WAEC · 2013 · May/June · General Maths · Paper 2

Question 1

(a)
Simplify, without using tables or a calculator: $\dfrac{\frac34\left(3\frac38 + 1\frac58\right)}{2\frac18 - 1\frac12}$ .
(b)
Given that $\log_{10}2 = 0.3010$ and $\log_{10}3 = 0.4771$ , evaluate, correct to 2 significant figures and without using tables or a calculator, $\log_{10}1.125$ .

Worked solution (try it first)

(a)

Work out the top and bottom separately.
Top: $3\frac38 + 1\frac58 = 5$ , so $\frac34 \times 5 = \frac{15}{4}$ .
Bottom: $2\frac18 - 1\frac12 = \frac{17}{8} - \frac{12}{8}$
$= \frac58$ .
Divide: $\frac{15}{4} \div \frac58 = \frac{15}{4} \times \frac85$
$= 6$ .

(b)

$1.125 = \frac98 = \frac{3^2}{2^3}$ .
So $\log 1.125 = 2\log 3 - 3\log 2$
$= 2(0.4771) - 3(0.3010)$
$= 0.9542 - 0.9030$
$= 0.0512$ , which is $0.051$ to 2 significant figures.

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Question 2

(a)
Solve: $7x + 4 < \frac12(4x + 3)$ .
Show the answer
$x < -\frac12$
(b)
Salem, Sunday and Shaka shared a sum of ₦1,100.00. For every ₦2.00 that Salem gets, Sunday gets fifty kobo, and for every ₦4.00 Sunday gets, Shaka gets ₦2.00. Find Shaka's share.

Worked solution (try it first)

(a)

Expand the bracket: $7x + 4 < 2x + \frac32$ .
Collect terms: $5x < \frac32 - 4 = -\frac52$ , so $x < -\frac12$ .

(b)

For every ₦2.00 Salem gets, Sunday gets ₦0.50: Salem : Sunday $= 2 : 0.5 = 4 : 1$ .
For every ₦4.00 Sunday gets, Shaka gets ₦2.00: Sunday : Shaka $= 4 : 2 = 2 : 1$ .
Make Sunday's parts match: $4 : 1 = 8 : 2$ and $2 : 1$ , so Salem : Sunday : Shaka $= 8 : 2 : 1$ (11 parts).
Shaka's share $= \frac{1}{11} \times 1100 = ₦100$ .

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Question 3

(a)
The present ages of a father and his son are in the ratio $10 : 3$ . If the son is 15 years old now, in how many years will the ratio of their ages be $2 : 1$ ?
(b)
The arithmetic mean of $x$ , $y$ and $z$ is 6 while that of $x$ , $y$ , $z$ , $t$ , $u$ , $v$ and $w$ is 9. Calculate the arithmetic mean of $t$ , $u$ , $v$ and $w$ .

Worked solution (try it first)

(a)

The ratio is $10 : 3$ and the son is 15, so the father is $\frac{10}{3} \times 15 = 50$ years old.
In $n$ years they will be $50 + n$ and $15 + n$ , in the ratio $2 : 1$ : $\frac{50 + n}{15 + n} = 2$ .
Cross-multiply: $50 + n = 30 + 2n$ , so $n = 20$ .
The ratio will be $2 : 1$ in 20 years.

(b)

A mean times the number of values gives their total.
$x + y + z = 3 \times 6 = 18$ .
All seven values add up to $7 \times 9 = 63$ .
So $t + u + v + w = 63 - 18 = 45$ , and their mean is $\frac{45}{4} = 11.25$ .

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Question 4

Plane mensuration

(a)
The area of a circle is $154\text{ cm}^2$ . It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

Find the radius: $\frac{22}{7}r^2 = 154$ , so $r^2 = 49$ and $r = 7$ cm.
Let the two equal sectors have angle $x$ each.
The third is three times the two together: $3 \times 2x = 6x$ .
All three fill the circle: $x + x + 6x = 360^\circ$ , so $x = 45^\circ$ and the third sector's angle is $270^\circ$ .
Arc of the third sector $= \frac{270}{360} \times 2 \times \frac{22}{7} \times 7$
$= 33$ cm.
Perimeter $= 33 + 7 + 7 = 47$ cm.

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Question 5

(a)
A boy $1.2\text{ m}$ tall stands $6\text{ m}$ away from the foot of a vertical lamp pole $4.2\text{ m}$ high. If the lamp is at the tip of the pole, (i) represent this information in a diagram; (ii) calculate the length of the shadow of the boy cast by the lamp; (iii) calculate the angle of elevation of the lamp from the boy, correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw the lamp pole (4.2 m) and the boy (1.2 m) upright on the ground, 6 m apart.
The light from the lamp just passes the top of the boy's head and reaches the ground at the end of his shadow.

(ii)

Let the shadow be $y$ m long.
The boy and the pole make two similar right-angled triangles with the tip of the shadow: $\frac{y}{1.2} = \frac{y + 6}{4.2}$ .
Cross-multiply: $4.2y = 1.2y + 7.2$ .
So $3y = 7.2$ and $y = 2.4\text{ m}$ .

(iii)

From the top of the boy's head, the lamp is $4.2 - 1.2 = 3$ m higher and 6 m away horizontally: $\tan\theta = \frac36 = 0.5$ .
So $\theta \approx 26.6^\circ$
$\approx 27^\circ$ .

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Question 6

(a)
Two positive whole numbers $P$ and $q$ are such that $P$ is greater than $q$ and their sum is equal to three times their difference. (i) Express $P$ in terms of $q$ . (ii) Hence, evaluate $\dfrac{P^2 + q^2}{Pq}$ .
(b)
A man sold 100 articles at 25 for ₦66.00 and made a gain of $32\%$ . Calculate his gain or loss percent if he sold them at 20 for ₦50.00.

Worked solution (try it first)

(a)(i)

Their sum is three times their difference: $P + q = 3(P - q)$ .
Expand: $P + q = 3P - 3q$ .
So $4q = 2P$ and $P = 2q$ .

(ii)

Substitute $P = 2q$ : $\frac{P^2 + q^2}{Pq} = \frac{4q^2 + q^2}{2q \times q}$
$= \frac{5q^2}{2q^2}$
$= \frac52$
$= 2\frac12$ .

(b)

At 25 for ₦66, the 100 articles sold for $\frac{100}{25} \times 66 = ₦264$ .
That was a $32\%$ gain, so $264 = 1.32 \times$ cost price, and the cost price is $\frac{264}{1.32} = ₦200$ .
At 20 for ₦50 they would sell for $\frac{100}{20} \times 50 = ₦250$ .
Gain $= 250 - 200 = ₦50$ , which is $\frac{50}{200} \times 100\% = 25\%$ .
It is a gain of $25\%$ .

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Question 7

(a)
Copy and complete the table of values for the relation $y = 3x^2 - 5x - 7$ .

$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$

$y$ $35$ $-7$ $-9$ $5$

Model answer

$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$

$y$ $35$ $15$ $1$ $-7$ $-9$ $-5$ $5$ $21$

For example $x = -2$ : $12 + 10 - 7 = 15$ , and $x = 4$ : $48 - 20 - 7 = 21$ .
(b)
Using scales of 2 cm to 1 unit on the $x$ -axis and 2 cm to 5 units on the $y$ -axis, draw the graph of $y = 3x^2 - 5x - 7$ for $-3 \le x \le 4$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 1 unit across, 2 cm to 5 units up.

For (c): (i) roots $x \approx −0.9$ and $2.6$ ; (ii) the minimum is about $−9.1$ (at $x \approx 0.8$ ); (iii) draw the tangent at $(2, -5)$ and measure its slope using two points on it far apart: the gradient is about 7.
(c)
From your graph: (i) find the roots of the equation $3x^2 - 5x - 7 = 0$ ; (ii) estimate the minimum value of $y$ ; (iii) calculate the gradient of the curve at the point $x = 2$ .
Separate values with commas, e.g. 3, −2

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$35$			$-7$	$-9$		$5$

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$35$	$15$	$1$	$-7$	$-9$	$-5$	$5$	$21$

Try it on a graph

The curve and its tangent at x = 2 (gradient 7).

Worked solution (try it first)

(a)

Substitute each $x$ into $y = 3x^2 - 5x - 7$ .
For example, $x = -2$ gives $12 + 10 - 7 = 15$ and $x = 4$ gives $48 - 20 - 7 = 21$ .
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$

$y$ $35$ $15$ $1$ $-7$ $-9$ $-5$ $5$ $21$

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$35$	$15$	$1$	$-7$	$-9$	$-5$	$5$	$21$

(b)

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

(c)(i)

The roots are where the curve crosses the $x$ -axis: $x \approx -0.9$ and $x \approx 2.6$ .

(ii)

The lowest point of the curve is a little right of $x = 1$ : the minimum value is $y \approx -9.1$ .

(iii)

Draw the tangent at $(2, -5)$ .
It passes through about $(1, -12)$ and $(3, 2)$ , so the rise is $2 - (-12) = 14$ and the run is $3 - 1 = 2$ .
Its gradient is $14 \div 2 = 7$ .
(Check: $\frac{dy}{dx} = 6x - 5 = 7$ at $x = 2$ .)

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Question 8

(a)
If $(3 - x)$ , $6$ , $(7 - 5x)$ are consecutive terms of a geometric progression (G.P.) with constant ratio $r > 0$ , find the: (i) value of $x$ ; (ii) constant ratio.
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $ABCD$ is a quadrilateral with $\angle ABC = 90^\circ$ , $|AB| = 3\text{ cm}$ , $|BC| = 4\text{ cm}$ , $|CD| = 6\text{ cm}$ and $|DA| = 7\text{ cm}$ . Calculate $\angle ADC$ , correct to the nearest degree.

Worked solution (try it first)

(a)(i)

In a G.P. the ratio of consecutive terms is constant: $\frac{6}{3 - x} = \frac{7 - 5x}{6}$ .
Cross-multiply: $36 = (3 - x)(7 - 5x) = 21 - 22x + 5x^2$ .
So $5x^2 - 22x - 15 = 0$ , which factorises as $(5x + 3)(x - 5) = 0$ .
So $x = -\frac35$ or $x = 5$ .
With $x = 5$ the terms are $-2, 6, -18$ , whose ratio $-3$ is not positive.
So $x = -\frac35$ .

(ii)

With $x = -\frac35$ the terms are $3.6, 6, 10$ , and $r = \frac{6}{3.6} = \frac53$ .

(b)

Join $AC$ .
Triangle $ABC$ is right-angled at $B$ , so $|AC| = \sqrt{3^2 + 4^2} = 5\text{ cm}$ .
In triangle $ACD$ all three sides are known, so use the cosine rule for the angle at $D$ : $\cos\angle ADC = \frac{7^2 + 6^2 - 5^2}{2 \times 7 \times 6}$
$= \frac{60}{84}$
$\approx 0.7143$ .
So $\angle ADC \approx 44^\circ$ .

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Question 9

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) a trapezium $WXYZ$ such that $|WX| = 10.2\text{ cm}$ , $|XY| = 5.6\text{ cm}$ , $|YZ| = 5.8\text{ cm}$ , $\angle WXY = 60^\circ$ and $\overline{WX}$ is parallel to $\overline{YZ}$ ; (ii) a perpendicular from $Z$ to meet $\overline{WX}$ at $N$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $WX = 10.2$ cm and construct $60^\circ$ at $X$ ; mark $Y$ with $XY = 5.6$ cm. Through $Y$ draw a line parallel to $WX$ and mark $Z$ with $YZ = 5.8$ cm. Join $WZ$ . Then drop the perpendicular from $Z$ to $WX$ to meet it at $N$ . Measured: $|WZ| \approx 5.1$ cm and $|ZN| \approx 4.8$ cm.
(b)
Measure: (i) $|WZ|$ ; (ii) $|ZN|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw $WX = 10.2$ cm.
Construct $60^\circ$ at $X$ and mark $XY = 5.6$ cm on the arm.
$YZ \parallel WX$ , so the angle at $Y$ is $180^\circ - 60^\circ = 120^\circ$ (co-interior angles): construct $120^\circ$ at $Y$ , towards $W$ , and mark $YZ = 5.8$ cm.
Join $WZ$ .

(ii)

With centre $Z$ , draw an arc cutting $WX$ twice.
From those points draw equal arcs crossing below $WX$ .
Join $Z$ to the crossing.
It meets $WX$ at $N$ .

(b)

Measure: (i) $|WZ| \approx 5.1$ cm.

(ii)

$|ZN| \approx 4.8$ cm.
Check: $|ZN|$ is the height, $5.6\sin 60^\circ \approx 4.85$ cm.
$|WN| = 10.2 - 5.6\cos 60^\circ - 5.8$
$= 1.6$ cm, so $|WZ| = \sqrt{1.6^2 + 4.85^2} \approx 5.1$ cm.

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Question 10

(a)
A segment of a circle is cut off from a rectangular board $22\text{ cm}$ by $12\text{ cm}$ as shown in the diagram, leaving $5\text{ cm}$ and $3\text{ cm}$ of the base on either side of the chord. If the radius of the circle is $1\frac12$ times the length of the chord, calculate, correct to 2 decimal places, the perimeter of the remaining portion. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
Evaluate, without using calculators or tables, $\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3} - \dfrac{\sqrt{12}}{6}\right)$ .

Worked solution (try it first)

(a)

The chord is what is left of the 22 cm base: $22 - 5 - 3 = 14$ cm.
The radius is $1\frac12 \times 14 = 21$ cm.
Find the angle $x$ at the centre: the perpendicular from the centre halves the chord, so $\sin\frac{x}{2} = \frac{7}{21} = \frac13$ , giving $\frac{x}{2} \approx 19.47^\circ$ and $x \approx 38.94^\circ$ .
Arc $= \frac{38.94}{360} \times 2 \times \frac{22}{7} \times 21$
$\approx 14.28$ cm.
The remaining board's edge is the top (22), the two sides (12 and 12), the two pieces of base (5 and 3) and the arc instead of the chord: $22 + 12 + 12 + 5 + 3 + 14.28 = 68.28$ cm.

(b)

$\frac{3}{\sqrt3} = \sqrt3$ and $\frac{\sqrt{12}}{6} = \frac{2\sqrt3}{6}$
$= \frac{\sqrt3}{3}$ .
So the expression is $\sqrt3\left(\frac{2}{\sqrt3} - \frac{\sqrt3}{3}\right) = 2 - 1$
$= 1$ .

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Question 11

Marks (%)	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Frequency	2	3	5	13	19	31	13	9	4	1

The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary ( $10.5, 20.5, \ldots, 100.5$ ), starting from $(0.5, 0)$ and ending at $(100.5, 100)$ , and join the points with a smooth S-shaped curve. Label both axes.

For (b): across from 60 the curve gives the 60th percentile, about 56; up from 34.5 it reads about 15, so about 85 of the 100 passed and the probability is about $0.85$ .
(b)
Use the graph to find the: (i) 60th percentile; (ii) probability that a student passed the test if the pass mark was fixed at $35\%$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive with the 60th-percentile reading.

Worked solution (try it first)

(a)

Make the cumulative frequency table, with the upper class boundaries:

Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Upper boundary	10.5	20.5	30.5	40.5	50.5	60.5	70.5	80.5	90.5	100.5
Cumulative frequency	2	5	10	23	42	73	86	95	99	100

Plot each cumulative frequency at its upper boundary, starting from $(0.5, 0)$ , and join the points with a smooth S-shaped curve.

(b)(i)

The 60th percentile is at $\frac{60}{100} \times 100 = 60$ on the cumulative frequency axis.
Go across to the curve and down: about 56.
(As a check, 60 lies between 42 at 50.5 and 73 at 60.5: $50.5 + \frac{60 - 42}{31} \times 10 \approx 56.3$ .)

(ii)

A pass is 35 or more.
Go up from 34.5 (the boundary below 35) to the curve and across: about 15 students scored less than 35.
So about $100 - 15 = 85$ passed, and $P(\text{passed}) \approx \frac{85}{100}$
$= 0.85$ .
A reading close to this from your own curve is fine.

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Question 12

Longitude & latitude

(a)
An aeroplane flies due north from a town $T$ on the equator at a speed of $950\text{ km}$ per hour for 4 hours to another town $P$ . It then flies eastwards to town $Q$ on longitude $65^\circ$ E. If the longitude of $T$ is $15^\circ$ E, (i) represent this information in a diagram; (ii) calculate the: (I) latitude of $P$ , correct to the nearest degree; (II) distance between $P$ and $Q$ , correct to 4 significant figures. $\left[\text{Take }\pi = \frac{22}{7}, \text{ radius of the earth} = 6400\text{ km}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw the earth with the equator and the meridian $15^\circ$ E.
Mark $T$ on the equator, $P$ due north of it on the same meridian, and $Q$ east of $P$ on the same latitude, at $65^\circ$ E.

(ii)

(I)** Flying north for 4 hours covers $950 \times 4 = 3800$ km along a meridian, a great circle of radius 6400 km.
If the latitude of $P$ is $\theta$ : $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400 = 3800$ , so $\theta \approx 34.0^\circ$ .
$P$ is at latitude $34^\circ$ N.
(II) $P$ and $Q$ are on latitude $34^\circ$ N, and the difference in longitude is $65^\circ - 15^\circ = 50^\circ$ .
$|PQ| = \frac{50}{360} \times 2 \times \frac{22}{7} \times 6400\cos 34^\circ$
$\approx 4632$ km (4 significant figures).

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Question 13

(a)
When one end of a ladder $LM$ is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of $37^\circ$ with the horizontal ground. (i) Represent this information in a diagram. (ii) Calculate, correct to 3 significant figures, the length of the ladder. (iii) If the foot of the ladder is pushed towards the wall by 2 metres, calculate, correct to the nearest degree, the angle which the ladder now makes with the ground.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw the wall vertical and the ground horizontal.
The ladder $LM$ runs from $L$ , 5 m up the wall, to $M$ on the ground, making $37^\circ$ with the ground at $M$ .

(ii)

The height, 5 m, is opposite the $37^\circ$ angle, and the ladder is the hypotenuse: $\sin 37^\circ = \frac{5}{|LM|}$ .
So $|LM| = \frac{5}{\sin 37^\circ}$
$= \frac{5}{0.6018}$
$\approx 8.31\text{ m}$ .

(iii)

First find how far the foot was from the wall: $\tan 37^\circ = \frac{5}{d}$ , so $d = \frac{5}{0.7536} \approx 6.635$ m.
Pushed 2 m closer, the foot is $6.635 - 2 = 4.635$ m from the wall.
The ladder is still 8.308 m long, and it is now the hypotenuse with 4.635 m adjacent to the new angle: $\cos\theta = \frac{4.635}{8.308}$
$\approx 0.5579$ .
So $\theta \approx 56^\circ$ .

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