Mathematics 2020 Past Questions | JAMB
Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
31. The chord ST of a circle is equal to the radius r of the circle. Find the length of arc ST
- A. πr/3
- B. πr/2
- C. πr/12
- D. πr/6
Correct Option: Answer is A
Solution:\(\frac{ \frac{r}{2}}{r}\) Sin \(\theta\) = \(\frac{1}{2}\)
\(\theta\) = sin\(^{-1}\) (\(\frac{1}{2}\)) = 30\(^o\) = 60\(^o\)
Length of arc (minor)
ST = \(\frac{\theta}{360}\) x 2\(\pi r\)
\(\frac{60}{360} \times 2 \pi \times r = \frac{\pi}{3}\)
32. A sector of circle of radius 7.2cm which subtends an angle of 300oat the centre is used to form a cone. What s the radius of the base of the cone?
- A. 8cm
- B. 9cm
- C. 6cm
- D. 7cm
Correct Option: Answer is C
Solution:
The length of the arc subtended by the sector of angle 300° = circumference of the base of the cone
\(\frac{300}{360} \times 2 \pi \times 7.2 = 2{\pi}r\)
5/6 x 2π x 7.2 = 2πr
5 x π x 2.4 = 2πr
5 x 2.4 = 2r
12 = 2r => r =6cm
33. A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m?
- A. 25m
- B. 23m
- C. 22m
- D. 20m
Correct Option: Answer is D
Solution:
V = 2080cm\(^3\), h = ?
r = 7cm
V = V\(\pi r^2h\)
h = \(\frac{V}{\pi r^2} = \frac{3080}{\frac{22}{7} \times 49}\)
\(\frac{3080}{54}\) = 20cm
34. The acres for rice, pineapple, cassava, cocoa and palm oil in a certain district are given respectively as 2, 5, 3, 11 and 9. What is the angle of the sector of cassava in a pie chart?
- A. 180o
- B. 36o
- C. 60o
- D. 108o
Correct Option: Answer is B
Total number of acres = 2 + 5 + 3 + 11 + 9 = 30
The angle of acres = 2 + 5 + 3 + 11 + 9 = 30
The angle of the sector for cassava in a pie chart = 3/30×360o=36o
35. Three consecutive terms of a geometric progression are give as n – 2, n and n + 3. Find the common ratio
- A. 3/2
- B. 2/3
- C. 1/2
- D. 1/4
Correct Option: Answer is A
Solution:
\(\frac{h}{n – 2} = \frac{n + 3}{n}\)
n\(^2\) = (n + 3) (n – 2)
n\(^2\) = n\(^2\) + n – 6
n\(^2\) + n – 6 – n\(^2\) = 0
n – 6 = 0
n = 6
Common ratio: \(\frac{n}{n – 2} = \frac{6}{6 – 2} = \frac{6}{4}\) = \(\frac{3}{2}\)
36. In a class of 40 students, 32 offer mathematics, 24 offer physics and 4 offer neither mathematics nor physics. How many offer both mathematics and physics?
- A. 4
- B. 8
- C. 16
- D. 20
Correct Option: Answer is A
Solution:
40 = 32 – x + x + 24 + 4
40 = 60 – x
x = 60 – 40
x = 20
37. The sum of the interior angle of pentagon is 6x + 6y. Find y in terms of x.
- A. y = 6 – x
- B. y = 90 – x
- C. y = 120 – x
- D. y = 150 – x
Correct Option: Answer is B
Solution:
Sum of interior angles = (2n – 4) 90o
For pentagon, n = 5
Sum of interior angles = 6 x 90o = 540o
6x + 6y = 540
6(x + y) = 540
x + y = 540/6 = 90
y = 90
y = 90 – x
38. The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group.
- A. 7
- B. 9
- C. 15
- D. 42
Correct Option: Answer is A
Solution:
39. A surveyor walks 500m up a hill which slopes at an angle of 30o. Calculate the vertical height through which he rises
- A. 252m
- B. 500m
- C. 250m
- D. 255m
Correct Option: Answer is C
h/500 = sin 30o
= 500 sin 30o
= 500 x 1/2
= 250m
40. Find the non-zero positive value of x which satisfies the equation
- A. 2
- B. √3
- C. √2
- D. 1
Correct Option: Answer is B
S.I = PRT/100
I = 1,240 – 1000 240
P = 1000; R = ? T = 3
∴ 240 = 240 x 100/ 1000 x 3
= 8%