WAEC · 2017 · Private · General Maths · Paper 2

(a)

1. Top: divide first, $1\frac34 \div 1\frac25 = \frac74 \times \frac57$
   $= \frac54$.
2. Then $2\frac12 + \frac54 = \frac{10}{4} + \frac54$
   $= \frac{15}{4}$.
3. Bottom: $2\frac14 - 1\frac12 = \frac94 - \frac64$
   $= \frac34$.
4. So the value is $\frac{15}{4} \div \frac34 = \frac{15}{4} \times \frac43$
   $= 5$.

(b)

1. In $\triangle PSR$, right-angled at $R$: $PR$ is opposite the $60^\circ$ angle at $S$ and $SR$ is adjacent, so $\tan 60^\circ = \frac{|PR|}{3\sqrt2}$.
2. Then $|PR| = 3\sqrt2 \times \sqrt3 = 3\sqrt6$ cm.
3. In $\triangle PRQ$, right-angled at $P$: $PR$ is opposite the $45^\circ$ angle at $Q$ and $QR = x$ is the hypotenuse, so $\sin 45^\circ = \frac{3\sqrt6}{x}$.
4. Then $x = \frac{3\sqrt6}{\frac{1}{\sqrt2}}$
   $= 3\sqrt6 \times \sqrt2$
   $= 3\sqrt{12}$
   $= 6\sqrt3\text{ cm}$.

(a)

1. Compound interest at $5\%$ for two years multiplies the money by $1.05^2 = 1.1025$.
2. So $1.1025P = 2\,205\,000$ and $P = ₦2,000,000.00$.

(b)

1. Height : base $= 3 : 2$, so the base is $\frac23$ of the height: $b = \frac{2h}{3}$.
2. Area: $\frac12 \times \frac{2h}{3} \times h = \frac{h^2}{3}$
   $= 30$, so $h^2 = 90$ and $h = \sqrt{90} \approx 9.49$ cm.

(a)

1. $|PR| = |PQ|$, so the triangle is isosceles and the base angles at $Q$ and $R$ are equal: $2y = 2x - \frac25y$.
2. Multiply by 5: $10y = 10x - 2y$, so $12y = 10x$ and $x = \frac65y$ (1).
3. The angles of a triangle add up to $180^\circ$: $\left(2\frac14x + \frac12y\right) + 2y + \left(2x - \frac25y\right) = 180$.
4. Collect terms: $\frac{17}{4}x + \frac{21}{10}y = 180$.
5. Multiply by 20: $85x + 42y = 3600$ (2).
6. Substitute (1) into (2): $85 \times \frac65y + 42y = 3600$, so $102y + 42y = 3600$, $144y = 3600$ and $y = 25$.
7. Then $x = \frac65 \times 25 = 30$.
8. Check: the angles are $80^\circ$, $50^\circ$ and $50^\circ$, which add up to $180^\circ$ ✓.

(a)

1. $OS = OP$ (radii), so triangle $OSP$ is isosceles and $\angle SPO = \angle PSO = 44^\circ$.
2. $\angle SOQ$ is an exterior angle of triangle $OSP$: $\angle SOQ = 44^\circ + 44^\circ$
   $= 88^\circ$.
3. $R$ is on the minor arc $SQ$, so $\angle SRQ$ stands on the major arc.
4. The reflex angle $SOQ$ is $360^\circ - 88^\circ = 272^\circ$, and $\angle SRQ = \frac12 \times 272^\circ$
   $= 136^\circ$.
5. $OS = OQ$ and $RS = RQ$, so $OSRQ$ is a kite, and $\angle OSR = \angle OQR$.
6. The angles of the quadrilateral add up to $360^\circ$: $88^\circ + 136^\circ + 2\angle OQR = 360^\circ$.
7. So $2\angle OQR = 136^\circ$ and $\angle OQR = 68^\circ$.

(a)

1. The marks are separate whole numbers, so draw a bar chart: one bar for each mark from 40 to 46, equal widths, with gaps between them, of heights $7, 4, 6, 2, 4, 2, 6$.
2. Label the axes and state the scale.

(b)

1. There are $7 + 4 + 6 + 2 + 4 + 2 + 6 = 31$ students, so the median is the $\frac{31 + 1}{2} = 16$th mark.
2. Running totals: $7, 11, 17, \ldots$.
3. The 12th to 17th students scored 42, so the median is 42.

(c)

1. Less than the median means 40 or 41: $7 + 4 = 11$ students.
2. The probability is $\frac{11}{31} \approx 0.355$.

(a)

1. Substitute each $x$ into $y = 3x^2 - x - 11$.
2. For example, $x = -3$ gives $27 + 3 - 11 = 19$ and $x = 3$ gives $27 - 3 - 11 = 13$.

3. $x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
   $y$	$19$	$3$	$-7$	$-11$	$-9$	$-1$	$13$

(b)

1. With 2 cm to 1 unit on the $x$-axis and 2 cm to 5 units on the $y$-axis, plot the seven points and join them with a smooth U-shaped curve.

(c)(i)

1. The roots are where the curve crosses the $x$-axis: $x \approx -1.8$ and $x \approx 2.1$.

(ii)

1. The lowest point is just right of $x = 0$: the minimum value is $y \approx -11.1$.

(iii)

1. Draw the tangent at $(1, -9)$.
2. It passes through about $(0, -14)$ and $(2, -4)$, so the rise is $-4 - (-14) = 10$ and the run is $2 - 0 = 2$.
3. Its gradient is $10 \div 2 = 5$.
4. (Check: $\frac{dy}{dx} = 6x - 1 = 5$ at $x = 1$.)

(a)

1. $2\log_6 x = \log_6 x^2$, so $\log_6 y + \log_6 x^2 = \log_6(yx^2) = 3$.
2. In index form, $yx^2 = 6^3 = 216$, so $y = \frac{216}{x^2}$.

(b)(i)

1. Draw three overlapping circles F, B and V in a rectangle (100%).
2. Put 5 in the centre.
3. Two sports only: F and B $45 - 5 = 40$, B and V $15 - 5 = 10$, F and V $25 - 5 = 20$.
4. One sport only: F $70 - 40 - 20 - 5 = 5$, B $60 - 40 - 10 - 5 = 5$, V $45 - 20 - 10 - 5 = 10$.

(ii)

1. (α)** Football and volleyball but not boxing: $20\%$.

(β)

1. Exactly two sports: $40 + 10 + 20 = 70\%$.

(γ)

1. The regions add up to $5 + 5 + 10 + 40 + 10 + 20 + 5 = 95\%$, so $x = 100 - 95 = 5\%$ liked none.

(a)

1. "Only Iniola passes" means Iniola passes and Akafi fails.
2. $P(\text{Akafi fails}) = 1 - \frac34$
   $= \frac14$, so the probability is $\frac14 \times \frac35 = \frac{3}{20}$.

(b)(i)

1. A radius meets a tangent at $90^\circ$, so $\angle OXP = \angle OYP = 90^\circ$.
2. The triangle is equilateral, so $\angle XPY = 60^\circ$.
3. The angles of quadrilateral $PXOY$ add up to $360^\circ$: $\angle XOY = 360^\circ - 90^\circ - 90^\circ - 60^\circ$
   $= 120^\circ$.

(ii)

1. Find the radius from the chord $XY = 10$ cm.
2. The perpendicular from $O$ to $XY$ halves both the chord and the angle, giving a right-angled triangle with half-chord 5 cm opposite an angle of $60^\circ$: $r = \frac{5}{\sin 60^\circ}$
   $\approx 5.7735$ cm.
3. The major sector $XZY$ has angle $360^\circ - 120^\circ = 240^\circ$.
4. Area $= \frac{240}{360} \times \frac{22}{7} \times 5.7735^2$
   $\approx 69.8$, which is $70\text{ cm}^2$ to the nearest whole number.

(a)(i)

1. Draw north at $A$ and draw $AB$, 100 km on $330^\circ$ ($30^\circ$ west of north).
2. Draw north at $B$ and draw $BC$, 300 km due west ($270^\circ$).
3. Join $C$ to $A$.

(ii)

1. (I)** At $B$, the direction back to $A$ is $330^\circ - 180^\circ = 150^\circ$ and the direction to $C$ is $270^\circ$, so $\angle ABC = 270^\circ - 150^\circ$
   $= 120^\circ$.
2. Cosine rule: $|AC|^2 = 100^2 + 300^2 - 2(100)(300)\cos 120^\circ$
   $= 10\,000 + 90\,000 + 30\,000$
   $= 130\,000$.
3. So $|AC| = \sqrt{130\,000} \approx 360.56$ km.
4. (II) Sine rule: $\sin\angle BAC = \frac{300\sin 120^\circ}{360.56}$
   $\approx \frac{259.81}{360.56}$
   $\approx 0.7206$, so $\angle BAC \approx 46.1^\circ$.
5. At $A$, $B$ is on $330^\circ$ and $C$ is $46.1^\circ$ further round anticlockwise.
6. Bearing of $C$ from $A$ $= 330^\circ - 46.1^\circ$
   $\approx 284^\circ$.

(a)

1. Mark $D_2$ and draw a north line there.
2. Draw $D_2D_1$, 6 cm (60 m) on $315^\circ$, and $D_2D_3$, 8 cm (80 m) on $060^\circ$.
3. The angle between them is $360^\circ - 315^\circ + 60^\circ = 105^\circ$.
4. $M$ is the same distance from all three dormitories, so it lies on the perpendicular bisector of $D_1D_2$ and on the perpendicular bisector of $D_2D_3$.
5. Construct both bisectors with compasses.
6. $M$ is where they cross.

(b)(i)

1. Measure $|D_1D_3| \approx 11.2$ cm, which is about 112 m.
2. (By the cosine rule: $|D_1D_3|^2 = 60^2 + 80^2 - 2(60)(80)\cos 105^\circ$
   $\approx 12\,485$, so $|D_1D_3| \approx 111.7$ m.) (ii) Measure $|MD_2| \approx 5.8$ cm, which is about 58 m.

1. List every set first.
2. $M = \{4, 8, 12, 16, 20, 24, 28\}$ (7 numbers).
3. $A$ is the members of $M$ between 3 and 13: $A = \{4, 8, 12\}$.
4. $B$ is those between 9 and 25: $B = \{12, 16, 20, 24\}$.

(a)

1. $P(A) = \frac{3}{7}$.

(b)

1. $B'$ is the members of $M$ not in $B$: $\{4, 8, 28\}$.
2. $P(B') = \frac37$.

(c)

1. $A \cup B = \{4, 8, 12, 16, 20, 24\}$ (12 counted once).
2. $P(A \cup B) = \frac67$.

(a)

1. Multiply the matrices (row by column): $\begin{pmatrix} 3x & 2 \\ 5 & 4 \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 5 & 2y \end{pmatrix} = \begin{pmatrix} 9x + 10 & -6x + 4y \\ 35 & -10 + 8y \end{pmatrix}$.
2. Match the entries with $\begin{pmatrix} 28 & 4 \\ 35 & 22 \end{pmatrix}$: $9x + 10 = 28$, so $x = 2$.
3. $8y - 10 = 22$, so $y = 4$.
4. (Check the other entry: $-6(2) + 4(4) = 4$ ✓.)

(b)

1. Draw the four boundary lines: $y = x + 1$ (dashed).
2. $2y = -2x + 3$, that is $y = -x + \frac32$ (solid).
3. $2x = 3$, that is $x = \frac32$ (dashed).
4. $y + 1 = 0$, that is $y = -1$ (dashed).
5. Test the origin in each: $0 < 1$ ✓, so keep the side of $y = x + 1$ below the line.
6. $0 \ge 3$ ✗, so keep the side of $y = -x + \frac32$ above the line.
7. $0 < 3$ ✓, so keep the left of $x = \frac32$.
8. $1 > 0$ ✓, so keep above $y = -1$.
9. The region satisfying all four is the triangle with corners $\left(\frac14, \frac54\right)$, $\left(\frac32, \frac52\right)$ and $\left(\frac32, 0\right)$.
10. The line $y = -1$ lies below it, so it doesn't cut any of it off.
11. Shade and label the triangle.

(a)(i)

1. Work out the bracket first: $2 * 3 = 2 + 3 + 2(2)(3) = 5 + 12 = 17$.
2. Then $17 * 5 = 17 + 5 + 2(17)(5)$
   $= 22 + 170$
   $= 192$.

(ii)

1. Write out each side with the rule.
2. $x * 7 = x + 7 + 14x = 15x + 7$, and $x * 5 = x + 5 + 10x = 11x + 5$.
3. For $(x * 5) * 2$, the first number is $11x + 5$: $(11x + 5) + 2 + 2(11x + 5)(2)$.
4. This is $11x + 7 + 44x + 20$, which is $55x + 27$.
5. Solve $15x + 7 = 55x + 27$: $-40x = 20$, so $x = -\frac12$.
6. The truth set is $\left\{-\frac12\right\}$.

(b)(i)

1. $2\overrightarrow{SR} = 3\overrightarrow{PQ}$, so $\overrightarrow{SR} = \frac32\begin{pmatrix} 8 \\ 6 \end{pmatrix}$
   $= \begin{pmatrix} 12 \\ 9 \end{pmatrix}$.
2. $\overrightarrow{RS}$ goes the other way: $\overrightarrow{RS} = \begin{pmatrix} -12 \\ -9 \end{pmatrix}$.

(ii)

1. Go round the trapezium from $P$: $\overrightarrow{PS} = \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS}$
   $= \begin{pmatrix} 8 - 4 - 12 \\ 6 + 3 - 9 \end{pmatrix}$
   $= \begin{pmatrix} -8 \\ 0 \end{pmatrix}$.