Specimen Paper 1 Higher

The two marked angles add up to $90°$ because the third angle is a right angle.

$7x + 5x + 18 = 90$, so $12x = 72$ and $x = 6$.

The angles are $7 \times 6 = 42°$, $5 \times 6 + 18 = 48°$ and $90°$, so the smallest is $42°$.

Rearrange $p = \frac{F}{A}$ to $A = \frac{F}{p}$.

$A = \frac{140}{35} = 4$

The area is $4$ m².

The probabilities add up to 1.

$1 - 0.24 - 0.32 = 0.44$

Blue and green share this equally: $0.44 \div 2 = 0.22$ each.

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

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

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

There are 4 tiles, each $4 \times 3 = 12$ cm², so the total area is $4 \times 12 = 48$ 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 the coin 160 times, more than anyone else, so his results are the most reliable.

(b) Altogether there were 300 heads and 100 tails, a ratio of 3 : 1. Heads is about three times as likely, not twice, so Paul is wrong.

(c) P(heads) $\approx \frac{300}{400} = \frac{3}{4}$

P(two heads) $= \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

(a) $\cos 30° = \frac{\sqrt{3}}{2}$

(b) $x$ is opposite the $30°$ angle and 12 is the hypotenuse.

$\sin 30° = \frac{x}{12}$, so $x = 12 \times 0.5 = 6$.

(a) Round $1.899$ to $2$: $2 \times 10^{27} \times 0.3 = 0.6 \times 10^{27} = 6 \times 10^{26}$ kg.

(b) $1.899$ was rounded up to $2$, so the estimate is an overestimate.

Total after 11 games: $33 \times 11 = 363$.

Mean after 10 games was $35$, so total: $35 \times 10 = 350$.

11th game: $363 - 350 = 13$ points, so Jordan is right.

(a) $30 \div 6 = 5$ red counters.

(b) For 1 : 2 with 30 yellow, there must be $30 \div 2 = 15$ red.

$15 - 5 = 10$ red counters were added.

The denominator 3 means cube root: $\sqrt[3]{125} = 5$.

The numerator 2 means square: $5^2 = 25$.

First part: $50 \times 3 = 150$ miles.

Second part: $150 \div 30 = 5$ hours.

Total: $300$ miles in $8$ hours.

Average speed $= 300 \div 8 = 37.5$ mph, so Sean is wrong.

Square both sides: $m^2 = \frac{k^3 + 1}{4}$

Multiply by 4: $4m^2 = k^3 + 1$

Subtract 1: $k^3 = 4m^2 - 1$

Cube root: $k = \sqrt[3]{4m^2 - 1}$

Multiply every term by $6x$:

$2(x + 2) + 3(x - 2) = 18x$

$2x + 4 + 3x - 6 = 18x$

$5x - 2 = 18x$, so $-2 = 13x$ and $x = -\frac{2}{13}$.

Numerator: $2x^2 - 3x - 5 = (2x - 5)(x + 1)$

Denominator: $x^2 + 6x + 5 = (x + 5)(x + 1)$

Cancel the common factor $(x + 1)$:

$\frac{2x - 5}{x + 5}$, so $a = 2$, $b = -5$, $c = 1$, $d = 5$.

Direct proportion is a straight line through the origin: D.

Inverse proportion $y = \frac{k}{x}$ has two curves in opposite quadrants: A.

$y = kx^2$ is a U-shaped parabola through the origin: B.

$y = \frac{k}{x^2}$ is always positive, so both branches are above the $x$-axis: C.

$PQ = PR$, so triangle $PQR$ is isosceles and angle $SQR$ = angle $TRQ$ (base angles).

$S$ and $T$ are midpoints of equal sides, so $QS = \frac{1}{2}PQ = \frac{1}{2}PR = RT$.

$QR$ is a side of both triangles.

So triangles $QTR$ and $RSQ$ have two sides and the included angle equal: congruent by SAS.

Hemisphere volume: $\frac{2}{3}\pi r^3 = \frac{250}{3}\pi$, so $r^3 = 125$ and $r = 5$.

Curved surface: $\frac{1}{2} \times 4\pi \times 5^2 = 50\pi$

Flat face: $\pi \times 5^2 = 25\pi$

Total: $50\pi + 25\pi = 75\pi$ cm²

Expand the numerator: $(6 - \sqrt{5})(6 + \sqrt{5}) = 36 + 6\sqrt{5} - 6\sqrt{5} - 5 = 31$

So the expression is $\frac{31}{\sqrt{31}}$.

Multiply top and bottom by $\sqrt{31}$: $\frac{31\sqrt{31}}{31} = \sqrt{31}$

Let the integers be $n$ and $n + 1$.

Difference of squares: $(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$

Sum: $n + (n + 1) = 2n + 1$

Both equal $2n + 1$, so the statement is true for any two consecutive integers.

P(red then blue) $= \frac{x}{10} \times \frac{10 - x}{9}$

P(blue then red) $= \frac{10 - x}{10} \times \frac{x}{9}$

Add: $\frac{2x(10 - x)}{90} = \frac{x(10 - x)}{45} = \frac{10x - x^2}{45}$

$\overrightarrow{AB} = -3\mathbf{a} + 6\mathbf{b}$

$\overrightarrow{AX} = \frac{1}{3}\overrightarrow{AB} = -\mathbf{a} + 2\mathbf{b}$

$\overrightarrow{CX} = 3\mathbf{a} + (-\mathbf{a} + 2\mathbf{b}) = 2\mathbf{a} + 2\mathbf{b}$

$\overrightarrow{CY} = 6\mathbf{b} + 5\mathbf{a} - \mathbf{b} = 5\mathbf{a} + 5\mathbf{b}$

$\frac{2}{5}\overrightarrow{CY} = 2\mathbf{a} + 2\mathbf{b} = \overrightarrow{CX}$

$A$ is $(-2, 0)$ and $B$ is $(0, 4)$, so the gradient of $AB = \frac{4}{2} = 2$.

Perpendicular gradient $= -\frac{1}{2}$.

Through $C(5, -1)$: $-1 = -\frac{1}{2}(5) + c$, so $c = \frac{3}{2}$.

$y = -\frac{1}{2}x + \frac{3}{2}$