Iteration, growth and decay – Year 11 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Increase: \(1+0.12=\) 1.12. Decrease: \(1-0.08=\) 0.92. For an increase add the percentage (as a decimal) to 1; for a decrease subtract it from 1.
   2. \(1+0.025=\) 1.025; \(1-0.15=\) 0.85. 2.5% as a decimal is 0.025 and 15% is 0.15.
   3. \(240\times1.35=\) £324. A 35% increase gives the multiplier 1.35, so one multiplication does the whole job.
   4. \(50\times0.76=\) £38. A 24% decrease leaves 76%, so the multiplier is 0.76.

2. Answer to Question 2

   1. 1.04. 4% as a decimal is 0.04, and we keep the original 100%, so the multiplier is \(1+0.04=1.04\).
   2. \(2000\times1.04^3=2000\times1.124864=\) £2 249.73. Compound interest multiplies by 1.04 once for each year, so we use the power 3. Round to the nearest penny.
   3. \(2249.73-2000=\) £249.73. The interest is the final amount minus the amount invested.
   4. Amount \(=2000\times1.04^n\). After \(n\) years the multiplier has been applied \(n\) times.

3. Answer to Question 3

   1. \(1-0.20=\) 0.8. A 20% fall leaves 80% of the value each year.
   2. \(15\,000\times0.8^3=15\,000\times0.512=\) £7 680. The multiplier is applied once for each of the 3 years.
   3. \(15\,000-7\,680=\) £7 320. Subtract the value after 3 years from the original price.
   4. \(\dfrac{7320}{15000}\times100=\) 48.8% (not \(3\times20=60\%\)). Divide the loss by the original value. The loss is not 60%, because each fall is worked out on a smaller value.

4. Answer to Question 4

   1. Year 1: 1.03, year 2: 1.025, year 3: 1.02. Add each rate (as a decimal) to 1.
   2. \(4000\times1.03\times1.025\times1.02=\) £4 307.46. Multiply by each year's multiplier in turn, and round to the nearest penny.
   3. Overall multiplier \(=1.03\times1.025\times1.02=1.076865\), so the increase is about 7.7%. Multiply the three multipliers together; the part above 1 is the overall increase.

5. Answer to Question 5

   1. Multiplier 1.2, so original \(=90\div1.2=\) £75. The price after the increase is 120% of the original, so we divide by 1.2. Check: \(75\times1.2=90\).
   2. Multiplier 0.85, so original \(=680\div0.85=\) £800. The value after the decrease is 85% of the original, so we divide by 0.85. Check: \(800\times0.85=680\).
   3. \(31\,200\div1.04=\) £30 000. The new salary is 104% of the old one, so we divide by 1.04. Check: \(30\,000\times1.04=31\,200\).

6. Answer to Question 6

   1. \(x_1=\sqrt[3]{4.5}=\) 1.651; \(x_2=\sqrt[3]{4.651}=\) 1.669; \(x_3=\sqrt[3]{4.669}=\) 1.671.
   2. Let \(f(x)=x^3-x-3\). \(f(1.6)=4.096-1.6-3=-0.504\lt 0\) and \(f(1.7)=4.913-1.7-3=0.213>0\). The sign changes, so there is a solution between 1.6 and 1.7.
   3. The values settle at about 1.672, so \(x=1.67\) to 2 d.p.

7. Answer to Question 7

   1. \(1500\times1.045^5=\) £1 869.27. 4.5% interest gives the multiplier 1.045, used 5 times.
   2. \(1869.27-1500=\) £369.27. Subtract the amount invested.
   3. \(1500\times1.045^7=1500\times1.36086=2041.29\), which is more than £2 000, so yes.
   4. With compound interest, interest is paid on the interest already earned as well as on the original amount.

8. Answer to Question 8

   1. \(3.20\times1.03^4=\) £3.60. 3% inflation gives the multiplier 1.03, used 4 times.
   2. Price \(=3.20\times1.03^n\).
   3. \(1.03^4=1.1255\), so the increase is 12.6%. Subtract 1 from the overall multiplier and change it into a percentage.
   4. Both rise by the same percentage, but the loaf costs more in pounds: \(1.40\times1.1255=£1.58\), which is less than £3.60.

9. Answer to Question 9

   1. \(80\,000\times0.98^{10}=\) 65 366. A 2% fall gives the multiplier 0.98, used 10 times.
   2. \(65\,366\div80\,000=0.817\), so it has fallen by about 18.3% (not 20%).
   3. After 6 years: \(80\,000\times0.98^6=70\,867\). After 7 years: \(80\,000\times0.98^7=69\,450\). So the population first falls below 70 000 in year 7.

10. Answer to Question 10

    1. \(200\times1.5^4=200\times5.0625=\) 1 012.5, about 1 013 bacteria. A 50% increase gives the multiplier 1.5, used 4 times. We round because we cannot have half a bacterium.
    2. \(N=200\times1.5^t\). The number starts at 200 and is multiplied by 1.5 each hour.
    3. \(t=5\): \(1519\). \(t=6\): \(2278\). So after 6 hours the number is first more than 2 000. Try whole numbers of hours until the number first passes 2000.

11. Answer to Question 11

    1. \(600\times0.7\times0.8\times0.8=\) £268.80. A 30% loss gives the multiplier 0.7 and each 20% loss gives 0.8. Multiply by each one in turn.
    2. \(\dfrac{600-268.80}{600}=\) 55.2%. Divide the total loss, £331.20, by the original £600 and multiply by 100.
    3. No: it has lost 55.2%, which is more than half. Half of £600 is £300, and the phone is now worth only £268.80, which is less than half.

12. Answer to Question 12

    1. After the rise: \(100\times1.1=\) £110. After the reduction: \(110\times0.9=\) £99.
    2. No: the jacket costs £99, which is £1 less than the start, because the 10% reduction applies to the higher price.
    3. \(1.1\times0.9=0.99\), so the overall change is a 1% decrease.

13. Answer to Question 13

    1. Simple: 5% of £1 000 is £50 a year, so \(1000+10\times50=\) £1 500.
    2. \(1000\times1.05^{10}=\) £1 628.89.
    3. \(1628.89-1500=\) £128.89.
    4. In compound interest each year's interest is also earned on the previous interest, so the yearly interest keeps growing, while the simple interest stays at £50 a year.

14. Answer to Question 14

    1. \(5000\times1.07^{10}=5000\times1.9672=\) £9 835.76.
    2. After 10 years it is \(1.967\times\) the start (not quite double). After 11 years it is \(1.07^{11}=2.105\times\), which is more than double. So 11 years.
    3. Simple interest at 7% gives £350 a year, so doubling needs £5 000 of interest: \(5000\div350=14.3\) years. So in 14 years it is not quite doubled; it takes about 14.3 years.

15. Answer to Question 15

    1. \(1.05^3=\) 1.157625. 5% interest gives the multiplier 1.05, so over 3 years the multiplier is \(1.05^3\).
    2. \(2315.25\div1.157625=\) £2 000. To reverse the growth we divide by the overall multiplier. The original sum was £2000.
    3. \(2000\times1.05=\) £2 100. Multiply the original sum by 1.05 once.
    4. \(2100\times1.05=2205\) and \(2205\times1.05=2315.25\). ✓ Growing the answer forward gives back £2315.25, so it is correct.

16. Answer to Question 16

    1. \(4494.40\div4000=1.1236\). Divide the final amount by the amount invested to get the overall multiplier.
    2. The yearly multiplier is the square root: \(\sqrt{1.1236}=1.06\), so the rate is 6%. The same multiplier is used twice, so the yearly multiplier is the square root of the overall one.
    3. \(4000\times1.06^2=4000\times1.1236=4494.40\). ✓ Growing £4000 by 6% for 2 years gives back the final amount, so the rate is correct.
    4. \(4000\times1.06^5=\) £5 352.90. Use the multiplier 1.06 five times: \(1.06^5\approx1.33823\).

17. Answer to Question 17

    1. \(80\times0.75^3=\) 33.75 g. A 25% loss gives the multiplier 0.75, used 3 times.
    2. \(M=80\times0.75^n\).
    3. After 7 days: \(80\times0.75^7=10.68\) g (still above 10). After 8 days: \(80\times0.75^8=8.01\) g. So after 8 days.
    4. \(0.75^2=0.5625\) and \(0.75^3=0.42\): the mass halves between 2 and 3 days, in about 2.4 days.

18. Answer to Question 18

    1. \(1000\times1.032^4=\) £1 134.28. 3.2% interest gives the multiplier 1.032, used 4 times.
    2. \(1000\times1.025\times1.035^3=\) £1 136.44. Use 1.025 for the first year and 1.035 for each of the next three years.
    3. Bank B is better by \(1136.44-1134.28=\) £2.16.
    4. Bank A: 3.2%. Bank B: \(\sqrt[4]{1.13644}=1.0325\), so about 3.25%.

19. Answer to Question 19

    1. \(x^2-x-1=0\) gives \(x^2=x+1\). Divide by \(x\): \(x=1+\dfrac1x\).
    2. \(x_1=1+\dfrac11=\) 2; \(x_2=1+\dfrac12=\) 1.5; \(x_3=1+\dfrac1{1.5}=\) 1.667; \(x_4=1+\dfrac1{1.667}=\) 1.6.
    3. \(f(x)=x^2-x-1\): \(f(1)=-1\lt 0\) and \(f(2)=1>0\), so there is a sign change and a root between 1 and 2.
    4. The values settle near 1.618, so \(x=1.62\). Check: \(1.618^2-1.618-1=0.000\). ✓

20. Answer to Question 20

    1. \(m^{12}=2\), so \(m=2^{1/12}=\) 1.0595.
    2. The rate is 5.9% (to 1 decimal place).
    3. At 6%: \(72\div6=12\) years. \(1.06^{12}=2.012\), so the money has just doubled. The rule is accurate.
    4. At 30%: the rule gives \(72\div30=2.4\) years, but \(1.3^{3}=2.2\) and \(1.3^2=1.69\), so the real doubling time is about 2.6 years. The rule becomes less accurate for large rates because it is based on small percentages.