Senior High School Certificate Examination …...2011/05/10 · Prepared by Peace Corps, Liberia The West African Examinations Council Senior High School Certificate Examination MATHEMATICS

Prepared by Peace Corps, Liberia

The West African Examinations Council Senior High School Certificate Examination

MATHEMATICS May 2011 Exam Solutions

2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 1 Peace Corps Liberia

Name:__________________________

SOLUTIONS THE WEST AFRICAN EXAMINATIONS COUNCIL

Senior High School Certificate Examination

May 2011 MATHEMATICS 1 ½ hours

PAPER 1

OBJECTIVE TEST

Exam questions are listed on the left and solutions are given to the right. Please note these solutions are

intended as a guide and a study tool. They are not designed to teach you how to solve each problem.

Independent research and study will still be necessary if a concept is unfamiliar.

1. If P = { factors of 42 } and

Q = { factors of 54 }, find 𝑃 ∩ 𝑄.

A. {1, 3, 6}

B. {1, 2, 3, 6}

C. {1, 2, 3}

D. {1, 2, 6}

This question requires you to understand elements of set

theory and factorization. In order to find the factors in each

set, use prime factorization and create a factor tree of the

given number in each set.

Start with a prime factor

Stop when all factors are prime.

Multiply the prime factors to find all the numbers in the set.

Remember that one is always a factor:

𝑃 = { 1, 2, 3, 6, 7, 14, 21 }

Using the same method for factors of 54, we get:

𝑄 = { 1, 2, 3, 6, 9, 18, 27, 54 }

The symbol ∩ means intersection. Intersection means having

the same numbers in common. Therefore:

𝑃 ∩ 𝑄 = { 1, 2, 3, 6 }

301 S. H. S. C. E.

MAY 2011

MATHEMATICS

Objective and Essay Tests

2 ½ hours

1&2


2. In a given regular polygon, the ratio

of the sum of the exterior angles to

the interior angles is 1:3. How many

sides does the polygon have?

A. 5

B. 6

C. 7

D. 8

3. Solve for x in the equation

2(2𝑥+1) − 9(2𝑥) + 4 = 0

A. (−1,2)

B. (1,−2)

C. (1, 2)

D. (−1,−2)

This question requires you to know the Polygon Angle-Sum

Theorem and the Polygon Exterior Angle-Sum Theorem.

The former theorem is that the sum of the measures of the

interior angles of an 𝑛 − 𝑔𝑜𝑛 is (𝑛 − 2)180𝑜.

The latter theorem is that the sum of the measures of the

exterior angles of a polygon, one at each vertex, is 360𝑜.

1

3=

360o

(n − 2)180o

Set the ratio of the sums equal to the ratio of the formulas

(𝑛 − 2)180o = 3(360o) Cross-multiply

(𝑛 − 2) = 6 Divide both sides by 180𝑜 and simplify

𝑛 = 8 Add 2 to each side and simplify

This question requires understanding of factoring polynomial

equations and rules of exponents.

The fastest way to solve this equation may be to do trial-and-

error with a table of values using the answers provided.

Method 1: Table of Values

x 2(2𝑥+1) − 9(2𝑥) + 4 -2 1.875 -1 0 1 -6 2 0

Therefore our solutions are x = (-1, 2)

Method 2: Factoring

2(2𝑥+1) − 9(2𝑥) + 4 = 0 Original equation

(2𝑥)2(𝑥+1) − 9(2𝑥) + 4 = 0 Extract a (2𝑥) from the first

term

(2𝑥)2(𝑥+1) − 8(2𝑥) − 1(2𝑥) + 4 = 0

(2𝑥)2(𝑥+1) − 23(2𝑥) − 1(2𝑥) + 4 = 0

(2𝑥)2(𝑥+1) − 22(2𝑥+1) − 1(2𝑥) + 4 = 0

Split the second term into 2 terms, and factor 2(𝑥+1) out

of one of the terms.

(2𝑥+1 − 1)(2𝑥 − 4) = 0 Factor the equation

2𝑥+1 = 1, 2𝑥 = 4 Split the equation

𝑥 = (−1,2) Solve


4. A television set was sold for $300

which indicated a loss of 20%. Find

the cost price of the TV.

A. $275

B. $350

C. $375

D. $450

5. A vertical pole is 6m high. If the pole

casts a shadow 9m long at the same

time a tree casts a shadow 30m

long, find the height of the tree.

A. 54 𝑚

B. 30 𝑚

C. 20 𝑚

D. 15 𝑚

6. Simplify: 4+3𝑖

3−4𝑖

A. 25𝑖

B. 𝑖

C. −𝑖

D. −25𝑖

This question requires understanding of percentages, and

can be solved by writing and solving an algebraic equation.

Relate Cost price minus the product of cost price and percent

loss equals the selling price.

Define Cost price = p

Percent loss = 0.20

Selling price = $300

Write 𝑝 − 0.20𝑝 = $300

Solve 𝑝 = $375

This question assumes a linear relationship between the

changing height of an object and the change in its shadow

length, and that the objects are located near enough to each

other that the sun causes a similar enough shading effect to

use a direct proportionality to solve.

Define: x = height of the tree

Solve: 6𝑚

9𝑚=

𝑥

30𝑚

(9𝑚)(𝑥) = 180𝑚2

𝑥 = 20𝑚

This question requires understanding of the variable “i,”

which represents an imaginary number equivalent to √−1.

The expression can be simplified by recognizing how to apply

the “difference of two squares” formula.

Simplification Process

4 + 3𝑖

3 − 4𝑖 Original equation

(4 + 3𝑖

3 − 4𝑖) ( 3 + 4𝑖

3 + 4𝑖)

Multiply by 3+4𝑖

3+4𝑖

12 + 16𝑖 + 9𝑖 − 12

9 + 12𝑖 − 12𝑖 + 16

Multiply the numerators and

denominators across using “FOIL”

25𝑖

25 Simplify

𝑖 Simplify


7. Given that 𝑡𝑎𝑛𝜃 =5

12 and that

0𝑜 < 𝜃 < 90𝑜 , what is the value of

𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃?

A. 7

13

B. 8

13

C. 9

13

D. 12

13

This question can be answered using the trigonometric

ratios for a right triangle and the Pythagorean Theorem.

We can construct an appropriate right triangle using the

information given in the problem and the following

relationship (recall that ∠ is a symbol meaning “angle”):

𝑡𝑎𝑛𝜃 =5

12=

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑒𝑔 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∠𝜃

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑒𝑔 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠𝜃

The length of the hypotenuse (h), the longest side, can be

solved using the Pythagorean Theorem, where the square of

the hypotenuse is equal to the sum of the squares of each of

the legs.

Relate ℎ2 = (12)2 + (5)2

Simplify ℎ2 = 169

Solve √ℎ2 = √169

ℎ = 13

The two remaining ratios we need to solve the equation in

the problem are as follows:

𝑐𝑜𝑠𝜃 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑒𝑔 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 ∠𝜃

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=12

13

𝑠𝑖𝑛𝜃 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑒𝑔 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 ∠𝜃

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=5

13

Therefore:

𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃 =12

13−5

13=7

13

5

12

𝜽

h


8. Marie has some quarters while

Kollie has some nickels. If the

number of quarters is four more

than three times the number of

nickels and the total value of their

money is $5.80, find the number of

quarters.

A. 6

B. 8

C. 20

D. 22

9. If M and N are two intersecting sets

such that 𝑛(𝑀) = 20, 𝑛(𝑁) = 30

and 𝑛(𝑀 ∪ 𝑁) = 40, find 𝑛(𝑀 ∩ 𝑁).

A. 10

B. 20

C. 50

D. 60

10. Solve for x and y in log𝑥 𝑦 = 2 and 𝑥𝑦 = 8

A. (−2,4)

B. (2,4)

C. (−2,−4)

D. (2,−4)

This question requires knowledge of the value of a quarter

($0.25) and of a nickel ($0.05). Begin by defining variables,

and then write equations based on the relationships defined

in the problem. These equations will form a linear system of

equations, which may then be solved using either the

substitution or elimination (not shown) methods.

Define Number of quarters = q

Number of nickels = n

Write Equation 1:

𝑞 = 3𝑛 + 4

Equation 2:

$0.25𝑞 + $0.05𝑛 = $5.80

Rewrite Equation 1

𝑛 =𝑞 − 4

3

Substitute into Equation 2

$0.25𝑞 + $0.05 (𝑞 − 4

3) = $5.80

Solve 3($0.25𝑞) + $0.05(𝑞 − 4) = 3($5.80)

$0.75𝑞 + $0.05𝑞 − $0.20 = $17.40

$0.80𝑞 = $17.60

𝑞 = 22

This question requires understanding of set theory

notation. 𝑛(𝑀) refers to the number of items in set M, ∪

represents the union of sets, 𝑛(𝑀 ∪ 𝑁) means the number of

items in the union of both sets, counting each commonality

only once, and ∩ represents the intersection of the sets, which

includes the items both sets have in common, counted once.

Identity: 𝑛(𝑀 ∩ 𝑁) = 𝑛(𝑀) + 𝑛(𝑁) − 𝑛(𝑀 ∪ 𝑁)

Substitute: 𝑛(𝑀 ∩ 𝑁) = 20 + 30 − 40

Simplify: 𝑛(𝑀 ∩ 𝑁) = 10

This question involves properties of logarithms.

Write Equation 1:

log𝑥 𝑦 = 2

Equation 2: 𝑥𝑦 = 8

Rewrite Equation 1 𝑥2 = 𝑦

Substitute into Equation 2 𝑥(𝑥2) = 8

Solve 𝑥3 = 8

√𝑥33

= √83

𝑥 = 2

Substitute into Equation 2 (2)𝑦 = 8

𝑦 = 4


Sheep, 700

Cattle, 300 Pigs,

500

Rabbits, 900

11. Express 21

4% as decimal.

A. 0.25

B. 0.0225

C. 2.25

D. 2.75

The pie chart below shows the number of rabbits,

sheep, cattle and pigs on a farm. Use it to answer

questions 12 and 13.

12. What angle represents the number of sheep

on the farm?

A. 45𝑜

B. 60𝑜

C. 80𝑜

D. 105𝑜

13. What is the probability that an animal

selected by the farmer for a feast will be a

pig?

A. 1

8

B. 5

24

C. 7

24

D. 3

8

This question requires understanding of the relationship

between percentages and decimals, as well as converting

mixed fractions to decimals.

21

4%

Begin with the mixed fraction percentage

9

4%

Convert the improper fraction to a mixed fraction

2.25% Divide

0.0225 Move the decimal 2 places

The pie chart in the problem is not to scale. The pie chart

below is a more accurate representation of the data.

This question requires knowing that the rotation around a

full circle is represented by convention as 360𝑜. After finding

the total number of animals on the farm (2400), a direct

proportionality may be used to determine the angle (θ).

Solve: 700 𝑠ℎ𝑒𝑒𝑝

2400 𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑖𝑚𝑎𝑙𝑠=

𝜃

360𝑜

7 𝑠ℎ𝑒𝑒𝑝


𝜃

360𝑜

𝜃 = 105𝑜

This question can be solved by simply taking the number of

pigs, dividing that number by the total number of animals,

and reducing that fraction.

500 𝑝𝑖𝑔𝑠


5 𝑝𝑖𝑔𝑠

24 𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑖𝑚𝑎𝑙𝑠


14. Solve for x in: 4𝑥 + 2 [3 4−2 1

] = [10 04 2

]

A. [−2 10 2

]

B. [1 −22 0

]

C. [1 −2−2 0

]

D. [−1 22 0

]

15. Evaluate log10 √35+ log10 √2− log10√7

A. 1

3

B. −1

2

C. 1

3

D. 1

2

16. Simplify: (√4𝑎2𝑥3

)(√2𝑎𝑥3

)

A. 2𝑎√4𝑥3

B. 2𝑎√𝑥23

C. 2 √𝑎𝑥23

D. 2 √𝑎2𝑥

3

This question requires understanding how to solve a matrix

equation that includes a scalar product, subtraction, and

a scalar quotient.

Solve:

4𝑥 + 2 [3 4−2 1

] = [10 04 2

] Original equation

4𝑥 + [6 8−4 2

] = [10 04 2

] Multiply the matrix on the left by its scalar factor

4𝑥 = [10 04 2

] − [6 8−4 2

] Subtract, getting the term with the variable alone

4𝑥 = [4 −88 0

] Simplify [Note: the solution can be

determined by the sign of the matrix elements at this point.

4𝑥

4=[4 −88 0

]

4 Divide both sides by 4

𝑥 = [1 −22 0

] Simplify

This question requires understanding the properties of

logarithms.

Evaluate:

log10√35 + log10√2 − log10√7 Original equation

1

2(log10 35 + log10 2 − log10 7)

Use the power property to extract one half from every term

1

2(log10

35 ∗ 2

7)

Use the product and quotient properties to rewrite the

expression. 1

2(log10 10) Multiply and divide

1

2 Simplify, recognizing log10 10 = 1

This question requires understanding on how to simplify

radical expressions.

Simplify:

(√4𝑎2𝑥3

)(√2𝑎𝑥3

) Original equation

(413𝑎23𝑥13)(2

13𝑎13𝑥13)

Distribute the radicals as fractional exponents

(223𝑎33𝑥23)(2

13)

Combine factors with the same bases and make the base 4 factor

into a base 2 factor

(233𝑎𝑥

23) Simplify

2𝑎√𝑥23

Simplify


17. How many pounds of cornmeal costing 25

cents should be mixed with 15 pounds of

flour costing 16 cents to produce a mixture

of 22 cents a pound?

A. 33

B. 30

C. 22

D. 20

18. If 7𝑛 × 73 = 712, what is the value of

n?

A. 36

B. 15

C. 9

D. 4

19. In the diagram below, if 𝑙 ∥ 𝑚,

and 𝒓 = 91, then what is the value

for 𝒕 + 𝒖?

A. 178

B. 179

C. 180

D. 181

This question requires an algebraic equation to be built and

solved.

Relate and Define

1) Pounds of cornmeal (c) times its price per

pound is equal to the total cost of cornmeal (Tc)

2) Pounds of flour (f) times its price per pound is

equal to the total cost of flour (Tf)

3) c plus f, all times the mixture price per pound,

is equal to the sum of Tc and Tf

Write $0.25𝑐 = 𝑇𝑐 $0.16(15) = 𝑇𝑓

$0.22(𝑐 + 15) = 𝑇𝑐 + 𝑇𝑓

Substitute and Solve

$0.22(𝑐 + 15) = $0.25𝑐 + $0.16(15)

$0.22𝑐 + $3.3 = $0.25𝑐 + $2.4

$0.9 = $0.03𝑐

𝑐 = 30

This question requires understanding how to solve an

equation when it includes multiplying powers with the same

base.

Solve:

7𝑛 × 73 = 712 Original equation

𝑛 + 3 = 12 Property of multiplying powers

with the same base

n = 9 Subtract 3 from each side

This question requires understanding of some of the

geometric theorems or postulates, including those for:

supplementary angles, vertical angles, corresponding

angles, alternate exterior angles, and same-side exterior

angles.

Solve:

𝑟𝑜 + 𝑡𝑜 = 180𝑜 Same-side exterior angles

91𝑜 + 𝑡𝑜 = 180𝑜 Substitute

𝑡 = 89 Solve

𝑡𝑜 = 𝑢𝑜 Vertical angles

𝑢 = 89 Relate

Evaluate: 𝑡 + 𝑢

89 + 89

178


20. Write 1.02616 correct to 3

significant figures.

A. 1.02

B. 1.026

C. 1.0262

D. 1.03

21. If 𝑥+𝑦

𝑎−𝑏=2

3, then

9𝑥+9𝑦

10𝑎−10𝑏 is equal to

A. 9

10

B. 20

27

C. 2

3

D. 3

5

22. Change 235(6) to a base ten numeral.

A. 95

B. 99

C. 102

D. 107

23. Find the sum of the first five terms in the

progression 7, 2, - 3, ...

A. 85

B. 25

C. −15

D. −35

This question requires understanding of the first two rules of

significant figures and the process of rounding. Always begin

counting significant figures from the left.

First rule of significant figures: All non-zero digits are significant.

Second rule of significant figures: All zeroes appearing between non-

zero digits are significant.

Counting three from the left, we reach 1.02. Using rules of rounding, we

then look at the fourth digit. If it is 5 or greater, the number in the

hundredth’s place increases by one (2 goes to 3). If the number is less

than 5, then the 2 will remain. The next digit after 1.02 is 6; 1.026

rounds to 1.03.

This question requires understanding of the distributive

property and algebraic substitution.

9𝑥 + 9𝑦

10𝑎 − 10𝑏 Original expression

9(𝑥 + 𝑦)

10(𝑎 − 𝑏)

Use the Distributive Property to

factor out a 9 from the numerator

and a 10 from the denominator

9(2)

10(3) Replace

𝑥+𝑦

𝑎−𝑏 with

2

3

18

30 Simplify

3

5 Reduce

This question requires understanding of how to change

number bases.

235(6) can be written 2 ∗ 62 + 3 ∗ 61 + 5 ∗ 60 which simplifies, when

added together using the base ten system, to 95.

This question can be solved by recognizing the given

progression is arithmetic, and that each number after the

first is 5 less than the previous number.

The first 5 terms are: 7, 2, -3, -8, and -13

The sum of these numbers is -15


24. How many sides are there in a

hexagon?

A. 9

B. 8

C. 7

D. 6

25. How many liters are there in a cubic

centimeter?

A. 10

B. 1

C. 0.1

D. 0.001

26. Express 2.35 × 10−4 as a decimal

number.

A. 23,500

B. 2,350

C. 0.000235

D. 0.0000235

27. An angle whose vertex is a point on

a circle is called

A. 𝑎 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑎𝑛𝑔𝑙𝑒.

B. 𝑎𝑛 𝑖𝑛𝑠𝑐𝑟𝑖𝑏𝑒𝑑 𝑎𝑛𝑔𝑙𝑒.

C. 𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑔𝑙𝑒.

D. 𝑎𝑛 𝑎𝑟𝑐.

This question requires the test taker to have memorized the

prefixes used to describe the number of sides on a polygon.

Hexa means 6

This question requires the test taker to have memorized unit

conversion between liters and volume measured in metric

cubic length.

There are 1000 cubic centimeters(𝑐𝑚3) in one liter(𝐿),

therefore 1 𝐿

1,000 𝑐𝑚3= 0.001 𝐿 𝑝𝑒𝑟 𝑐𝑚3

This question requires understanding how to transition

between scientific and standard notations.

2.35 × 10−4 Original expression

. ⏟ ⏟ ⏟ 2⏟ . 35 × 10−4 The exponent is negative four, so

move the decimal left 4 places.

0.000235 Fill the blank moved spaces with

zeroes, put a zero before the

decimal, and remove the power.

This question requires understanding geometric vocabulary

given a definition. The definition of each of the answers is as

follows:

Central angle: an angle whose vertex is the center of a circle

Inscribed angle: an angle whose vertex is a point on a circle

Vertical angle: an angle whose sides are opposite rays of

another angle

Arc: part of a circle


28. . If x is a positive integer satisfying 𝑥7 = 𝑘

and 𝑥9 = 𝑚, which of the following must

be equal to 𝑥11?

A. 𝑚2

𝑘

B. 𝑚2 − 𝑘

C. 𝑚2 − 7

D. 2𝑘 −𝑚

3

29. Find the distance between the points (-3, 2)

and (5, -4).

A. 100

B. 64

C. 36

D. 10

30. Solve for x in √4𝑥 − 11= 2√x − 1.

A. 9

B. 5

C. −5

D. −9

This problem requires understanding of multiplication and

division of powers with the same base. 𝑥9 becomes 𝑥11 by

multiplying it by 𝑥2. 𝑥2 can be obtained by dividing 𝑥9 by 𝑥7.

Evaluate:

𝑥9

𝑥7=𝑚

𝑘= 𝑥2

𝑥11 = 𝑥9 ∗ 𝑥2 = 𝑚 ∗𝑚

𝑘=𝑚2

𝑘

This problem requires deriving or having memorized the

distance formula for finding the distance between two points

located on a coordinate plane. The distance formula is based

on the Pythagorean Theorem.

Relate 𝐷 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

Define 𝑥1 = −3, 𝑦1 = 2, 𝑥2 = 5, 𝑦2 = −4

Write 𝐷 = √[5 − (−3)]2 + [(−4) − 2]2

Solve 𝐷 = √[8]2 + [−6]2

𝐷 = √100 = 10

This problem requires understanding how to solve a single

variable algebraic equation involving radicals. Remember,

FOIL is for multiplying two two-term expressions together. It

stands for first, outer, inner, and last.

Solve:

√4𝑥 − 11= 2√x − 1 Original equation

(√4𝑥 − 11)2= (2√x − 1)

2 Square both sides

4𝑥 − 11= (2√x − 1)(2√x − 1) Expand

4𝑥 − 11 = 4x− 2√x − 2√x + 1 Use FOIL

4√x= 12 Bring each term with an x to

the left, and constants right

√x= 3 Divide both sides by 4 and simplify

x = 9 Square both sides and simplify


The diagram below shows the scores of a group

of 12th

graders on a Mathematics test. Use it to

answer questions 31 to 34.

Score 0 1 2 3 4 5

Frequency 1 2 4 4 7 2

31. Find the mode of the scores.

A. 2

B. 3

C. 4

D. 5

32. Find the median score.

A. 2

B. 3

C. 4

D. 5

33. How many students took this test?

A. 20

B. 15

C. 6

D. 5

34. What is the mean score?

A. 5

B. 4

C. 3

D. 2

These questions require understanding of the statistical

measures of central tendency given a frequency table

which shows how many times the same data point occurs in

the data set.

The mode is the most frequently occurring value, which is a

score of 4 in this data set.

The median score is the middle value or mean of the two

middle values.

The total number of data points, given the sum of the numbers in the

frequency chart, is 20. For even totals, there will be two middle values,

in this case the 10th and 11th scores, counting up from the lowest test

scores. These two values will be 3 and 3; the mean of these is 3.

See the previous answer.

The mean score is found by taking the sum of the data values

and dividing by the number of data values. The sum is

calculated by adding together each score multiplied by its

frequency.

0(1) + 1(2) + 2(4) + 3(4) + 4(7) + 5(2)

20= 3


35. Which of the following is not a factor of 80?

A. 5

B. 8

C. 12

D. 16

Use the equation below to answer questions 36

and 37.

𝑀 =𝐾

√1 −𝑉2

𝑃2

36. Find 𝑀 if 𝐾 = 9, 𝑃 = 5 and 𝑉 = 4.

A. 14

B. 15

C. 16

D. 17

37. Make V the subject of the equation.

A. 𝑉 =𝑃

𝑀√𝑀2 + 𝑃2

B. 𝑉 =𝑃

𝑀√𝐾2 +𝑀2

C. 𝑉 =𝑃

𝑀√𝑀2 −𝐾2

D. 𝑉 =𝑃

𝑀√𝑃2 − 𝐾2

This question requires understanding of multiplication tables

up to 12 (by process of elimination, 12 does not multiply by

any whole numbers to equal 80), or of how to make a factor

tree, which is shown in problem 1.

This question requires understanding of the order of

operations. Replace K, P, and V in the equation with their

given values, then simplify.

𝑀 =9

√1 −42

52

Original equation with values

𝑀 =9

√1 −1625

Evaluate exponents

𝑀 =9

√0.36

Divide, then subtract, under

the radical

𝑀 =9

0.6 Evaluate the radical

𝑀 = 15 Divide

This question requires understanding of algebraic equation

manipulation that includes exponents.

(𝑀)2 =

(

𝐾

√1−𝑉2

𝑃2)

2

Square both sides

𝑀2 (1 −𝑉2

𝑃2) = 𝐾2

Simplify and multiply both sides by the

denominator on the right side

−𝑃2 +𝑉2 =𝐾2

𝑀2(−𝑃2)

Divide both sides by 𝑀2 and multiply

both sides by −𝑃2

𝑉2 =𝐾2

𝑀2(−𝑃2) + 𝑃2 Add 𝑃2 to each side

√𝑉2 = √𝑃2

𝑀2(−𝐾2 +𝑀2)

Factor out a 𝑃2

𝑀2 and take the square root

of each side

𝑉 =𝑃

𝑀√𝑀2 − 𝐾2 Simplify


38. Solve |−2𝑥 + 1| ≤ 1.

A. −1 ≤ 𝑥 ≤ 1

B. −1 ≤ 𝑥 ≤ 0

C. 0 ≤ 𝑥 ≤ 1

D. 0 ≤ 𝑥 ≤ −1

39. Of the 6 courses offered by the music

department in her school, Musulyn must

choose exactly two of them. How many

different combinations of two courses are

possible for Musulyn if there are no

restrictions on which two courses she can

choose?

A. 12

B. 15

C. 24

D. 30

40. If the function of 𝑓 is defined as

𝑓(𝑥) = 𝑥 2 − 7𝑥 + 10 and 𝑓(𝑡 + 1) = 0,

what are the possible values of 𝑡 ?

A. (1,4)

B. (−1,4)

C. (1,−4)

D. (−1,−4)

This question requires understanding how to solve an

algebraic equation containing an absolute value and an

inequality.

Properties of Absolute Value Inequalities

Let k represent a positive real number. |𝑥| ≥ 𝑘 is equivalent to 𝑥 ≤ −𝑘 𝑜𝑟 𝑥 ≥ 𝑘. |𝑥| ≤ 𝑘 is equivalent to −𝑘 ≤ 𝑥 ≤ 𝑘.

Solve:

−1 ≤ −2𝑥 + 1 ≤ 1

−2 ≤ −2𝑥 ≤ 0

1 ≥ 𝑥 ≥ 0

Rewrite as compound inequality

Subtract 1 from each side

Divide each side by -2, switching the inequalities

This problem requires understanding of combination

selection and factorials. The number of combinations of n

items chosen r at a time can be calculated using the

following formula:

𝐶𝑛 𝑟 =𝑛!

𝑟! (𝑛 − 𝑟)! 𝑓𝑜𝑟 0 ≤ 𝑟 ≤ 𝑛

For 6 courses chosen 2 at a time, Musulyn’s number of

choices is calculated as follows:

𝐶6 2 =6!

2! (6 − 2)!=720

2(24)= 15

This question requires understanding how to factor a

quadratic expression.

𝑓(𝑥) = 𝑥 2 − 7𝑥 + 10 Original function

𝑓(𝑥) = (𝑥 − 5)(𝑥 − 2) Factored function using FOIL

𝑓(𝑡 + 1) = (𝑡 + 1 − 5)(𝑡 + 1 − 2) Replace 𝑥 with 𝑡 + 1

0 = (𝑡 − 4)(𝑡 − 1) Replace 𝑓(𝑡 + 1) with 0 and

simplify.

𝑡 = 4, 1 Solve


41. What is the area of the trapezoid below?

A. 80

B. 120

C. 240

D. 360

42. What is the least common multiple of 6, 8

and 12?

A. 24

B. 48

C. 72

D. 144

43. Find the simple interest on $10,000.00 at

an interest rate of 20% for 6 months.

A. $1,000.00

B. $2,000.00

C. $3,000.00

D. $4,000.00

The formula for the area of a trapezoid is

𝐴 =1

2ℎ(𝑏1 + 𝑏2) ,

where h is the height, and 𝑏1 and 𝑏2 are the top and bottom

lengths, respectively, which are parallel line segments.

Assuming the side of length 20 (𝑏2) and the side of length 15

(h) have an interior angle of 90 degrees, we can use

geometry to split the trapezoid into a right triangle and a

rectangle to determine the length of the base.

The length of the base can be calculated using

𝑏1 = 20 + √172 − 152 = 28.

Plugging our values into the area equation, we get

𝐴 =1

2(15)(28 + 20) = 360

This question requires understanding of multiplication. The

least common multiple (LCM) can be determined by

comparing a table of multiples.

6 12 18 24

8 16 24

12 24

This question requires understanding how to calculate simple

interest. The formula is as follows:

𝐼 = 𝑝𝑟𝑡, where

I is the simple interest,

p is the principal, or initial amount,

r is the interest rate, and

t is the time in years (note that the time given is in months).

Putting our data in the formula yields:

𝐼 = $10,000.00 ∗ 0.20 ∗6 𝑚𝑜𝑛𝑡ℎ𝑠

12𝑚𝑜𝑛𝑡ℎ𝑠𝑦𝑒𝑎𝑟

= $1,000.00


44. If −3 is one of the roots of 2𝑟2 + 𝒃𝑟 − 3 = 0, what is the value

for 𝒃?

A. 5

B. 4

C. −4

D. −5

45. In a class of 80 students, every student had

to study Economics, Geography or both. If

65 students studied Economics and 50

studied Geography, how many students

studied both subjects?

A. 15

B. 30

C. 35

D. 45

46. In the diagram below, ST and QR are

parallel. /PS/ = 6 𝑐𝑚, /SQ/ = 8 𝑐𝑚 and

/PR/ = 182

3𝑐𝑚 Find /𝑷𝑻/.

A. 7 𝑐𝑚

B. 8 𝑐𝑚

C. 9 𝑐𝑚

D. 10 𝑐𝑚

This question requires understanding of quadratic

factorization.

2𝑟2 + 𝒃𝑟 − 3 = 0 Original equation

1. (2𝑟 − 1)(𝑟 + 3) = 0

2. (2𝑟 + 6) (𝑟 −1

2) = 0

The two possible factorizations

where -3 is a root

2𝑟2 + 6𝑟 − 𝑟 − 3 = 0 Both equations simplified

using FOIL

2𝑟2 + 5𝑟 − 3 = 0 Simplify

𝑏 = 5

This question is most easily solved by setting variables

solving a system of equations (∴ means “therefore”).

Define Students that study Economics only = E

Students that study Geography only = G

Students that study both = EG

Write 𝐸 + 𝐺 + 𝐸𝐺 = 80

𝐸 + 𝐸𝐺 = 65, ∴ 𝐸 = 65 − 𝐸𝐺

𝐺 + 𝐸𝐺 = 50, ∴ 𝐺 = 50 − 𝐸𝐺

Solve 65 − 𝐸𝐺 + 50 − 𝐸𝐺 + 𝐸𝐺 = 80

−𝐸𝐺 + 115 = 80

𝐸𝐺 = 35

This question requires understanding of similar triangles and

their proportionalities according to the Side-Splitter

Theorem.

Side-Splitter Theorem: If a line is parallel to one side of a

triangle and intersects the other two sides, then it divides

those sides proportionally.

Therefore, based on the triangle, 𝑃𝑆

𝑆𝑄=𝑃𝑇

𝑇𝑅 𝑎𝑛𝑑

𝑃𝑄

𝑃𝑆=𝑃𝑅

𝑃𝑇.

Plugging in our values (where 182

3𝑐𝑚 =

56

3𝑐𝑚),

6 𝑐𝑚 + 8 𝑐𝑚

6 𝑐𝑚=

563 𝑐𝑚

𝑃𝑇

14 𝑐𝑚(𝑃𝑇) =6 ∗ 56

3 𝑐𝑚2

𝑃𝑇 = 8 𝑐𝑚


47. If y varies directly as the square of x and

inversely as z and y = 2, when x = 5 and z =

100, find y when x =3 and z = 4.

A. 18

B. 16

C. 10

D. 8

48. Simplify log√3 27√3

A. 5

B. 3

C. 7

D. 9

49. A sack of grain can feed 80 chickens for 18

days. How long will this sack of grain last for

120 chickens?

A. 9 𝑑𝑎𝑦𝑠

B. 12 𝑑𝑎𝑦𝑠

C. 13 𝑑𝑎𝑦𝑠

D. 28 𝑑𝑎𝑦𝑠

50. If 𝑐𝑜𝑠 𝑥 is negative and 𝑠𝑖𝑛 𝑥 is also negative,

which of the following is true?

A. 0𝑜 < 𝒙 < 90𝑜

B. 90𝑜 < 𝒙 < 180𝑜

C. 180𝑜 < 𝒙 < 270𝑜

D. 270𝑜 < 𝒙 < 360𝑜

END OF OBJECTIVE TEST

This question would be better written: If y varies directly as the square of x and inversely with z, and y = 2

when x = 5 and z = 100, find y when x =3 and z = 4.

𝑦 =𝑘𝑥2

𝑧

Plugging in our values yields: 2 =𝑘(5)2

100, meaning k=8, so

𝑦 =8(3)2

4= 18

This question requires understanding of logarithms.

Evaluate:

log√3 27√3 Original equation

log√3 33312 Rewrite the term inside the logarithm

log√3 372 = 𝑥

Simplify the product inside the logarithm and set the whole expression equal to 𝑥

312𝑥 =3

72

Use the definition of the logarithm to rewrite the equation

1

2𝑥 =

7

2 Equate the exponents

𝑥 = 7 Solve

This question requires a direct proportionality to solve. Let x

equal the number of days the grain sack will last.

80 𝑐ℎ𝑖𝑐𝑘𝑒𝑛𝑠 ∗ 18 𝑑𝑎𝑦𝑠 = 120 𝑐ℎ𝑖𝑐𝑘𝑒𝑛𝑠 ∗ 𝑥 𝑑𝑎𝑦𝑠

𝑥 = 12 𝑑𝑎𝑦𝑠

Using the polar coordinate system, 𝑐𝑜𝑠 𝑥 gives the equivalent

x-component of an ordered pair and 𝑠𝑖𝑛 𝑥 gives the

equivalent y-coordinate. If both are negative, the point lies in

the third quadrant of the polar coordinate plane,

where 180𝑜 < 𝒙 < 270𝑜.


PAPER 2 1 ½ hours

ESSAY

[60 Marks]

Paper 2 consists of seven questions divided into two sections, A & B. Section A has four compulsory questions and section B has

three questions for which you are required to answer any two.

Write your answers in ink (blue or black) only.

For each question, all necessary details of working including diagrams must be shown with the answer.

Credit will be given for clarity of expression and orderly presentation of material.

NOTE: Make sure you understand these directions! The majority of students perform poorly on this section because they

do not answer enough questions. Solutions are provided here with the amount of detail and ‘clarity of expression’ you

should include on the actual exam.

SECTION A

COMPULSORY

[36 marks]

Answer all the questions in this section.

1. (a) Solve for x and y in 2𝑥 + 6𝑦 = 10 and

3𝑥−2𝑦 =1

27

This question requires you to use tools of algebra

and substitutions to solve for x and y.

3𝑥−2𝑦 = 3−3 Rewrite the second equation

𝑥 − 2𝑦 = −3 Equate the exponents

3𝑥 − 6𝑦 = −9 2𝑥 + 6𝑦 = 10

Multiply the previous equation by 3

and write it above the first equation

5𝑥 = 1 Add the equations (elimination)

𝑥 =1

5 Solve

2 (1

5) + 6𝑦 = 10

Substitute the value for x into the first equation.

6𝑦 =50

5−2

5=48

5

Subtract 2

5 from each side and

simplify.

𝑦 =8

5 Solve


(b) Find P if

2

𝑥−3−

3

𝑥−2 is expressed as

𝑃

(𝑥−3)(𝑥−2).

2. (a) Write the quadratic equation whose roots

have a sum of 3

2 and a product of 3.

(b) The probability that an event A occurs is 1

5 and

the probability that event B occurs is 1

3. If the

two events are independent, find the probability that at least one of the events occurs.

3. (a) If (𝑎 − 3) is one of the factors of

𝑎2+ 14𝑎 − 51, find the other factor.

The only way to combine the two fractions is by

giving them the same base. This can be done as

follows:

This question is impossible.

For two independent events, the probability of at

least one of the events occurring is the sum of the

probabilities of occurrence of each event.

𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) =1

5(3

3) +

1

3(5

5) =

8

15

This second term in the unknown factor, when

added to the second term in the first factor, gives

14, and when multiplied by the second term in the

first factor, gives -51.

Using x as the second term in the unknown factor:

𝑥 − 3 = 14

𝑥 = 17

Multiplying 17 by -3 gives -51, which is correct,

therefore, our other factor is

(𝑎 + 17)

2

𝑥 − 3−

3

𝑥 − 2 Original equation

2

𝑥 − 3(𝑥 − 2

𝑥 − 2)−

3

𝑥 − 2(𝑥 − 3

𝑥 − 3)

Multiply the left and right

fractions by fractions equal to

one that will give them the

denominator of the second

expression in the question.

2𝑥 − 4

(𝑥 − 3)(𝑥 − 2)−

3𝑥 − 9

(𝑥 − 3)(𝑥 − 2) Simplify

−𝑥 + 5

(𝑥 − 3)(𝑥 − 2) Subtract

𝑃 = −𝑥 + 5 𝑜𝑟 5 − 𝑥 Relate


(b) A doctor has 20 ounces of a 30% Argyrols solution. How many ounces of a 40% solution should he add to produce a 32% Argyrols solution?

4. (a) Find the value of 3𝑛+2 given that 3𝑛 = 𝑝

(b) Simplify 3√28 − 5√63+ 4√112

This question can be solved by creating an equation

based on the information provided.

If x is equal to the ounces (oz) of 40% Argyrols

solution, then

20 𝑜𝑧 ∗ 0.30 + 𝑥 ∗ 0.40 = (20 + 𝑥 𝑜𝑧) ∗ 0.32

6 + 𝑥 ∗ 0.40 = 6.4 + 0.32 ∗ 𝑥 𝑜𝑧

𝑥 ∗ 0.08 = 0.4

𝑥 = 5 𝑜𝑧

This problem requires understanding the properties

of exponents.

3𝑛+2 can be written 3𝑛 ∗ 32.

Since 3𝑛 = 𝑝, we can rewrite the previous expression as

𝑝 ∗ 32, so

3𝑛+2 = 9𝑝.

This question requires understanding how to

simplify radical expressions.

Simplify:

3√28 − 5√63+ 4√112 Original equation

3√4 ∗ 7 − 5√9 ∗ 7 + 4√16 ∗ 7 Factor a 7 from under each radical

3 ∗ 2√7− 5 ∗ 3√7 + 4 ∗ 4√7 Take the square root of each of the numbers that form a

perfect square

6√7 − 15√7+ 16√7 Simplify the multiplication

(6 − 15 + 16)√7 Factor out a √7

7√7 Simplify


SECTION B

[24 marks]

Answer any two questions in this section.

5. In the diagram, /PS/ = 20cm, and ∠𝑆𝑃𝑅 = 60𝑜 .

(a) Calculate /SR/.

(b) Find the value of QR.

(Note: The provided drawing is not to scale)

This question requires determining or having

memorized the ratio of the sides generated by

putting the given angle into one of the primary

trigonometric functions.

Knowing that sin(60𝑜) =√3

2, and that

sin(60𝑜) =𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 60𝑜

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 , we can set

an equation:

√3

2=

/SR/

20 𝑐𝑚

∴ /SR/ = 10√3 𝑐𝑚

Using the Pythagorean Theorem, /PR/ can be

determined. Subtracting PQ yields the value for QR.

(20 𝑐𝑚)2 = (10√3 𝑐𝑚)2 + (/𝑃𝑅/)2

(/𝑃𝑅/)2 = 400 𝑐𝑚2 − 300 𝑐𝑚2

/𝑃𝑅/ = 10 𝑐𝑚

𝑄𝑅 =/𝑃𝑅/−𝑃𝑄 = 10 𝑐𝑚 − 6 𝑐𝑚 = 4 𝑐𝑚


(c) Find the area of triangle PRS.

6. (a) The first term of an arithmetic progression is −8. If the ratio of the 7𝑡ℎ and 9𝑡ℎ term is 5:8, find the common difference of the progression.

(b) Mrs. Peters invested $8,000.00 in bonds and savings. If the bonds yielded an interest of 4% and the savings an interest of 5%, how much was invested in savings if the income from both investments was $380.00?

This question requires knowing how to determine

the area of a triangle given the length of the sides.

The formula for the area of a triangle is:

𝐴𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =1

2𝑏 ∗ ℎ,

where b is the length of the base, in this case /PR/,

and h is the height, in this case /SR/.

𝐴𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑃𝑅𝑆 =1

210 𝑐𝑚 ∗ 10√3 𝑐𝑚 = 50√3 𝑐𝑚2

This question can most easily be solved by using the

formula for the last term of an arithmetic

progression, which is

𝐴(𝑛) = 𝑎 + (𝑛 − 1)𝑑,

where A(n) is the nth term,

a is the first term,

n is the term number, and

d is the common difference.

Based on the question, 𝐴(7) = −8 + (7 − 1)𝑑, and

𝐴(9) = −8 + (9 − 1)𝑑, and 𝐴(7)

𝐴(9)=5

8 .

∴ 5

8=−8 + 6𝑑

−8 + 8𝑑

Solve:

−40 + 40𝑑 = −64 + 48𝑑 Cross multiply

24 = 8𝑑 Collect variable terms on one side and constants on the other

𝑑 = 3 Divide

This question can be solved by formulating and

solving an appropriate equation.

Define Bond investment = b

Savings investment = s

Write 𝑠 + 𝑏 = $8,000.00, or 𝑏 = $8,000.00 − 𝑠 0.04𝑏 + 0.05𝑠 = $380.00 Combined: 0.04($8,000.00 − 𝑠) + 0.05𝑠 =$380.00

Solve 0.04($8,000.00 − 𝑠) + 0.05𝑠 = $380.00

$320.00 − 0.04𝑠 + 0.05𝑠 = $380.00

$320.00 + 0.01𝑠 = $380.00

0.01𝑠 = $60.00

𝑠 = $6,000.00


7. R is the midpoint of 𝑃𝑇̅̅ ̅̅ , and Q is the midpoint

of 𝑃𝑅̅̅ ̅̅ . If S is a point between R and T such that

the length of 𝑄𝑆̅̅ ̅̅ is 10 and the length of 𝑃𝑆̅̅̅̅ is 19,

find the length of 𝑆𝑇̅̅̅̅ with the help of a diagram not drawn to scale.

The diagram may be drawn as follows:

Given the information provided in the problem,

𝑃𝑅̅̅ ̅̅ =1

2𝑃𝑇̅̅̅̅ and 𝑃𝑄̅̅ ̅̅ ̅ =

1

2𝑃𝑅̅̅ ̅̅ , so

𝑃𝑄̅̅ ̅̅ =1

4𝑃𝑇̅̅̅̅ .

The problem gives us PS͞ and QS͞, from which we can

determine PQ͞.

𝑃𝑄̅̅ ̅̅ = 𝑃𝑆̅̅̅̅ − 𝑄𝑆̅̅̅̅ =1

4𝑃𝑇,

Substituting the values in the problem yields

19 − 10 =1

4𝑃𝑇̅̅̅̅ .

Solve:

9 =1

4𝑃𝑇̅̅ ̅̅ Simplify the left side

𝑃𝑇̅̅̅̅ = 36 Multiply both sides by 4

Finally, ST͞ can be calculated as follows:

𝑆𝑇̅̅̅̅ = 𝑃𝑇̅̅̅̅ − 𝑃𝑆̅̅̅̅ = 36 − 19 = 17

END OF PAPER

R P T Q S

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Senior High School Certificate Examination …...2011/05/10 · Prepared by Peace Corps, Liberia The West African Examinations Council Senior High School Certificate Examination MATHEMATICS