Problem solving and GCSE exam practice – Year 11 GCSE maths worksheet answers to download as PDF

← Back to the worksheet

1. Answer to Question 1

   1. \(2+3+7=12\) parts, so one part is \(240\div12=\) £20. Add the numbers in the ratio to find the total number of equal parts.
   2. \(2\times20=\) £40, \(3\times20=\) £60, \(7\times20=\) £140. Check: \(40+60+140=240\). Multiply each number in the ratio by the value of one part.
   3. \(0.35\times140=\) £49. 35% as a decimal is 0.35. The friend spends £49.

2. Answer to Question 2

   1. \(3\times4=12\) and \(10^5\times10^{-2}=10^3\), so \(12\times10^3=\) \(1.2\times10^4\). Multiply the numbers and add the powers of 10. 12 is not between 1 and 10, so we write \(12=1.2\times10^1\).
   2. \(7.2\times10^{-4}\). Move the decimal point 4 places to the right to get 7.2, so the power is \(-4\).
   3. \(\dfrac63=2\) and \(10^{8-3}=10^5\): \(2\times10^5\). Divide the numbers and subtract the powers of 10.
   4. \(\dfrac{1.5\times10^8}{3\times10^5}=0.5\times10^3=\) 500 seconds. Time \(=\) distance \(\div\) speed. 0.5 is not between 1 and 10, so \(0.5\times10^3=5\times10^2=500\).

3. Answer to Question 3

   1. \(6x-15=x+10\), so \(5x=25\) and \(x=5\). Expand the bracket, then subtract \(x\) and add 15 to both sides. Check: \(3\times5=15\) and \(5+10=15\).
   2. Add the equations: \(3x=12\), so \(x=4\). Then \(y=4-1=3\). \(x=4\), \(y=3\). The \(y\) terms have opposite signs, so adding the equations eliminates \(y\).
   3. \(2\times4+3=11\) ✓ and \(4-3=1\) ✓. Both equations are true, so the solution is correct.

4. Answer to Question 4

   1. \(h^2=6.5^2-2.5^2=42.25-6.25=36\), so \(h=\) 6 m. The ladder is the hypotenuse of a right-angled triangle, so we subtract the squares. The ladder reaches 6 m up the wall.
   2. New foot distance 3.5 m: \(h^2=42.25-12.25=30\), \(h=5.48\) m. The top slides down \(6-5.48=\) 0.52 m. Use Pythagoras again with the new distance 3.5 m, then subtract the new height from the old height.

5. Answer to Question 5

   1. \(80\times0.85\times0.9=\) £61.20. A 15% reduction gives the multiplier 0.85 and a 10% reduction gives 0.9. Multiply by each in turn.
   2. \(0.85\times0.9=0.765\), so the reduction is 23.5%. Multiply the multipliers; the part below 1 is the overall reduction.
   3. He added 15% and 10%, but the second reduction is 10% of a smaller price. The overall reduction is 23.5%, not 25%.

6. Answer to Question 6

   1. Let the numbers be \(n\), \(n+1\), \(n+2\). The sum is \(3n+3=3(n+1)\), which is a multiple of 3. Consecutive numbers differ by 1, so we call them \(n\), \(n+1\) and \(n+2\). Since \(n+1\) is a whole number, the sum is 3 times a whole number.
   2. \((n+1)^2-(n-1)^2=(n^2+2n+1)-(n^2-2n+1)=4n\). Expand both squares; the \(n^2\) terms and the 1s cancel, leaving \(2n+2n\).
   3. \(n=50\): \(4\times50=\) 200 (check: \(2601-2401=200\)). \(51=50+1\) and \(49=50-1\), so this is the expression from part (b) with \(n=50\).

7. Answer to Question 7

   1. A: \(60+24\times15=60+360=\) £420. B: \(24\times22=\) £528.
   2. \(528-420=\) £108.
   3. \(60+15m=22m\), so \(60=7m\) and \(m=8.57\): the totals are equal after about 8.6 months, so from month 9 A is cheaper.
   4. At 6 months A costs \(60+90=150\) and B costs \(132\), so B is cheaper.

8. Answer to Question 8

   1. Multiply by \(\dfrac{10}4=2.5\): flour \(300\times2.5=\) 750 g; eggs \(2\times2.5=\) 5 eggs. 10 people is \(\tfrac{10}{4}=2.5\) times as many as 4 people, so every amount is multiplied by 2.5.
   2. \(0.75\times1.20=\) 90p. Change 750 g into 0.75 kg before multiplying by the price per kilogram.
   3. \(90+5\times30=90+150=\) £2.40. Add the cost of the flour and the cost of 5 eggs.
   4. \(2000\div300\times4=26.7\), so 26 people. Each person needs \(300\div4=75\) g of flour, and \(2000\div75=26.7\). We round down because the flour would run out.

9. Answer to Question 9

   1. \(12\times8-\pi\times2^2=96-12.57=\) 83.43 m². Subtract the area of the circular pond from the area of the rectangle.
   2. \(83.43\div20=4.17\), so 5 bags. We round up because 4 bags cover only 80 m².
   3. \(5\times6.50=\) £32.50. Multiply the number of bags by the price of one bag.
   4. Circumference \(=2\pi\times2=12.57\) m. Cost \(=12.57\times4.20=\) £52.78. The edging goes round the circumference of the pond.

10. Answer to Question 10

    1. \(90\div60=\) 1.5 h and \(60\div40=\) 1.5 h.
    2. \(1.5+1.5=\) 3 hours.
    3. \(150\div3=\) 50 km/h.
    4. The two speeds were used for different distances. Here the distances differ, but the times happen to be equal, so the answer is 50; if the times had differed, simply averaging the speeds would be wrong.

11. Answer to Question 11

    1. \(\dfrac7{10}\times\dfrac69=\dfrac{42}{90}=\) \(\dfrac7{15}\). Multiply along the branches; the second fraction has denominator 9 because the counter is not replaced.
    2. \(\dfrac3{10}\times\dfrac79+\dfrac7{10}\times\dfrac39=\dfrac{21}{90}+\dfrac{21}{90}=\dfrac{42}{90}=\) \(\dfrac7{15}\). Different colours can happen in two orders, so we add the two products.
    3. \(1-\dfrac7{15}=\) \(\dfrac8{15}\). 'At least one red' is the complement of 'both blue', so subtract from 1.

12. Answer to Question 12

    1. \(5\times12=\) 60. The total is the mean times the number of values.
    2. New total \(=6\times13=78\), so the sixth number is \(78-60=\) 18. The sixth number is the difference between the two totals.
    3. The total for seven numbers is \(7\times13=91\), so the seventh number is \(91-78=\) 13, equal to the mean.

13. Answer to Question 13

    1. \(5\sin12^\circ=\) 1.04 m. The ramp is the hypotenuse and the height is opposite the \(12^\circ\) angle, so we use sine.
    2. \(5\cos12^\circ=\) 4.89 m. The horizontal distance is adjacent to the angle, so we use cosine.
    3. \(\sin5^\circ=\dfrac{1.04}{L}\), so \(L=\dfrac{1.04}{0.0872}=\) 11.9 m. The height is opposite the angle and the ramp length is the hypotenuse. Rearrange to make \(L\) the subject.

14. Answer to Question 14

    1. \(2500\times1.03^4=\) £2 813.77.
    2. After 6 years \(2500\times1.03^6=2985.13\); after 7 years \(=3074.68\). He needs 7 years.
    3. Simple interest gives £75 a year, so £2 800 after 4 years. After 1 year both give £2 575, but from year 2 the compound interest is also paid on earlier interest, so compound gives more (£2 813.77).

15. Answer to Question 15

    1. \(w(w+5)=84\), so \(w^2+5w-84=0\). The area of a rectangle is length times width; expand and subtract 84.
    2. \((w+12)(w-7)=0\). Width cannot be negative, so \(w=7\). We need two numbers with product \(-84\) and sum 5: 12 and \(-7\).
    3. Length \(=12\). Perimeter \(=2\times(7+12)=\) 38 cm. The length is \(7+5=12\) cm.
    4. Square side \(=38\div4=9.5\) cm. Area \(=9.5^2=\) 90.25 cm². A square has four equal sides, so each side is a quarter of the perimeter.

16. Answer to Question 16

    1. Side \(=10\) cm, so the area is 100 cm². The circle just fits, so the side of the square equals the diameter, 10 cm.
    2. \(100-\pi\times5^2=100-78.54=\) 21.46 cm². Subtract the area of the circle from the area of the square.
    3. \(\dfrac{21.46}{100}=\) 0.2146. The probability is the shaded area divided by the area of the square.
    4. \(200\times0.2146=\) 43. Multiply the number of darts by the probability: \(200\times0.2146=42.9\), which is about 43.

17. Answer to Question 17

    1. \(\dfrac{15}{10}=\) 1.5. Divide the corresponding heights: large \(\div\) small.
    2. Volume scale factor \(=1.5^3=3.375\). \(675\div3.375=\) 200 ml. For similar shapes the volume scale factor is the cube of the length scale factor. Divide because we are going from large to small.
    3. Area scale factor \(=1.5^2=2.25\). \(160\times2.25=\) 360 cm². The area scale factor is the square of the length scale factor. Multiply because we are going from small to large.

18. Answer to Question 18

    1. \(3a+2c=37\) and \(2a+3c=33\).
    2. Multiply the first by 3 and the second by 2: \(9a+6c=111\) and \(4a+6c=66\). Subtract: \(5a=45\), so \(a=9\). Then \(27+2c=37\) so \(c=5\). Adult £9, child £5.
    3. \(4\times9+5\times5=36+25=\) £61. Substitute \(a=9\) and \(c=5\).

19. Answer to Question 19

    1. \((2n+1)^2-1=4n^2+4n+1-1=4n^2+4n=4n(n+1)\).
    2. Of two consecutive whole numbers one is even, so the product is even.
    3. Since \(n(n+1)=2k\) for some whole number \(k\), \(4n(n+1)=8k\), a multiple of 8. For example, \(7^2-1=48=8\times6\).

20. Answer to Question 20

    1. \(\pi\times3^2\times10=\) \(90\pi\) cm³.
    2. \(\dfrac43\pi\times1.5^3=\dfrac43\pi\times3.375=\) \(4.5\pi\) cm³.
    3. \(90\pi\div4.5\pi=\) 20 spheres.
    4. Spheres: \(20\times4\pi\times2.25=180\pi\). Cylinder: \(2\pi\times3\times10+2\pi\times9=60\pi+18\pi=78\pi\). The spheres have more than twice the surface area.