Specimen Paper 2 Higher

Take 11 from both sides: $3t = w - 11$

Divide both sides by 3: $t = \dfrac{w - 11}{3}$

Put every price in pounds.

Paris: $1980 \div 1.34 = £1477.61$

New York: $2250 \div 1.52 = £1480.26$

London: £1480

The lowest is Jardins of Paris.

Readings: 110, 84, 78, 94, 90, 120

Total $= 576$

Mean $= 576 \div 6 = 96$

The mean equals the target, so yes, the target was met.

(a) The largest frequency is 10, in the class $160 < h \le 170$.

(b) The points are plotted at the start of each class instead of at the midpoints (135, 145, …).

The last point has been joined back to the first, but a frequency polygon is not a closed shape.

(a) 36 minutes $= 0.6$ hours, so distance $= 15 \times 0.6 = 9$ km.

Plot (9 am, 0) to (9.36 am, 9), then on to (10.45 am, 17).

(b) 15 minutes $= 0.25$ hours

Distance $= 18 \times 0.25 = 4.5$ km

Multiplier for 4% growth: $1.04$

$7500 \times 1.04^2 = 7500 \times 1.0816 = £8112$

Let Becky have $x$. Chris has $2x$ and Dan has $2x + 7$.

$x + 2x + 2x + 7 = 57$, so $5x = 50$ and $x = 10$

Becky 10, Chris 20, Dan 27

Equal shares would be $57 \div 3 = 19$ each, but Chris keeps his 20, so Dan is wrong.

Rectangle: $19 \times 16 = 304$ cm²

Radius $= 8$, semicircle $= \frac{1}{2} \pi \times 8^2 = 100.53$ cm²

Shaded $= 304 - 100.53 = 203.47$ cm²

$\frac{203.47}{304} \times 100 = 66.9\%$

Triangle $ABC$ is isosceles with $AB = CB$, so angles at $A$ and $C$ are equal.

$\angle BAC = (180 - 50) \div 2 = 65°$

Bearing of $C$ from $A$ $= 70 + 65 = 135°$

(a) The line goes from (0, 210) to (140, 0).

Gradient $= \frac{0 - 210}{140 - 0} = -1.5$

(b) The depth of water goes down by 1.5 cm every second: it is the rate the tank is emptying.

(a) $0.7 \times 0.7 = 0.49$

(b) Losing at least one is everything except winning both.

$1 - 0.49 = 0.51$

(a) $1 \div 2.5 = 0.4$

(b) Top: $4.3 \times \tan 39° = 3.4821$

Bottom: $23.4 - 6.06 = 17.34$

$3.4821 \div 17.34 = 0.20081$

$\sqrt[3]{0.20081} = 0.586$ (3 s.f.)

$(3x - 1)(x + 5) = 3x^2 + 15x - x - 5 = 3x^2 + 14x - 5$

$(3x^2 + 14x - 5)(4x - 3)$

$= 12x^3 - 9x^2 + 56x^2 - 42x - 20x + 15$

$= 12x^3 + 47x^2 - 62x + 15$

Triangles $ABD$ and $CBA$ are similar (both right-angled and they share angle $B$).

So $\frac{CB}{AB} = \frac{AB}{DB}$

$CB = \frac{13 \times 13}{5} = \frac{169}{5} = 33.8$ cm

$T = k\sqrt{L}$

New length $= 1.4L$, so new $T = k\sqrt{1.4L} = \sqrt{1.4} \times k\sqrt{L}$

$\sqrt{1.4} = 1.1832$

Increase $= 18.3\%$

The £300k–£400k bar holds 20 houses over a width of 100, so its height stands for 0.2 per £1000.

That sets the scale: the bar is 2 large squares tall, so each large square up is 0.1.

0–100: $0.14 \times 100 = 14$

100–150: $0.6 \times 50 = 30$

150–200: $0.8 \times 50 = 40$

Total $= 14 + 30 + 40 = 84$

First differences: 2, 4, 6, 8; second difference 2, so start with $n^2$.

$n^2$: 1, 4, 9, 16, 25

Sequence minus $n^2$: 0, −1, −2, −3, −4, which is $-n + 1$

nth term $= n^2 - n + 1$

$f(x + 2) = 3(x + 2)^2 - 2(x + 2) - 8$

$= 3(x^2 + 4x + 4) - 2x - 4 - 8$

$= 3x^2 + 12x + 12 - 2x - 12$

$= 3x^2 + 10x$

$\frac{1}{2}x(x - 2) = 2.5 \Rightarrow x^2 - 2x - 5 = 0$

$x = \frac{2 \pm \sqrt{4 + 20}}{2} = 1 \pm \sqrt{6}$, so $x = 3.449$ (positive)

Sides: 3.449 and 1.449

Hypotenuse $= \sqrt{3.449^2 + 1.449^2} = 3.742$

Perimeter $= 3.449 + 1.449 + 3.742 = 8.64$ cm

(a) Draw a tangent at $t = 6$ and find rise ÷ run, e.g. from about (2, 32) to (10, 60): $28 \div 8 = 3.5$ m/s².

(b) Strips of width 4: $v(0) = 0,\ v(4) = 35,\ v(8) = 51,\ v(12) = 54$

$\frac{4}{2}(0 + 35) = 70$

$\frac{4}{2}(35 + 51) = 172$

$\frac{4}{2}(51 + 54) = 210$

Total $= 70 + 172 + 210 = 452$ m

2016: $1.05(9500 - 250) = 9712.5$

2017: $1.05(9712.5 - 250) = 9935.625$

2018: $1.05(9935.625 - 250) = 10169.90…$

About 10170 bees.

$x$: upper bound 99.75

$y = 67$ to 2 s.f.: lower bound 66.5

Upper bound of $D = \frac{99.75}{66.5} = 1.5$

Gradient of radius $OP = \frac{3}{4}$

The tangent is perpendicular, so its gradient is $-\frac{4}{3}$

$3 = -\frac{4}{3}(4) + c \Rightarrow c = \frac{25}{3}$

$y = -\frac{4}{3}x + \frac{25}{3}$

Join $AO$ and extend it to $D$. Let $\angle OAB = x$ and $\angle OAC = y$.

$OA = OB$ (radii), so $\angle OBA = x$ (base angles of an isosceles triangle).

Exterior angle of triangle $AOB$: $\angle BOD = x + x = 2x$

Similarly $\angle COD = 2y$

$\angle BOC = 2x + 2y = 2(x + y) = 2 \times \angle BAC$