WAEC · 2011 · May/June · General Maths · Paper 2

(a)

1. “Of” means multiply, and it comes before the division: $\frac12 \text{ of } \frac14 = \frac18$.
2. Dividing by $\frac13$ is multiplying by 3: $\frac18 \times 3 = \frac38$.
3. This is the numerator.
4. Denominator over the LCM 12: $\frac{2}{12} - \frac{9}{12} + \frac{6}{12} = -\frac{1}{12}$.
5. Divide: $\frac38 \div \left(-\frac{1}{12}\right) = \frac38 \times (-12)$
   $= -\frac92$.
6. So the value is $-4\frac12$.

(b)

1. Take logs of both sides: $\log\sqrt{x} = \bar1.6741$, that is $\frac12\log x = \bar1.6741$.
2. Multiply by 2, doubling the characteristic and the mantissa separately: $2 \times (-1) = -2$ and $2 \times 0.6741 = 1.3482$.
3. So $\log x = -2 + 1.3482 = \bar1.3482$.
4. Read the antilog of $\bar1.3482$ from the tables: $x = 0.2229$.

(a)

1. $q$ is inside a square root, so square both sides: $t^2 = \frac{pq}{r} - r^2 q$.
2. Multiply every term by $r$ to clear the fraction: $rt^2 = pq - r^3 q$.
3. Both terms on the right contain $q$, so take it out as a common factor: $rt^2 = q(p - r^3)$.
4. Divide by the bracket: $q = \frac{rt^2}{p - r^3}$.

(b)

1. Write both sides as powers of 3: $9 = 3^2$ and $27 = 3^3$.
2. So $3^{2(1 - x)} = 3^{3y}$, and the powers must be equal: $2 - 2x = 3y$, which is $2x + 3y = 2$ (1).
3. The second equation is $x - y = -\frac32$ (2).
4. Multiply (2) by 3: $3x - 3y = -\frac92$ (3).
5. Add (1) and (3): $5x = -\frac52$, so $x = -\frac12$.
6. From (2), $y = x + \frac32 = 1$.
7. So $x + y = -\frac12 + 1 = \frac12$.

(a)

1. Find the angle first: $\frac{\theta}{360} \times \frac{22}{7} \times 21^2 = 280$, so $\theta = \frac{280 \times 360 \times 7}{22 \times 441}$
   $\approx 72.73^\circ$.
2. Arc $= \frac{72.73}{360} \times 2 \times \frac{22}{7} \times 21$
   $\approx 26.67$ cm.
3. (Or use area $= \frac12 \times$ arc $\times\ r$: arc $= \frac{2 \times 280}{21} \approx 26.67$ cm.)
4. Perimeter $= 26.67 + 21 + 21 \approx 68.7$ cm.

(b)

1. When the sector is bent into a cone, its radius (21 cm) becomes the slant height and its arc becomes the base circumference: $2\pi r = 26.67$, so $r = \frac{26.67}{2 \times 22/7}$
   $\approx 4.24$ cm.
2. The vertical angle is the angle at the tip of the cone.
3. Half of it is in a right-angled triangle with hypotenuse 21 (the slant height) and opposite side 4.24 (the radius): $\sin\frac{y}{2} = \frac{4.24}{21}$
   $\approx 0.2020$, so $\frac{y}{2} \approx 11.66^\circ$ and $y \approx 23^\circ$.

(a)(i)

1. $PT \parallel QS$, and $TQ$ crosses both, so $\angle TQS = \angle PTQ = 42^\circ$ (alternate angles).
2. In triangle $QTS$ the angles add up to $180^\circ$: $x + 42^\circ + 38^\circ = 180^\circ$, so $x = 100^\circ$.

(ii)

1. At $Q$, the angles along the straight line are $y$, $\angle TQS = 42^\circ$ and $\angle RQS$, which the diagram gives as $60^\circ$.
2. So $y + 42^\circ + 60^\circ = 180^\circ$ and $y = 78^\circ$.

(b)

1. $\angle PSQ$ stands on the same arc $PQ$ as $\angle POQ$ at the centre, so $\angle PSQ = \frac12 \times 150^\circ$
   $= 75^\circ$.
2. Then $\angle PSR = \angle PSQ + \angle QSR$
   $= 75^\circ + 40^\circ$
   $= 115^\circ$.
3. $PQRS$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$: $\angle PQR = 180^\circ - 115^\circ$
   $= 65^\circ$.
4. So $\angle RQS = \angle PQR - \angle SQP$
   $= 65^\circ - 45^\circ$
   $= 20^\circ$.

(a)

1. The whole circle, $360^\circ$, stands for the whole grant, $100\%$.
2. So each $1\%$ is $3.6^\circ$: magazines $10 \times 3.6^\circ = 36^\circ$, new books $35 \times 3.6^\circ = 126^\circ$, repairs $54^\circ$, furniture $108^\circ$ and training $36^\circ$.
3. Check: $36 + 126 + 54 + 108 + 36 = 360$.
4. Draw a circle with compasses, draw one radius as a starting line, then measure each angle in turn with a protractor.
5. Label each sector with the item and its angle.

(b)

1. New books get $35\%$ of $1,300, which is $455.
2. Furniture gets $30\%$, which is $390.
3. The increase is $65, so the percentage increase over furniture is $\frac{65}{390} \times 100 \approx 16.7\%$, which is $17\%$ to the nearest whole number.
4. (Straight from the percentages: $\frac{35 - 30}{30} \times 100$ gives the same.)

1. Draw a Venn diagram with three overlapping circles, M, A and E, inside a rectangle for the 40 students.
2. Fill in the regions you are given: M only 6, A only 9, M and A only 5, A and E only 2.
3. Let the centre (all three) be $x$.
4. Find the centre. Accounts has 19 students: $9 + 5 + 2 + x = 19$, so $x = 3$.
5. M and E only: Mathematics has 18: $6 + 5 + 3 + (\text{M and E only}) = 18$, so M and E only is 4.
6. E only: Economics has 16: $4 + 3 + 2 + (\text{E only}) = 16$, so E only is 7.

(a)

1. The students who passed at least one subject are $6 + 5 + 9 + 3 + 4 + 2 + 7 = 36$.
2. So $40 - 36 = 4$ students failed all three.

(b)

1. Failing both means passing neither.
2. Passed Mathematics or Economics (or both): $18 + 16 - (4 + 3) = 27$.
3. So $40 - 27 = 13$ failed both, and $\frac{13}{40} \times 100 = 32\frac12\%$.

(c)

1. 19 passed Accounts, so $40 - 19 = 21$ failed it.
2. $P(\text{failed Accounts}) = \frac{21}{40}$.

(a)

1. To divide by a fraction, multiply by it turned upside down: $\frac{x^2 - 4}{x^2 + x} \times \frac{x + 1}{x^2 - 4x + 4}$.
2. Factorise each part: $x^2 - 4 = (x + 2)(x - 2)$, $x^2 + x = x(x + 1)$ and $x^2 - 4x + 4 = (x - 2)^2$.
3. So the product is $\frac{(x + 2)(x - 2)}{x(x + 1)} \times \frac{x + 1}{(x - 2)^2}$.
4. Cancel $(x + 1)$ and one $(x - 2)$ to get $\frac{x + 2}{x(x - 2)}$.

(b)(i)

1. The roots of $ax^2 + bx + c = mx + k$ are the $x$-values where the line meets the curve.
2. From the graph, these are $x = -1.5$ and $x = 1.5$.

(ii)

1. $M(0, 2)$ is on the curve: put $x = 0$, $y = 2$ to get $c = 2$.
2. $L(-1, 0)$ gives $a - b + 2 = 0$, so $a - b = -2$.
3. $N(2, 0)$ gives $4a + 2b + 2 = 0$, so $2a + b = -1$.
4. Add the two equations: $3a = -3$, so $a = -1$ and then $b = 1$.
5. The curve is $y = -x^2 + x + 2$.

(iii)

1. The line of symmetry is halfway between the points where the curve crosses the $x$-axis, $x = -1$ and $x = 2$: $x = \frac{-1 + 2}{2} = 0.5$.

(a)

1. $\sin x = 0.6 = \frac35$.
2. Draw a right-angled triangle with opposite 3 and hypotenuse 5.
3. The adjacent side is $\sqrt{25 - 9} = 4$.
4. So $\cos x = \frac45$.
5. Then $2\cos x + 3\sin x = \frac85 + \frac95$
   $= \frac{17}{5}$.

(b)(i)

1. $OC$ is the line of symmetry of the isosceles triangle, so it cuts $\angle ACB$ in half: $\angle OCA = 65^\circ$.
2. Let the semicircle touch $AC$ at $X$.
3. A tangent is perpendicular to the radius, so $\angle OXC = 90^\circ$ and triangle $OXC$ is right-angled with hypotenuse $OC = 30$.
4. The radius $OX$ is opposite the $65^\circ$ angle: $r = 30\sin 65^\circ$
   $\approx 27.19$
   $\approx 27.2\text{ cm}$.

(ii)

1. The shaded part is the triangle minus the semicircle.
2. $OC$ is perpendicular to $AB$, so in triangle $OAC$, $\angle AOC = 90^\circ$ and $|OA| = 30\tan 65^\circ \approx 64.34$ cm.
3. Area of triangle $ABC = \frac12 \times |AB| \times |OC|$
   $= |OA| \times 30$
   $\approx 1930.1\text{ cm}^2$.
4. Area of the semicircle $= \frac12\pi r^2$
   $= \frac12 \times \frac{22}{7} \times 27.19^2$
   $\approx 1161.7\text{ cm}^2$.
5. Shaded area $\approx 1930.1 - 1161.7$
   $= 768.4\text{ cm}^2$.

(a)

1. Draw $QR = 7$ cm.
2. Construct $60^\circ$ at $Q$ and mark $QP = 7$ cm on the arm.
3. With centres $P$ and $R$ and radius 7 cm, draw arcs meeting at $S$.
4. Join $PS$ and $RS$.

(b)

1. Points equidistant from the lines $PQ$ and $QR$ lie on the bisector of $\angle PQR$: construct it (in a rhombus it is the diagonal $QS$).
2. Points equidistant from the points $Q$ and $R$ lie on the perpendicular bisector of $QR$: construct it.
3. $X$ is where the two lines cross.

(c)

1. Measure $|XR| \approx 4.0$ cm.
2. (Check: $X$ is above the midpoint of $QR$ and $\angle XQR = 30^\circ$, so $|XR| = |XQ|$
   $= \frac{3.5}{\cos 30^\circ}$
   $\approx 4.04$ cm.)

(a)

1. Surface areas are in the ratio of the squares of the radii: $\frac{4\pi(12)^2}{4\pi R^2} = \frac{9}{49}$.
2. So $\frac{12^2}{R^2} = \frac{9}{49}$, and taking square roots, $\frac{12}{R} = \frac37$.
3. So $R = 28\text{ cm}$.
4. Volume of the bigger sphere $= \frac43\pi R^3$
   $= \frac43 \times \frac{22}{7} \times 28^3$
   $= \frac43 \times \frac{22}{7} \times 21\,952$
   $\approx 91\,989\text{ cm}^3$.

(b)

1. Draw the diagram.
2. From $X$, go 3 km due west to $Y$.
3. At $Y$, draw north.
4. North-west is the bearing $315^\circ$, so $YZ$ goes up and to the left, 5 km long.
5. At $Y$, the direction back to $X$ is due east ($090^\circ$) and the direction to $Z$ is $315^\circ$, so the angle inside the triangle is $\angle XYZ = 360^\circ - (315^\circ - 90^\circ)$
   $= 135^\circ$.

(i)

1. Cosine rule: $|XZ|^2 = 3^2 + 5^2 - 2(3)(5)\cos 135^\circ$
   $= 34 + 30(0.7071)$
   $\approx 55.21$.
2. So $|XZ| \approx 7.43$ km, which is 7 km to the nearest km.

(ii)

1. Sine rule for the angle at $X$: $\frac{\sin\angle YXZ}{5} = \frac{\sin 135^\circ}{7.431}$, so $\sin\angle YXZ = \frac{5 \times 0.7071}{7.431}$
   $\approx 0.4758$ and $\angle YXZ \approx 28.4^\circ$.
2. At $X$, $Y$ is due west ($270^\circ$) and $Z$ is $28.4^\circ$ further round clockwise, towards north.
3. Bearing of $Z$ from $X$ $= 270^\circ + 28.4^\circ$
   $\approx 298^\circ$.

1. Find $x$ first. The total number of throws is $2 + 5 + x + 11 + 9 + 10 = 37 + x$, and $P(3) = \frac{x}{37 + x} = 0.26$.
2. So $x = 0.26(37 + x) = 9.62 + 0.26x$, which gives $0.74x = 9.62$ and $x = 13$.
3. The die was thrown 50 times.

(a)

1. With 50 scores, the median is halfway between the 25th and 26th.
2. The running totals are $2, 7, 20, 31, \ldots$: the 21st to 31st scores are all 4, so the 25th and 26th are both 4.
3. Median $= 4$.

(b)

1. $\sum fx = 2 + 10 + 39 + 44 + 45 + 60$
   $= 200$, so the mean is $\frac{200}{50} = 4$.
2. Then $\sum f(x - 4)^2 = 2(9) + 5(4) + 13(1) + 11(0) + 9(1) + 10(4)$
   $= 18 + 20 + 13 + 0 + 9 + 40$
   $= 100$.
3. Standard deviation $= \sqrt{\frac{100}{50}}$
   $= \sqrt 2$
   $\approx 1.41$.

(a)(i)

1. Area of a trapezium $= \frac12 \times (\text{sum of the parallel sides}) \times \text{height}$.
2. So $60 = \frac12 \times (15 + 25) \times h$
   $= 20h$, and $h = 3\text{ cm}$.

(ii)

1. Drop a perpendicular from $P$ to $SR$.
2. It is the height, 3 cm, and it is opposite the $30^\circ$ angle at $S$, with $PS$ as the hypotenuse: $\sin 30^\circ = \frac{3}{|PS|}$.
3. So $|PS| = \frac{3}{\sin 30^\circ}$
   $= \frac{3}{\frac12}$
   $= 6\text{ cm}$.

(b)

1. Let the sum be ₦$y$.
2. Ade got $\frac35y$, leaving $\frac25y$.
3. Nelly got $\frac13$ of the remainder: $\frac13 \times \frac25y = \frac{2}{15}y$.
4. Austin got the rest of the remainder: $\frac25y - \frac{2}{15}y = \frac{4}{15}y$.
5. Austin's share is ₦3,000 more than Nelly's: $\frac{4}{15}y - \frac{2}{15}y = 3000$.
6. So $\frac{2}{15}y = 3000$ and $y = 22\,500$.
7. Ade received $\frac35 \times 22\,500 =$ ₦13,500.

(a)

1. $P$ varies directly as $Q$ and inversely as $R^2$: $P = \frac{kQ}{R^2}$.
2. With $P = 1$, $Q = 8$, $R = 2$: $1 = \frac{8k}{4}$, so $k = \frac12$ and $P = \frac{Q}{2R^2}$.
3. When $P = 3$ and $R = 5$: $3 = \frac{Q}{50}$, so $Q = 150$.

(b)(i)

1. $A$ and $B$ are both on latitude $20^\circ$N, and the difference in longitude is $60^\circ - 20^\circ = 40^\circ$.
2. Distance along the parallel $= \frac{40}{360} \times 2 \times 3.142 \times 6400\cos 20^\circ$
   $\approx 4199.1$ km.
3. Average speed $= \frac{4199.1}{6} \approx 700$ km/h.

(ii)

1. Due north is along a meridian, a great circle of radius 6400 km.
2. If the latitude changes by $y^\circ$: $\frac{y}{360} \times 2 \times 3.142 \times 6400 = 420$, so $y \approx 3.76^\circ$.
3. $C$ is at $20^\circ + 3.76^\circ \approx 24^\circ$N.