June 2018 Paper 3 Higher

(a) As age increases, time decreases, so the correlation is negative.

(b) An 11-year-old's times on the graph are about 14.5 to 16 seconds; 12 seconds is far faster than the trend suggests.

(c) The oldest girl on the graph is under 14, so predicting for 15 means going beyond the data, which is unreliable.

$5(p + 3) = 5p + 15$

$-2(1 - 2p) = -2 + 4p$

$5p + 15 - 2 + 4p = 9p + 13$

The parallel sides are 2 cm and 7 cm, and the height is 4 cm.

Area $= \frac{1}{2}(2 + 7) \times 4 = 18$ cm².

A triangle needs $\frac{1}{2} \times b \times h = 18$, so $b \times h = 36$, e.g. base 9 cm, height 4 cm.

The probability of landing on 4 is 0.65 on every throw, so the probability of not landing on 4 is $1 - 0.65 = 0.35$.

1. On the first throw, $0.65 + 0.25 \ne 1$: the 'not land on 4' branch should be 0.35, not 0.25.

2. On the second throw after landing on 4, 'land on 4' is labelled 0.35 and 'not land on 4' 0.65; these are the wrong way round.

(a) $AB$ is the hypotenuse and $BC$ is adjacent to angle $ABC$.

$\cos ABC = \frac{7}{11}$, so $ABC = \cos^{-1}(\frac{7}{11}) = 50.5^\circ$.

(b) New $\cos ABC = \frac{7}{10} = 0.7$, which is more than $\frac{7}{11} \approx 0.636$, so it increases.

(a) Total counters: $18 \div 0.45 = 40$.

Red and white together: $1 - 0.45 - 0.25 = 0.3$.

Red is twice white, so $P(\text{white}) = 0.1$ and $P(\text{red}) = 0.2$.

Red counters: $0.2 \times 40 = 8$.

(b) The number of silver marbles is $0.5 \times$ the total. That is only a whole number if the total is even.

Multiply both sides by 2: $5 - x = 4x - 14$.

Add $x$ and 14 to both sides: $19 = 5x$.

$x = \frac{19}{5} = 3.8$

Angle sum of a pentagon: $(5 - 2) \times 180 = 540^\circ$.

Known angles: $125 + 115 + 90 = 330$, so $540 - 330 = 210^\circ$ remains.

If $ABC = x$, then $x + 2x = 210$, so $x = 70$.

$BCD = 2 \times 70 = 140^\circ$.

(a) $d^3 = (1.4 \times 10^{-4})^3 = 2.744 \times 10^{-12}$

$\frac{w}{d^3} = \frac{5.6 \times 10^{-5}}{2.744 \times 10^{-12}} = 2.0408 \times 10^{7}$

$T = \sqrt{2.0408 \times 10^7} = 4517.5... = 4.52 \times 10^3$

(b) $w$ is multiplied by 1.1 and $d^3$ by $1.05^3 = 1.1576$. $T$ is multiplied by $\sqrt{1.1 \div 1.1576} = 0.975$, which is less than 1, so $T$ falls (to about $4.40 \times 10^3$).

$20 = 2^2 \times 5$, $45 = 3^2 \times 5$, $120 = 2^3 \times 3 \times 5$

LCM $= 2^3 \times 3^2 \times 5 = 360$ seconds

One hour $= 3600$ seconds, and $3600 \div 360 = 10$

So they flash together 10 times (11 if you count the start as well).

Mia's sale was 90% of her price: $162\,000 \div 0.9 = 180\,000$

Jerry's sale was 120% of his price: $180\,000 \div 1.2 = 150\,000$

Jerry paid £150 000.

(a) From $(0, 4)$ to $(8, 16)$: gradient $= \frac{16-4}{8-0} = \frac{12}{8} = 1.5$

(b) The volume rises 1.5 litres every second, so the gradient is the rate the container fills, in litres per second.

(c) At $t = 0$ there are already 4 litres, so the intercept is the starting volume.

Area ratio $3 : 4$, so length ratio $\sqrt{3} : \sqrt{4} = \sqrt{3} : 2$

Length scale factor from B to A: $\frac{\sqrt{3}}{2} = 0.8660$

Volume scale factor: $0.8660^3 = 0.6495$

Volume of A $= 10 \times 0.6495 = 6.50$ cm³

Each of the 16 teams plays the other 15 teams twice (once as each of the two sides), so every ordered pair of teams is one match.

Number of ordered pairs: $16 \times 15 = 240$

(Check: $\frac{16 \times 15}{2} = 120$ pairs, each playing twice, gives 240.)

(a) Read speeds at 0, 5, 10, 15, 20 s: 0, 22, 28, 32, 35 m/s.

Each strip is a trapezium of width 5:

$2.5(0+22) + 2.5(22+28) + 2.5(28+32) + 2.5(32+35)$

$= 55 + 125 + 150 + 167.5 = 497.5$ m

(b) The curve bulges above the straight lines joining the points, so the trapezia leave out some area. It is an underestimate.

(a) $n = 2$: $4a + 2b = -2$

$n = 4$: $16a + 4b = 12$

Double the first: $8a + 4b = -4$. Subtract: $8a = 16$, so $a = 2$, then $b = -5$.

6th term: $2(36) - 5(6) = 72 - 30 = 42$

(b) Second differences are 2, so start with $n^2$: 1, 4, 9, 16, 25.

Sequence minus $n^2$: $-1, -2, -3, -4, -5$, which is $-n$.

nth term $= n^2 - n$

In triangle $BCD$: $\frac{BD}{\sin 34°} = \frac{12.5}{\sin 109°}$

$BD = \frac{12.5 \sin 34°}{\sin 109°} = 7.393$ cm

In triangle $ABD$: $AD^2 = 11.4^2 + 7.393^2 - 2(11.4)(7.393)\cos 86°$

$AD^2 = 129.96 + 54.65 - 11.76 = 172.85$

$AD = 13.1$ cm

(a) $1^3 + 1 = 2 < 7$ and $2^3 + 2 = 10 > 7$. Since 7 is between 2 and 10, a solution lies between 1 and 2.

(b) $x^3 + x = 7 \Rightarrow x^3 = 7 - x \Rightarrow x = \sqrt[3]{7 - x}$

(c) $x_1 = \sqrt[3]{5} = 1.70998$

$x_2 = \sqrt[3]{5.29002} = 1.74242$

$x_3 = \sqrt[3]{5.25758} = 1.73885 \approx 1.74$

$\tan e = \frac{x}{4x-1}$ and $\tan f = \frac{6x+5}{12x+31}$

$x(12x+31) = (6x+5)(4x-1)$

$12x^2 + 31x = 24x^2 + 14x - 5$

$12x^2 - 17x - 5 = 0 \Rightarrow (3x-5)(4x+1) = 0$

$x = \frac{5}{3}$ (reject $x = -\frac{1}{4}$, since $x$ is a length)

Fill the Venn diagram from the centre out:

All three: 2. French and Spanish only: 4. German and Spanish only: $7 - 2 = 5$.

German only: 0 (all German speakers speak another). French and German only: $10 - 2 - 5 = 3$. French only: $31 - 2 - 4 - 3 = 22$.

Spanish only: $50 - 8 - 22 - 3 - 2 - 4 - 5 = 6$

P(both) $= \frac{6}{50} \times \frac{5}{49} = \frac{30}{2450} = \frac{6}{490}$

(a) $AD = CB$: opposite sides of a parallelogram are equal.

Angle $DAP$ = angle $BCQ$: opposite angles of a parallelogram are equal.

Angle $ADP$ = angle $CBQ = 90°$ (given).

So triangles $ADP$ and $CBQ$ are congruent (ASA).

(b) From (a), $AP = CQ$. $AP$ lies on $AB$ and $CQ$ lies on $DC$, which are parallel. A quadrilateral with one pair of sides equal and parallel is a parallelogram, so $APCQ$ is a parallelogram and $AQ$ is parallel to $PC$.