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
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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
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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
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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
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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
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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
24 𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑖𝑚𝑎𝑙𝑠=
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
2400 𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑖𝑚𝑎𝑙𝑠=
5 𝑝𝑖𝑔𝑠
24 𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑖𝑚𝑎𝑙𝑠
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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
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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
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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
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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
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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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 12 Peace Corps Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 13 Peace Corps Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 14 Peace Corps Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 15 Peace Corps Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 16 Peace Corps Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 17 Peace Corps Liberia
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.
Prepared by Peace Corps, Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 19 Peace Corps Liberia
(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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 20 Peace Corps Liberia
(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
Prepared by Peace Corps, Liberia
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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 22 Peace Corps Liberia
(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
2012 Liberia WAEC Mathematics SOLUTIONS Version 1.0 Page | 23 Peace Corps Liberia
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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