June 2019 Paper 2 Higher

(a) $14n - 11n > 6$, so $3n > 6$ and $n > 2$.

(b) Subtract 3 from every part: $-2 - 3 < x \le 4 - 3$, so $-5 < x \le 1$.

Draw an open circle at $-5$ (not included), a filled circle at $1$ (included) and join them.

Make a table of values using $y = 2x - 3$:

| $x$ | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| $y$ | −7 | −5 | −3 | −1 | 1 | 3 | 5 |

Plot the points and join them with a single ruled line from $(-2,-7)$ to $(4,5)$.

(i) $\frac{10}{30} = \frac{1}{3}$ of the sample chose the theme park.

$\frac{1}{3} \times 195 = 65$

(ii) Assume the 30 students represent the whole group fairly. If they do not, the true number could be quite different from 65.

Volume $= 30 \times 19 \times 6 = 3420\text{ cm}^3$

Water $= \frac{2}{3} \times 3420 = 2280\text{ cm}^3 = 2280\text{ ml}$

$2280 \div 275 = 8.29...$

Only 8 cups can be completely filled.

AC is the hypotenuse (16 cm) and AB is opposite the 38° angle.

$\sin 38^\circ = \frac{AB}{16}$

$AB = 16 \sin 38^\circ = 9.850... = 9.85\text{ cm}$

The display began 8.3, so the digits after 8.3 have been cut off (truncated), not rounded.

The smallest value is 8.3 and $y$ can get as close as you like to 8.4 without reaching it.

$8.3 \le y < 8.4$

Abby : Ben = 2 : 7. Chloe and Denesh each get $1.5 \times 2 = 3$ parts.

Total parts $= 2 + 7 + 3 + 3 = 15$

One part $= 360 \div 15 = £24$

Ben gets $7 \times 24 = £168$

(a) Move the decimal point 3 places right to get 5.62, so $0.00562 = 5.62 \times 10^{-3}$.

(b) $1.452 \times 10^3$: move the point 3 places right to get $1452$.

(a) Radius of B is $0.9 \times$ radius of A, so area of B is $0.9^2 = 0.81$ times area of A.

Area A : area B $= 1 : 0.81 = 100 : 81$

(b) $e^2 = 1.44 f^2$, so $e = \sqrt{1.44} f = 1.2 f$.

$e : f = 1.2 : 1 = 6 : 5$

(a) Not late $= 1 - 0.15 = 0.85$. Each Friday pair of branches is late 0.15, not late 0.85.

(b) P(not late on both days) $= 0.85 \times 0.85 = 0.7225$

P(late at least once) $= 1 - 0.7225 = 0.2775$

(a) Add the frequencies as you go: $5, 5+30=35, 55, 70, 78, 80$.

(b) Plot each total at the top of its interval, e.g. $(40, 35)$, and join with a smooth curve.

(c) At $t = 90$ the graph reads about 74, so $80 - 74 = 6$ workers take longer.

$\frac{6}{80} \times 100 = 7.5\%$

Sector angle: $\frac{\theta}{360} \times \pi \times 7^2 = 40$, so $\theta = \frac{40 \times 360}{49\pi} = 93.54...^\circ$

Arc $= \frac{93.54...}{360} \times 2\pi \times 7 = 11.43$ cm

Perimeter $= 7 + 7 + 11.43 = 25.43 = 25.4$ cm (3 s.f.)

$x^2 + 3x - 10 = (x + 5)(x - 2)$

$(x + 5) \div \frac{(x + 5)(x - 2)}{x - 1} = (x + 5) \times \frac{x - 1}{(x + 5)(x - 2)} = \frac{x - 1}{x - 2}$

$6 + \frac{x - 1}{x - 2} = \frac{6(x - 2) + x - 1}{x - 2} = \frac{7x - 13}{x - 2}$

(a)(i) Draw a tangent at $t = 15$. Reading two points on it, e.g. it rises 7.5 over 9 s:

gradient $\approx 7.5 \div 9 = 0.83$

(ii) This is the car's acceleration in m/s² at 15 seconds.

(b) Speeds at $t = 0, 5, 10, 15, 20$: $0, 4, 12, 18, 20$.

Area $= \frac{5}{2}(0+4) + \frac{5}{2}(4+12) + \frac{5}{2}(12+18) + \frac{5}{2}(18+20) = 10 + 40 + 75 + 95 = 220$ m

$f(m - 1) = 3m + 4$

$fm - f = 3m + 4$

$fm - 3m = f + 4$

$m(f - 3) = f + 4$

$m = \frac{f + 4}{f - 3}$

$3y = 4x + 7$ gives $y = \frac{4}{3}x + \frac{7}{3}$, gradient $\frac{4}{3}$.

Perpendicular gradient $= -\frac{3}{4}$.

Through $(3, -5)$: $-5 = -\frac{3}{4}(3) + c$, so $c = -5 + \frac{9}{4} = -\frac{11}{4}$.

$y = -\frac{3}{4}x - \frac{11}{4}$

(a) Small : large $= 4 : 7$ has 11 parts; red : yellow $= 3 : 5$ has 8 parts. The total must divide by both 11 and 8, and LCM(11, 8) $= 88$.

(b) With 88 cubes: small $= 4 \times 8 = 32$, yellow $= 5 \times 11 = 55$.

All small cubes are yellow, so large yellow $= 55 - 32 = 23$.

Triangle ABD is isosceles, so $\angle BAD = \angle BDA = \frac{180 - 40}{2} = 70^\circ$ (base angles of an isosceles triangle are equal; angles in a triangle sum to 180°).

$\angle BCD = 180 - 70 = 110^\circ$ (opposite angles of a cyclic quadrilateral sum to 180°).

Triangle BCD is isosceles: $\angle CDB = \frac{180 - 110}{2} = 35^\circ$.

$\angle ADE = 180 - 70 - 35 = 75^\circ$ (angles on a straight line sum to 180°).

$BE = 15 \tan 35^\circ = 10.50$ cm

$MA = \frac{3}{5} \times 15 = 9$ cm, so $MB = \sqrt{9^2 + 15^2} = \sqrt{306} = 17.49$ cm

The angle is at M in triangle EMB, right-angled at B:

$\tan \theta = \frac{10.50}{17.49}$, so $\theta = 30.98... = 31.0^\circ$

(a) $\vec{FE} = \vec{FC} + \vec{CD} + \vec{DE} = \mathbf{a} - \mathbf{b} + \mathbf{a} + \mathbf{b} = 2\mathbf{a}$

(b) $\vec{FM} = 2\mathbf{a} - \frac{1}{2}\mathbf{b}$, so $\vec{FX} = \frac{n}{n+1}(2\mathbf{a} - \frac{1}{2}\mathbf{b})$.

Also $\vec{FX} = \vec{FC} + k\vec{CE} = \mathbf{a} - \mathbf{b} + k(\mathbf{a} + \mathbf{b})$.

Compare coefficients: $1 + k = \frac{2n}{n+1}$ and $1 - k = \frac{n}{2(n+1)}$.

Add them: $2 = \frac{5n}{2(n+1)}$, so $4n + 4 = 5n$ and $n = 4$.