June 2019 Paper 1 Foundation

There are 60 minutes in an hour.

$180 \div 60 = 3$ hours

Multiply by 100 to turn a decimal into a percentage.

$0.73 \times 100 = 73\%$

Work out the bracket first: $3 + 5 = 8$

Then $10 \times 8 = 80$

Check each odd number from 21 to 29.

21 = 3 × 7, 25 = 5 × 5, 27 = 3 × 9, so these are not prime.

23 and 29 have no factors other than 1 and themselves, so either is correct.

Add the numbers and halve: $(7 + 15) \div 2 = 22 \div 2 = 11$

Hotel for one person: $7 \times £50 = £350$

Cost for one person: $£150 + £350 + £250 = £750$

For 4 people: $4 \times £750 = £3000$

(a) From the chart: red 8, white 10, yellow 5.

Blue $= 30 - 8 - 10 - 5 = 7$

(b) White has the most flowers (10), so the mode is white.

Write each with denominator 12:

$\frac{1}{3} = \frac{4}{12}$, $\frac{3}{4} = \frac{9}{12}$, $\frac{1}{4} = \frac{3}{12}$, $\frac{7}{12}$, $\frac{1}{2} = \frac{6}{12}$

Order: $\frac{3}{12}, \frac{4}{12}, \frac{6}{12}, \frac{7}{12}, \frac{9}{12}$, i.e. $\frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{7}{12}, \frac{3}{4}$

(a) 9 am to 10 30 am is 1.5 hours.

Distance $= 4 \times 1.5 = 6$ miles

(b) Leaves library: 10 30 + 50 min = 11 20 am

$1\frac{1}{4}$ hours is 1 hour 15 minutes: 11 20 + 1 h 15 min = 12 35 pm

(a) $3t = 12$, so $t = 4$

(b) Add 2 to both sides: $x = 8$

(c) $6w + 2 = 20$

$6w = 18$

$w = 3$

Grid method:

$70 \times 50 = 3500$, $70 \times 8 = 560$, $4 \times 50 = 200$, $4 \times 8 = 32$

$3500 + 560 + 200 + 32 = 4292$

(a) The angle at B is a right angle, so $x = 90 - 25 - 25 = 40$

(b)(i) $b = 125°$ because vertically opposite angles are equal (or $d = 125°$ because corresponding angles are equal).

(ii) Angles around a point add up to $360°$, so $a + b + c = 360 - 125 = 235°$

There are 10 mm in 1 cm, so $x$ cm is $10x$ mm.

(a) $70 \div 5 = 14$

(b) Each whole contains 2 halves, so 48 wholes contain $48 \times 2 = 96$ halves. Dividing by $\frac{1}{2}$ is the same as multiplying by 2, so the answer should be 96, not 24.

(a) $8 \times 8 = 64$, so $\sqrt{64} = 8$

(b) $5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$

(a) $5 \times 2m = 10m$ and $5 \times (-3) = -15$, so $10m - 15$

(b) 3 is a factor of both terms: $3n + 12 = 3(n + 4)$

(i) Maxine: she made more throws (50 against 10), and more trials give a more reliable estimate.

(ii) Combine all the results: $7 + 30 = 37$ Tails out of $10 + 50 = 60$ throws.

Estimate $= \frac{37}{60}$

Along the length: $600 \div 30 = 20$ stones

Across the width: $120 \div 30 = 4$ stones

Number needed: $20 \times 4 = 80$

Cost: $80 \times £2.50 = £200$

$£200 < £220$, so he has enough money.

(a) $\frac{2}{3} - \frac{1}{5} = \frac{10}{15} - \frac{3}{15} = \frac{7}{15}$

(b) $\frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$

Let B have side 2, so A has side 1.

Area of B $= 4$, area of A $= 1$

The diagonal halves A, so the shaded part $= \frac{1}{2}$

$\frac{0.5}{4} \times 100 = 12.5\%$

Boys: $40 - 22 = 18$

Girls on the bus: $10 - 6 = 4$

Boys who walk: $18 - 7 - 6 = 5$

Students who walk: $5 + 9 = 14$

(a) Probabilities add to 1: $1 - 0.2 = 0.8$

Red and yellow are equal: $0.8 \div 2 = 0.4$ each.

(b) 0.2 of the cubes are blue, so $12 \div 0.2 = 60$ cubes.

(a) $60 \div 15 = 4$, so she needs $4 \times 50 = 200$ g of sugar.

Flour: $3 \times 200 = 600$ g

(b) Butter: $2 \times 200 = 400$ g

$400 \div 250 = 1.6$, so she must buy 2 packs.

$72 = 2^3 \times 3^2$ and $90 = 2 \times 3^2 \times 5$

Shared primes: $2 \times 3^2 = 18$

The plan is a circle, so the solid is round when seen from above.

Both elevations are rectangles, so the solid is a cylinder.

From the grid: diameter 4 cm (radius 2 cm) and height 5 cm.

Reflect A in the $x$-axis: $(3, 2) \to (3, -2)$, $(5, 2) \to (5, -2)$, $(3, 5) \to (3, -5)$

B has matching corners at $(-3, -3)$, $(-1, -3)$, $(-3, -6)$.

Move from $(3, -2)$ to $(-3, -3)$: 6 left and 1 down.

So $c = -6$ and $d = -1$

Pens in the ratio $7 \times 2 : 3 \times 5 : 4 \times 6 = 14 : 15 : 24$

Total parts $= 53$, and $212 \div 53 = 4$

Green pens $= 24 \times 4 = 96$

Area of PQRS: $PQ \times 10 = 45$, so $PQ = 4.5$ cm, so $BC = 4.5$ cm.

Perimeter of ABCD: $2 \times AB + 2 \times 4.5 = 26$

$2 \times AB = 17$, so $AB = 8.5$ cm

(a) The lowest point of the curve is at $(1, -4)$.

(b) The roots are where the curve crosses the $x$-axis ($y = 0$): $x = -1$ and $x = 3$.

Check: $(-1)^2 - 2(-1) - 3 = 0$ and $3^2 - 6 - 3 = 0$