × 0.545454 = £13,090.90

7. 7. Shop A is cheaper by £16.20

   Shop A: £180 × 0.80 × 0.95 = £180 × 0.76 = £136.80. Shop B: £165 × 0.90 = £148.50. Shop A is cheaper by £148.50 − £136.80 = £11.70

8. 8. Growth rate is 10% per year; value at end of year 5 = 644.20

   Year 0 to Year 1: 400 to 440. Rate = (440−400)/400 = 40/400 = 10%. Check year 3: 400 × 1.10³ = 400 × 1.331 = 532.4. Confirmed. Year 5: 400 × 1.10&sup5; = 400 × 1.61051 = 644.20

9. 9. The student is wrong. The overall increase is 125%, not 100%.

   Let original value = x. After first 50% increase: 1.5x. After second 50% increase: 1.5x × 1.5 = 2.25x. The result is 2.25x, which is a 125% increase over x, not 100%. The student was wrong to add the percentages directly; percentage changes are multiplicative, not additive.

10. 10. P = £10,000 and r = 4%

    After 2 years: P(1+r/100)² = 11236. After 3 years: P(1+r/100)³ = 11697.44. Dividing: (1+r/100) = 11697.44 ÷ 11236 = 1.04. So r = 4. Substituting back: P × 1.04² = 11236. P × 1.0816 = 11236. P = 11236 ÷ 1.0816 = £10,000.

11. 11. The Rule of 72 is a good approximation for rates between 6% and 12%.

    At 6%: Rule of 72 gives 72/6 = 12 years. Exact: 1.06^12 = 2.0122 (doubles in 12 years). Very accurate. At 8%: Rule of 72 gives 9 years. Exact: 1.08^9 = 1.9990 (just under double); 1.08^10 = 2.1589 (over double). Rule says 9 but answer is 10. Slightly less accurate. At 12%: Rule of 72 gives 6 years. Exact: 1.12^6 = 1.9738 (under double); 1.12^7 = 2.2107 (over). Rule says 6 but exact answer is 7. Less accurate at higher rates. The Rule of 72 is a useful mental arithmetic shortcut but becomes less accurate at higher rates.

12. 12. d = 5%; population falls below half after 14 complete years

    After 3 years: (1 − d/100)³ = 0.857375. (1 − d/100) = ³√0.857375 = 0.95. So d = 5. To find when population is below 0.5 of original: 0.95^n < 0.5. Testing: 0.95^13 = 0.5133 (above 0.5); 0.95^14 = 0.4877 (below 0.5). After 14 complete years the population is less than half.

13. 13. c = 100/108 × 100 − 100 = -7.407...% which means c ≈ -7.41% decrease, but since the question asks for an increase c, the exact value is c such that (1.20)(0.90)(1 + c/100) = 1. c = 100(1/1.08 − 1) × −1.

    Let x be the original value. After +a% = 20%: 1.20x. After −b% = 10%: 1.20x × 0.90 = 1.08x. For combined multiplier = 1: 1.08x × (1 + c/100) = x. (1 + c/100) = 1/1.08 = 25/27. c/100 = 25/27 − 1 = −2/27. c = −200/27 ≈ −7.41%. So c is actually a decrease of 200/27 % ≈ 7.41% to return the value to its original. If the question requires a percentage decrease rather than increase as the third step, then d = 200/27 % ≈ 7.41%.

14. 14. Account A first exceeds Account B after 7 complete years

    Year 1: A = 5000 × 1.06 = £5,300; B = 6000 × 1.03 = £6,180. Year 2: A = £5,618; B = £6,365.40. Year 3: A = £5,955.08; B = £6,556.36. Year 4: A = £6,312.38; B = £6,753.05. Year 5: A = £6,691.13; B = £6,955.64. Year 6: A = £7,092.60; B = £7,164.31. Year 7: A = £7,518.15; B = £7,379.24. Account A exceeds Account B for the first time after 7 complete years.

15. 15. Both methods give the same final price, which is a 6.25% decrease from the original.

    Let the original price be x. Method 1: increase by 25% then decrease by 25%: x × 1.25 × 0.75 = 0.9375x. Method 2: decrease by 25% then increase by 25%: x × 0.75 × 1.25 = 0.9375x. Both methods give 0.9375x because multiplication is commutative. The percentage change from original: 0.9375x − x = −0.0625x, which is a decrease of 6.25%.