WAEC · 2018 · Private, 2nd series · General Maths · Paper 2

1. Write each fact as an equation.
2. The sum of the first ten terms: $S_{10} = \frac{10}{2}[2a + 9d]$
   $= 5(2a + 9d)$
   $= 130$, so $2a + 9d = 26$.
3. The fifth term is 3 times the first: $a + 4d = 3a$, so $4d = 2a$ and $a = 2d$.

(a)

1. Put $a = 2d$ into $2a + 9d = 26$: $4d + 9d = 26$, so $13d = 26$ and the common difference is $d = 2$.

(b)

1. The first term is $a = 2d = 4$.
2. (Check: $5(8 + 18) = 130$ ✓ and $T_5 = 4 + 8 = 12 = 3 \times 4$ ✓.)

(c)

1. The last term is 28: $a + (n - 1)d = 28$, so $4 + 2(n - 1) = 28$, $2(n - 1) = 24$, $n - 1 = 12$ and there are $n = 13$ terms.

(a)

1. $\sin X = \frac35$ gives a right-angled triangle with sides 3, 4 and 5, so $\cos X = \frac45$.
2. Then $5(\cos X)^2 - 3 = 5 \times \frac{16}{25} - 3$
   $= \frac{16}{5} - 3$
   $= \frac15$.

(b)

1. Volume of the pyramid $= \frac13 \times (42 \times 11) \times 12$
   $= 1848\text{ cm}^3$.
2. The cone has the same height (12 cm) and volume: $\frac13 \times \frac{22}{7} \times r^2 \times 12 = 1848$.
3. Simplify: $\frac{88}{7}r^2 = 1848$, so $r^2 = 147$.
4. In surd form, $r = \sqrt{147}$
   $= \sqrt{49 \times 3}$
   $= 7\sqrt3$ cm.

(a)

1. The prices are in classes, so draw a histogram.
2. The class 1.0–1.9 really runs from 0.95 to 1.95, so mark the boundaries $0.95, 1.95, 2.95, 3.95, 4.95, 5.95$ on the horizontal axis.
3. Draw touching bars of heights $23, 48, 107, 90, 32$.
4. Label the axes "Price (₦ million)" and "Number of vehicles".

(b)

1. The most preferred price is the one bought most often: the mode.
2. It's in the tallest bar, 2.95–3.95.
3. Join the top-left corner of that bar to the top-left corner of the next bar, and the top-right corner to the top-right corner of the bar before.
4. Read down from where the lines cross: about 3.7, so about ₦3,700,000.
5. By formula, as a check: $2.95 + \frac{107 - 48}{(107 - 48) + (107 - 90)} \times 1 = 2.95 + \frac{59}{76}$
   $\approx 3.73$ million.

(a)

1. The perimeter is 28 cm: $(2y + x) + 4y + (6y - 2x + 1) = 28$.
2. Collect like terms: $12y - x + 1 = 28$, so $12y - x = 27$ (1).
3. The two equal sides give a second equation: $6y - 2x + 1 = 4y$, so $2y - 2x = -1$ (2).
4. From (1), $x = 12y - 27$.
5. Substitute into (2): $2y - 2(12y - 27) = -1$.
6. So $2y - 24y + 54 = -1$, $-22y = -55$ and $y = \frac52$.
7. Then $x = 12 \times \frac52 - 27 = 3$.

(b)

1. $|PQ| = 2y + x = 5 + 3 = 8\text{ cm}$.
2. $|QR| = 4y = 10\text{ cm}$.
3. $|PR| = 6y - 2x + 1$
   $= 15 - 6 + 1$
   $= 10\text{ cm}$.
4. Check: $8 + 10 + 10 = 28$ ✓.

(a)

1. First inequality: $5 - x > 1$, so $4 > x$, that is $x < 4$.
2. Second: $9 + x \ge 8$, so $x \ge -1$.
3. Both at once: $x$ is at least $-1$ and less than 4, so $-1 \le x < 4$.

(b)

1. $PQRS$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$: $\angle PQR = 180^\circ - 56^\circ$
   $= 124^\circ$.
2. $|PQ| = |QR|$, so triangle $PQR$ is isosceles: $\angle QRP = \frac{180^\circ - 124^\circ}{2}$
   $= 28^\circ$.
3. $PS$ is a diameter, so $\angle PRS = 90^\circ$ (angle in a semicircle).
4. Then $\angle QRS = \angle QRP + \angle PRS$
   $= 28^\circ + 90^\circ$
   $= 118^\circ$.

(a)(i)

1. Put $y = 20\,000$: $\frac{x^2}{8} + 5x = 20\,000$.
2. Multiply by 8: $x^2 + 40x - 160\,000 = 0$.

(ii)

1. Formula with $a = 1$, $b = 40$, $c = -160\,000$: $x = \frac{-40 \pm \sqrt{40^2 + 4 \times 160\,000}}{2}$
   $= \frac{-40 \pm \sqrt{641\,600}}{2}$
   $= \frac{-40 \pm 801.0}{2}$.
2. The number sold can't be negative, so $x \approx 380.5$ (thousand): about 380,500 books.

(b)(i)

1. Draw the tree $QT$ vertical, with $R$ halfway up, and $P$ on the ground.
2. The angle of elevation of $T$ from $P$ is $28^\circ$.

(ii)

1. Let $|PQ| = d$.
2. Then $|QT| = d\tan 28^\circ$, and $|QR| = \frac12 d\tan 28^\circ$.
3. So $\tan\angle QPR = \frac{|QR|}{d}$
   $= \frac12\tan 28^\circ$
   $\approx 0.2659$, and the angle of elevation of $R$ is about $14.9^\circ \approx 15^\circ$.

(a)

1. Integrate term by term, adding 1 to each power and dividing by the new power: $\int (2x^3 - 4x + 3)\,dx = \frac{x^4}{2} - 2x^2 + 3x$.
2. At $x = 2$: $\frac{16}{2} - 8 + 6 = 6$.
3. At $x = 1$: $\frac12 - 2 + 3 = \frac32$.
4. Top limit minus bottom limit: $6 - \frac32 = 4\frac12$.

(b)

1. The inverse of $P^{-1}$ is $P$, so find the inverse of $\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}$.
2. Its determinant is $(-1)(-3) - (1)(4) = 3 - 4 = -1$.
3. Swap $-1$ and $-3$, change the signs of $1$ and $4$, and divide by $-1$: $P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix}$
   $= \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}$.
4. Check: $\begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ ✓.

(a)(i)

1. Draw the cone upside down (vertex at the bottom), with base radius 21 cm at the top and height $h$ from the vertex up to the centre of the base.

(ii)

1. $4.158$ litres $= 4158\text{ cm}^3$.
2. Volume $= \frac13 \times \frac{22}{7} \times 21^2 \times h$
   $= 462h$
   $= 4158$, so $h = 9$ cm.

(b)

1. The water left is a smaller cone with the same shape.
2. Its surface radius is 14 cm, so by similar triangles its depth is $9 \times \frac{14}{21} = 6$ cm.
3. Its volume is $\frac13 \times \frac{22}{7} \times 14^2 \times 6 = 1232\text{ cm}^3$.
4. Water drawn out $= 4158 - 1232 = 2926\text{ cm}^3$.

(a)

1. Running totals of the frequencies, against the upper class boundaries:

2. Marks	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
   Upper boundary	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
   Cumulative frequency	4	11	23	41	61	75	84	88	90

(b)

1. Plot each cumulative frequency at its upper boundary, starting from $(9.5, 0)$, and draw a smooth S-shaped curve through the points.

(c)(i)

1. There are 90 students, so the median is at 45 on the cumulative frequency axis.
2. Go across to the curve and down: about 51.5.
3. (Check: 45 lies between 41 at 49.5 and 61 at 59.5, and $49.5 + \frac{45 - 41}{20} \times 10 = 51.5$.)

(ii)

1. A distinction is 75 or more.
2. Go up from 74.5 to the curve and across: about 80 students scored less than 75.
3. So about $90 - 80 = 10$ got a distinction, and the probability is about $\frac{10}{90} = \frac19$.
4. (A careful reading gives about 79.5 below, so 10.5 above and $\frac{10.5}{90} \approx 0.12$.
5. Either is a fair reading from a curve.)

(a)(i)

1. Draw $SR = 12$ cm.
2. Construct $90^\circ$ at $S$ and mark $SP = 8$ cm on the arm.
3. With centre $P$ and radius 6 cm, and centre $R$ and radius 11 cm, draw arcs meeting at $Q$, on the side of $PR$ away from $S$.
4. Join $PQ$ and $QR$.

(ii)

1. With centre $Q$, draw an arc cutting $SR$ twice.
2. From those points draw equal arcs crossing below $SR$.
3. Join $Q$ to the crossing.
4. It meets $SR$ at $K$.

(b)

1. Measure: (i) $|QK| \approx 9.1$ cm.

(ii)

1. $\angle QRS \approx 56^\circ$.
2. Check: with $S$ at the origin, $R$ at $(12, 0)$ and $P$ at $(0, 8)$, the two arcs meet at $Q \approx (5.89, 9.15)$, so $|QK| \approx 9.15$ cm and $\angle QRS = \tan^{-1}\frac{9.15}{12 - 5.89}$
   $\approx 56^\circ$.

(a)(i)

1. Add or multiply, then take the remainder on dividing by 5.
2. For example $3 \oplus 4 = 7 \equiv 2$ and $3 \otimes 4 = 12 \equiv 2$.
3. The $\oplus$ table (rows 2, 3, 4) is $4, 0, 1$ / $0, 1, 2$ / $1, 2, 3$.
4. The $\otimes$ table is $4, 1, 3$ / $1, 4, 2$ / $3, 2, 1$.

(ii)

1. (α)** $(4 \otimes e) \oplus 2 \equiv 1$, so $4 \otimes e \equiv 1 - 2 \equiv 4$.
2. Then $4e \equiv 4 \pmod 5$, so $e = 1$.

(β)

1. $(4 \oplus n) \otimes 2 \equiv 2$.
2. From the $\otimes$ table, the number that gives 2 when multiplied by 2 is 1, so $4 \oplus n \equiv 1$, and from the $\oplus$ table $n = 2$.

(b)

1. From $p$
   $\Rightarrow q$, only the contrapositive $\sim q$
   $\Rightarrow \sim p$ follows.

(i)

1. Is the converse: not valid.

(ii)

1. Is the contrapositive: valid.

(iii)

1. Is the inverse: not valid.

(a)(i)

1. Draw north at $X$ and $XY$, 285 m on $078^\circ$.
2. From $Y$ draw a line due south to $Z$, and join $Z$ to $X$ (307 m).

(ii)

1. The north line at $X$ and the south line at $Y$ are parallel, so $\angle XYZ = 78^\circ$ (alternate angles).
2. Sine rule for the angle at $Z$: $\sin\angle YZX = \frac{285\sin 78^\circ}{307}$
   $= \frac{278.77}{307}$
   $\approx 0.9081$, so $\angle YZX \approx 65.2^\circ$ (acute, because it is opposite the shorter of the two known sides).
3. (I) At $Z$, $Y$ is due north, and $X$ is $65.2^\circ$ round anticlockwise (towards the west).
4. Bearing of $X$ from $Z$ $= 360^\circ - 65.2^\circ$
   $= 294.8^\circ$
   $\approx 295^\circ$.
5. (II) The third angle is $\angle YXZ = 180^\circ - 78^\circ - 65.2^\circ$
   $= 36.8^\circ$.
6. Sine rule: $|YZ| = \frac{307\sin 36.8^\circ}{\sin 78^\circ}$
   $\approx \frac{183.9}{0.9781}$
   $\approx 188$ m.

(a)

1. With 2 cm to 2 units, draw both axes from $-10$ to $10$, crossing at $O$, and number them.

(b)(i)

1. Plot $P(3, 2)$, $Q(-1, 5)$, $R(0, 8)$ and $S(3, 7)$ and join them in order.

(ii)

1. A rotation of $90^\circ$ anticlockwise about the origin sends $(x, y)$ to $(-y, x)$: $P_1(-2, 3)$, $Q_1(-5, -1)$, $R_1(-8, 0)$ and $S_1(-7, 3)$.

(iii)

1. $y - x = 0$ is the line $y = x$.
2. A reflection in it swaps the coordinates, $(x, y) \to (y, x)$: $P_2(3, -2)$, $Q_2(-1, -5)$, $R_2(0, -8)$ and $S_2(3, -7)$.

(c)

1. Compare each point with its final image: $P(3, 2) \to P_2(3, -2)$ and $Q(-1, 5) \to Q_2(-1, -5)$.
2. Each $(x, y)$ goes to $(x, -y)$, so $T$ is a reflection in the $x$-axis.

(d)

1. $P_1(-2, 3)$ and $Q_1(-5, -1)$: the gradient of $P_1Q_1$ is $\frac{-1 - 3}{-5 - (-2)} = \frac43$.
2. The line is $y - 3 = \frac43(x + 2)$, and $y = 0$ gives $x + 2 = -\frac94$, so $W$ is $(-4.25, 0)$.
3. $P_1S_1$ lies on $y = 3$ and $R_1W$ on $y = 0$, so they are parallel.
4. The other two sides are not parallel (gradients 3 and $\frac43$), so $P_1S_1R_1W$ is a trapezium.