Specimen Paper 3 Higher

(a) The point at 4 hours with a mark of 10 is far below the pattern of the rest: $(4, 10)$.

(b) Ignore the outlier and draw one straight line through the middle of the other points, e.g. from about $(2, 22)$ to $(13, 86)$. As hours go up, marks go up: positive correlation.

(c) Go up from 9 hours to your line and across: about 65.

(d) The data only goes up to about 13 hours. Extending the line to 18 hours is extrapolation, so the prediction is unreliable (and marks can't keep rising past 100).

To the nearest cm, the true length is within half a centimetre either side of 13.

$13 - 0.5 = 12.5$ and $13 + 0.5 = 13.5$

So $12.5 \le L < 13.5$.

The line crosses the $y$-axis at $(0, 1)$, so $c = 1$.

It also passes through $(2, 5)$. Gradient $= \frac{5 - 1}{2 - 0} = 2$

So $y = 2x + 1$

(a) Midpoints: 30, 34, 38, 42.

$30 \times 24 + 34 \times 12 + 38 \times 8 + 42 \times 6 = 720 + 408 + 304 + 252 = 1684$

$1684 \div 50 = 33.68$ inches

(b) $\frac{24}{50} = 0.48$, less than half. Three quarters would be 0.75, far more than the share sold, so the manager is correct.

(a) Area of wall $= \frac{1.8 + 2.7}{2} \times 0.8 = 1.8\text{ m}^2 = 18\,000\text{ cm}^2$

Area of a tile $= 15 \times 7.5 = 112.5\text{ cm}^2$

Tiles $= 18\,000 \div 112.5 = 160$; packs $= 160 \div 9 = 17.8$, so 18 packs.

(b) $160 \times 1.1 = 176$ tiles; $176 \div 9 = 19.6$, so 20 packs.

10% more packs would be $18 \times 1.1 = 19.8$. She actually needs 2 more packs, which is about 11%, so her assumption is not correct.

Find two numbers that multiply to $-4$ and add to $3$: these are $4$ and $-1$.

$x^2 + 3x - 4 = (x + 4)(x - 1)$

Rewrite each line as $y = mx + c$:

A: $y = 2x + 4$ (gradient 2)

B: $y = \frac{1}{2}x + 2$ (gradient $\frac{1}{2}$)

C: $y = -x + 2$ (gradient $-1$)

D: $y = 2x - 4$ (gradient 2)

Parallel lines have equal gradients, so A and D.

(a) Each year multiplies the amount by $1.03$, so after 2 years it is multiplied by $1.03^2 = 1.0609$.

Original $= 2652.25 \div 1.0609 = £2500$

(b) Saver Account: $1.04^5 = 1.2167$, so the interest is about $21.67\%$.

Investment Account: $21\%$.

$21.67\% > 21\%$, so the Saver Account is best.

$CD = 1.5 \times 1.7 = 2.55$ m, so $C$ is $2.55 - 1.7 = 0.85$ m above $A$.

In the right-angled triangle from $A$ across to below $C$: $\tan 52^\circ = \frac{0.85}{BD}$

$BD = \frac{0.85}{\tan 52^\circ} = 0.6640... = 0.664$ m (3 s.f.)

Draw $x + y = 4$ (through $(0, 4)$ and $(4, 0)$), $y = x - 1$ (through $(0, -1)$ and $(1, 0)$) and $y = 3x$ (through $(0, 0)$ and $(1, 3)$). All are strict, so dashed lines are fine.

Test $(1, 1)$: $2 < 4$, $1 > 0$, $1 < 3$. All true, so the region containing $(1, 1)$ is wanted.

R is the triangle with corners $(1, 3)$, $(2.5, 1.5)$ and $(-0.5, -1.5)$.

Halve the $x$ coefficient: $x^2 + 2x = (x + 1)^2 - 1$

So $x^2 + 2x - 8 = (x + 1)^2 - 1 - 8 = (x + 1)^2 - 9$

$m = 1$, $n = -9$

Let the unknown edge be $d$. The space diagonal satisfies $7^2 + 5^2 + d^2 = 15^2$

$d^2 = 225 - 49 - 25 = 151$, so $d = \sqrt{151} = 12.288...$ cm

Volume $= 7 \times 5 \times 12.288... = 430.09... = 430\text{ cm}^3$ (3 s.f.)

Each of the 14 boys can be paired with any of the 12 girls.

$14 \times 12 = 168$ ways

$x^2 + 5x + 6 = (x + 2)(x + 3)$

$(x + 3) \div \frac{(x + 2)(x + 3)}{x - 2} = (x + 3) \times \frac{x - 2}{(x + 2)(x + 3)} = \frac{x - 2}{x + 2}$

$4 - \frac{x - 2}{x + 2} = \frac{4(x + 2) - (x - 2)}{x + 2} = \frac{4x + 8 - x + 2}{x + 2} = \frac{3x + 10}{x + 2}$

(a) $10^6 \times 2^{-8} = \frac{1\,000\,000}{256} = 3906.25$, so about 3906 errors.

(b) In the model the number halves every hour, and halving a positive number never gives 0, so the model never reaches zero. (You could also argue that in reality the count must eventually drop below 1, and there can't be a fraction of an error.)

$f(x + 3)$ shifts the graph 3 units left: $(9, 1) \to (6, 1)$

$-f(\ldots)$ reflects in the $x$-axis: $(6, 1) \to (6, -1)$

So $P = (6, -1)$

Radius $= 12x$, height $= 16x$, so slant height $l = \sqrt{(12x)^2 + (16x)^2} = 20x$

Curved surface area: $\pi \times 12x \times 20x = 240x^2\pi = 2160\pi$, so $x^2 = 9$ and $x = 3$

So $r = 36$ and $h = 48$

Volume $= \frac{1}{3}\pi \times 36^2 \times 48 = 20\,736\pi$, so $V = 20\,736$

$P(\text{H})^2 = 0.09$, so $P(\text{H}) = \sqrt{0.09} = 0.3$

$P(\text{T}) = 1 - 0.3 = 0.7$

$P(\text{TT}) = 0.7^2 = 0.49$

(a) Move the decimal point 4 places right to get 4.23, so $0.000423 = 4.23 \times 10^{-4}$

(b) $4.5 \times 10^4 = 4.5 \times 10\,000 = 45\,000$

Area scale factor $= 253.5 \div 6 = 42.25$

Length scale factor $= \sqrt{42.25} = 6.5$

Volume scale factor $= 6.5^3 = 274.625$

Clay needed $= 2 \times 274.625 = 549.25$ kg

$549.25 \div 10 = 54.925$, so he must buy 55 bags.

(a) $3x^2 - x^3 + 3 = 0 \Rightarrow x^3 = 3x^2 + 3$. Divide every term by $x^2$: $x = 3 + \frac{3}{x^2}$

(b) $x_1 = 3 + \frac{3}{3.2^2} = 3.29296...$

$x_2 = 3 + \frac{3}{3.29296...^2} = 3.27666...$

$x_3 = 3 + \frac{3}{3.27666...^2} = 3.27942...$

(c) Each value is a closer estimate of a solution of $3x^2 - x^3 + 3 = 0$.

The $n$th term is $6n + 1$. Take two terms $6a + 1$ and $6b + 1$.

$(6a + 1)^2 - (6b + 1)^2 = 36a^2 + 12a - 36b^2 - 12b$

$= 12(3a^2 - 3b^2 + a - b) = 12(a - b)(3a + 3b + 1)$

If $a$ and $b$ are both odd or both even, $a - b$ is even.

If one is odd and one even, $a + b$ is odd, so $3(a + b)$ is odd and $3(a + b) + 1$ is even.

Either way $(a - b)(3a + 3b + 1)$ is even, so the difference is $12 \times$ even $= 24k$: a multiple of 24.