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2020 Mathematics WAEC Theory Past Questions

WAEC: MATHEMATICS 2020 Mathematics WAEC Theory Past Questions

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Question 1a

If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of μ = {x: 1 ≤ x ≤ 10}, find \( \scriptsize A’ \cap B’ \cap C’ \)

Solution

μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 4, 6, 8, 10}

B = {3, 6, 9}

C = {1, 2, 3, 6}

A’ = {1, 3, 5, 7, 9}

B’ = {1, 2, 4, 5, 7, 8, 10}

C’ = {4, 5, 7, 8, 9, 10}

Hence, 

:- \( \scriptsize A’ \cap B’ \cap C’ \)

= {1, 3, 5, 7, 9} \( \scriptsize \cap \) {1, 2, 4, 5, 7, 8, 10} \( \scriptsize \cap \) {4, 5, 7, 8, 9, 10}

= {5, 7}

Question 1b

Tickets for a movie premiere cost $18.50 each while a bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?

Solution

Retail price for 1 dics = $18.50

Pack of 5 discs = $80.00

Total cost for 5 persons = $18.50 x 5

= $92.50

Amount saved by 5 persons

= $(92.50 – 80)

= $12.50

Amount saved by each person =  \(  \frac{12.50}{5} \)

= $2.50

Alternative Solution

Price for each person = \( \frac{80}{5} \)

= $16.00

Amount saved by each person = $(18.00 – 16.00) = $2.50

Question 2a

Given that P = \( \left (  \frac{rk}{Q}  \: – \: ms\right)^{\frac{2}{3}} \)

(i) Make Q the subject of the relation;

(ii) Find, correct to two decimal places the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10

Solution

(i) \( \scriptsize P = \normalsize \left (  \frac{rk}{Q}  \: – \: ms\right)^{\frac{2}{3}} \)

\( \scriptsize P^{\frac{3}{2}} = \normalsize \frac{rk}{Q}  \: – \: ms \) \( \scriptsize P^{\frac{3}{2}} \: + \: ms =  \normalsize \frac{rk}{Q}   \)

Cross multiply

\( \scriptsize Q \left( P^{\frac{3}{2}} \: + \: ms \right) = rk \)

∴\( \scriptsize Q  = \normalsize \frac{rk}{P^{\frac{3}{2}} \: + \: ms}\)

 

(ii) Given that k = 4 , m = 15, P = 3, r = 10 and s = 0.2

∴  \( \scriptsize Q  = \normalsize \frac{(10)(4)}{3^{\frac{3}{2}} \: + \: (15)(0.2)}\)

= \(   \normalsize \frac{40}{\sqrt{3}^{3} \: + \: 3}\)

= \(   \normalsize \frac{40}{\sqrt{27} \: + \: 3}\)

= \(   \normalsize \frac{40}{5.1962 \: + \: 3}\)

= \(   \normalsize \frac{40}{8.192}\)

= 4.8803

= 4.88

Question 2b

Given that \( \frac{x \: + \: 2y}{5} \scriptsize = x \: – \: 2y \)

find x: y

x + 2y = 5x – 10y

Collect like terms

x – 5x = -10y – 2y

-4x = -12y

Divide both sides by 4

x = 3y

Divide both sides by y

\( \frac{x}{y} = \frac{3}{1} \)

Hence x : y = 3 : 1

Question 3a

In the diagram, O is the centre of the circle ABCDE, \( \scriptsize \bar{|BC|} = \bar{|CD|} \) and ∠BCD = 108º. Find ∠CDE

Solution

∠CBD = ∠CDB (Base angles of an iscosceles triangle)

∴ 2∠CBD  + 108º = 180º (Sum of angles in a triangle)

2∠CBD   = 180º –  108º

2∠CBD   = 72º

∠CBD = \( \frac{72}{2} \)

∠CBD = 36º

Also, ∠BDE = 90º

Hence,

∠CDE = ∠CDB + ∠BDE

 = 36º + 90º

= 126º

Question 3b

Given that tan x = √3, \( \scriptsize  \; \; \; \; 0^o \leq  x \leq 90^o \)

Evaluate \( \frac{(cosx)^2 \: – \: sinx}{(sinx)^2 \: + \: cosx} \)

Solution

tan x = √3 for  \( \scriptsize  \; \; \; \; 0^o \leq  x \leq 90^o \)

∴ tanx = \(\frac{opp}{adj}  = \frac{\sqrt{3}}{1} \)

By Pythagoras’ theorem

\( \scriptsize a^2 = ( \sqrt{3})^2 \: + \: 1^2 \) \( \scriptsize a^2 = 3 \: + \: 1 \) \( \scriptsize a^2 = 4\) \( \scriptsize a = \sqrt{4}\) \( \scriptsize a = 2\)

∴ sinx = \( \frac{\sqrt{3}}{2} \)

cosx = \( \frac{1}{2} \)

Substitute sin x & cosx  into

\( \frac{(cosx)^2\: – \:  sinx}{(sinx)^2 \: + \: cosx} \) \( \frac{(\frac{1}{2} )^2 \: – \: \frac{\sqrt{3}}{2}  }{(\frac{\sqrt{3}}{2}  )^2 \: + \: \frac{1}{2}} \)

= \( \frac{\frac{1}{4}  \: – \: \frac{\sqrt{3}}{2} }{\frac{3}{4}  \: + \: \frac{1}{2}} \)

Finding the L.C.M of the numerator and denominator, we have;

\( \frac{\frac{1 \: – \: 2\sqrt{3}}{4}}{\frac{3 \: + \: 2}{4}}\)

= \( \frac{\frac{1 \: – \: 2\sqrt{3}}{4}}{\frac{5}{4}}\)

Solving the fraction we have

\( \frac{1 \: – \: 2\sqrt{3}}{4} \: \times \: \frac{4}{5}\) \( \frac{1 \: – \: 2\sqrt{3}}{5}\)

Question 4a

The total surface area of a cone of slant height 1 cm and base radius r cm is 224𝛑 cm2.

If r : l = 2 : 5, find: (a) Correct to one decimal place, the value of r,

Solution:

Ratio of the radius to slant height of a right circular cone is

r : l = 2 : 5

\( \frac{r}{l} = \frac{2}{5} \)

∴ 2l = 5r

l = \( \frac{5r}{2} \)

Total surface area = πr² + πrl

224π = πr(r + l)

224 = \( \scriptsize r \left[ r \: + \: \normalsize \frac{5r}{2} \right] \)

224 = \( \scriptsize r \left[ \normalsize \frac{2r \: + \: 5r}{2} \right] \)

224 = \( \scriptsize r \left[ \normalsize \frac{7r}{2} \right] \)

Crosss multiply

224 x 2 = 7r²

448 = 7r²

r² = \( \frac{448}{7} \)

r² = 64

r = √64

r = 8cm

Hence,

Slantheight(l) = \( \frac{5r}{2} \)

= \( \frac{5 \: \times \: 8}{2} \)

= \( \frac{40}{2} \)

= 20cm

Question 4b

Correct to the nearest whole number, the volume of the cone

Take π = \( \frac{22}{7} \)

Solution

By Pythagoras’ theorem

\( \scriptsize h^2 = 20^2 – 8^2 \) \( \scriptsize h = \sqrt{20^2 – 8^2} \) \( \scriptsize h = \sqrt{336}\) \( \scriptsize h = 18.33cm\)

∴vol = \( \frac{1}{3} \scriptsize \: \times \: \pi r^2h \)

= \( \frac{1}{3} \: \times \: \frac{22}{7} \scriptsize \: \times \: 8^2 \: \times \: 18.33\)

= \( \frac{25808.64}{21} \)

Volume = 1228.98

= 1229cm³

Question 5

A die was rolled a number of times. The outcomes are as shown in the table.

Number

1

2

3

4

5

6

outcomes

32

m

25

40

28

45

If the probability of obtaining 2 is 0.15, find the:

 

(a) Value of m;

Solution:

From the table

∑f = 32 + m + 25 + 40 + 28 + 45

= 170 + m

Pr(2) = 0.15

\( \frac{m}{170 \: + \: m} \scriptsize = 0.15 \)

m = 0.15(170 + m)

m = 25.5 + 0.15m

Collect like terms

m – 0.15m = 25.5

0.85m = 25.5

m = \( \frac{25.5}{0.85} \)

m = 30

 

(b) Number of times the die was rolled;

Solution:

Total number of times the die was rolled

= 170 + m

= 170 + 30

= 200

 

(c) Probability of obtaining an even number

Solution:

Pr(obtain an even number)

= \( \frac{m \: + \: 45 \: + \: 40}{170 \: + \: m} \)

= \( \frac{30 \: + \: 45 \: + \: 40}{170 \: + \: 30} \)

= \( \frac{115}{200} \)

or

0.575

Question 6a and 6b

(a) Copy and complete the table of values for the relation y = 3sin2x

Solution:


(b) Using a  scale of 2 cm to 15ᵒ on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of y = 3 sin 2x for 0º ≤ x ≤ 150º

Question 6c

(c) Use the graph to find the truth set of

(i) 3 sin 2x + 2 = 0

3 sin2x + 2 = 0

3 sin2x = -2

∴ y = -2

The truth is x = 111º

 

(ii) \( \frac{3}{2} \scriptsize sin2x \: = \: 0.25 \)

cross multiply

3sin2x = 0.25 x 2

3sin2x = 0.5

∴ y = 0.5

The truth set is x = 4.5 and 85.5

Question 7a

(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48m and the base radius is 14m. Calculate, correct to three significant figures, the surface area of the structure.

Take π = \( \frac{22}{7} \)

Solution

Slant height l is given by

l² = 48² + 14²

l² = 2304 + 196

l² = 2500

l = √2500

= 50m

Hence,

Surface area of the structure

= surface area of cone + surface area of hemisphere

Surface area of the structure

= πrl + 2πr²

\( \scriptsize = \left (  \frac{22}{7} \scriptsize \: \times \: 14 \: \times \: 50 \right) \: + \: \left (\scriptsize 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 14^2 \right) \)

= 2200 + 1232

= 3432

= 3430m² (3.s.f)

Question 7b

Five years ago, Musah was twice as old as Sesay’s. If the sum of their ages is 100. Find Sesay’s present age.

Let Sesay’s age = x

Musah’s age = y

From the first statement,

x + y = 100………(i)

From the second statement,

y – 5 = 2(x – 5)

y – 5 = 2x – 10

-5 + 10 = 2x – y

5 = 2x – y

∴ 2x – y = 5 …….(ii)

Adding equation (i) and (ii)

x + y = 100………(i)

2x – y = 5 …….(ii)

3x = 105

x = \( \frac{105}{3} \)

x = 35

put x = 35 into (i)

35 + y = 100

y = 100 – 35

y = 65

∴ Sesay’s age = x = 35 years

Musah’s ae = y = 65 years

Aliter

Let Musah’s age = M

M – 5 = 2(100 – M – 5)

M – 5 = 2(95 – M)

M + 2M = 190 + 5

3M = 195

M = \( \frac{195}{3} \)

M = 65

∴ Musah’s age = M = 65 years

Sesay’s age = 100 – 65 = 35 years

Question 8a

Ms. Maureen spent \( \frac{1}{4}\) of her monthly income at a shopping mall. \( \frac{1}{3}\)  at an open market and \( \frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had ₦225,000.00 left, find:

(i) her monthly income;

(ii) the amount spent at the open market.

Solution:

(i) Let woman’s monthly income = x

Food = \(\normalsize  \frac{1}{4} \scriptsize \: \times \: x = \normalsize  \frac{x}{4} \) 

Open market = \(\normalsize  \frac{1}{3} \scriptsize \: \times \: x = \normalsize  \frac{x}{3} \) 

The remainder = \( \scriptsize x  \: – \: \left( \normalsize  \frac{x}{4} \: + \: \frac{x}{3} \right) \)

= \( \scriptsize x  \: – \:  \normalsize  \frac{x}{4} \: + \: \frac{x}{3}  \)

= \(  \frac{12x \: – \: 3x \: – \: 4x}{12}  \)

Remainder = \(  \frac{5x}{12}  \)

Fraction of the remainder spent at the mechanic workshop

= \( \frac{2}{5} \: \times \:  \frac{5x}{12}  \)

= \(  \frac{x}{6}  \)

New remainder = \(  \frac{5x}{12} \: – \: \frac{x}{6}  \)

=\( \frac{5x \: – \: 2x}{12}  \)

=\( \frac{3x}{12}  \)

=\( \frac{x}{4}  \)

Her monthly income is

\( \frac{x}{4} \scriptsize  = 225,000  \)

x = ₦225,000 x 4

x = ₦900,000

 

(ii) Amount spent on the open market

= \( \frac{x}{3} \)

= \( \frac{900,000}{3} \)

= 300,000

Question 8b

The third term of an Arithmetic Progression (A.P) is 4m – 2n. If the ninth term of the progression is 2m – 8n. Find the common difference in terms of m and n

Solution

\( \scriptsize U_3 = 4m – 2n \)

a + 2d = 4m – 2n ………..(1)

\( \scriptsize U_9 = 2m – 8n \)

a + 8d = 2m – 8n ………..(2)

Subtract equation (2) from (1)

-6d = 2m + 6n

-6d = 2(m + 3n)

Divide both sides by -6

d = \( \frac{-2}{6} \scriptsize (m \: + \: n) \)

d = \( \frac{-1}{3} \scriptsize (m \: + \: n) \)

Question 9

Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km h on a bearing of 049ᵒ and cyclist Y travels at a rate of 9 km h on a bearing of 319ᵒ.

(a) Illustrate the information on a diagram.

Solution

(b) After travelling for two hours, calculate correct to the nearest whole number. The:

(i) Distance between cyclists X and Y

(ii) Bearing of cyclists X and Y.

 

(c) Find the average speed at which cyclist X will get to Y in 4 hours.

Solution:

Using cosine rule we have

|XY|² = |QY|² + |QX|²  – 2|QY| |QX| cos Q

= 9² + 5² – 2(9)(5) cos 90º

= 81 + 25 – 0

|XY| = √106

= 10.296kmh-1

Recall,

Speed = \( \frac{distance}{time} \)

10.296kmh-1  = \( \frac{distance}{2hrs} \)

distance = 10.296 x 2

= 20.592

= 21km

Aliter

Since the diagram is a right angled ΔXYQ

|XY|² = |QY|²  + |QX|²  (pythagoras theorem)

But,

|QY| = 9 x 2 = 18km

|QX| = 5 x 2 = 10km

Hence,

|XY|² = 18² + 10²

|XY| = √424

= 20.591

= 21km

 

(ii) since, we are dealing with a right-angled ∆XYQ, we can use

ta Y = \( \frac{|QX|}{|QY|} = \frac{10}{18} \)

Y = \( \scriptsize tan^{-1} 0.556 \)

Y = 29.06

Hence, the bearing of X from Y 

= 180º – (29.06º + 41º)

= 109.94º

=110º

 

(c) Find the average speed at which cyclist X will get to Y in 4 hours.

Average speed  = \( \frac{distance}{time} \)

= \( \frac{20.59km}{4hr} \)

= 5.148

= 5.15kmh-1

Question 10

The table shows the distribution of marks obtained by students in an examination

Solution:

Marks

Upper Class
Boundary

Frequency

Cumulative
Frequency

0-9

9.5

7

7

10-19

19.5

11

18

20-29

29.5

17

35

30-39

39.5

20

55

40-49

49.5

29

84

50-59

59.5

34

118

60-69

69.5

30

148

70-79

79.5

25

173

80-89

89.5

21

194

90-99

99.5

6

200

 

 

(b) Draw the cumulative frequency curve for the distribution.

Solution:

 

(c) Using the curve. Find correct to one decimal place, the:

(i) Median mark.

Solution:

To get the median score, we use

Q2 =  \( \frac{1}{2}\scriptsize  (N)th \)

=  \( \frac{1}{2}\scriptsize  (200)th \)

= 100th

=54.5 (from the curve)

 

(ii) Lowest mark for distinction if 5% of the students passed with distinction.

If 5% made distinction, then

⇒ 100% – 5% = 95%

It means 95% did not make distinction

Hence,

95 percentile = \( \frac{95}{100} \scriptsize \: \times \: 200th \)

= 190th

= 87.5% (from the curve)

Question 11a

In the diagram, MNPQ is a circle with center O, |MN|=|NP| and ∠OMN=50ᵒ. Find: (i) ∠MNP   (ii) ∠POQ

Solution

∠NPQ = 180º – 50º (cyclic quadrilateral)

∠MPQ = 90º (angle in a semi-circle)

∴∠MPN = 130º – 90º 

∠MPN = 40º

If ∠MPN = 40º, then ∠PMN = 40º (Base angles of an isosceles triangle)

∠MNP + ∠PMN + ∠MPN =  180º

∠MNP + 40º + 40º = 180º (Sum of angles in a triangle)

∠MNP = 180º – (40º + 40º)

∠MNP = 180º – 80º

∠MNP = 100º

 

(ii) ∠PMQ = ∠OMN – ∠PMN

= 50º – 40º

= 10º

∠POQ = 2 x ∠PMQ (angle at centre)

= 2 x 10º = 20º

 

Question 11b

Find the equation of the line which has the same gradient as 8y + 4x = 24 and passes through the point (-8, 12)

Solution:

8y + 4x = 24

Put the above equation in the form y = mx + c to get gradient (m)

∴ 8y = -4x + 24

divide through by 8

y = \( – \: \frac{1}{2} \scriptsize x  \: + \: 3 \)

∴ m = \( – \: \frac{1}{2} \)

Hence, the equation of the line at (-8, 12)

\( \scriptsize y \: – \: y_1 = m(x \: – \: x_1) \) \( \scriptsize y \: – \: 12 = \normalsize – \:  \frac{1}{2} \scriptsize (x \: – \: (-8)) \) \( \scriptsize y \: – \: 12 = \normalsize – \:  \frac{1}{2}\scriptsize (x \: +\: 8) \)

Multiply both sides by the L.C.M (2)

2y – 24 = -x – 8

2y = -x – 8 + 24

2y = -x + 16

Divide both sides by 2

y = \( – \: \frac{1}{2}\scriptsize \: + \: 8 \)

Question 12a

In the diagram, AB is a tangent to the circle with centre O and COB is a straight line. If CD//AB and ∠ABE = 40ᵒ, find ∠ODE

Solution

∠OCD = ∠ABC = 40ᵒ (Alternate angles)

∠CDE = 90ᵒ (∠s in a semicircle)

∠OCD =∠OCD = 40ᵒ (base ∠s of isosceles ∆)

Hence,

∠ODE = ∠OCD − ∠ODC

∠ODE = 90ᵒ − 40ᵒ

= 50ᵒ

Question 12b

Solution

(i)

\( \scriptsize \angle BCD = 180^o \: – \: 125^o = 55^o \)

∴ |BE| = 5 x sin55º

= 4.0958cm

Area of parallelogram

= 7 x 4.0988

= 28.6706

= 28.7cm²

Question 12c

(c) If \( \scriptsize x = \normalsize \frac{1}{2} \left( \scriptsize1 \: – \: \sqrt{2}  \right) \)

Evaluate \( \left( \scriptsize 2x^2 \: – \: 2x \right ) \)

Solution

\( \scriptsize x = \normalsize \frac{1}{2} \left( \scriptsize1 \: – \: \sqrt{2}  \right) \)

Substitute the value of x into \( \scriptsize 2x^2 \: – \: 2x \)

 = \( \scriptsize 2\left [ \normalsize \frac{1}{2} \left(\scriptsize1 \: – \: \sqrt{2} \right)  \right ]^2 \scriptsize \: – \: 2\left [ \normalsize \frac{1}{2} \left(\scriptsize 1 \: – \: \sqrt{2} \right)  \right ] \)

= \( \scriptsize 2\left [ \normalsize \frac{1}{4} \left(\scriptsize1 \: – \: \sqrt{2} \right)^2  \right ]\scriptsize  \: – \:  \left( 1 \: – \: \sqrt{2} \right) \)

Expanding

\( \left( \scriptsize 1 \: – \: \sqrt{2} \right)^2 \\ \scriptsize = (1 \: – \: \sqrt{2})(1 \: – \: \sqrt{2}) \\ \scriptsize =  1 \: – \: \sqrt{2} \: – \: \sqrt{2} \: + \: 2 \\ \scriptsize = 1 \: – \: 2\sqrt{2} \: + \: 2 \)

Hence, \( \scriptsize 2x^2 \: – \: 2x \)

= \( \scriptsize 2 \left [ \normalsize \frac{1}{4} \left( \scriptsize 1 \: – \: 2\sqrt{2} \: + \: 2 \right) \right] \scriptsize  \: – \:  1 \: + \: \sqrt{2} \)

= \( \frac{1}{2} \left ( \scriptsize 1  \: – \: 2\sqrt{2} \: + \: 2 \right) \scriptsize  \: – \:  1 \: + \: \sqrt{2} \)

= \( \frac{1 \: – \: 2\sqrt{2}\: + \: 2 \: – \: 2 \: + \: 2 \sqrt{2}}{2} \)

= \( \frac{1}{2} \)

Question 13

(a) Using a ruler and a pair of compasses only, construct: (i) (𝞪) ∆ABC with |AB| = 7.5cm. |AC| = 13.5 cm and ∠ABC = 120ᵒ: (𝞫) locus l1 of points equidistant from A and B. (𝞬) locus l2 of points equidistant from B and C.

Solution

 

(b) Using the method of completing the square, solve \( \scriptsize 4x^2 \: -\: 4\sqrt{3x} \: + \: 3 = 0 \)

Leaving the answer in surd form.

Solution

Using completing the square method

\( \scriptsize 4x^2 \: -\: 4\sqrt{3x} \: + \: 3 = 0 \)

Move constant to the R.H.S

\( \scriptsize 4x^2 \: -\: 4\sqrt{3x} = -3 \)

Divide through by 4

\( \scriptsize x^2 \: – \: \sqrt{3x} = \normalsize \frac{-3}{4} \)

Add the square of half of the coefficient of x to both sides

\( \scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-3}{2} \right)^2 = \normalsize \frac{-3}{4} \: + \: \left ( \normalsize \frac{-3}{2} \right)^2\) \( \scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-3}{2} \right)^2 = \normalsize \frac{-3}{4} \: + \: \frac{3}{4} \) \( \scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-3}{2} \right)^2 =0\)

Combine the squares on the L.H.S

\( \left(\scriptsize x \: – \:   \normalsize \frac{\sqrt{3}}{2}\right) \scriptsize = 0 \)

Square root both sides.

\( \scriptsize x  \: – \: \normalsize \frac{\sqrt{3}}{2} \scriptsize = 0 \) \( \scriptsize x = \normalsize \frac{\sqrt {3}}{2} \)

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