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

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1. Answer to Question 1 (4 marks)

   1. \(\dfrac{15}{24}+\dfrac{14}{24}=\) \(\dfrac{29}{24}\) or \(1\dfrac5{24}\) (1 for common denominator, 1 for the answer) The lowest common denominator of 8 and 12 is 24. The answer is \(\tfrac{29}{24}=1\tfrac5{24}\).
   2. \(\dfrac{10}{3}\div\dfrac53=\dfrac{10}{3}\times\dfrac35=\) 2 (1 for turning into improper fractions and inverting, 1 for the answer) Change the mixed numbers into improper fractions, then multiply by the reciprocal of \(\tfrac53\). The answer is 2.

   Check (b) by multiplying back: \(2\times1\tfrac23=3\tfrac13\), the number we started with.

2. Answer to Question 2 (4 marks)

   1. \(85\times1.12=\) £95.20 (1 for the multiplier, 1 for the answer) A 12% increase gives the multiplier 1.12. The new amount is £95.20.
   2. Decrease \(=63\) (1). \(\dfrac{63}{840}=\) 7.5% (1) The decrease is \(840-777=63\), and we divide by the original roll, 840. The percentage decrease is 7.5%.

   A common error in (b) is to divide 63 by 777, the new roll, which gives about 8.1% and scores only the first mark. Check (a): 12% of £85 is £10.20.

3. Answer to Question 3 (4 marks)

   1. \(6500\times0.82\times0.88=\) £4690.40 (1 for the multipliers, 1 for the answer) An 18% fall gives the multiplier 0.82 and a 12% fall gives 0.88. The motorbike is worth £4690.40 after two years.
   2. \(0.82\times0.88=0.7216\) (1), so the decrease is 27.8% (1) The bike keeps 72.16% of its value, so the overall decrease is \(100-72.16=27.84\), which is 27.8% to 1 d.p.

   Do not add the percentages: 30% is wrong.

4. Answer to Question 4 (4 marks)

   1. \(456\div1.2=\) £380 (1 for 120%, 1 for the answer) The price including VAT is 120% of the price before VAT, so we divide by 1.2. The price before VAT is £380.
   2. \(59.50\div0.7=\) £85 (1 for 70%, 1 for the answer) The sale price is 70% of the original price, so we divide by 0.7. The original price was £85.

   Taking 20% off £456 gives £364.80, which is wrong.

5. Answer to Question 5 (4 marks)

   1. \(100x=36.3636\ldots\), \(x=0.3636\ldots\), subtract: \(99x=36\) (1), \(x=\dfrac{36}{99}=\dfrac4{11}\) (1) Two digits repeat, so we multiply by 100 and subtract to remove the recurring part. Then divide 36 and 99 by 9.
   2. \(25\,000\times1.04^6=31\,633\), so 31 600 (1) 4% growth gives the multiplier 1.04, used 6 times. The population is about 31 600.
   3. \(1.04^{12}=1.601\) gives \(40\,026\); \(1.04^{11}=1.539\) gives \(38\,490\). So 12 years (1) Try powers of 1.04 until \(25\,000\times1.04^n\) first passes 40 000. It takes 12 years.

Total : 20 marks