WAEC · 2018 · May/June · General Maths · Paper 2

1. After the first year the car has lost $30\%$, so it is worth $70\%$ of ₦900,000: $0.7 \times 900\,000 = 630\,000$, which is ₦630,000.
2. In each later year it loses $22\%$ of its value at the start of that year, so each year multiplies the value by $0.78$.
3. 1st March 2011 to 28th February 2015 is 4 full years: the first year, then 3 more.
4. Value $= 630\,000 \times 0.78^3$
   $= 630\,000 \times 0.474552$
   $\approx 298\,967.76$, which is ₦299,000 to the nearest hundred naira.

(a)

1. The point $(3, 10)$ is on the graph, so $x = 3$ and $y = 10$ satisfy the equation: $10 = 2p(9) - p^2(3) - 14$.
2. So $10 = 18p - 3p^2 - 14$, which rearranges to $3p^2 - 18p + 24 = 0$.
3. Divide by 3: $p^2 - 6p + 8 = 0$.
4. Factorise: $(p - 2)(p - 4) = 0$.
5. So $p = 2$ or $p = 4$.

(b)

1. $Q$ is on both lines, so solve the equations simultaneously.
2. Arrange them with $x$ first: $-2x + 3y = 21$ (1) and $5x + 4y = 5$ (2).
3. Multiply (1) by 4 and (2) by 3 so the $y$ terms match: $-8x + 12y = 84$ and $15x + 12y = 15$.
4. Take the first from the second: $23x = -69$, so $x = -3$.
5. From (1): $6 + 3y = 21$, so $y = 5$.
6. $Q$ is the point $(-3, 5)$.

(a)

1. The diagonals of a rhombus cut each other in half at right angles.
2. So each side is the hypotenuse of a right-angled triangle with legs $\frac{10.2}{2} = 5.1$ cm and $\frac{9.3}{2} = 4.65$ cm.
3. Side $= \sqrt{5.1^2 + 4.65^2}$
   $= \sqrt{26.01 + 21.6225}$
   $= \sqrt{47.6325}$
   $\approx 6.902$ cm.
4. The four sides are equal, so the perimeter is $4 \times 6.902 \approx 27.6$ cm.

(b)

1. $\sin x = \frac35$ gives a right-angled triangle with sides 3, 4 and 5, so $\cos x = \frac45$ and $\tan x = \frac34$.
2. Then $5\cos x - 4\tan x = 5 \times \frac45 - 4 \times \frac34$
   $= 4 - 3$
   $= 1$.

(a)(i)

1. $QOS$ is a diameter, so $\angle QRS = 90^\circ$ (angle in a semicircle).
2. $\angle QST$ is an exterior angle of triangle $QRS$, so it equals the sum of the two opposite interior angles: $3x + 15 = x + 90$.
3. So $2x = 75$ and $x = 37.5$.

(ii)

1. $\angle QST = 3(37.5) + 15 = 127.5^\circ$.
2. $RST$ is a straight line: $\angle RSQ = 180^\circ - 127.5^\circ$
   $= 52.5^\circ$.

(b)

1. Change both to base ten: $2N4_{\text{seven}} = 2 \times 49 + 7N + 4$
   $= 102 + 7N$ and $15N_{\text{nine}} = 81 + 45 + N = 126 + N$.
2. So $102 + 7N = 126 + N$, $6N = 24$ and $N = 4$.
3. (A digit 4 is allowed in both base seven and base nine.)

(a)

1. Turn the mean into a total: $m + n + s + p + q = 5 \times 12 = 60$.
2. The new numbers add up to $(m + n + s + p + q) + 4 - 3 + 6 - 2 + 8 = 60 + 13$
   $= 73$, so their mean is $\frac{73}{5} = 14.6$.

(b)

1. A quarter of the people are below the 25th percentile (15 years) and three quarters below the 75th percentile (65 years).
2. So between 15 and 65 years there are $75\% - 25\% = 50\%$ of the people: $0.5 \times 500 = 250$.
3. (That's $375 - 125 = 250$.)

(a)

1. $60\%$ passed, so $40\%$ failed: $\frac{40}{100} \times 240 = 96$ cars had faults.
2. Draw three overlapping circles C (clutch), B (brakes) and S (steering) in a rectangle of 96.
3. Put 8 in the centre.
4. The pair totals include it: C and S only $= 14 - 8 = 6$.
5. C and B only $= 20 - 8 = 12$.
6. B and S only is given as 6, and C only as 28.
7. Let Brakes only be $x$, so Steering only is $2x$.
8. All the regions add up to 96: $28 + 6 + 12 + 8 + 6 + x + 2x = 96$, so $60 + 3x = 96$ and $x = 12$.
9. Brakes only 12, Steering only 24.

(b)(i)

1. Faulty brakes: every region inside B, $12 + 12 + 8 + 6 = 38$.

(ii)

1. Only one fault: $28 + 12 + 24 = 64$.

(a)

1. Gradient $= \frac{-7 - 5}{-4 - 2}$
   $= \frac{-12}{-6}$
   $= 2$.
2. So $y = 2x + c$.
3. Using $(2, 5)$: $5 = 4 + c$, so $c = 1$.
4. The line is $y = 2x + 1$.

(b)(i)

1. Draw north at $P$.
2. $Q$ is 5 km from $P$ on $030^\circ$.
3. The bearing of $P$ from $R$ is $300^\circ$, so the bearing of $R$ from $P$ is $300^\circ - 180^\circ = 120^\circ$: $R$ is 8 km from $P$ on $120^\circ$.
4. The angle between the lines at $P$ is $120^\circ - 30^\circ = 90^\circ$.

(ii)

1. (I)** Triangle $QPR$ is right-angled at $P$: $|QR| = \sqrt{5^2 + 8^2}$
   $= \sqrt{89}$
   $\approx 9.43$ km.
2. (II) At $Q$: $\tan\angle PQR = \frac{8}{5} = 1.6$, so $\angle PQR \approx 58.0^\circ$.
3. At $Q$, the direction back to $P$ is $030^\circ + 180^\circ = 210^\circ$, and $R$ is $58.0^\circ$ further round anticlockwise.
4. Bearing of $R$ from $Q$ $= 210^\circ - 58.0^\circ = 152^\circ$.

(a)

1. Each sale multiplies the price by $1 + \frac{x}{100}$.
2. Bola pays $300\left(1 + \frac{x}{100}\right) = 300 + 3x$.
3. James pays $(300 + 3x)\left(1 + \frac{x}{100}\right) = 300 + 3x + 3x + \frac{3x^2}{100}$
   $= 300 + 6x + \frac{3x^2}{100}$.
4. James paid $6x + \frac34$ more than ₦300: $\frac{3x^2}{100} = \frac34$, so $x^2 = 25$ and $x = 5$ (a percentage profit is positive).

(b)

1. Split the double inequality.
2. $3x - 2 < 10 + x$ gives $2x < 12$, $x < 6$.
3. $10 + x < 2 + 5x$ gives $8 < 4x$, $x > 2$.
4. So $2 < x < 6$.

(a)(i)

1. In triangle $PTQ$, two sides and the angle between them are known, so use the cosine rule: $|TQ|^2 = 6^2 + 4^2 - 2(6)(4)\cos 30^\circ$
   $= 52 - 41.57$
   $\approx 10.43$, so $|TQ| \approx 3.23$ cm.
2. $TQ \parallel SR$, so triangle $PTQ$ is similar to triangle $PSR$, with every length multiplied by $\frac{|PS|}{|PT|} = \frac{10}{4} = 2.5$.
3. So $|SR| = 2.5 \times 3.23 \approx 8.07$, which is 8 cm to the nearest whole number.
4. (Also $|PR| = 2.5 \times 6 = 15$ cm.)

(ii)

1. Area of a triangle $= \frac12ab\sin C$.
2. Area of $PSR = \frac12 \times 10 \times 15 \times \sin 30^\circ$
   $= 37.5\text{ cm}^2$.
3. Area of $PTQ = \frac12 \times 4 \times 6 \times \sin 30^\circ$
   $= 6\text{ cm}^2$.
4. So area of $TQRS = 37.5 - 6 = 31.5$, which is 32 cm² to the nearest whole number.

(a)

1. $\angle SPQ = 90^\circ$, so $|SQ| = \sqrt{5^2 + 12^2} = 13$ cm.
2. $PR$ is perpendicular to $SQ$, so it is the height of the triangle on the base $SQ$.
3. Work out the area two ways: $\frac12 \times 5 \times 12 = \frac12 \times 13 \times |PR|$.
4. So $|PR| = \frac{60}{13} \approx 4.62$ cm.

(b)(i)

1. Draw the wall vertical and the two ladders leaning against it at the same angle to the ground, $L$ (10 m) with its foot 8 m from the wall and $M$ (12 m).

(ii)

1. Ladder $L$ reaches $\sqrt{10^2 - 8^2} = 6$ m up the wall.
2. Both ladders make the same angle with the ground, so the two triangles are similar, and everything scales by $\frac{12}{10}$.
3. So $M$ reaches $6 \times \frac{12}{10} = 7.2$ m up the wall.

(a)

1. Put each $x$ into $y = 2x^2 + x - 10$.
2. For $x = -5$: $50 - 5 - 10 = 35$.
3. For $x = -4$: $32 - 4 - 10 = 18$.
4. For $x = -2$: $8 - 2 - 10 = -4$.
5. For $x = 1$: $2 + 1 - 10 = -7$.
6. For $x = 3$: $18 + 3 - 10 = 11$.
7. For $x = 4$: $32 + 4 - 10 = 26$.
8. The row is $35, 18, 5, -4, -9, -10, -7, 0, 11, 26$.

(b)

1. With 2 cm to 1 unit across and 2 cm to 5 units up, plot the ten points and join them with a smooth U-shaped curve.

(c)(i)

1. $2x^2 + x = 10$ is the same as $2x^2 + x - 10 = 0$, that is $y = 0$.
2. The curve crosses the $x$-axis at $x = -2.5$ and $x = 2$.

(ii)

1. $2x^2 + x - 10 = 2x$ means the curve meets the line $y = 2x$.
2. Draw the line through $(-3, -6)$, $(0, 0)$ and $(3, 6)$.
3. It meets the curve at $x = -2$ and $x = 2.5$.
4. (As a check, the equation is $2x^2 - x - 10 = 0$, which factorises as $(x + 2)(2x - 5) = 0$.)

(a)

1. Write $p\mathbf x + q\mathbf y = \mathbf z$ in columns: $p\begin{pmatrix} 2 \\ 3 \end{pmatrix} + q\begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}$.
2. The tops give $2p + 5q = -4$ and the bottoms give $3p - 2q = 13$.
3. Multiply the first by 2 and the second by 5: $4p + 10q = -8$ and $15p - 10q = 65$.
4. Add them: $19p = 57$, so $p = 3$.
5. Then $2(3) + 5q = -4$, so $5q = -10$ and $q = -2$.

(b)

1. With 2 cm to 2 units, draw both axes from $-8$ to $8$.

(i)

1. Plot $W(2, 3)$, $X(4, -1)$, $Y(-3, -4)$ and $Z(-3, 2)$ and join them in order.

(ii)

1. A rotation of $90^\circ$ anticlockwise about the origin sends $(x, y)$ to $(-y, x)$: $W_1(-3, 2)$, $X_1(1, 4)$, $Y_1(4, -3)$ and $Z_1(-2, -3)$.
2. Plot and join them, labelling each vertex.

(a)

1. There are 30 students: $1 + 1 + x + 5 + y + 1 + 4 + 3 + 1 = 30$, so $x + y = 14$.
2. The mean is 52, so the marks add up to $30 \times 52 = 1560$: $10 + 20 + 30x + 200 + 50y + 60 + 280 + 240 + 90 = 1560$, which gives $30x + 50y = 660$, or $3x + 5y = 66$.
3. From the first equation $x = 14 - y$.
4. Substitute: $3(14 - y) + 5y = 66$, so $42 + 2y = 66$, $y = 12$ and $x = 2$.

(b)

1. Classes of width 10 starting at 1: each mark goes in the class that ends with it (10 is in 1–10, 20 in 11–20, and so on).

2. Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90
   Frequency	1	1	2	5	12	1	4	3	1

(c)

1. Draw the histogram on the class boundaries $0.5, 10.5, 20.5, \ldots, 90.5$, with touching bars of heights $1, 1, 2, 5, 12, 1, 4, 3, 1$.

(d)

1. The tallest bar is 40.5–50.5.
2. Join its top-left corner to the top-left corner of the next bar, and its top-right corner to the top-right corner of the bar before.
3. Read down from where the lines cross: about 44.
4. By formula: $40.5 + \frac{12 - 5}{(12 - 5) + (12 - 1)} \times 10 = 40.5 + \frac{7}{18} \times 10$
   $\approx 44.4$.