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2019 Mathematics WAEC Theory Past Questions

WAEC: MATHEMATICS 2019 Mathematics WAEC Theory Past Questions

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Question 1a

(a) Given that 110x – 40five, find the value of x.

Solution

Step 1: convert all numbers on both sides to numbers in base ten

110x = 40five

\( \scriptsize (1 \: \times \: x^2) \: + \: (1 \: \times \: x^1) = 4 \: \times \: 5^1 \)

x2 + x = 20

x2 + x – 20 = 0

Step 2: solve the quadratic equation

x2 + 5x – 4x – 20 = 0

x(x + 5) – 4(x + 5) = 0

(x + 5) (x – 4) = 0

x = 4

Question 1b

Simplify \( \frac{15}{\sqrt{75}}\scriptsize  \: + \: \sqrt{108} \: + \: \sqrt{432} \)

Leaving the answer in the form \( \scriptsize a \sqrt{b} \), where a and b are positive integers.

Solution

b) Step 1: write each surd in the base form, i.e.

i.e \( \frac{15}{\sqrt{25 \: \times \: 3}}\scriptsize  \: + \: \sqrt{36 \: \times \: 3} \: + \: \sqrt{36 \: \times \: 12} \)

\( \frac{15}{5\sqrt{3}}\scriptsize  \: + \: 6 \sqrt{3} \: + \:  \sqrt{36} \: \times \: \sqrt{4 \: \times \: 3} \)

= \( \frac{15}{5\sqrt{3}}\scriptsize  \: + \: 6 \sqrt{3} \: + \:  6 \: \times \:  2 \sqrt{ 3} \)

= \( \frac{15}{5\sqrt{3}}\scriptsize  \: + \: 6 \sqrt{3} \: + \:  12\sqrt{ 3} \)

= \( \frac{3}{\sqrt{3}}\scriptsize  \: + \: 18 \sqrt{3}  \)

Step 2: simplifying further we have

\( \frac{3}{\sqrt{3}}  \: + \: \frac{18 \sqrt{3}}{1}  \)

= \( \frac{3 \: + \: 18 \sqrt{9}}{\sqrt{3}} \)

= \( \frac{3 \: + \: 54}{\sqrt{3}} \)

= \( \frac{57}{\sqrt{3}} \)

Step 3: rationalize the denominator

\( \frac{57}{\sqrt{3}} \: \times \: \frac{\sqrt{3}}{\sqrt{3}} \)

= \( \frac{57 \sqrt{3}}{3}  \)

= \( \scriptsize 19 \sqrt{3} \)

Question 2a

find the equation of the line which passes through the points A(-2, 7) and B(2, -3)

Solution

Step 1: the equation of the line passing through the points (x1, y1) and (x2, y2) is given as

\( \frac{y_2 \: – \: y_1}{x_2 \: – \: x_1}  = \frac{y \: – \: y_1}{x \: – \: x_1} \)

Step 2: substitute A (-2, 7) and B (2, -3) into the formula, we have

\( \frac{-3 \: – \: 7}{2 \: –  \: (-2)}  = \frac{y \: – \: 7}{x \: – \:(-2)} \) \( \frac{-3 \: – \: 7}{2 \:  + \: 2}  = \frac{y \: – \: 7}{x \: + \: 2 } \)

4(y – 7) =  -10(x + 2)

Step 3: divide through by 2 and open the brackets

4y – 28 = -10x – 20

2y – 14 = -5x – 10

2y + 5x – 4 = 0

Question 2b

Given that \( \frac{5b \: – \: a}{8b \: + \: 3a} = \frac{1}{5}\), find correct to two decimal places, the value of \(\frac{a}{b} \)

Solution

Step 1: Using equivalent fractions, i.e.

\( \frac{5b \: – \: a}{8b \: + \: 3a} = \frac{1}{5}\)

  5b – a = 1 ———– (1)

8b + 3a = 5 ———- (2)

Step 2: from equation (1)

a  = 5b – 1 ————— (*)

Substitute for a in equation (2)

8b + 3(5b – 1)  = 5

8b + 15b – 3 = 5

23b = 8

b = \( \frac{8}{23} \)

Step 3: substitute b = \( \frac{8}{23} \) in equation (*)

⇒ a = \( \scriptsize 5 \left( \normalsize \frac{8}{23} \right) \scriptsize \: – \: 1 \)

a = \( \frac{40}{23} \: – \: \frac{1}{1} \)

a = \( \frac{40 \: – \: 23}{23}  \)

a = \( \frac{17}{23}  \)

Step 4: hence \( \frac{a}{b} \) is equal to

\( \frac{17}{23} \: \div \: \frac{8}{23} \) \( \frac{17}{23} \: \times\: \frac{23}{8} \)

= \( \frac{17}{8}  \)

= 2.13 (Correct to 2 decimal places)

Question 3a

Ali, Musah and Yusuf shared N420,000.00 in the ratio 3:5:8 respectively. Find the sum of Ali and Yusuf’s shares.

Solution

Step 1: Given that Ali, Musah and Yusuf shares are in the ratio 3:5:8, that is

Ali ———— 3

Musah ——– 5

Yusuf ——— 8

Total ——— 16

Step 2: the sum of Ali and Yusuf’s ratios is 3 + 8 = 11

Step 3: sum of their shares is

\( \frac{11}{16} \: \times \: \frac{420,000}{1} \)

= N288,750

Question 3b

(b) Solve: \( \scriptsize 2 \left(\normalsize \frac{1}{8} \right)^x \scriptsize  = 32^{x \: – \: 1} \)

Solution

Step 1:

Writing this in index form we have

\( \scriptsize 2(2^{-3})^x = 2^{(5(x \: – \: 1)} \) \( \scriptsize 2^1 \: \times \: 2^{-3x} = 2^{(5(x \: – \: 1)} \) \( \scriptsize 2^{1 \: – \: 3x}  = 2^{(5(x \: – \: 1) }\)

Step 2: dividing through by 2

\( \frac{ 2^{1 \: – \: 3x}}{2}  = \frac{2^{(5(x \: – \: 1)}}{2} \)

1 – 3x = 5x – 5

1 + 5 = 5x + 3x

6 = 8x

x = \( \frac{6}{8} \)

x = \( \frac{3}{4} \)

Question 4

In the diagram, PQRS is a quadrilateral, <PQR = <PRS = 900, |PQ| = 3cm, |QR| = 4cm and |PS| = 13cm. Find the area of the quadrilateral.

Solution

Step 1: the quadrilateral is made up of two right-angled triangles. PQR and PRS

Step 2: From ΔPQR

PR2 = PQ2 + QR2 (By Pythagoras)

PR2 = 32 + 42

PR2 = 9 + 16

PR2 = 25

PR =  √25

PR = 5cm

Step 3: From ΔPQR, by Pythagoras

PS2 = PR2 + RS2

132 = 52 + RS2

RS2 = 169 – 25

RS2 = 144

RS = √144

RS = 12cm

Step 4: Area of  PQR

= ½  x base x height

= ½ x 4 x 3

= 6cm²

Step 5: Area of  PRS

= ½ x base x height

= ½ x 12 x 5

= 30cm²

Step 6: Hence the area of the quadrilateral 

= 6 + 30

= 36cm2

Question 5

Three red balls, five, green balls and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \( \frac{1}{6} \), find:

a. the number of blue balls in the sack

b. the probability of picking a green ball

Solution

Step 1: Let the following be the number of balls as specified

Red balls ———– 3

Green balls ——— 5

Blue balls ———– x

Total balls ———– 8 + x

Step 2: given that the probability of picking a red ball is \( \frac{1}{6} \), then

\( \frac{1}{6} = \frac{3}{8 \: + \: x} \)

Step 3: cross multiply

8 + x = 3 x 6

8 + x = 18

x = 18 – 8

x = 10

Step 4: hence the number of balls in the bag is  8 + 10 = 18

Step 5: the probability of picking a green ball is \( \frac{5}{18}\)

Question 6a

(a) The force of attraction, F, between two bodies varies directly as the product of their masses, m1 and m2 and inversely as the square of the distance, d, between them.

Given that F = 20N, when m1 = 25kg, m1 = 10kg and d = 5m, find:

(i) An expression F in terms of m1, m2 and d;

(ii) The distance, d when F = 30N, m1 = 7.5kg and m2 = 4kg

Solution

(i) Step 1: translating this information into a mathematical equation we have

\( \scriptsize F \propto \normalsize \frac{m_1 m_2}{d^2} \)

Step 2: if K = constant then we have that

\( \scriptsize F = \normalsize \frac{m_1 m_2}{d^2}\scriptsize k \)—– (*)

Step 3: substituting the given parameters into equation (*)

⇒ \( \scriptsize 20 = \normalsize \frac{25 \: \times \: 10}{5 \: \times \: 5}\scriptsize k \)

100 = 50k

k = \( \frac{100}{50} \)

k = 2

Step 4: substituting the value of the constant k = 2 into equation (*), then

\( \scriptsize F = \normalsize \frac{2 m_1 m_2}{d^2} \)

Hence the expression for F in terms of m1, m2 and d;

(ii) Step 5: if F = 30, m1 = 7.5kg and m2 = 4kg, then we have

30 = \( \frac{2 \: \times \: 7.5 \: \times \: 4}{d^2} \)

30d² = 60

d² = \( \frac{60}{30} \)

d² = 2

d = √2

d = 1.44

Question 6b

(b)

Find the value of x in the diagram

Solution

Step 1: from the diagram x + (x+200) + (x + 400) + (x + 800) + x + 600 = 5400

(Sum of the interior angels of a pentagon)

Step 2: 5x + 200 = 5400

5x = 5400 – 2000

5x = 3400

x = \(\frac{340}{5}\)

x = 680

Question 7

The data show the marks obtained by students in a Biology test.

(a) construct a frequency distribution table using the class interval 0 – 9, 10 – 19, 20 – 29, ……..

Solution

Step 1: the frequency table is constructed as follows:

Class interval

Tally

Frequency

Cumulative Frequency

0 – 9

I

1

1

10 – 19

III

3

4

20 – 29

IIII

3

7

30 – 39

IIII I

6

13

40 – 49

IIII III

8

21

50 – 59

IIII IIII II

12

33

60 – 69

IIII I

6

39

70 – 79

III

3

42

80 – 89

IIII

4

46

90 – 99

IIII

4

50

 

(b) Draw a cumulative frequency curve for the distribution

Solution

The cumulative frequency curve is the graph obtained when the cumulative frequency is plotted against the upper-class boundaries of the distribution.

 

 

(c) Use the graph to estimate the:

(i) Median

Solution

 Step 1: the median is the 50th percentile

i.e. \( \left(\frac{50}{100} \scriptsize \: \times \: 50 \right)^{th} \scriptsize = 25^{th} \: observation \)

Step 2: the median is traced out from the cumulative frequency curve, the median is 

= 52.5

(ii) Percentage of students who scored at least 66 marks, correct to the nearest whole number.

Solution

Step 1: the number of students that scored 66 marks correct to the nearest whole number is

= 50 – 37

= 13

Step 2: the percentage of students that scored at least 66 is

\( \frac{13}{50} \: \times \: \frac{100}{1} \)

= 26%

Question 8a

(a) Solve the inequality \( \frac{1}{3} \scriptsize x \: – \: \frac{1}{4} \scriptsize (x \: + \: 2) \geq 3x \: – \: 1 \frac{1}{3} \)

Solution

Step 1: multiply inequality through by the LCM of the denominators which is 12

\( \frac{1}{3} \scriptsize x \: \times \: 12 \: – \: \frac{1}{4} \scriptsize (x \: + \: 2) \: \times \: 12 \geq 3x \: \times \: 12 \: – \: 1 \frac{1}{3} \: \times \: 12 \) \( \scriptsize 4x \: – \: 3(x\: + \: 2) \geq 36x \: – \: 16 \)

Step 2: Collecting like terms

\( \scriptsize 4x \: – \: 3(x\: + \: 2) \geq 36x \: – \: 16 \) \( \scriptsize x \: – \: 36x \geq -16 \: + \: 6 \) \( \scriptsize -35x \geq -16 \: + \: 6 \) \( \scriptsize -35x \geq -10 \) \( \scriptsize x  \leq \normalsize \frac{-10}{-35} \)

Note: (Sign changes when wwe divide by a negative number)

\( \scriptsize x  \leq \normalsize \frac{2}{7} \)

Question 8b

In the diagram, ABC is a right-angled triangle on horizontal ground. |AD| is a vertical tower <BAC = 900, <ACB = 350, <ABD = 520 and |BC| = 66m.

Find, correct to two decimal places.

(i) The height of the tower

Step 1: from ΔABC

Sin 350 = \( \frac{AB}{BC} \)

Sin 350 = \( \frac{AB}{66} \)

AB = 66 x sin350

AB = 37.856

Step 2: from ΔDAB

Tan 520 = \( \frac{AD}{AB} \)

Tan 520 = \( \frac{AD}{37.856} \)

AD = 37.856 x Tan 520

AD = 48.453

|AD| = 48.45 (to 2 decimal places)

 

(ii) The angle of elevation of the top of the tower from C

Step 1: from ΔBAC

Cos 350 = \( \frac{AC}{BC} \)

Cos 350 = \( \frac{AC}{66} \)

|AC| = 66 x cos 35

|AC| = 66 x 0.81915

|AC| = 54.06

Step 2: Let angle \( \scriptsize \hat{DCA} = \theta \). be the angle of elevation of the top of the Tower, then

Tan θ = \( \frac{AD}{AC} \)

Tan θ = \( \frac{48.45}{54.06} \)

Tan θ = 0.89622

θ = \( \scriptsize tan^{-1} (0.89622) \)

θ = 41.870 (to 2 decimal places)

Question 9

(a) Copy and complete the table values of y = 2cosx + 3sinx for \( \scriptsize 0^o \leq x \leq 360^o \)

x

00

600

1200

1800

2400

3000

3600

y

2.0

      

-3.60

    

Solution

Step 1: the table can be completed as shown below:

y = 2cosx + 3sinx

X

00

600

1200

1800

2400

3000

3600

2cosx

2.0

0.1

-1.0

-2.0

-1.0

1.0

2.0

3sinx

0.0

2.59

2.59

0.0

-2.59

-2.59

0.0

Y

2.0

3.6

1.6

-2.0

-3.6

-1.6

2.0

(b) Using a scale of 2cm to 600 the x-axis and 2cm to 1unit on the y-axis, draw the graph of y = 2cosx + 3sinx for \( \scriptsize 0^o \leq x \leq 360^o \)

 

(c) Using the graph:

(i) Solve 2cosx + 3sinx = -1

Solution

Step 1: 2cosx + 3sinx = -1

Compare with y = 2cosx + 3sinx

  y = -1

Step 2: the solution is obtained from the graph at the points of intersection of the graph and the line y = -1

Step 3: from the graph, the solution is labeled at points A and B,

x = 1350 or 3090

(ii) Find, correct to one decimal place, the value of y when x = 3420

Solution

From the graph the value of y when x = 3420 is 1.0

Question 10

A woman bought 130kg of tomatoes for N52,000.00. She sold half of the tomatoes at a profit of 30%. The rest of the tomatoes began to go bad, she then reduced the selling price per kg by 12%.

Calculate:

(a) The new selling price per kg

Solution

Step 1: Half of 130kg of tomatoes is

\( \frac{1}{2}\scriptsize  \: \times \: 130 =65kg \)

Step 2: 65kg of tomatoes was bought for

\( \frac{1}{2}\scriptsize  \: \times \: N52,000.00 = 65kg \)

= N26,000.00

Step 3: The selling price of the 65kg of tomatoes is:

100% = N26,000.00

1% = \( \frac{26,000}{100} \) 

130% = \( \frac{26,000}{100}  \scriptsize \: \times \: \frac{130}{1}\) 

= N33,800

Step 4: the amount per kg is \( \frac{33,800}{65} \) 

= N520

Step 5: the selling price of the remaining tomatoes per kg is

(100 – 12) = 88%

= 88% of N520

= \( \frac{88}{100}  \: \times \: \frac{520}{1}\) 

= N457.60

 

(b) The percentage profit on the entire sales if she threw away 5kg of bad tomatoes.

Step 1: the remaining bad tomatoes sold is (65.5)kg = 60kg

Step 2: the selling price of the 60kg tomatoes is

60 x N457.60

= N27,456.00

Step 3: the total amount realize for selling the tomatoes is

N33,800 + N27,456

= N61,256.00

Step 4: the profit realize on the sale

= N61,256 – N52,000

= N9,256.00

Step 5: the percentage profit on the sales of the tomatoes if 5kg of bad ones was thrown away.

= \( \frac{9,256}{52,000} \: \times \: \frac{100}{1} \)

= 17.8%

Question 11a

(a) the third and sixth terms of a geometric progression (G.P) are \( \frac{1}{4} \)   and \( \frac{1}{32} \)   respectively.

Find:

(i) the first term and the common ratio

Solution

Step 1: the nth term of a G.P is given as

Tn = arn-1

T3 = ar3-1

\( \frac{1}{4} \scriptsize =  ar^2\) …………….. (1)

Step 2: similarly,

T6 = ar6-1

\( \frac{1}{32} \scriptsize =  ar^5\)  …………….. (2)   

Step 3: Divide equation (2) by (1)

\( \frac{ar^5}{ar^2} = \frac{1}{32} \: \times \: \frac{4}{1} \) \( \scriptsize r^3 = \frac{1}{8}  \) \( \scriptsize r =  \sqrt[3]{\frac{1}{8}}  \)

Step 4: substitute r = \( \frac{1}{2} \) in equation (1)

\( \frac{1}{4} \scriptsize =  a \: \times \: \left( \normalsize \frac{1}{2} \right)^2\) \( \frac{1}{4} \scriptsize =  a \: \times \: \frac{1}{4}\)

a = \( \frac{1}{4} \: \times \: \frac{4}{1} \)

a = 1

Therefore the first term is 1 and the common ratio is \( \frac{1}{2} \)

(ii) the seventh term.

Step 5: the seventh term is obtained from

T7 = ar7-1

T7 = \( \scriptsize 1 \: \times \: \left( \normalsize \frac{1}{2}\right)^6 \)

T7 = \( \frac{1}{64} \)

Question 11b

Given that 2 and -3 are the roots of the equations ax2 + bx + c = 0, find the values of a, b and c.

Solution

Step 1: sum of the roots

= 2 + (-3) = 2 – 3 = -1

Step 2: product of roots

= 2 x -3 = -6

Step 3: substitute the sum and product into the formula

x2 – (sum of roots)x + product of roots = 0

x2 – (-1)x + (-6) = 0

x2 + x – 6 = 0

Step 4: compare x2 + x – 6 = 0 with ax2 + bx + c = 0

⇒ a = 1, b = 1 and c = -6

Question 12a

Given that sin y = \( \frac{8}{17} \), find the value of  \( \frac{tany}{1 \: + \: 2tany} \)

Solution

Step 1: given that Siny = \( \frac{8}{17} \)

By Pythagoras

\( \scriptsize 17^2 = 8^2 \: + \: a^2 \) \( \scriptsize 289 = 64 \: + \: a^2 \) \( \scriptsize a^2 = 289 – 64  \) \( \scriptsize a^2 = 225 \) \( \scriptsize a = \sqrt{225} \)

a = 15

Step 2: tan y = \( \frac{8}{15} \)

Step 3: Hence the value of \( \frac{tany}{1 \: + \: 2tany}   \)

= \( \frac{8}{15} \scriptsize \: \div \: \left(\scriptsize 1 \: + \: 2 \: \times \: \normalsize \frac{8}{15} \right) \)

= \( \frac{8}{15} \scriptsize \: \div \: \left(\scriptsize1 \: + \:  \normalsize \frac{16}{15} \right) \)

= \( \frac{8}{15} \scriptsize \: \div \: \left(\normalsize \frac{31}{15} \right) \)

= \( \frac{8}{15} \scriptsize \: \times \: \normalsize \frac{15}{31}  \)

= \( \frac{8}{31}   \)

Question 12b

An amount of N300,000.00 was shared among Otobo, Ada, and Adeola. Otobo received N60,000.00, Ada received \( \frac{5}{12}\) of the remainder, while the rest went to Adeola. In what ratio was the money shared?

Solution

Step 1: the remainder is N300,000 – N60,000

= N240,000

Step 2: Ada’s share is \( \frac{5}{12} \scriptsize \: \times \: 240,000\)

= N100,000

Hence Ada received N100,000

Step 3: Adeolu’s share is N240,000 – N100,000

= N140,000

Step 4: therefore the money was shared in the ratio

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

Question 13a

In the diagram \( \scriptsize \bar{RS} \) and \( \scriptsize \bar{RT} \) are tangent to the circle with centre O. <TUS = 680, <SRT = x and  <UTO = y. Find the value of x.

Solution

Step 1: Obtuse \( \scriptsize \hat{TOS} = 2 \: \times \: 68\) {angle subtends at the centre of a circle}

\( \scriptsize \hat{TOS} =136^o \)

Step 2: But \( \scriptsize \hat{OTR} = \hat{OSR} = 90^o\)  

(radius perpendicular to tangent at the points of contact)

Step 3: similarly,

\(\scriptsize \hat{OTR} \: + \: \hat{OSR} \: + \: \hat{TOS} \: + \: \hat{TRS} = 360^o \)

(sum of angles in a quadrilateral)

⇒ 900 + 900 + 1360 + x = 3600

1800 + 1360 + x = 3600

316 + x = 360

x = 360 – 316

x = 440

Question 13b

Two tanks A and B are filled to capacity with diesel. Tank A holds 600 litres of diesel more than tank B. If 100 litres of diesel was pumped out of each tank, tank A would then contain 3 times as much diesel as tank B. Find the capacity of each tank.

Solution

Step 1: Let the litres of diesel in tank B = x litres.

⇒ then the litres of diesel in tank A = (x + 600)litres

Step 2: If 100litres of diesel was pumped from each tank then we have that

Tank A holds = (x + 600 – 100) litres

= (x + 500)litres

Tank B holds = (x – 100)litres

Step 3: from the information given in the question, it implies that

(x + 500) = 3(x – 100)

Step 4: opening the brackets and solving for x, we have

x + 500 = 3x – 300

500 + 300 = 3x – x

2x = 800

x = \( \frac{800}{2} \)

x = 400litres

Step 5: thus tank A’s capacity is (x + 600) = 400 + 600

= 1000 litres

Therefore the capacity of each tank is as follows

Tank A = 1000 litres

Tank B = 400 litres.

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