November 2019 Paper 1 Foundation

The digits of 1074 are thousands, hundreds, tens, ones: 1, 0, 7, 4.

The 7 is in the tens column, so it is worth $7 \times 10 = 70$.

Look at the second decimal place: it is 8.

8 is 5 or more, so round the 5 up: $4.58 \approx 4.6$

Multiplying by 100 moves every digit two places to the left.

$31.7 \times 100 = 3170$

The highest common factor of 28 and 70 is 14.

$\frac{28}{70} = \frac{28 \div 14}{70 \div 14} = \frac{2}{5}$

Per cent means out of 100, so divide by 100.

$15\% = 15 \div 100 = 0.15$

(a) January has 3 whole squares: $3 \times 8 = 24$

(b) $12 = 8 + 4$, so draw one whole square and a half square for April.

(c) February $3.5 \times 8 = 28$, March $2.5 \times 8 = 20$

Total $= 24 + 28 + 20 + 12 = 84$

$1\frac{1}{4}$ hours $= 60 + 15 = 75$ minutes

1 hour 25 minutes $= 60 + 25 = 85$ minutes

Difference $= 85 - 75 = 10$ minutes

Four blocks: $4 \times 650 = 2600$ g

Total: $3$ kg $= 3000$ g

Last block: $3000 - 2600 = 400$ g

Angles on a straight line add up to $180°$.

$x = 180 - 100 - 35 = 45°$

(a) Go 3 right from $O$ then 2 up, and mark $A$.

(b) $B = (-3, 2)$ and $C = (1, -2)$.

Midpoint $= \left(\frac{-3 + 1}{2}, \frac{2 + (-2)}{2}\right) = (-1, 0)$

Work systematically. Fix the first throw as H, then list the four ways for the other two:

HHH, HHT, HTH, HTT

Then repeat with T first:

THH, THT, TTH, TTT

There are $2 \times 2 \times 2 = 8$ outcomes.

(a) T-shirts: $3 \times 25 = 75$

Total with the jacket: $60 + 75 + 80 = 215$

$215 > 200$, so no, he is 15 dollars short.

(b) Round the rate down: $0.749 \approx 0.7$

$60 \times 0.7 = £42$. This is already an underestimate and is more than £40, so he is wrong.

(a) Multiply the numbers and write the letters together: $2 \times 5 = 10$, so $10ab$.

(b) $x$ terms: $3x + 5x = 8x$

$y$ terms: $2y - y = y$

Answer: $8x + y$

Split 15 into 10 and 5:

$23 \times 10 = 230$

$23 \times 5 = 115$

$230 + 115 = 345$

(a) Males: $120 - 75 = 45$

Males sitting: $45 - 29 = 16$

Females sitting: $30 - 16 = 14$

Females standing: $75 - 14 = 61$

(b) 29 of the 120 people are males who stand, so the probability is $\frac{29}{120}$.

(a) The flat part runs from 12:45 to 13:30, which is 45 minutes.

(b) He drives 25 km home between 13:30 and 14:00, which is half an hour.

Speed $= 25 \div 0.5 = 50$ km/h

$x - 1 = 2$, so $x = 3$

$2x^2 = 2 \times 3^2 = 2 \times 9 = 18$

School A: $x = (360 - 60) \div 2 = 150°$

Tigers at A $= \frac{150}{360} \times 480 = 200$

School B: tigers angle $= 360 - 160 - 110 = 90°$

Tigers at B $= \frac{90}{360} \times 760 = 190$

$200 \ne 190$, so Henry is not correct.

The filled circle at $-3$ means $-3$ is included: $p \ge -3$

The open circle at $1$ means $1$ is not included: $p < 1$

Together: $-3 \le p < 1$

$108 = 2^2 \times 3^3$

$120 = 2^3 \times 3 \times 5$

Take the highest power of each prime:

LCM $= 2^3 \times 3^3 \times 5 = 8 \times 27 \times 5 = 1080$

Women: $60 \div 2 = 30$

Men: $30 \div 3 = 10$

Children: $60 - 30 - 10 = 20$

Children : men $= 20 : 10 = 2 : 1$, so $n = 2$

$1\frac{3}{4} = \frac{7}{4}$ and $1\frac{1}{3} = \frac{4}{3}$

$\frac{7}{4} \times \frac{4}{3} = \frac{28}{12} = \frac{7}{3} = 2\frac{1}{3}$

1. With centre $P$, draw an arc cutting $CD$ at two points.
2. From each of these points, draw arcs of equal radius that cross on the other side of $CD$.
3. Join $P$ to where the arcs cross. This line meets $CD$ at right angles.

Leave all arcs showing.

Angle $ACB = 180 - 75 - 51 = 54°$

Split 54 in the ratio 2 : 1: one part $= 54 \div 3 = 18°$

So $ACD = 18°$ and $DCB = 36°$

In triangle $BDC$: angle $BDC = 180 - 51 - 36 = 93°$

Red: $4 \times 5 = 20$ kg

Blue: $5 \times 9 = 45$ kg

Total $= 20 + 45 + 6 = 71$ kg

Mean $= 71 \div 10 = 7.1$ kg

$7.1 > 7$, so Donna is not correct.

(a) Multiply the powers: $(p^2)^5 = p^{2 \times 5} = p^{10}$

(b) Divide the numbers and subtract the powers:

$12 \div 6 = 2$, $x^{7-3} = x^4$, $y^{3-1} = y^2$

Answer: $2x^4y^2$

Distance $PQ = 12 \times 1.5 = 18$ km

On the drawing: $18 \div 4 = 4.5$ cm

From $P$, measure $70°$ clockwise from north and mark $Q$ 4.5 cm along that line.

(i) Measure $QL$ in cm and multiply by 4 (about 20 to 23 km).

(ii) At $L$, measure clockwise from north round to $Q$: about $317°$ to $330°$.

Angle $AOB$ is not obtuse, so $2x + 3x + 10 \le 90$

$5x \le 80$

$x \le 16$, so the greatest value is $x = 16$

(a) $BC = 15$ matches $QR = 10$, so the scale factor is $\frac{10}{15} = \frac{2}{3}$

$PR = 9 \times \frac{2}{3} = 6$ cm

(b) $GK = 10 - 4 = 6$ cm

The triangles are congruent, so $EG = GK = 6$ and $GF = HG = 4$

$EF = 6 - 4 = 2$ cm