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Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?
A
N16,200
B
N1,4,200
C
12,000
D
8000.
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
D = P(I - \(\frac{R}{100}\))n where P = N20,000; R = 10%, n = 2
D = 20,000(I - \(\frac{10}{100}\))2
= 20000(0.9)2
20,000 x 0.81 = N16,200
Source: Myschool Past Questions Archive (Record #36176)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
If x + y = 12 and 3x - y = 20, find the value of 2x - y
A
8
B
10
C
12
D
15
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x + y = 12
+ 3x - y = 20
---------------
4x = 32
x = 8
x + y = 12; 8 + y = 12; y = 12 - 8
= 4
2x - y = 2(8) - 4; 16 - 4 = 12
Source: Myschool Past Questions Archive (Record #36179)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
If loga270 - loga10 + loga \(\frac{1}{3}\) = 2, what is the value of a?
A
3
B
4
C
5
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
loga270 - loga10 + loga\(\frac{1}{3}\) = 2
loga270 + loga\(\frac{1}{3}\) - loga10 = 2
loga(270\(\frac{\frac{1}{3}}{10}\)) = 2
loga9 = 2
aa = 9
a = \(\sqrt{9}\) = 3
Source: Myschool Past Questions Archive (Record #36185)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
a conical water-jug is 7cm in diameter and 6cm deep. find the volume of water it can hold. [Take \(\pi = \frac{22}{7}\)]
A
22cm2
B
44cm3
C
77cm3
D
3082
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Volume = \(\frac{1}{3} \pi r^2h = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{6}{1}\)
V = \(\frac{6468}{84}\)
= 77cm3
Source: Myschool Past Questions Archive (Record #36187)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
If p = [\(\frac{Q(R - T)}{15}\)]\(^ \frac{1}{3}\), make T the subject of the relation
A
T = R + \(\frac{P^3}{15Q}\)
B
T = R - \(\frac{15P^3}{Q}\)
C
T = R + \(\frac{P^3}{15Q}\)
D
T = 15R - \(\frac{Q}{P^3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Cubic both sides; P3 = \(\frac{Q(R - T)}{15}\)
(cross multiplication) Q(R - T) = 15P3
(divide both sides by Q); R - T = 15\(\frac{1}{Q}\)
(subtract r from both sides) - T = \(\frac{15P^3}{Q - R}\)
T = R - \(\frac{15P^3}{Q}\)
Source: Myschool Past Questions Archive (Record #36189)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
A seller allows 20% discount for cash payment on the marked price of his goods. What is the ratio of the cash payment to the marked price
A
1:4
B
1:5
C
3:4
D
4:5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
(1 - \(\frac{20}{100}\)) = cash payment fraction= \(\frac{4}{5}\)
The ratio of the cash payment to the marked price = 4:5
Source: Myschool Past Questions Archive (Record #36190)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
If p = {1, 3, 5, 7, 9} and Q = {2, 4, 6, 8, 10} are subsets of a universal set. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What are the elements of P1 \(\cap\) Q1?
A
{1, 3, 5, 7, 9}
B
[2, 4, 6, 8, 10}
C
\(\varnothing\)
D
{1, 2, 6, 8, 10}
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
P1 = {2, 4, 6, 8, 10}; {1, 3, 5, 7, 9}
P1 \(\cap\) q1 = {} or \(\varnothing\)
Source: Myschool Past Questions Archive (Record #36192)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Which of the following is not a measure of centre tendency?
A
mean
B
mode
C
range
D
median
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #36197)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
The temperature in a chemical plant was -5oC at 2.00 am. The temperature fell by 6oC and then rose again by 7oC. What was the final temperature?
A
-6oC
B
-4oC
C
4oC
D
8oC
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Temperature = -5oC
It fell by 6oC = -5oC - 6oC
= -11oC
Then rose by 7oC = -11 + 7
= -4oC (final temperature)
Source: Myschool Past Questions Archive (Record #36199)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
Simplify \(\frac{20}{5\sqrt{28} - 2 \sqrt{63}}\)
A
\(\frac{5\sqrt{7}}{7}\)
B
\(\frac{5\sqrt{5}}{7}\)
C
\(\frac{7\sqrt{5}}{7}\)
D
\(\frac{7\sqrt{7}}{5}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{20}{5 \sqrt{4 \times 7} - 2\sqrt{9 \times 7}}\)
\(\frac{20}{10 \sqrt{7} - 6\sqrt{7}}\)
= \(\frac{20 \sqrt{7}}{4 \sqrt{7}}\)
= \(\frac{5\sqrt{7}}{7}\)
Source: Myschool Past Questions Archive (Record #36200)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Diagram
Solution
In the diagram, |QR| = 5cm, PQR = 60o and PSR = 45o. Find |PS|, leaving your answe in surd form.
Attached Figure / Diagram
A
4\(\sqrt{5}\)cm
B
3\(\sqrt{7}\)cm
C
4\(\sqrt{6}\)cm
D
5\(\sqrt{6}\)cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
tan 6o = \(\frac{|PR|}{|QR|}\)
\(\sqrt{3} = \frac{|PR|}{5}\)
= |PR| = \(5 \sqrt{3}\)cm
sin 45 = \(\frac{|PR|}{|PS|}\)
\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)
|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)
= 5\(\sqrt{6}\)cm
Source: Myschool Past Questions Archive (Record #36203)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Diagram
Solution
How many pupils took the test?
Attached Figure / Diagram
A
36
B
38
C
39
D
40
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No. of pupils = 1 + 3 + 4 + 7 + 8 + 6 + 6 + 4 + 1 = 40
Source: Myschool Past Questions Archive (Record #36211)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Diagram
Solution
If the pass mark was 4. What percentage of the pupils failed the test?
Attached Figure / Diagram
A
20.0%
B
37%
C
62.5%
D
80.0%
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No. of pupils that failed = 1 + 3 + 4 = 8
% failed = \(\frac{8}{40} \times 100\)% = 20.0%
Source: Myschool Past Questions Archive (Record #36212)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Diagram
Solution
In the diagram |XY| = 12cm, |XZ| = 9cm, |ZN| = 3cm and ZY||NM, calculate |MY|
Attached Figure / Diagram
A
3cm
B
4cm
C
5cm
D
6cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{|YM|}{|YX|} = \frac{|ZM|}{|ZX|}\)
\(\frac{|YM|}{12} = \frac{3}{9}\)
|YM| = 3 x \(\frac{12}{9}\)
|MY| = 4cm
Source: Myschool Past Questions Archive (Record #36215)
Mathematics
WAEC / WASSCE 2006
Number and Numeration
Solution
(a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
(b) By rationalising the denominator, simplify : \(\frac{7\sqrt{5}}{\sqrt{7}}\), leaving your answer in surd form.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
\(4\frac{2}{9} - 1\frac{13}{15} = \frac{38}{9} - \frac{28}{15}\)
= \(\frac{190 - 84}{45}\)
= \(\frac{106}{45}\)
\(2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3} = \frac{11}{5} + (\frac{4}{7} \times \frac{7}{3})\)
= \(\frac{11}{5} + \frac{4}{3}\)
= \(\frac{33 + 20}{15}\)
= \(\frac{53}{15}\)
\(\therefore \frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}} = \frac{106}{45} \div \frac{53}{15}\)
= \(\frac{106}{45} \times \frac{15}{53}\)
= \(\frac{2}{3}\)
(b) \(\frac{7\sqrt{5}}{\sqrt{7}}\)
= \(\frac{7\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\)
= \(\frac{7\sqrt{35}}{7}\)
= \(\sqrt{35}\)
Source: Myschool Past Questions Archive (Record #53010)