June 2019 Paper 3 Foundation

478 lies between 400 and 500.

The tens digit is 7, which is 5 or more, so round up: 500.

The 8 times table: 40, 48, 56, 64, ...

The ones between 41 and 60 are 48 and 56; either will do.

1 km = 1000 m

$1.5 \times 1000 = 1500$ metres.

Powers of 3: $3^1 = 3$, $3^2 = 9$, $3^3 = 27$, $3^4 = 81$

From the list, the powers of 3 are 9 and 27.

Per cent means out of 100, so $19\% = \frac{19}{100}$.

$10\%$ of 80 is 8, so $20\%$ is $2 \times 8 = 16$.

Or: $80 \times 0.2 = 16$.

Green: $26 + 7 = 33$

Red: $16 + 11 = 27$

Difference: $33 - 27 = 6$

A quarter of 52 is $52 \div 4 = 13$ litres already in the tank.

Fuel needed: $52 - 13 = 39$ litres (or $\frac{3}{4} \times 52 = 39$).

$4e + 6f + 7e - f$

$e$ terms: $4e + 7e = 11e$

$f$ terms: $6f - f = 5f$

Answer: $11e + 5f$

Given away: $35 + 50 + 75 = 160$

Left: $400 - 160 = 240$

$\frac{240}{400} = \frac{3}{5}$

(a) Difference per parcel: $35.38 - 15.25 = 20.13$

$12 \times 20.13 = £241.56$

(b) He used 20 parcels instead of 21 and £15 instead of £15.25. Both are smaller than the real values, so the real cost is more than £300.

Total parts: $6 + 9 + 10 = 25$

Ben gets 9 of the 25 parts, so $\frac{9}{25}$.

Terms: 24, 32, 40, 48, 56, 64, 72, 80, 88, ...

(a) 40 is in the sequence and is in the 5 times table.

(b) No. 80 and 88 are terms, so 85 is skipped. Also, every term is even and 85 is odd.

Top: $5.27 + 3.5 = 8.77$

Bottom: $7.9 - 4.36 = 3.54$

$8.77 \div 3.54 = 2.477401129...$

(a) $4 \times 70 + 50 = 280 + 50 = £330$

(b) Undo the rule: $680 - 50 = 630$, then $630 \div 70 = 9$ weeks.

One rectangle stands upright (4 by 6), the other lies flat (6 by 4) beside it.

The step on top is $6 - 4 = 2$ cm.

Perimeter: $4 + 2 + 6 + 4 + 6 + 4 + 6 = 32$ cm

Or: $20 + 20 - 2 \times 4 = 32$ cm (the joined 4 cm edge is lost from both).

Measure: the box is about 1.8 cm wide and the mast about 10.8 cm tall.

The mast is about $10.8 \div 1.8 = 6$ times the box width.

Height $\approx 6 \times 1.5 = 9$ m

(a) The highest frequency is 7, for 2 points, so the mode is 2.

(b) $0\times4 + 1\times3 + 2\times7 + 3\times5 + 4\times6 + 5\times5$

$= 0 + 3 + 14 + 15 + 24 + 25 = 81$

$y = 2x + 4$

Subtract 4: $y - 4 = 2x$

Divide by 2: $x = \frac{y - 4}{2}$

EB is a diagonal of the square, so it cuts the right angle at B in half: angle $EBD = 45^\circ$.

The triangle is equilateral, so angle $DBC = 60^\circ$.

Angle $EBC = 45 + 60 = 105^\circ$

$850 \times 18.53 = 15\,750.5$ rand

$15\,750.5 \div 200 = 78.7525$

Only whole notes count, so 78 notes.

Gallons: $560 \div 34.5 = 16.231...$

Litres: $16.231... \times 4.55 = 73.855...$

Cost: $73.855... \times 1.08 = £79.76$

Alan: $100 - 32 - 40 = 28\%$, and $0.28 \times 400 = 112$

Beryl: $1 - \frac{3}{10} - \frac{1}{10} = \frac{6}{10}$, and $\frac{6}{10} \times 500 = 300$

Charlie: $150 \div 3 = 50$, so medium $= 4 \times 50 = 200$

Total: $112 + 300 + 200 = 612$

(a) In both A and B: 6, 9

A only: 1, 5, 8. B only: 2. Neither: 3, 4, 7.

(b) $A \cap B = \{6, 9\}$, 2 numbers out of 9.

$P = \frac{2}{9}$

Multiplier for 1.5% interest: 1.015

$200\,000 \times 1.015^4 = 212\,272.71$

Interest: $212\,272.71 - 200\,000 = £12\,272.71$

(a) 80 plants, so the median is about the 40th–41st value.

Running totals: 7, 20, 34, 46, ... The 40th value is in $40 < h \le 50$.

(b) Plot each frequency at the middle of its class: (15, 7), (25, 13), (35, 14), (45, 12), (55, 16), (65, 18). Join with straight lines.

Things wrong with the graph:

1. The points are joined with curves; a time series should use straight lines.
2. The vertical axis starts at 6, which exaggerates the changes.

Other valid points: the first quarter of each year is missing, and it is not clear what 2, 3, 4 on the horizontal axis mean.

Interior angles of a hexagon: $(6 - 2) \times 180 = 720^\circ$

By symmetry $A = B = 117^\circ$; let $D = E = x$ and $C = F = 2x$.

$117 + 117 + 2x + 2x + x + x = 720$, so $6x = 486$ and $x = 81$

Angle $AFE = 2 \times 81 = 162^\circ$

Radius $= 0.8$ m

Top and bottom: $2 \times \pi \times 0.8^2 = 4.021...$

Curved surface: $2 \times \pi \times 0.8 \times 1.8 = 9.047...$

One tank: $13.069...$ m²; three tanks: $39.21$ m²

Paint covers $7 \times 5 = 35$ m² $< 39.21$, so No, he does not have enough.

$3x - 4y = 11$ and $9x + 2y = 5$

Double the second: $18x + 4y = 10$

Add to the first: $21x = 21$, so $x = 1$

$9(1) + 2y = 5$, so $2y = -4$ and $y = -2$