Theory paper · 9 questions · partial

WAEC · 2021 · Private, 2nd series · General Maths · Paper 2

Topics include Commercial arithmetic, Sequences & series (AP, GP), Linear & simultaneous equations, Solid mensuration, Quadratics & their graphs, Circle geometry.

Our copy of this paper is missing questions 10, 11, 12, 13.

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Answer every question in order, timed if you like (suggested 2 h 15 min). You're marked when you hand in, then you see where to focus and the working for each question.

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Question 1

(a)
Kojo and Alidu invested $4,000.00 and $6,000.00 respectively in a business venture. After one year, a profit of $4,800.00 was realised. If after deduction of expenses of $1,280.00, the remainder is shared in proportion to their investment, find the amount received by each person.
Separate values with commas, e.g. 3, −2
(b)
The 8th term of an Arithmetic Progression (A.P.) is $-8$ and the 3rd term is 12. Find the: (i) common difference; (ii) first term.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Remove the expenses first: $4800 - 1280 = 3520$ dollars to share.
Investments $4000 : 6000 = 2 : 3$ (5 parts), so one part is $\frac{3520}{5} = 704$ .
Kojo: $2 \times 704 = 1408$ , that is $1,408.00.
Alidu: $3 \times 704 = 2112$ , that is $2,112.00.

(b)

$T_8 = a + 7d = -8$ and $T_3 = a + 2d = 12$ .
Subtract: $5d = -20$ , so (i) $d = -4$ and (ii) $a = 12 - 2(-4) = 20$ .

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The distance between two ports $M$ and $N$ is $2100\text{ km}$ . Two ships travel towards each other, one leaving port $M$ at $20\text{ km/h}$ and, at the same time, another from port $N$ at $15\text{ km/h}$ . How:

(a)
long will it take the two ships to meet?
(b)
far will they be from port $M$ when they meet?

Worked solution (try it first)

(a)

The ships move towards each other, so the gap between them closes at $20 + 15 = 35\text{ km/h}$ .
Let them meet after $t$ hours: $35t = 2100$ , so $t = 60$ .
They meet after 60 hours.

(b)

In 60 hours the ship from $M$ travels $20 \times 60 = 1200\text{ km}$ .
They meet 1200 km from port $M$ .
(Check: the other ship travels $15 \times 60 = 900$ km, and $1200 + 900 = 2100$ ✓.)

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Question 3

The curved surface area of a cone is $242\text{ cm}^2$ . If the slant height is $4\text{ cm}$ more than the radius, calculate, correct to one decimal place, the: $\left[\text{Take }\pi = \frac{22}{7}\right]$

(a)
radius;
(b)
height;
(c)
volume of the cone.

Worked solution (try it first)

(a)

Let the radius be $r$ .
Then $l = r + 4$ .
Curved surface: $\frac{22}{7} \times r(r + 4) = 242$ , so $r^2 + 4r = 77$ and $r^2 + 4r - 77 = 0$ .
Factorise: $(r + 11)(r - 7) = 0$ .
The radius is positive, so $r = 7.0$ cm.

(b)

$l = 11$ cm, so $h = \sqrt{11^2 - 7^2}$
$= \sqrt{72}$
$\approx 8.5$ cm.

(c)

Volume $= \frac13 \times \frac{22}{7} \times 7^2 \times \sqrt{72}$
$\approx 435.6\text{ cm}^3$ .

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Question 4 ✱✱

(a)
$\overline{RU}$ and $\overline{RS}$ are tangents to the circle with centre $O$ at $U$ and $S$ , and $T$ is a point on the major arc $US$ . If $\angle URS = 80^\circ$ , find $\angle UTS$ .
(b)
$P$ is a point $270\text{ m}$ from $Q$ on a bearing of $328^\circ$ and $R$ is $420\text{ m}$ from $Q$ on a bearing of $058^\circ$ . Find the bearing of $P$ from $R$ .

Worked solution (try it first)

(a)

A tangent is perpendicular to the radius, so $\angle OUR = \angle OSR = 90^\circ$ .
The angles of quadrilateral $OURS$ add up to $360^\circ$ : $\angle UOS = 360^\circ - 90^\circ - 90^\circ - 80^\circ$
$= 100^\circ$ .
$T$ is on the major arc, so $\angle UTS$ is half the angle at the centre: $\angle UTS = 50^\circ$ .

(b)

Draw north at $Q$ .
$P$ is 270 m on $328^\circ$ and $R$ is 420 m on $058^\circ$ .
The angle between them is $(360^\circ - 328^\circ) + 58^\circ = 90^\circ$ , so triangle $PQR$ is right-angled at $Q$ .
At $R$ : $\tan\angle QRP = \frac{270}{420}$ , so $\angle QRP \approx 32.7^\circ$ .
At $R$ , the direction back to $Q$ is $058^\circ + 180^\circ = 238^\circ$ , and $P$ is $32.7^\circ$ further round clockwise.
Bearing of $P$ from $R$ $= 238^\circ + 32.7^\circ$
$\approx 271^\circ$ .

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Question 5

Probability

(a)
There are 25 boys and 15 girls in a class, all of them equally likely to be chosen for a contest. If two students are chosen one after the other without replacement for the contest, find the probability that: (i) two boys are chosen; (ii) a boy and a girl are chosen; (iii) two boys or two girls are chosen.
Separate values with commas, e.g. 3, −2
(b)
A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability that a girl takes away two red beads, one after the other, from the box.

Worked solution (try it first)

(a)

There are 40 students.
Without replacement, the second is chosen from the 39 left.

(i)

Two boys: $\frac{25}{40} \times \frac{24}{39} = \frac{600}{1560}$
$= \frac{5}{13}$ .

(ii)

A boy and a girl, in either order: $\frac{25}{40} \times \frac{15}{39} + \frac{15}{40} \times \frac{25}{39} = \frac{375 + 375}{1560}$
$= \frac{750}{1560}$
$= \frac{25}{52}$ .

(iii)

Two girls: $\frac{15}{40} \times \frac{14}{39} = \frac{210}{1560}$ .
Two boys or two girls: $\frac{600 + 210}{1560} = \frac{810}{1560}$
$= \frac{27}{52}$ .
(Check: (ii) and (iii) cover every case, and $\frac{25}{52} + \frac{27}{52} = 1$ .)

(b)

12 beads, 5 red.
After one red bead is taken away, 11 remain with 4 red: $\frac{5}{12} \times \frac{4}{11} = \frac{5}{33}$ .

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Question 6 ✱✱

(a)
The sum of two numbers is 91. If one-quarter of one of the numbers added to one-fifth of the other is 21, find the numbers.
Separate values with commas, e.g. 3, −2
(b)
If $2$ and $-\frac58$ are the roots of the quadratic equation $px^2 + qx + r = 0$ , find the values of $p$ , $q$ and $r$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let the numbers be $x$ and $y$ .
Their sum is 91: $x + y = 91$ (1).
A quarter of one plus a fifth of the other is 21: $\frac x4 + \frac y5 = 21$ .
Multiply by 20: $5x + 4y = 420$ (2).
From (1), $y = 91 - x$ , so $5x + 4(91 - x) = 420$ .
Then $5x + 364 - 4x = 420$ and $x = 56$ .
So $y = 35$ .
The numbers are 56 and 35.

(b)

The root $x = 2$ gives the factor $(x - 2)$ .
The root $x = -\frac58$ gives $8x = -5$ , so the factor $(8x + 5)$ .
The equation is $(x - 2)(8x + 5) = 0$ .
Expand: $8x^2 + 5x - 16x - 10 = 8x^2 - 11x - 10$
$= 0$ .
So $p = 8$ , $q = -11$ and $r = -10$ .

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Question 7

(a)
The marked price for an item with a profit of $15\%$ is ₦60,000.00. If the item was sold at a discount of $5\%$ , calculate, correct to one decimal place, the percentage profit made on the transaction.
(b)
Sarah has ₦15.00 and Bobo has ₦39.00. How much must Sarah give to Bobo so that Bobo shall have five times as much as Sarah?

Worked solution (try it first)

(a)

The marked price includes a $15\%$ profit, so it is $115\%$ of the cost price.
Cost price $= \frac{60\,000}{1.15} \approx$ ₦52,173.91.
With a $5\%$ discount, the selling price is $0.95 \times 60\,000 =$ ₦57,000.00.
Profit $= 57\,000 - 52\,173.91 =$ ₦4,826.09.
Percentage profit $= \frac{4826.09}{52\,173.91} \times 100\%$
$= 9.25\%$ .
Correct to one decimal place: $9.3\%$ .

(b)

Let Sarah give Bobo ₦ $x$ .
Sarah then has $15 - x$ and Bobo has $39 + x$ .
Bobo has five times as much as Sarah: $39 + x = 5(15 - x)$ .
So $39 + x = 75 - 5x$ , $6x = 36$ and $x = 6$ .
Sarah must give Bobo ₦6.00.
Check: Sarah has ₦9 and Bobo ₦45, and $45 = 5 \times 9$ ✓.

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Question 8 ✱✱

(a)
A boat sails $42\text{ km}$ from $M$ to $N$ on a bearing of $072^\circ$ , then to $P$ on a bearing of $342^\circ$ . $P$ is on a bearing of $038^\circ$ from $M$ . (i) Illustrate the information in a diagram. (ii) Calculate, correct to four significant figures: I. $|MP|$ ; II. $|NP|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw north at $M$ , and $MN$ , 42 km on $072^\circ$ .
Draw north at $N$ and $NP$ on $342^\circ$ .
Draw $MP$ on $038^\circ$ from $M$ to meet it at $P$ .

(ii)

At $N$ , the direction back to $M$ is $252^\circ$ and the direction to $P$ is $342^\circ$ , so $\angle MNP = 342^\circ - 252^\circ$
$= 90^\circ$ .
At $M$ , $\angle NMP = 72^\circ - 38^\circ$
$= 34^\circ$ .
I. $MP$ is the hypotenuse: $\cos 34^\circ = \frac{42}{|MP|}$ , so $|MP| = \frac{42}{0.8290} \approx 50.66$ km.
II. $\tan 34^\circ = \frac{|NP|}{42}$ , so $|NP| = 42 \times 0.6745 \approx 28.33$ km.

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Question 9 ✱✱

(a)
A tailor had a piece of cloth in the shape of a trapezium. The perpendicular distance between the two parallel edges was $38\text{ cm}$ . The lengths of the two parallel edges are $46\text{ cm}$ and $60\text{ cm}$ . The tailor cut off a semi-circular piece of the cloth of radius $18\text{ cm}$ from the $60\text{ cm}$ edge. Calculate, correct to one decimal place, the area of the remaining piece of cloth. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

Area of the trapezium $= \frac12(\text{sum of the parallel sides}) \times \text{height}$
$= \frac12(46 + 60) \times 38$
$= 53 \times 38$
$= 2014\text{ cm}^2$ .
Area of the semicircle cut off $= \frac12 \pi r^2$
$= \frac12 \times \frac{22}{7} \times 18^2$
$= \frac{3564}{7}$
$\approx 509.14\text{ cm}^2$ .
Remaining cloth $= 2014 - 509.14 = 1504.86\text{ cm}^2$ .
Correct to one decimal place: $1504.9\text{ cm}^2$ .

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