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Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
Simplify \(\frac{2}{2 + x} + \frac{2}{2 - x}\)
A
\(\frac{4}{4 - x^3}\)
B
\(\frac{8}{4 - x^2}\)
C
\(\frac{4x}{4 - x^2}\)
D
\(\frac{8 - 4x}{4 - x^2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2}{2 + x} + \frac{2}{2 - x}\)
\(\frac{2(2 - x) + 2(2 + x)}{(2 + x)(2 - x)} = \frac{4 - 2x + 4 + 2x}{4 - 2x + 2x - x^2}\)
= \(\frac{8}{4 - x62}\)
Source: Myschool Past Questions Archive (Record #36553)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
If y = \(\frac{(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular
A
\(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\)
B
\(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\)
C
\(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\)
D
\(\frac{\sqrt{9y^2 N - 3m}}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
y = \(\frac{(2\sqrt{x^2 + m})}{3N}\)
3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
Source: Myschool Past Questions Archive (Record #36558)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a
A
profit of 13%
B
loss of 12%
C
Profit of 10%
D
loss of 5%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
s.p = N6900
%profit = 15%
%profit = \(\frac{s.p - c.p}{c.p}\) x 100%
15% = \(\frac{6900 - c.p}{c.p}\) x 100%
\(\frac{15}{100}\)c.p = N6900 - c.p
0.15 c.p = N6900 - c.p
1.15c.p + c.p = N6900
c.p = \(\frac{6900}{1.15}\)
= 6000.00
Now new S.P = N6600
profit = s.p - c.p = 6000 - 6600
= 600
%profit = \(\frac{600}{6600}\) x 100%
= 10%
Source: Myschool Past Questions Archive (Record #36563)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
What is the value of x when y = 5?
y = \(\frac{1}{2}\) x + 1
A
8
B
9
C
10
D
11
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
when y = 5; x = ?; y = \(\frac{1}{2}\)x + 1
5 = \(\frac{1}{2}\)x + 1
5 - 1 = \(\frac{1}{2}\)x
4 = \(\frac{1}{2}\)x
x = 4 x 2
x = 8
Source: Myschool Past Questions Archive (Record #36584)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
The subtraction below is in base seven. Find the missing number.
5 1 6 2seven
-2 6 4 4seven
--------
2 * 1 5
--------
A
2
B
3
C
4
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
5 1 6 2seven
-2 6 4 4seven
--------
2 2 1 5
--------
the missing number is 2
Source: Myschool Past Questions Archive (Record #36585)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k
A
\(\frac{8}{3}\)
B
\(\frac{7}{3}\)
C
\(\frac{5}{3}\)
D
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x2 + kx + \(\frac{16}{9}\); Perfect square
But, b2 - 4ac = 0, for a perfect square
where a - 1; b = k; c = \(\frac{16}{9}\)
k2 - 4(1) x \(\frac{16}{9}\) = 0
k2 - \(\frac{64}{9}\) = 0
k2 = \(\frac{64}{9}\)
k = \(\sqrt{\frac{64}{9}}\)
k = \(\frac{8}{3}\)
Source: Myschool Past Questions Archive (Record #36591)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
If x km/h = y m/s, then y =
A
\(\frac{7}{18}\)x
B
\(\frac{11}{20}\)x
C
\(\frac{4}{15}\)x
D
\(\frac{5}{18}\)x
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x kmh-1 = y ms-1
\(\frac{x km}{1 hr}\) = y ms-1
\(x \times \frac{1km}{1hr}\) = y ms-1
\(x \times \frac{1000m}{60 \times 60s}\) = y ms-1
\(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms-1
\(x \times \frac{5}{18} ms^{-1}\)
\(x \times \frac{5}{18} ms^{-1}\) = y ms-1
y = \(\frac{5}{18}\)x
Source: Myschool Past Questions Archive (Record #36592)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Diagram
Solution
The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test?
Attached Figure / Diagram
A
100
B
85
C
80
D
70
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Pass mark = 50%
No. of students that passed = f50 + f65 + f80
= 45 + 25 + 15
= 85
Source: Myschool Past Questions Archive (Record #36599)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Diagram
Solution
The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50?
Attached Figure / Diagram
A
55\(\frac{1}{3}\)%
B
60%
C
65%
D
66\(\frac{2}{3}\)%
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\)
= \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100%
= \(\frac{120}{180}\) x 100%
= 66\(\frac{2}{3}\)%
Source: Myschool Past Questions Archive (Record #36600)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Diagram
Solution
In the diagram, QR//ST, /PQ/ = /PR/ and o. Find the value of y
Attached Figure / Diagram
A
105o
B
110o
C
130o
D
150o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\bigtriangleup\) PQR,
Q = S = 75o (Corresponding angle)
R = Q = 75o (Base angles of an isosceles \(\bigtriangleup\))
But, y + 75o = 180o (Sum of angles in a straight line)
y = 180 - 75
y = 105o
Source: Myschool Past Questions Archive (Record #36602)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Diagram
Solution
In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\)
Attached Figure / Diagram
A
\(\frac{KL}{JI}\)
B
\(\frac{HK}{JK}\)
C
\(\frac{JI}{KL}\)
D
\(\frac{HK}{LK}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ
o
Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\)
\(\frac{LH}{JH} = \frac{KL}{JI}\)
Source: Myschool Past Questions Archive (Record #36603)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Diagram
Solution
In the diagram, o and < POQ = 3x. some construction lines are shown. Calculate the value of x.
Attached Figure / Diagram
A
10o
B
11o
C
22o
D
35o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the diagram, OP bisects o
3x = 33o
x = \(\frac{33^o}{3}\)
= 11o
Source: Myschool Past Questions Archive (Record #36605)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) \(A \cap (B' \cap C')\) ;
(b) \((A \cup B) \cap (B \cup C)\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
\(U = {2, 3, 4, 5, 6, 7, 8, 9}\)
\(A = {2, 4, 6, 8}\)
\(B = {2, 3, 7, 9}\)
\(C = {4, 5, 6, 7, 8}\)
(a) \(A \cap (B' \cap C')\) = \({2, 4, 6, 8} \cap ({4, 5, 6, 8} \cap {2, 3, 9})\)
B' = 4,5,6,8
C' = 2,3
B' n C' = Ø
\(A \cap (B' \cap C')\) = \({2, 4, 6, 8} \cap { }\) Ø
= Ø
(b) \((A \cup B) \cap (B \cup C)
= {2, 3, 4, 6, 7, 8, 9} \cap {2, 3, 4, 5, 6, 7, 8, 9}\)
= \({2, 3, 4, 6, 7, 8, 9}\)
Source: Myschool Past Questions Archive (Record #52794)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
(a)
Curved Surface Area = \(\pi rl\)
\(115.5 = \frac{22}{7} \times r \times 10.5\)
\(115.5 = 33r\)
\(r = \frac{115.5}{33} = 3.5 cm\)
(b)
\(\therefore h^{2} + (3.50)^{2} = (10.5)^{2}\)
\(h^{2} = 10.5^{2} - 3.5^{2}\)
\(h^{2} = 98 \implies h = \sqrt{98}\)
\(h = 9.8994 cm \approxeq 9.90 cm\)
(c) Volume of a cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 9.90\)
= \(\frac{23.1 \times 11}{2}\)
= \(127.05 cm^{3} \approxeq 127 cm^{3}\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option .
Source: Myschool Past Questions Archive (Record #52811)
Mathematics
WAEC / WASSCE 2010
Number and Numeration
Solution
(a) Madam Kwakyewaa imported a quantity of frozen fish costing GH¢ 400.00. The goods attracted an import duty of 15% of its cost. She also paid a sales tax of 10% of the total cost of the goods including the import duty and then sold the goods for GH¢ 660.00. Calculate the percentage profit.
(b) In a school, there are 1000 boys and a number of girls. The 48% of the total number of students that were successful in an examination was made up of 50%of the boys and 40% of the girls. Find the number of girls in the school.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Import duty : \(\frac{15}{100} \times 400 = GH¢ 60.00\)
New cost price : GH¢400.00 + GH¢ 60 = GH¢ 460.00
Sales tax : \(\frac{10}{100} \times 460 = GH¢ 46.00\)
Total cost price : \(GH¢(460 + 46) = GH¢ 506.00\)
Profit = GH¢(660 - 506)
= GH¢ 154.00
\(% profit = \frac{154}{506} \times 100%\)
= \(30.43%\).
(b) Let the number of girls be t.
Therefore the total number of students = (1000 + t) students.
The number that passed the exam = \(\frac{48}{100} \times (1000 + t)\)
= \(0.48(1000 + t)\)
\(\implies 0.48(1000 + t) = (\frac{50}{100} \times 1000 + \frac{40}{100} \times t)\)
\(0.48(1000 + t) = 500 + 0.4t\)
\(480 + 0.48t = 500 + 0.4t \implies 0.48t - 0.4t = 500 - 480\)
\(0.08t = 20\)
\(t = \frac{20}{0.08} = 250\)
There are 250 girls in the school.
Source: Myschool Past Questions Archive (Record #52826)