Fractions, decimals and percentages – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(\dfrac{8}{12}+\dfrac{9}{12}=\dfrac{17}{12}=\) \(1\dfrac5{12}\) The lowest common denominator of 3 and 4 is 12, so we change both fractions into twelfths before adding. The answer is \(1\tfrac5{12}\).
   2. \(\dfrac{11}{2}-\dfrac{11}{4}=\dfrac{22}{4}-\dfrac{11}{4}=\dfrac{11}{4}=\) \(2\dfrac34\) Change the mixed numbers into improper fractions, then use the common denominator 4. The answer is \(2\tfrac34\).
   3. \(\dfrac{3\times10}{5\times21}=\dfrac{30}{105}=\) \(\dfrac27\) Multiply the numerators and the denominators, then simplify by dividing by 15. We could also cancel first: 5 into 10 and 3 into 21. The answer is \(\tfrac27\).
   4. \(\dfrac94\div\dfrac32=\dfrac94\times\dfrac23=\dfrac{18}{12}=\) \(1\dfrac12\) Change to improper fractions, then multiply by the reciprocal of the second fraction. The answer is \(1\tfrac12\).

2. Answer to Question 2

   1. \(7\div40=0.175\) and \(0.175\times100=\) 17.5% A fraction is a division, so we divide the numerator by the denominator, then multiply by 100 for the percentage.
   2. \(0.36=\dfrac{36}{100}=\) \(\dfrac9{25}\), and as a percentage 36% The decimal 0.36 means 36 hundredths, and we simplify by dividing top and bottom by 4.
   3. \(62.5\%=0.625=\dfrac{625}{1000}=\) \(\dfrac58\) Divide the percentage by 100 to get the decimal, then simplify \(\tfrac{625}{1000}\) by dividing by 125.
   4. \(3\div11=0.272727\ldots=\) \(0.\dot2\dot7\), which is 27.3% (1 d.p.) The digits 27 repeat for ever, so we put a dot over the 2 and over the 7.

3. Answer to Question 3

   1. 10% \(=8.40\), 5% \(=4.20\), so 15% \(=\) £12.60 Find 10% by dividing by 10, halve it for 5%, then add. 15% of £84 is £12.60.
   2. 12.5% is \(\tfrac18\): \(640\div8=\) 80 12.5% is half of 25%, which is \(\tfrac18\). So 12.5% of 640 is 80.
   3. 1% \(=3.5\), so 7% \(=\) 24.5 Find 1% by dividing by 100, then multiply by 7. 7% of 350 is 24.5.
   4. 10% \(=0.24\), so 30% \(=0.72\) and 5% \(=0.12\); total \(=\) 0.84 kg Build 35% as 30% plus 5%. 35% of 2.4 kg is 0.84 kg.

4. Answer to Question 4

   1. Multiplier \(1.08\): \(60\times1.08=\) £64.80 An 8% increase means we keep 100% and add 8%, so the multiplier is \(1+0.08=1.08\). The new value is £64.80.
   2. Multiplier \(0.85\): \(250\times0.85=\) 212.5 kg A 15% decrease leaves \(100\%-15\%=85\%\), so the multiplier is 0.85. The new value is 212.5 kg.
   3. Multiplier \(1.025\): \(840\times1.025=\) 861 m An increase of 2.5% gives the multiplier \(1+0.025=1.025\). The new value is 861 m.
   4. Multiplier \(0.996\): \(12\,500\times0.996=\) £12 450 A decrease of 0.4% leaves 99.6%, so the multiplier is 0.996. The new value is £12 450.

5. Answer to Question 5

   1. \(\dfrac{20}{80}=0.25=\) 25% increase Percentage change \(=\dfrac{\text{change}}{\text{original}}\times100\). The change is \(100-80=20\), so this is a 25% increase.
   2. \(\dfrac{60}{250}=0.24=\) 24% decrease The value falls by \(250-190=60\). Always divide by the original value, 250. This is a 24% decrease.
   3. \(\dfrac{9}{36}=0.25=\) 25% increase The change is \(45-36=9\), and we divide by the original value 36. This is a 25% increase.
   4. \(\dfrac{0.5}{2.5}=0.2=\) 20% decrease The change is \(2.5-2=0.5\), divided by the original 2.5. This is a 20% decrease.

6. Answer to Question 6

   1. \(54\div1.2=\) £45 The new price is 120% of the original, so we divide by the multiplier 1.2 to reverse the change. The original price was £45.
   2. \(68\div0.85=\) £80 After a 15% decrease the price is 85% of the original, so we divide by 0.85. The original price was £80.
   3. \(175.50\div1.08=\) £162.50 The value is 108% of the original, so we divide by 1.08. The original value was £162.50.
   4. \(91\div0.65=\) 140 g After a 35% decrease the mass is 65% of the original, so we divide by 0.65. The original mass was 140 g.

7. Answer to Question 7

   1. \(480\div1.2=\) £400 The price including VAT is 120% of the price without VAT, so we divide by 1.2.
   2. \(36\div1.2=30\) without VAT, so VAT \(=36-30=\) £6
   3. \(27\times0.85=\) £22.95 A 15% reduction leaves 85% of the price. The lamp costs £22.95 in the sale.
   4. The 20% is added to the price without VAT (£400), so the VAT is \(20\%\) of £400 \(=\) £80. 20% of £480 is wrong because £480 already includes VAT.

8. Answer to Question 8

   1. \(120\times0.75=\) £90 A 25% reduction leaves 75% of the price, so the multiplier is 0.75. The sale price is £90.
   2. \(90\times0.9=\) £81 The voucher takes 10% off the sale price, not the original price, so the multiplier is 0.9. The coat costs £81.
   3. \(0.75\times0.9=0.675\), so the overall reduction is \(1-0.675=\) 32.5%. Multiplying the two multipliers gives one multiplier for both changes. The overall saving is 32.5%, not 35%.
   4. \(120\times0.65=£78\). The second customer pays £3 less (\(81-78\)). The second customer pays £78, so the second customer pays less, by £3.

9. Answer to Question 9

   1. Taxable income \(=28\,000-12\,570=15\,430\). Tax \(=0.2\times15\,430=\) £3086 Only the income above the personal allowance is taxed, so we subtract £12 570 first. Priya pays £3086 tax a year.
   2. \(28\,000-3086=\) £24 914 a year, and \(24\,914\div12=\) £2076.17 a month. Subtract the tax from her salary, then divide by 12 for one month.
   3. \(28\,000\times1.035=\) £28 980 A 3.5% rise gives the multiplier 1.035. Her new salary is £28 980.
   4. New taxable income \(=28\,980-12\,570=16\,410\). New tax \(=3282\). Increase \(=3282-3086=\) £196. The new tax is \(0.2\times16\,410=3282\). Her yearly tax increases by £196.

10. Answer to Question 10

    1. \(18\,000\times0.85=\) £15 300 A 15% fall leaves 85% of the value, so the multiplier is 0.85. The car is worth £15 300 after 1 year.
    2. \(18\,000\times0.85^3=\) £11 054 (11 054.25) The value falls by 15% each year, so we use the multiplier 0.85 three times. The car is worth about £11 054.
    3. Year 4: \(18\,000\times0.85^4=9396\). Year 5: \(18\,000\times0.85^5=7987\). So 5 years. Keep multiplying by 0.85 until the value drops below £8000. The value is first below £8000 after 5 years.
    4. Multiplier \(0.85^3=0.614125\). Loss \(=1-0.614125=0.3859\), so 38.6%. The car keeps 61.4% of its value, so it loses about 38.6%. This is not \(3\times15\%=45\%\), because each fall is worked out on a smaller value.

11. Answer to Question 11

    1. \(42\,000\times1.024^5=47\,288\), so 47 300.
    2. \(42\,000\times1.024^7=49\,585\), which is below 50 000, but \(42\,000\times1.024^8=50\,775\). So 8 years.
    3. \(42\,000\times0.024=1008\approx1000\). The first year's increase is 2.4% of 42 000.
    4. Compare the two populations year by year. Year 14: \(58\,539\) and \(58\,265\) (first town still bigger). Year 15: \(59\,944\) and \(60\,071\) (second town bigger). So the second town overtakes after 15 years.

12. Answer to Question 12

    1. Ayo \(\dfrac{54}{72}=75\%\). Bella \(\dfrac{45}{72}=62.5\%\). Carlos \(\dfrac{63}{72}=87.5\%\). Divide each score by 72 and multiply by 100.
    2. Ayo \(\dfrac{63}{90}=70\%\). Bella \(\dfrac{72}{90}=80\%\). Carlos \(\dfrac{54}{90}=60\%\). Divide each score by 90 and multiply by 100.
    3. Ayo \(0.6\times75+0.4\times70=45+28=\) 73%. Bella \(0.6\times62.5+0.4\times80=37.5+32=\) 69.5%. Carlos \(0.6\times87.5+0.4\times60=52.5+24=\) 76.5%. The final grade is a weighted mean: Paper 1 counts more because its weight is 60%.
    4. Carlos has the highest final percentage (76.5%). Carlos has the highest final percentage, because the paper with the bigger weight was his best paper.

13. Answer to Question 13

    1. Walk: \(\tfrac25\times360=\) 144. Bus: \(\tfrac14\times360=\) 90.
    2. \(1-\tfrac25-\tfrac14=\tfrac{20-8-5}{20}=\) \(\tfrac{7}{20}\) \(=35\%\). The fractions must add up to 1 for the whole group, so subtract the other two fractions.
    3. Car or bike: \(360-144-90=126\). Cycling: \(0.4\times126=\) 50.4, so about 50 pupils (the data must be rounded).
    4. Walk \(144^\circ\), bus \(90^\circ\), car or bike \(126^\circ\). Check: \(144+90+126=360\).

14. Answer to Question 14

    1. Interest each year \(=0.03\times2000=60\). Over 4 years \(=240\). Total \(=\) £2240. Simple interest is the same every year because it is always worked out on the original £2000.
    2. \(2000\times1.03^4=\) £2251.02 Compound interest uses the multiplier 1.03 once for each of the 4 years.
    3. \(2251.02-2240=\) £11.02 So Jay has £11.02 more than Sam.
    4. With compound interest the interest earned is added to the account, so the next year’s interest is calculated on a larger amount.

15. Answer to Question 15

    1. \(0.9\times0.75=\) 0.675 Taking 10% off gives the multiplier 0.9 and taking 25% off gives 0.75. Multiply them to get one multiplier for both discounts.
    2. \(87.75\div0.675=\) £130 Dev's price is 0.675 of the original, so we divide by 0.675 to reverse the change. The marked price was £130.
    3. \(1-0.675=0.325\), so 32.5% off. The overall multiplier leaves 67.5% of the price, so the reduction is 32.5%.
    4. 35% off would give \(130\times0.65=£84.50\), which is less than £87.75. The 25% is taken from the already reduced price, so the overall discount is only 32.5%.

16. Answer to Question 16

    1. \(x=0.777\ldots\), \(10x=7.777\ldots\). Subtract: \(9x=7\), so \(x=\dfrac79\). The repeating part has one digit, so we multiply by 10 to line up the repeats. Subtracting removes the endless decimal part.
    2. \(x=0.4545\ldots\), \(100x=45.4545\ldots\). Subtract: \(99x=45\), so \(x=\dfrac{45}{99}=\) \(\dfrac5{11}\). The repeating part has two digits, so we multiply by 100. Then divide 45 and 99 by their HCF, 9.
    3. \(x=0.2111\ldots\), \(10x=2.111\ldots\), \(100x=21.111\ldots\). Subtract: \(90x=19\), so \(x=\dfrac{19}{90}\). Only the 1 repeats, so we use \(100x\) and \(10x\), which both end in \(.111\ldots\). The fraction \(\tfrac{19}{90}\) is already in its simplest form.
    4. \(x=0.999\ldots\), \(10x=9.999\ldots\), so \(9x=9\) and \(x=1\). So \(0.\dot9\) and 1 are two ways of writing the same number.

17. Answer to Question 17

    1. \(1000\times1.07^{10}=\) £1967.15 7% compound interest gives the multiplier 1.07, used 10 times. The account is worth £1967.15.
    2. \(1.07^{10}=1.967\lt 2\) and \(1.07^{11}=2.105>2\). So 11 years. After 10 years the multiplier is 1.967, just under 2, and after 11 years it is \(1.07^{11}=2.105\), more than 2. So the amount first doubles after 11 years.
    3. \(72\div7=10.3\) years, close to the exact answer of about 10.2 years (between 10 and 11). The rule of 72 is a quick estimate, and here it agrees well with the exact calculation.
    4. \(1.04^{17}=1.948\) and \(1.04^{18}=2.026\). So 18 years. Try powers of 1.04 until the result first passes 2. It takes 18 years to double at 4%.

18. Answer to Question 18

    1. \(1-\tfrac13=\) \(\tfrac23\) After spending \(\tfrac13\), Lena has \(\tfrac23\) of her money left.
    2. \(\tfrac25\times\tfrac23=\) \(\tfrac4{15}\) She spends \(\tfrac25\) of what is left, so we multiply \(\tfrac25\) by \(\tfrac23\).
    3. \(\tfrac23-\tfrac4{15}=\tfrac{10-4}{15}=\tfrac6{15}=\) \(\tfrac25\) Subtract the lunch from what was left after the book, using the common denominator 15.
    4. \(\tfrac25\) of the original is £12, so the original is \(12\div\tfrac25=12\times\tfrac52=\) £30. The book cost \(\tfrac13\times30=\) £10. So Lena started with £30 and the book cost £10. Check: \(30-10=20\), lunch \(=\tfrac25\times20=8\), and \(20-8=12\).

19. Answer to Question 19

    1. Profit \(=330-240=90\). \(\dfrac{90}{240}=0.375=\) 37.5% Percentage profit is the profit divided by the cost price, times 100. The percentage profit is 37.5%.
    2. \(275\times1.4=\) £385 A 40% profit means selling at 140% of the cost price, so the multiplier is 1.4. The selling price is £385.
    3. \(399\div1.33=\) £300 The selling price is 133% of the cost price, so we divide by 1.33. The cost price was £300.
    4. \(192\div0.8=\) £240 A 20% loss means the selling price is 80% of the cost price, so we divide by 0.8. The cost price was £240.

20. Answer to Question 20

    1. \(\left(1+\dfrac x{100}\right)\left(1-\dfrac x{100}\right)=1-\dfrac{x^2}{10\,000}\) (difference of two squares).
    2. \(1-\dfrac{x^2}{10\,000}=0.91\), so \(x^2=900\) and \(x=\) 30. A price 9% lower means the multiplier is 0.91, so \(\tfrac{x^2}{10\,000}=0.09\). We take the positive root, so the percentage is 30%.
    3. \(200\times1.3=260\) and \(260\times0.7=182\). \(\dfrac{18}{200}=9\%\) lower ✓.
    4. The multiplier \(1-\dfrac{x^2}{10\,000}\) is always less than 1 for \(x>0\), so the final price is always lower.