Specimen Paper 1 Foundation

There are 100 cm in 1 m, so divide by 100.

$530 \div 100 = 5.3$ m

3 hours is $3 \times 60 = 180$ minutes.

$\frac14$ of an hour is $15$ minutes.

$180 + 15 = 195$ minutes

Look at the third decimal place: it is 5.

5 or more rounds up, so $4.43|54$ becomes $4.44$.

Multiply by 100 to change a decimal to a percentage.

$0.9 \times 100 = 90$%

$(-3)^3 = (-3) \times (-3) \times (-3)$

$(-3) \times (-3) = 9$, then $9 \times (-3) = -27$

(a) An even number must end in 2 or 4. Put the biggest digits first and keep an even digit for the end: $5412$.

(b) Take each first digit in turn and pair it with the other three:

1: 12, 14, 15; 2: 21, 24, 25; 4: 41, 42, 45; 5: 51, 52, 54 (12 numbers).

A dual bar chart works well.

- Put the four sports along the horizontal axis.
- Use a frequency scale up to 16 on the vertical axis, going up in equal steps (e.g. 2s).
- For each sport draw a boys bar and a girls bar side by side: Hockey 3, 6; Tennis 8, 14; Football 15, 7; Golf 9, 1.
- Add a key for boys and girls, and label both axes.

(a) Try 18: $18 \div 2 = 9$, which ends in 9, not 4. (38 → 19 works too.)

(b) Alice is wrong. 27 ends in 7 but $27 = 3 \times 9$, so it is not prime. (57 = 3 × 19 also works.)

Split 63 into 60 and 3.

$247 \times 60 = 14820$

$247 \times 3 = 741$

$14820 + 741 = 15561$

Convert each limit to cm ($\times 2.54$):

- height: $22 \times 2.54 = 55.88$ cm (bag 50 cm, fine)
- width: $14 \times 2.54 = 35.56$ cm (bag 40 cm, too wide)
- depth: $9 \times 2.54 = 22.86$ cm (bag 20 cm, fine)

The bag is too wide, so Chris cannot take it.

Boys brown: $12 - 5 - 4 = 3$

Girls total: $30 - 12 = 18$

Girls green: $9 - 4 = 5$

Girls blue: $18 - 7 - 5 = 6$

Column totals: blue $5 + 6 = 11$, brown $3 + 7 = 10$ (check: $11 + 10 + 9 = 30$).

Red : black $= 28 : 84$

The HCF of 28 and 84 is 28, so divide both by 28:

$28 : 84 = 1 : 3$

Pattern $n$ has $n + 2$ grey tiles in a column and $n$ white tiles beside it.

(a) Pattern 4: 6 grey tiles in a column, 4 white tiles next to the middle ones.

(b) Total tiles $= (n + 2) + n = 2n + 2$. For $n = 20$: $2 \times 20 + 2 = 42$.

(c) Grey tiles go 3, 4, 5, ... so the expression is $n + 2$.

(a) Round: $4.2 \to 4$ and $50.1 \to 50$.

$4 \times 50 = 200$p a week, and $200 \times 13 = 2600$p $= £26$.

(b) Both 4.2 and 50.1 were rounded down, so the estimate is an underestimate. (The exact cost is about £27.35.)

Measure the plan: about 12 cm by 4 cm.

Scale 1 cm : 2 m, so the floor is $24$ m by $8$ m.

Area $= 24 \times 8 = 192$ m². Each board covers $1.2 \times 1 = 1.2$ m².

$192 \div 1.2 = 160$ boards are needed ($20$ along $\times$ $8$ across), which is more than 150. So she does not have enough.

Split it into two rectangles.

Tall part: $10 \times 2 = 20$ cm²

Bottom right part: $(8 - 2) \times 2 = 12$ cm²

Total $= 20 + 12 = 32$ cm²

Rotating 90° clockwise about $(0,0)$ sends $(x, y)$ to $(y, -x)$.

$(1,2) \to (2,-1)$

$(1,4) \to (4,-1)$

$(2,2) \to (2,-2)$

Plot and join these three points.

Men: $\frac{7}{20} \times 500 = 175$

Women: $40\% \times 500 = 200$

Children: $500 - 175 - 200 = 125$

(a) Milk needed: $7 \times \frac34 = 5.25$ pints, so he must buy at least 6 pints.

3 two-pint bottles: $3 \times 93 = 279$p $= £2.79$

(2 two-pint + 2 one-pint $= 290$p costs more, and 5 pints is not enough.)

(b) 6 pints lasts 7 days only if he uses up to $6 \div 7 \approx 0.857$ pints a day. If he uses more than that he will need more bottles and pay more.

The two unknown angles add to $90°$.

$7x + 5x + 18 = 90$

$12x = 72$, so $x = 6$

Angles: $7 \times 6 = 42°$ and $5 \times 6 + 18 = 48°$. The smallest is $42°$.

$p = \frac{F}{A}$, so $A = \frac{F}{p}$.

$A = \frac{140}{35} = 4$ m²

The probabilities add up to 1.

$1 - 0.24 - 0.32 = 0.44$

Blue and green are equal, so each is $0.44 \div 2 = 0.22$.

Let each tile be $l$ long and $w$ wide.

Across the top: $2l + w = 11$. Down the side: $l + w = 7$.

Subtracting: $l = 4$, so $w = 3$.

One tile $= 4 \times 3 = 12$ cm², and there are 4 tiles: $4 \times 12 = 48$ cm².

Volume $= 100 \times 60 \times 40 = 240000$ cm³

$1$ litre $= 1000$ cm³, so that is $240$ litres.

Bags: $240 \div 8 = 30$

Cost: $30 \times £2.50 = £75$, which is more than £70, so Sally is wrong.

(a) Sharif threw 160 times, more than anyone else, so his results are the most reliable.

(b) Paul got $80 : 40 = 2 : 1$, but all results together give $300$ heads and $100$ tails, $3 : 1$. So heads is about three times as likely, not twice: Paul is not correct.

(c) P(heads) $\approx \frac{300}{400} = \frac34$.

P(two heads) $= \frac34 \times \frac34 = \frac{9}{16}$

(a) $\cos 30° = \frac{\sqrt3}{2}$

(b) $x$ is opposite the $30°$ angle and 12 cm is the hypotenuse, so use sine.

$\sin 30° = \frac{x}{12}$

$x = 12 \times 0.5 = 6$

Multiply each term in the first bracket by each term in the second:

$x \times x = x^2$, $x \times (-1) = -x$, $3 \times x = 3x$, $3 \times (-1) = -3$

$x^2 - x + 3x - 3 = x^2 + 2x - 3$

$x^2 - 16 = x^2 - 4^2$, a difference of two squares.

So $x^2 - 16 = (x + 4)(x - 4)$

Multiply the first equation by 3: $12x + 3y = 75$

Add the second: $12x + 3y + x - 3y = 75 + 16$, so $13x = 91$ and $x = 7$.

Substitute: $4(7) + y = 25$, so $y = -3$.

Check: $7 - 3(-3) = 16$ ✓