WAEC · 2015 · May/June · General Maths · Paper 2

(a)

1. Brackets first.
2. Change to improper fractions: $5\frac13 - 2\frac34 = \frac{16}{3} - \frac{11}{4}$
   $= \frac{64 - 33}{12}$
   $= \frac{31}{12}$.
3. Then divide: $3\frac49 \div \frac{31}{12} = \frac{31}{9} \times \frac{12}{31}$
   $= \frac{12}{9}$
   $= \frac43$.
4. Finally add: $\frac43 + 5\frac{9}{10} = \frac{40}{30} + \frac{177}{30}$
   $= \frac{217}{30}$
   $= 7\frac{7}{30}$.

(b)

1. List the sample space: each of 2, 3, 4 with each of 1, 3, 5 gives $3 \times 3 = 9$ equally likely pairs.
2. Their sums:

3. 	1	3	5
   2	3	5	7
   3	4	6	8
   4	5	7	9

4. "Greater than 3 and less than 7" means a sum of 4, 5 or 6.
5. There are 4 such pairs: $(2, 3)$, $(3, 1)$, $(3, 3)$ and $(4, 1)$.
6. So the probability is $\frac49$.

(a)

1. The denominators are 4 and 8, so multiply every term by 8: $32 + 6(x + 2) \le 3x + 8$.
2. Expand: $32 + 6x + 12 \le 3x + 8$, so $6x + 44 \le 3x + 8$.
3. Collect: $3x \le -36$, so $x \le -12$.

(b)

1. The rectangle is 20 cm high and $10 + x + 10 = 20 + x$ cm wide.
2. The shaded area is the rectangle without the square: $20(20 + x) - x^2 = 484$.
3. Expand: $400 + 20x - x^2 = 484$, so $x^2 - 20x + 84 = 0$.
4. Factorise: $(x - 6)(x - 14) = 0$, so $x = 6$ or $x = 14$.

(a)

1. At each corner, interior angle + exterior angle $= 180^\circ$.
2. Sharing $180^\circ$ in the ratio $5 : 2$ (7 parts), the exterior angle is $\frac27 \times 180^\circ = \frac{360^\circ}{7}$.
3. The number of sides is $360^\circ$ divided by the exterior angle: $360 \div \frac{360}{7} = 7$ sides.

(b)

1. $\angle QPR$ and $\angle QSR$ stand on the same arc $QR$, so $\angle QPR = \angle QSR = 40^\circ$ (angles in the same segment).
2. $\angle UQR = 68^\circ$ is an exterior angle of triangle $PQR$ (as $PQU$ is a straight line), so it equals the two opposite interior angles: $68^\circ = 40^\circ + \angle PRQ$, giving $\angle PRQ = 28^\circ$.
3. $\angle TPS = 74^\circ$ is an exterior angle of the cyclic quadrilateral $PQRS$, so it equals the interior opposite angle: $\angle SRQ = 74^\circ$.
4. So $\angle PRS = \angle SRQ - \angle PRQ$
   $= 74^\circ - 28^\circ$
   $= 46^\circ$.

1. Draw the trapezium with $SR$ (60 cm) at the bottom and $PQ$ (20 cm) at the top.
2. Drop perpendiculars from $P$ and $Q$ to $SR$, meeting it at $N$ and $M$.
3. Both are 40 cm long.

(a)

1. Area $= \frac12(20 + 60) \times 40$
   $= 1600\text{ cm}^2$.

(b)

1. In the right-angled triangle $PNS$: $|SN| = \sqrt{50^2 - 40^2} = 30$ cm.
2. $NM = PQ = 20$ cm, so $|MR| = 60 - 30 - 20 = 10$ cm.
3. In triangle $QMR$, the height 40 is opposite $\angle QRS$ and $MR = 10$ is adjacent: $\tan\angle QRS = \frac{40}{10} = 4$.
4. So $\angle QRS \approx 76^\circ$.

(a)(i)

1. Use the table entries that make the working easiest.
2. When $x = 0$, $y = -12$: $0 - 0 + q = -12$, so $q = -12$.
3. When $x = 4$, $y = 0$: $16p - 20 - 12 = 0$, so $16p = 32$ and $p = 2$.
4. The relation is $y = 2x^2 - 5x - 12$.

(ii)

1. Substitute each missing $x$: $x = -1$ gives $2 + 5 - 12 = -5$.
2. $x = 1$ gives $2 - 5 - 12 = -15$.
3. $x = 2$ gives $8 - 10 - 12 = -14$.
4. $x = 3$ gives $18 - 15 - 12 = -9$.
5. The full row is $21, 6, -5, -12, -15, -14, -9, 0, 13$.

(b)

1. Plot the points with the scales given and join them with one smooth curve.

(c)(i)

1. Read up from $x = 1.8$ to the curve and across: $y \approx -14.5$.
2. (By calculation, $2(1.8)^2 - 9 - 12 = -14.52$.) (ii) Draw the line $y = -8$ and read down from where it meets the curve: $x \approx -0.6$ and $x \approx 3.1$.

(a)(i)

1. Draw $WX = 8$ cm.
2. Construct $60^\circ$ at $X$ and mark $XY = 5.5$ cm on the arm.
3. Through $Y$ construct a line parallel to $WX$.
4. With centre $X$ and radius 8.3 cm, cut it at $Z$.
5. Join $WZ$ and $YZ$.

(ii)

1. Drop perpendiculars from $Y$ and $Z$ to $WX$, meeting it at $Q$ and $P$.
2. $PQYZ$ is the rectangle.

(b)

1. Measure: (i) $|QX| \approx 2.8$ cm.

(ii)

1. $\angle XWZ \approx 76^\circ$.
2. Check: $|QX| = 5.5\cos 60^\circ = 2.75$ cm and the height is $5.5\sin 60^\circ \approx 4.76$ cm.
3. $Z$ is $\sqrt{8.3^2 - 4.76^2} \approx 6.80$ cm along from $X$, so $|WP| \approx 1.20$ cm and $\angle XWZ = \tan^{-1}\frac{4.76}{1.20}$
   $\approx 76^\circ$.

(a)

1. The $n$th term of an A.P. is $a + (n - 1)d$.
2. With $a = -8$: the 7th term is $-8 + 6d$ and the 9th term is $-8 + 8d$.
3. Their ratio is $5 : 8$: $\frac{-8 + 6d}{-8 + 8d} = \frac58$.
4. Cross-multiply: $8(-8 + 6d) = 5(-8 + 8d)$.
5. So $-64 + 48d = -40 + 40d$, $8d = 24$ and $d = 3$.

(b)

1. Let a basket of pawpaw cost ₦$p$ and a basket of mangoes cost ₦$m$.
2. Total cost: $30p + 100m = 2450$ (1).
3. Profit on pawpaw is $40\%$ of $30p$ and on mangoes $30\%$ of $100m$: $0.4 \times 30p + 0.3 \times 100m = 855$, which is $12p + 30m = 855$ (2).
4. Multiply (1) by 0.3: $9p + 30m = 735$ (3).
5. Take (3) from (2): $3p = 120$, so $p = 40$.

(i)

1. A basket of pawpaw cost ₦40.00.

(ii)

1. From (1), $100m = 2450 - 1200 = 1250$, so the mangoes cost ₦1,250.00.
2. They sold at a $30\%$ profit: $1.3 \times 1250 =$ ₦1,625.00.

(a)

1. Use the exact values $\tan 60^\circ = \sqrt3$, $\cos 30^\circ = \frac{\sqrt3}{2}$ and $\sin 60^\circ = \frac{\sqrt3}{2}$: $\frac{2\sqrt3 + \frac{\sqrt3}{2}}{\frac{\sqrt3}{2}}$.
2. Multiply the top and bottom by 2: $\frac{4\sqrt3 + \sqrt3}{\sqrt3} = \frac{5\sqrt3}{\sqrt3}$
   $= 5$.

(b)

1. Draw the aeroplane $A$ above and 1050 m horizontally from the tower.
2. The angles of depression, $36^\circ$ to the top and $41^\circ$ to the base, are measured down from the horizontal through $A$, and equal the angles of elevation from the ground level lines (alternate angles).

(i)

1. The drop from the aeroplane's height to the base of the tower is $1050\tan 41^\circ \approx 912.75$ m, and to the top of the tower is $1050\tan 36^\circ \approx 762.87$ m.
2. The tower is the difference: $912.75 - 762.87 \approx 150$ m.

(ii)

1. The shortest distance from the aeroplane to the base is the straight line, the hypotenuse of the bigger triangle: $\cos 41^\circ = \frac{1050}{d}$, so $d = \frac{1050}{0.7547} \approx 1391$ m.

(a)

1. Multiply both sides by $d(m + p)$ to clear the fraction: $hd(m + p) = mt$.
2. Expand: $hdm + hdp = mt$.
3. Collect the $m$ terms on one side: $hdp = mt - hdm$.
4. Take out $m$: $hdp = m(t - hd)$.
5. So $m = \frac{hdp}{t - hd}$.

(b)

1. Join $XM$.
2. $WM$ passes through the centre $O$, so it is a diameter and $\angle WXM = 90^\circ$ (angle in a semicircle).
3. In $\triangle WXM$: $\angle WMX = 180^\circ - 90^\circ - 48^\circ$
   $= 42^\circ$ (angles in a triangle).
4. $XYZM$ is a cyclic quadrilateral and $W, M, Z$ are in a straight line, so $\angle WMX$ is its exterior angle at $M$.
5. An exterior angle of a cyclic quadrilateral equals the interior opposite angle: $\angle XYZ = 42^\circ$.
6. Hence $\angle WYZ = 42^\circ$.

(c)(i)

1. Work out $m + n + 2$, then take the remainder when dividing by 7.
2. For example, $5 \oplus 6 = 13 = 7 + 6$, so $5 \oplus 6 = 6$.

3. $\oplus$	1	3	5	6
   1	4	6	1	2
   3	6	1	3	4
   5	1	3	5	6
   6	2	4	6	0

(ii)

1. For I, look along the row for 3: the entry 3 appears only in the column for 5, so the truth set is $\{5\}$.
2. For II, look at the diagonal, where $n \oplus n$ gives $4, 1, 5, 0$.
3. None of them is 3, so the truth set is the empty set, $\{\}$ or $\varnothing$.

(a)

1. The cumulative frequencies are the running totals of the frequencies.
2. Plot them at the upper class boundaries:

3. Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
   Upper boundary	9.5	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
   Cumulative frequency	7	18	35	55	84	118	148	173	194	200

4. Plot $(9.5, 7), (19.5, 18), \ldots, (99.5, 200)$, starting from $(-0.5, 0)$, and draw a smooth S-shaped curve through them.

(b)(i)

1. If the top $5\%$ got a distinction, then $95\%$ scored below the lowest distinction mark.
2. $95\%$ of 200 is 190.
3. Go across from 190 to the curve and down: about 87.6.
4. (Check: 190 lies between 173 at 79.5 and 194 at 89.5, and $79.5 + \frac{190 - 173}{21} \times 10 \approx 87.6$.)

(ii)

1. "At most $45\%$" means 45 or less, so read the curve at 45.5: go up from 45.5 and across, about 72 candidates.
2. The probability is about $\frac{72}{200} = 0.36$.
3. A reading close to this from your own curve is fine.