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Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
If 3x\(^o\) 4(mod 5), find the least value of x
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
3x \(\equiv\) 4(mod 5)
In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.
Thus; 3x = 4 + 5 = 9
x = \(\frac{9}{3}\)
x = 3
Source: Myschool Past Questions Archive (Record #49402)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16
A
6
B
7
C
8
D
9
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(Q_1 = \frac{1}{4}\) (N + 1)th
\(\frac{1}{4} \times 12^{th}\) no.
= 3rd no (\(\cong\) 4)
\(Q_3 = \frac{3}{4}\) (N + 1)th
= \(\frac{3}{4}\) x 12th no.
= 9th no. (\(\cong\) 12)
Hence, interquartile range
= \(Q_3 - Q_1\)
= 12 - 4
= 8
Source: Myschool Past Questions Archive (Record #49403)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0
A
\(\frac{153}{250}\)
B
\(\frac{68}{111}\)
C
\(\frac{61}{100}\)
D
\(\frac{21}{33}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)
= \(\frac{612}{1000}\)
= \(\frac{153}{250}\)
Source: Myschool Past Questions Archive (Record #49425)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Diagram
Solution
In the diagram, which of the following ratios is equal to \(\frac{|PN|}{|PQ|}\)?
Attached Figure / Diagram
A
\(\frac{|PN|}{|PR|}\)
B
\(\frac{|PM|}{|PQ|}\)
C
\(\frac{|PM|}{|PR|}\)
D
\(\frac{|PR|}{|PQ|}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{|PN|}{|PQ|}\) = \(\frac{|PM|}{|PR|}\)
Source: Myschool Past Questions Archive (Record #49448)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
If P and Q are two statements, under what condition would p|q be false?
A
If p is true and q is true
B
If p is true and q is false
C
If p is false and q is false
D
If p is false and q is true
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the truth table above, p \(\to\) q would be false if p is true and q is false
Source: Myschool Past Questions Archive (Record #49455)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Diagram
Solution
The solution of x + 2 \(\geq\) 2x + 1 is illustrated
Attached Figure / Diagram
A
i
B
ii
C
iii
D
iv
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x + 2 \(\geq\) 2x + 1
x - 2x \(\geq\) 1 - 2
-x \(\geq\) -1
\(\frac{-x}{-1}\) \(\geq\) \(\frac{-1}{-1}\)
x \(\leq\) 1
Source: Myschool Past Questions Archive (Record #49535)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Diagram
Solution
The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m
Attached Figure / Diagram
A
90\(^o\)
B
60\(^o\)
C
45\(^o\)
D
30\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram, < WOZ = 180\(^o\) (angle on a straight line)
< WOX = < XOY = < YOZ
(|WX| = |XY| = |YZ|)
\(\frac{180^o}{3}\) = 60\(^o\)
= 60\(^o\)
M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)
< YOZ + 2m (base angles of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\)
2m = 180\(^o\) - 60\(^o\)
2m = 120\(^o\)
m = \(\frac{120^o}{2}\)
= 60\(^o\)
Source: Myschool Past Questions Archive (Record #49549)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Diagram
Solution
In the diagram, PQ is a straight line, (m + n) = 110\(^o\) and (n + r) = 130\(^o\) and (m + r) = 120\(^o\). Find the ratio of m : n : r
Attached Figure / Diagram
A
2 : 3 : 4
B
3 : 4 : 5
C
4 : 5 : 6
D
5 : 6 : 7
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
m + n = 110\(^o\), (n + r) = 130\(^o\)
(m + n) = 120\(^o\)
then, r = 130\(^o\) - n
and;
m + (130^o - n) = 120\(^o\)
m - n = -10\(^o\)
2m + (n + r) = 110 + 120 = 230
2m + 130 = 230
2m = 230 - 130
m = \(\frac{100}{2}\) = 50\(^o\)
n = 110\(^o\) - 50\(^o\)
= 60\(^o\)
r = 130\(^o\) - 60\(^o\) = 70\(^o\)
Hence, the ratio m : n : r
= 50 : 60 : 70
= 5 : 6 : 7
Source: Myschool Past Questions Archive (Record #49580)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Diagram
Solution
Donations during the launching of a church project were sent in sealed envolopes. The table shows the distribution of the amount of money in the envelope. How much was the donation?
Attached Figure / Diagram
A
N26,792.00
B
N26,972.00
C
N62.792.00
D
N62,972.00
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10
= 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20
= N62,972
Source: Myschool Past Questions Archive (Record #49584)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
M and N are two subsets of the universal set (U). If n(U) = 48, n(M) = 20, n(N) = 30 and n(MUN) = 40, find n(M \(\cap\) N)'
A
18
B
20
C
30
D
38
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let n(M \(\cup\) N \) = x
Then 20 - x + x + 30
- x = n(M \(\cup\) N)
50 - x = 40
50 - 40 = x
10 = x
x = 10
Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10)
= 8 + 10 + 20
= 38
Source: Myschool Past Questions Archive (Record #49632)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
The graph of y = x\(^2\) and y = x intersect at which of these points?
A
(0,0), (1,1)
B
(0,0), (0,1)
C
(1, 0), (0, 0)
D
(0, 0) (0, 0)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
y = x\(^2\) ....(1)
y = x ......(2)
y = y
x\(^2\) - x
x\(^2\) - x = 0
x(x - 1) = 0
x = 0 or x - 1 = 0
x = 0 or x = 1
when x = 0, y = 0\(^2\) = 0
when x = 1, y = 1\(^2\) = 1
Hence; the two graphs interest at (0, 0) and (1, 1)
Source: Myschool Past Questions Archive (Record #49641)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Beginning of 1st year (March 2011 - February 2012) - N900,000
Depreciation (30%) = \(\frac{30}{100} \times N900,000\)
= N270,000
Price at the end of 1st year = N(900,000 - 270,000)
= N630,000.
Beginning of 2nd year (March 2012 - February 2013) - N630,000
Depreciation (22%) = \(\frac{22}{100} \times N630,000\)
= N138,600
Price at the end of 2nd year = N(630,000 - 138,600)
= N491,400.
Beginning of 3rd year (March 2013 - February 2014) - N491,400
Depreciation (22%) = \(\frac{22}{100} \times N491,400\)
= N108,108
Price at the end of the 3rd year = N(491,400 - 108,108)
= N383,292.
Beginning of 4th year (March 2014 - February 2015) - N383,292
Depreciation (22%) = \(\frac{22}{100} \times N383,292\)
= N84,324.24
Price at the end of the 4th year = N(383,292 - 84,324.24)
= N298,967.76 \(\approxeq\) N299,000 (to the nearest hundred naira).
Source: Myschool Past Questions Archive (Record #52238)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).
(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{m + n + s + p + q}{5} = 12 \)
\(\implies m + n + s + p + q = 12 \times 5 = 60\)
\(\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\)
= \(\frac{60 + 13}{5} = \frac{73}{5}\)
= 14.6.
(b) In the community, there are 500 people.
The 75th percentile age = 65 years
Number of people in the 75th percentile = \(\frac{75 \times 500}{100} = 375\)
The 25th percentile = 15 years
Number of people in the 25th percentile = \(\frac{25 \times 500}{100} = 125\)
Number of people between 15 years and 65 years = 375 - 125 = 250 people.
Source: Myschool Past Questions Archive (Record #52250)
Mathematics
WAEC / WASSCE 2018
Number and Numeration
Solution
In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.
(a) Draw a Venn Diagram to illustrate this information.
(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
(b) From the Venn Diagram above,
60% passed implies that 40% passed
: 40% of 240 = 96.
28 + 2x + x + 12 + 6 + 6 + 8 = 96
60 + 3x = 96
3x = 36
x = 12
(i) Faulty brakes = 12 + 12 + 8 + 6 =38
(ii) Cars with only one fault = 28 + 2x + x
= 28 + 3x = 28 + 3(12)
= 28 + 36 = 64
Source: Myschool Past Questions Archive (Record #52253)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Diagram
Solution
Find the value of m in the diagram
Attached Figure / Diagram
A
72\(^o\)
B
68\(^o\)
C
44\(^o\)
D
34\(^o\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2x + m = 180
x + m = 112
x = 122 - m
2(112 - m) + m = 180
224 - 2m + m = 180
224 - m = 180
224 - 180 = m
m = 44\(^o\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #50188)