Specimen Paper 3 Foundation

The thousands either side are 5000 and 6000.

5689 is past the halfway point of 5500, so it rounds up to 6000.

Work out the top and bottom first: $30 + 12 = 42$ and $5 + 3 = 8$.

$42 \div 8 = 5.25$

The reciprocal of a number is 1 divided by it.

$1 \div 0.125 = 8$ (because $0.125 = \frac{1}{8}$).

(i) $12 = 4 \times 3$, so 12 is a multiple of 4.

(ii) 2 and 5 each have exactly two factors, so either is prime. (1 is not prime.)

1 litre = 1000 ml, so 1.5 litres = 1500 ml.

$1500 + 250 = 1750$ ml, which is 1750 ml (or 1.75 litres).

(a) Put the 0 of the ruler on P and read the mark at Q: about 8 cm.

(b) Place the protractor centre on Q with the base line along QP and count up from 0: about **$35^\circ$**.

(a) $f + 2f + f = 4f$, so $4f = 20$ and $f = 5$.

(b) $18 - m = 6$ means $m = 18 - 6 = 12$.

(c) Add the powers: $d^2 \times d^3 = d^{2+3} = d^5$.

Round each number: $3.4 \approx 3$ and $5.3 \approx 5$.

$3 \times 5 = 15$, and even $4 \times 6 = 24$, so the answer must be less than 24. 180.2 is far too big.

From A to B there are 6 equal gaps, so each gap is $48 \div 6 = 8$.

A is 2 gaps left of 0: $A = -2 \times 8 = -16$.

B is 4 gaps right of 0: $B = 4 \times 8 = 32$.

Top row: $3 \times 12 + 2 = 36 + 2 = 38$.

Bottom row: $3n + 2 = 47$, so $3n = 45$ and $n = 15$.

(a) We need the cube root of 343: $7 \times 7 \times 7 = 343$, so the number is 7.

(b) $4^4 = 4 \times 4 \times 4 \times 4 = 256$

Square numbers: 1, 4, 9, 16, 25, 36.

$29 = 25 + 4 = 5^2 + 2^2$

$37 = 36 + 1 = 6^2 + 1^2$

So yes, Nadia is correct.

The terms go down by 6 each time.

32, 26, 20, 14, 8, 2, −4, −10

The first two terms below zero are −4 and −10.

(a) In angle CBA the middle letter, B, is where the angle is. Mark y inside the triangle at B.

(b) Find the four corners C, D, E and G and shade the face they enclose.

(c) A cuboid has 4 edges on top, 4 on the bottom and 4 upright: $4 + 4 + 4 = 12$.

One slice weighs $400 \div 10 = 40$ g.

Fibre is 5 g per 100 g, so in 40 g: $5 \times \frac{40}{100} = 2$ g.

The rectangle is 5 by 3, so its perimeter is 16.

(i) Cut a triangle off a corner: the slanted cut is shorter than the two edges it replaces, so the perimeter drops.

(ii) Cut a 1 by 1 square from a corner: two edges of 1 are removed and two new edges of 1 appear, so the perimeter stays 16.

(iii) Cut a 1 by 1 notch from the middle of a side: 1 unit of edge goes but 3 units appear, so the perimeter becomes 18.

(a) If A is one of the two equal angles and B is the other: $B = 70^\circ$.

If A and C are equal: $B = 180 - 70 - 70 = 40^\circ$.

If A is the odd angle: $B = (180 - 70) \div 2 = 55^\circ$.

(b) Two angles of $120^\circ$ would add to more than $180^\circ$, so 120° must be the single different angle. Then $B = (180 - 120) \div 2 = 30^\circ$ only.

(a) Fruit is 40%, so oats are $100 - 40 = 60\%$.

$40 : 60$, divide both by 40: $1 : 1.5$

(b) $1 + 3 = 4$ parts, 3 of them bran, so $\frac{3}{4}$.

Each deal of 3 boxes costs $2 \times 3.69 = £7.38$.

$50 \div 7.38 = 6.77...$, so 6 deals = 18 boxes, costing $6 \times 7.38 = £44.28$.

Left over: $50 - 44.28 = £5.72$, enough for 1 more box at £3.69.

Total: $18 + 1 = 19$ boxes.

A (even numbers) = {2, 4, 6, 8, 10}.

Overlap: 2 and 6. So A only: 4, 8, 10.

$A \cup B$ adds 1, 3, 9, which must be in B only.

Numbers in neither set: 5 and 7, written outside the ovals but inside the rectangle.

(a) The point (4, 10) sits well below the pattern of the others.

(b) Draw a straight ruler line through the middle of the other nine points, roughly from (2, 22) to (13, 86). As hours go up, marks go up: positive correlation.

(c) Go up from 9 hours to the line and across: about 65.

(d) The data only go up to 13 hours. Using the line at 18 hours is extrapolation, so we cannot be sure she would get full marks.

To the nearest cm, the true length is within half a centimetre either side of 13.

$13 - 0.5 = 12.5$ and $13 + 0.5 = 13.5$

So $12.5 \le L < 13.5$.

The line crosses the y-axis at 1, so $c = 1$.

It passes through (0, 1) and (2, 5): up 4 for 2 across, so the gradient is $4 \div 2 = 2$.

$y = 2x + 1$

(a) Midpoints: 30, 34, 38, 42.

$30 \times 24 + 34 \times 12 + 38 \times 8 + 42 \times 6 = 720 + 408 + 304 + 252 = 1684$

$1684 \div 50 = 33.68$ inches

(b) $\frac{24}{50} = 0.48$, less than half. Three quarters would be 0.75, far more than the share sold, so the manager is correct.

(a) Wall area $= \frac{1.8 + 2.7}{2} \times 0.8 = 1.8\text{ m}^2 = 18000\text{ cm}^2$

Tile area $= 15 \times 7.5 = 112.5\text{ cm}^2$

Tiles: $18000 \div 112.5 = 160$. Packs: $160 \div 9 = 17.8$, round up to 18.

(b) $160 \times 1.1 = 176$ tiles, $176 \div 9 = 19.6$, so 20 packs.

That is 2 more packs, but 10% of 18 is only 1.8, so the assumption is not correct.

Find two numbers that multiply to −4 and add to 3: $4$ and $-1$.

$x^2 + 3x - 4 = (x + 4)(x - 1)$

Rearrange each into $y = mx + c$:

A: $y = 2x + 4$ (gradient 2)

B: $y = \frac{1}{2}x + 2$ (gradient $\frac{1}{2}$)

C: $y = -x + 2$ (gradient −1)

D: $y = 2x - 4$ (gradient 2)

A and D have the same gradient, so they are parallel.

Density of B is 1.1 g/cm³, so density of A $= \frac{19}{22} \times 1.1 = 0.95$ g/cm³.

Mass of C $= 5 \times 0.95 + 15 \times 1.1 = 4.75 + 16.5 = 21.25$ g.

Density of C $= 21.25 \div 20 = 1.0625$ g/cm³.