Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 586–600 of 4910 verified questions
Sort by:
Mathematics
UTME / JAMB 2021
Probability
Solution
Find the probability that a number selected at random from 41 to 56 is a multiple of 9
A
\(\frac{1}{8}\)
B
\(\frac{2}{15}\)
C
\(\frac{3}{16}\)
D
\(\frac{7}{8}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given from 41 to 56
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56
The nos multiple of 9 are: 45, 54
P(multiple of 9) = \(\frac{2}{16}\)
= \(\frac{1}{8}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #60704)
Mathematics
WAEC / WASSCE 2021
Probability
Solution
A fair die is tossed twice what is the probability of get a sum of at least 10.
A
\(\frac{5}{36}\)
B
\(\frac{2}{3}\)
C
\(\frac{5}{18}\)
D
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)
From the table above, event space, n(E) = 6
sample space, n(S) = 36
Hence, probability sum of scores is at least 10, is;
\(\frac{n(E)}{n(S)}\)
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #61463)
Mathematics
UTME / JAMB 2021
Number and Numeration
Solution
In how many ways can 2 students be selected from a group of 5 students in a debating competition?
A
25 ways
B
10 ways
C
15 ways
D
20 ways
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(In\hspace{1mm} ^{5}C_{2}\hspace{1mm}ways\hspace{1mm}\)
=\(\frac{5!}{(5-2)!2!}\)
=\(\frac{5!}{3!2!}\)
=\(\frac{5\times4\times3!}{3!\times2\times1}\)
=10 ways.
Source: Myschool Past Questions Archive (Record #60619)
Mathematics
UTME / JAMB 2021
Number and Numeration
Solution
What is the rate of change of the volume V of a hemisphere with respect to its radius r when r = 2?
A
8π
B
16π
C
2π
D
4π
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(V = \frac{2}{3} \pi r^{3}\)
\(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\)
\(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\)
= \(8\pi\)
Source: Myschool Past Questions Archive (Record #60623)
Mathematics
UTME / JAMB 2021
Geometry and Trigonometry
Solution
determine the maximum value of y=3x\(^2\) + 5x - 3
A
6
B
0
C
2
D
No correct option
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option
Source: Myschool Past Questions Archive (Record #60628)
Mathematics
UTME / JAMB 2021
Statistics and Probability
Solution
Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?
A
\(\frac{2}{3}\)
B
\(\frac{1}{2}\)
C
\(\frac{1}{3}\)
D
\(\frac{1}{4}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Numbers: 1 (odd), 2 (even), 3 (odd), 4 (even)
Total ways to remove 2 numbers out of 4 = C(4,2) = 6 possible pairs:
- (1,2) → odd + even = odd
- (1,3) → odd + odd = even
- (1,4) → odd + even = odd
- (2,3) → even + odd = odd
- (2,4) → even + even = even
- (3,4) → odd + even = odd
Favourable outcomes (even sum): (1,3) and (2,4) → 2 ways
Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #60702)
Mathematics
WAEC / WASSCE 2021
Number and Numeration
Solution
Factorize 6pq-3rs-3ps+6qr
A
3(r -p)(2q + s)
B
3(p + r)( 2q - 2q - s)
C
3(2q - s)(p + r)
D
3(r - p)(s - 2q)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
6pq-3rs-3ps+6qr = 3 (2pq - rs - ps + 2qr)
= 3 ({2pq + 2qr} {-ps - rs})
= 3 (2q{ p + r} -s{p + r})
= 3 ({2q - s}{p + r})
Source: Myschool Past Questions Archive (Record #61143)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The height of an equilateral triangle of side is 10\(\sqrt{3}\)cm. Calculate its perimeter.
A
20cm
B
60cm
C
40cm
D
30cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Height of an equilateral triangle, h = a\(\frac{\sqrt{3}}{2}\), where a is the side of the equilateral triangle.
10\(\sqrt{3}\) = a\(\frac{\sqrt{3}}{2}\)
cross multiply--> 2 x 10\(\sqrt{3}\) = a\(\sqrt{3}\)
\(\sqrt{3}\) cancels out \(\sqrt{3}\) on both sides
20 = a
The perimeter of an equilateral triangle is: P = 3a
P = 3 * 20 = 60cm
Source: Myschool Past Questions Archive (Record #61207)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h.
He spent 2hours for a meeting and returned through town M by bus at an average speed of 40KM/H.
If the distance covered by the bus was 2km longer than that of the car and he arrived at town M at 1 :55PM.
calculate distance from M to N.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Time spent traveling = 3 hours 55 minutes - 2 hours
= 1hr 55mins
Let x be the distance from M - N
and t , time spent from M - N
Then t = x / 72
Return time = (23 / 12) - t
(23 / 12) - t = (x + 2) / 40
(23 / 12) - ( x / 72) = (x + 2) / 40
\(\frac{138 - x}{9}\) = \(\frac{x + 2}{5}\)
690 - 5x = 9x + 18
14x = 672
x = 48km
Source: Myschool Past Questions Archive (Record #61469)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270".
Calculate, correct: (a) two significant figures, |YZ|
(b) The nearest degree, the bearing of Y from Z
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
|YZ|\(^2\) = 15\(^2\) + 20\(^2\)
= 225 + 400 = 625
YZ = \sqrt625
YZ = 25
(b) tan \theta = \(\frac{15}{20}\)
tan \theta = 0.75
\theta = tan -1 (0.75)
= 36.87^\circ
Bearing of Y from Z = (90 + 36.87)^\circ = 126.87^\circ
= 127^\circ
Source: Myschool Past Questions Archive (Record #61470)
Mathematics
WAEC / WASSCE 2021
Number and Numeration
Diagram
Solution
In the diagram, \(\overline{AD}\) is a diameter of a circle with Centre O. If ABD is a triangle in a semi-circle ∠OAB=34",
find: (a) ∠OBD (b) ∠OCB
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
∠OAB = ∠OBA=34^\circ (base
Source: Myschool Past Questions Archive (Record #61485)
Mathematics
WAEC / WASSCE 2021
Number and Numeration
Solution
(a) A man shared his property among his children as follows:
Child's name
Ann
Afia
Kojo
Nuno
Akom
Percentage share
5
15
10
45
25
Represent the information on a pie chart
(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability th a girl takes away two red beads, one after the other, from the box.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) pie chart above.
Child's Name
Percentage Share (%)
Angle(^\circ)
Ann
5
5/100 * 360" = 18
Afia
15
15/100 x 360^\circ = 54
Kojo
10
10/100 x 360^\circ = 36
Nuno
45
45/100 x 360^\circ = 162
Akosua
25
25/100 x 360^\circ = 90
Total
100
360
(b) P(1st red) = 5/12
P(2nd red) = 4/11
Probability (two red beads) = 5/12 * 4/11
= 5/33 (Accept 0.1515)
Source: Myschool Past Questions Archive (Record #61489)
Mathematics
WAEC / WASSCE 2021
Number and Numeration
Solution
.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics.
If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information
(ii) how many students study both subjects
(iii) find the fraction of the class that study Biology but not Physics.
(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15m by 8 m which was to be carpeted.
The cost of carpeting was Gh¢ 890.00 per square metre. If a total of GH 216,120.00 was spent on painting and carpeting, how much was the cost of painting?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x = students that study both
(ii) 60- x + x + 48-x=80
x = 108-80
= 28
(iii) Fraction --> (60-28)/80
= 32/80 or 2/5
(b) Area of floor= 15x8= 120m\(^2\)
Cost of carpeting= 120 x GHC 890
= GHC106,800.00
Cost of paining = GHC(216,120 - 106,800)
= GHC109,320.00
Source: Myschool Past Questions Archive (Record #61495)
Mathematics
WAEC / WASSCE 2021
Geometry and Trigonometry
Solution
The table shows the distribution of the number of hours per day spent in studying by 50 students.
Number of hours per day
4
5
6
7
8
9
10
11
Number of students
5
7
5
9
12
4
3
5
Calculate, correct to two decimal places,
the: (a) mean; (b) standard deviation.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x
f
fx
fx\(^2\)
4
5
20
80
5
7
35
175
6
5
30
180
7
9
63
441
8
12
96
768
9
4
36
324
10
3
30
300
11
5
55
605
£f = 50
£fx = 365
£fx\(^2\) = 2,873
(a) Mean(\(\overline{x}\)) = \(\frac{365}{50}\) = 7.30 (2 d.p.)
(b) Standard deviation = \sqrt(\(\frac{2873}{50}\)) - (7.3)\(^2\)
= \sqrt4.17
= 2.0421
= 2.04 (2 d.p.)
Source: Myschool Past Questions Archive (Record #61562)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Diagram
Solution
In the diagram, \(\overline{MN}\)//\(\overline{PQ}\), < MNP = 2x, and < NPQ = (3x - 50)°. Find the value of < NPQ
Attached Figure / Diagram
A
200\(^o\)
B
100\(^o\)
C
120\(^o\)
D
90\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2x = 3x - 50 alternate angle
50\(^o\) = 3x - 2x
50\(^o\) = x
< NPQ = (3x - 50)\(^o\)
= 3(50)\(^o\) - 50
= 150 - 50 = 100\(^o\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #58325)