June 2018 Paper 3 Foundation

$\frac{9}{10}$ means 9 tenths.

Nine tenths is written $0.9$.

To change a decimal to a percentage, multiply by 100.

$0.3 \times 100 = 30$, so the answer is $30\%$.

2538 lies between 2500 and 2600.

The tens digit is 3, which is less than 5, so round down to $2500$.

Each term is 7 more than the one before, so the next term is $23 + 7 = 30$.

The nth term is $7n - 5$.

10th term: $7 \times 10 - 5 = 65$.

(a) Use the three largest digits, biggest first: $974$.

(b) Put the two smallest digits, 1 and 2, in the tens places. Put 6 and 8 in the units places.

$16 + 28 = 44$ (or $18 + 26 = 44$), the smallest possible total.

Find the factor pairs of 30:

$1 \times 30$, $2 \times 15$, $3 \times 10$, $5 \times 6$.

So the factors are $1, 2, 3, 5, 6, 10, 15, 30$.

Becky has $6 \times 2 = 12$ cousins.

David has $12 \times 2 = 24$ cousins.

(a) $1.44 \times 3.61 = 5.1984$ and $\sqrt{5.1984} = 2.28$.

(b) $3.54 - 0.96 = 2.58$

$2.58^2 = 6.6564$

$6.6564 - 4.096 = 2.5604$

(a) 08 25 to 09 05 is $40$ minutes.

(b) He reaches the Whitefield stop at 8:45 + 17 min = 9:02.

The next bus leaves Whitefield at 09 04 and reaches Manchester at 09 35.

9:35 + 15 min = 9:50, which is before 10 am, so yes.

Weekday hours: $20 - 8 = 12$.

Weekday pay: $12 \times 8.40 = £100.80$.

Weekend pay: $8 \times 11.20 = £89.60$.

Total: $100.80 + 89.60 = £190.40$, which is less than £200.

Increase: $600 - 560 = 40$.

As a fraction of last year's cost: $\frac{40}{560} = \frac{1}{14}$.

Measure the drawing: length about 11.5 cm, width about 5.8 cm.

Real length: $11.5 \times 200 = 2300$ cm $= 23$ m.

Real width: $5.8 \times 200 = 1160$ cm $= 11.6$ m.

Perimeter: $2 \times 23 + 2 \times 11.6 = 69.2$ m.

$y = 2$ is a horizontal line through 2 on the $y$-axis: graph D.

$y = x$ goes through the origin sloping upwards: graph F.

$x + y = 2$ gives $y = 2 - x$, crossing both axes at 2 and sloping down: graph A.

(a) Put the tens digits as stems and the units in order as leaves:

6 | 4 7 9 9
7 | 0 0 1 5 6 6 7
8 | 0 0 1 1 2 4 7
9 | 1 4

Key: 6 | 4 means 64 marks.

(b) Marks below 71: 64, 67, 69, 69, 70, 70, so 6 students failed.

$P(\text{fail}) = \frac{6}{20} = \frac{3}{10}$, but $\frac{1}{4} = \frac{5}{20}$, so Omar is wrong.

(a) Multiplication comes before subtraction: $12 - 2 \times 4 = 12 - 8 = 4$. Jenny subtracted first.

(b) The range is the largest value minus the smallest: $8 - 1 = 7$. Rehan used the last and first numbers in the list.

(a) Alan's £2.50 is $\frac{1}{8}$, so the total is $2.50 \times 8 = £20$.

Bispah gets $\frac{1}{2} \times 20 = £10$.

(b) Chan gets $20 - 2.50 - 10 = £7.50$.

Alan : Chan $= 2.5 : 7.5 = 1 : 3$.

Area $= \frac{1}{2} \times 3x \times 3x = \frac{9x^2}{2}$.

$\frac{9x^2}{2} = 162$, so $9x^2 = 324$ and $x^2 = 36$.

$x = 6$.

Divide the numbers: $2.645 \div 1.15 = 2.3$.

Subtract the powers: $10^9 \div 10^3 = 10^6$.

Answer: $2.3 \times 10^6$.

(a) As age increases, time decreases, so the correlation is negative.

(b) An 11-year-old's times on the graph are about 14.5 to 16 seconds; 12 seconds is far faster than the trend suggests.

(c) The oldest girl on the graph is under 14, so predicting for 15 means going beyond the data, which is unreliable.

$5(p + 3) = 5p + 15$

$-2(1 - 2p) = -2 + 4p$

$5p + 15 - 2 + 4p = 9p + 13$

The parallel sides are 2 cm and 7 cm, and the height is 4 cm.

Area $= \frac{1}{2}(2 + 7) \times 4 = 18$ cm².

A triangle needs $\frac{1}{2} \times b \times h = 18$, so $b \times h = 36$, e.g. base 9 cm, height 4 cm.

The probability of landing on 4 is 0.65 on every throw, so the probability of not landing on 4 is $1 - 0.65 = 0.35$.

1. On the first throw, $0.65 + 0.25 \ne 1$: the 'not land on 4' branch should be 0.35, not 0.25.

2. On the second throw after landing on 4, 'land on 4' is labelled 0.35 and 'not land on 4' 0.65; these are the wrong way round.

(a) $AB$ is the hypotenuse and $BC$ is adjacent to angle $ABC$.

$\cos ABC = \frac{7}{11}$, so $ABC = \cos^{-1}(\frac{7}{11}) = 50.5^\circ$.

(b) New $\cos ABC = \frac{7}{10} = 0.7$, which is more than $\frac{7}{11} \approx 0.636$, so it increases.

(a) Total counters: $18 \div 0.45 = 40$.

Red and white together: $1 - 0.45 - 0.25 = 0.3$.

Red is twice white, so $P(\text{white}) = 0.1$ and $P(\text{red}) = 0.2$.

Red counters: $0.2 \times 40 = 8$.

(b) The number of silver marbles is $0.5 \times$ the total. That is only a whole number if the total is even.

Multiply both sides by 2: $5 - x = 4x - 14$.

Add $x$ and 14 to both sides: $19 = 5x$.

$x = \frac{19}{5} = 3.8$

Angle sum of a pentagon: $(5 - 2) \times 180 = 540^\circ$.

Known angles: $125 + 115 + 90 = 330$, so $540 - 330 = 210^\circ$ remains.

If $ABC = x$, then $x + 2x = 210$, so $x = 70$.

$BCD = 2 \times 70 = 140^\circ$.

$AB = 8.4$ matches $DE = 12.6$.

Scale factor: $12.6 \div 8.4 = 1.5$.

(a) $DF = 6.4 \times 1.5 = 9.6$ cm.

(b) $CB = 15 \div 1.5 = 10$ cm.

Square both sides: $T^2 = \frac{g + 6}{2}$.

Multiply by 2: $2T^2 = g + 6$.

Subtract 6: $g = 2T^2 - 6$.