WAEC · 2021 · May/June · General Maths · Paper 2

(a)

1. The loan was agreed at $21\%$ simple interest for 3 years.
2. Interest $= \frac{25\,000 \times 21 \times 3}{100}$
   $= 15\,750$, that is $15,750.
3. So the total to repay is $25\,000 + 15\,750 = 40\,750$, that is $40,750.
4. He repaid it in 2 equal yearly instalments: $\frac{40\,750}{2} = 20\,375$.
5. He paid $20,375.00 each year.

(b)

1. Let the smaller number be $x$.
2. The next consecutive number is $x + 1$.
3. Thrice the smaller plus twice the larger is 17: $3x + 2(x + 1) = 17$.
4. So $5x + 2 = 17$, $5x = 15$ and $x = 3$.
5. The numbers are 3 and 4, with sum 7.
6. The smaller as a percentage of the sum: $\frac37 \times 100\% = 42.857\ldots\%$.
7. Correct to three significant figures: $42.9\%$.

(a)

1. From 10:00 a.m. to 1:55 p.m. is 3 hours 55 minutes.
2. Take off the 2-hour meeting: he spent 1 hour 55 minutes travelling, which is $1\frac{55}{60} = \frac{23}{12}$ hours.
3. Let the car's distance from $M$ to $N$ be $x$ km.
4. The bus travelled $x + 2$ km.
5. Time $= \frac{\text{distance}}{\text{speed}}$, so the travelling time is $\frac{x}{72} + \frac{x + 2}{40} = \frac{23}{12}$.
6. Multiply every term by 360: $5x + 9(x + 2) = 690$.
7. So $14x + 18 = 690$, $14x = 672$ and $x = 48$.
8. The distance from $M$ to $N$ is 48 km.

1. Draw north at $X$.
2. $Y$ is 15 km due south of $X$.
3. $Z$ is 20 km from $X$ on $270^\circ$, which is due west.
4. South and west are $90^\circ$ apart, so $\angle YXZ = 90^\circ$.

(a)

1. $|YZ| = \sqrt{15^2 + 20^2}$
   $= \sqrt{625}$
   $= 25$ km.

(b)

1. At $Z$: the opposite side is $XY = 15$ and the adjacent side is $ZX = 20$, so $\tan\angle XZY = \frac{15}{20}$ and $\angle XZY \approx 36.87^\circ$.
2. At $Z$, $X$ is due east ($090^\circ$), and $Y$ is $36.87^\circ$ further round clockwise.
3. Bearing of $Y$ from $Z$ $= 090^\circ + 36.87^\circ$
   $\approx 127^\circ$.

(a)

1. $|OA| = |OB|$ (radii), so triangle $OAB$ is isosceles.
2. Its base angles are equal: $\angle OBA = \angle OAB = 34^\circ$.

(b)

1. $\angle BOC$ is an exterior angle of triangle $OAB$, so it equals the sum of the two opposite interior angles: $34^\circ + 34^\circ = 68^\circ$.
2. A tangent is perpendicular to the radius at the point of contact, so $\angle OBC = 90^\circ$.
3. Angles in triangle $OBC$: $\angle OCB = 180^\circ - 90^\circ - 68^\circ$
   $= 22^\circ$.

(a)

1. The whole circle, $360^\circ$, is $100\%$, so $1\% = 3.6^\circ$.
2. Ann $5 \times 3.6^\circ = 18^\circ$, Afia $54^\circ$, Kojo $36^\circ$, Nuno $162^\circ$, Akosua $90^\circ$.
3. Check: $18 + 54 + 36 + 162 + 90 = 360$.
4. Draw the sectors with a protractor and label each with the child's name and angle.

(b)

1. There are $5 + 3 + 4 = 12$ beads.
2. The first bead is red with probability $\frac{5}{12}$.
3. It is taken away, so 11 beads remain, 4 of them red: the second is red with probability $\frac{4}{11}$.
4. Both red: $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$
   $= \frac{5}{33}$.

(a)(i)

1. Biology: $\frac34 \times 80 = 60$ students.
2. Physics: $\frac35 \times 80 = 48$.
3. Draw two overlapping circles B and P in a rectangle of 80, with nobody outside (each studies at least one).
4. Let $x$ study both: B only $60 - x$, P only $48 - x$.

(ii)

1. $(60 - x) + x + (48 - x) = 80$, so $108 - x = 80$ and $x = 28$.

(iii)

1. Biology but not Physics: $60 - 28 = 32$ students, which is $\frac{32}{80} = \frac25$ of the class.

(b)

1. Floor area $= 15 \times 8 = 120\text{ m}^2$.
2. Carpeting: $120 \times 890 =$ GH¢ 106,800.
3. Painting $= 216\,120 - 106\,800 =$ GH¢ 109,320.

(a)

1. Substitute each $x$ into $y = 2x^2 - x - 2$.
2. $x = -4$: $32 + 4 - 2 = 34$.
3. $x = -2$: $8 + 2 - 2 = 8$.
4. $x = -1$: $2 + 1 - 2 = 1$.
5. $x = 1$: $2 - 1 - 2 = -1$.
6. $x = 2$: $8 - 2 - 2 = 4$.
7. $x = 3$: $18 - 3 - 2 = 13$.
8. The full row is $34, 19, 8, 1, -2, -1, 4, 13, 26$.

(b)

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

(c)

1. $y = 2x + 3$ is a straight line: plot two or three points, for example $(-2, -1)$, $(0, 3)$ and $(3, 9)$, and join them with a ruler.

(d)(i)

1. Rearrange: $2x^2 - 3x - 5 = 0$ is the same as $2x^2 - x - 2 = 2x + 3$.
2. So its roots are the $x$-values where the curve meets the line: $x = -1$ and $x = 2.5$.

(ii)

1. $2x^2 - x - 2 < 0$ where the curve is below the $x$-axis, between the points where it crosses the axis: about $-0.8 < x < 1.3$.

(a)

1. $|PQ| : |QR| = 3 : 4$, so let $|PQ| = 3k$ and $|QR| = 4k$.
2. The right angle is at $Q$, so these two sides are the base and height: area $= \frac12 \times 3k \times 4k$
   $= 6k^2$
   $= 216$.
3. So $k^2 = 36$ and $k = 6$.
4. The sides are 18 cm and 24 cm.
5. By Pythagoras, $|PR| = \sqrt{18^2 + 24^2}$
   $= \sqrt{900}$
   $= 30\text{ cm}$.

(b)

1. In $n$ years the man will be $47 + n$ and his son $17 + n$.
2. The man will be twice as old: $47 + n = 2(17 + n)$.
3. So $47 + n = 34 + 2n$ and $n = 13$.
4. In 13 years.

1. $QU$ is perpendicular to $PS$, so it is the height of the trapezium: 12 cm.
2. Triangle $PQR$ has base $QR$ and the same height: $\frac12 \times |QR| \times 12 = 20$, so $|QR| = \frac{10}{3} \approx 3.33$ cm, and $|UT| = |QR| = 3.33$ cm.
3. $|PQ| = \sqrt{12^2 + 5^2} = 13$ cm.
4. In right-angled triangle $RTS$, $RT = 12$ is opposite the $50^\circ$ angle: $|RS| = \frac{12}{\sin 50^\circ}$
   $\approx 15.665$ cm and $|TS| = \frac{12}{\tan 50^\circ}$
   $\approx 10.069$ cm.
5. So $|PS| = 5 + 3.33 + 10.069 \approx 18.40$ cm.

(a)

1. Perimeter $\approx 13 + 3.33 + 15.665 + 18.40$
   $\approx 50.4$, which is 50 cm to the nearest whole number.

(b)

1. Area $= \frac12 \times (3.33 + 18.40) \times 12$
   $\approx 130.4$, which is 130 cm² to the nearest whole number.

(a)

1. The bearings given are from the farms to the cottage, so turn them round: from the cottage $C$, Dogbe's farm $D$ is on $200^\circ - 180^\circ = 020^\circ$ (5 km) and Mamu's farm $M$ is on $110^\circ + 180^\circ = 290^\circ$ (3 km).
2. The angle between these at $C$ is $360^\circ - 290^\circ + 20^\circ = 90^\circ$.

(i)

1. $|DM| = \sqrt{5^2 + 3^2}$
   $= \sqrt{34}$
   $\approx 5.8$ km.

(ii)

1. At $D$: $\tan\angle CDM = \frac35$, so $\angle CDM \approx 31.0^\circ$.
2. At $D$, the cottage is on $200^\circ$ and Mamu's farm is $31.0^\circ$ further round clockwise: $200^\circ + 31.0^\circ \approx 231^\circ$.

(b)

1. The wall ($x$ m), the ground ($x + 2$ m) and the ladder (10 m) make a right-angled triangle: $x^2 + (x + 2)^2 = 10^2$.
2. Expand: $2x^2 + 4x + 4 = 100$, so $x^2 + 2x - 48 = 0$.
3. Factorise: $(x - 6)(x + 8) = 0$.
4. A height can't be negative, so $x = 6$.

1. Use an assumed mean of 8 hours, with $d = x - 8$:

2. $x$	4	5	6	7	8	9	10	11	Total
   $f$	5	7	5	9	12	4	3	5	50
   $d$	$-4$	$-3$	$-2$	$-1$	0	1	2	3
   $fd$	$-20$	$-21$	$-10$	$-9$	0	4	6	15	$-35$
   $fd^2$	80	63	20	9	0	4	12	45	233

(a)

1. Mean $= 8 + \frac{-35}{50} = 8 - 0.7 = 7.30$ hours.

(b)

1. Standard deviation $= \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}$
   $= \sqrt{\frac{233}{50} - (-0.7)^2}$
   $= \sqrt{4.66 - 0.49}$
   $= \sqrt{4.17}$
   $\approx 2.04$ hours.

(a)

1. Triangle $PQS$ has angles in the ratio $2 : 3 : 3$ (8 parts of $180^\circ$): $45^\circ$, $67.5^\circ$ and $67.5^\circ$.
2. $|PQ| = |QS|$, so the equal angles are at $P$ and $S$: $\angle QPS = \angle QSP = 67.5^\circ$ and $\angle PQS = 45^\circ$.

(i)

1. $\angle SQR$ and $\angle SPR$ stand on the same arc $SR$, so $\angle SQR = 26^\circ$ (angles in the same segment).
2. So $\angle PQR = 45^\circ + 26^\circ$
   $= 71^\circ$.

(ii)

1. $\angle RPQ = \angle QPS - \angle SPR$
   $= 67.5^\circ - 26^\circ$
   $= 41.5^\circ$.

(iii)

1. In triangle $PQR$: $\angle PRQ = 180^\circ - 41.5^\circ - 71^\circ$
   $= 67.5^\circ$.
2. (Check: $\angle PRQ$ and $\angle PSQ$ stand on the same arc $PQ$, and $\angle PSQ = 67.5^\circ$ ✓.)

(b)

1. By the distance formula, $|PQ|^2 = (7 - 5)^2 + (3 - x)^2 = 29$, so $4 + (3 - x)^2 = 29$ and $(3 - x)^2 = 25$.
2. So $3 - x = 5$ or $3 - x = -5$, which gives $x = -2$ or $x = 8$.

(a)

1. From his first birthday to when he is 4 is 3 years.
2. Compound interest at $4\%$: $1000 \times 1.04^3 = 1000 \times 1.124864$
   $\approx 1124.86$, so the account holds $1,124.86.

(b)

1. Join $AC$.
2. $AD$ is a diameter, so $\angle ACD = 90^\circ$ (angle in a semicircle) and $\angle CAD = 180^\circ - 90^\circ - 50^\circ$
   $= 40^\circ$.
3. $ABCD$ is cyclic, so $\angle ABC = 180^\circ - \angle ADC$
   $= 130^\circ$.
4. $|AB| = |BC|$, so triangle $ABC$ is isosceles: $\angle BAC = \frac{180^\circ - 130^\circ}{2}$
   $= 25^\circ$.
5. $\angle BAD = \angle BAC + \angle CAD$
   $= 25^\circ + 40^\circ$
   $= 65^\circ$.