Mathematics 2021 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.
16. P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
- A. 6.5 units
- B. 13.0 units
- C. 3.5 units
- D. 7.0 units
Correct Option: Answer is B
Solution:
17. The angle of a sector of s circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector
- A. 8.8cm
- B. 25.4cm
- C. 25.6cm
- D. 29.8cm
Correct Option: Answer is D
Solution:
18. Find the length of a side of a rhombus whose diagonals are 6cm and 8cm
- A. 8cm
- B. 5cm
- C. 4cm
- D. 3cm
Correct Option: Answer is B
Solution:
A rhombus has two diagonals that bisect each other at right angles.
i.e this splits 6cm into 3cm each AND 8cm to 4cm
Using Hyp2 = adj2 + opp2
Hyp2 = 32 + 42
Hyp2 = 25
Hyp = 5
19. Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?
- A. 9
- B. 8
- C. 7
- D. 5
Correct Option: Answer is A
Each interior angle = (n-2)180 /n
140 = (n-2)180 /n
Cross multiply:
140n = 180n – 360
40n = 360
n = 9 sides
20. A cylinder pipe, made of metal is 3cm thick. If the internal radius of the pope is 10cm. Find the volume of metal used in making 3m of the pipe.
- A. 153πcm3
- B. 207πcm3
- C. 15300πcm3
- D. 20700πcm3
Correct Option: Answer is D
Solution:
Volume of a cylinder = πr2h
First convert 3m to cm by multiplying by 100
Volume of External cylinder = π x times 132 x 300
Volume of Internal cylinder = π x 102 x 300
Hence; Volume of External cylinder – Volume of Internal cylinder
Total volume (v) = π (169 – 100) x 300
V = π x 69 x 300
V = 20700πcm3
21. The locus of a point which moves so that it is equidistant from two intersecting straight lines is the?
- A. perpendicular bisector of the two lines
- B. angle bisector of the two lines
- C. bisector of the two lines
- D. line parallel to the two lines
Correct Option: Answer is B
Solution:
Angle bisector of the two lines.
What is an Angle Bisector? Angle bisector of two lines i.e. the line which bisects the angle between the two lines is the locus of a point which is equidistant from the two lines. In other words, an angle bisector has equal perpendicular distance from the two lines
22. 4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16
- A. 48o
- B. 84o
- C. 92o
- D. 276o
Correct Option: Answer is D
Solution:
Given that 4, 16, 30, 20, 10, 14 and 26
total = 120
numbers equals to or greater than 16 are:
16 + 30 + 20 + 26 = 92
The requires sum of angles = 92/120 x 360 = 276o
23. The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number
- A. 3
- B. 16
- C. 38
- D. 30
Correct Option: Answer is C
Solution:
Mean of 10 numbers = 16
The total sum of numbers = 16 x 10 = 160
Mean of 11 numbers = 18
Total sum of numbers = 11 x 18 = 198
The 11th number = 198 – 160 = 38
24. 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. 2/3
- B. 2/15
- C. 1/2
- D. 1/4
Correct Option: Answer is C
Solution:
sample space = 16
sum of numbers removed are (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even number = 8
Pr(even sum) = 8/16 = 1/22
25. Find the probability that a number selected at random from 41 to 56 is a multiply of 9
- A. 1/8
- B. 2/15
- C. 3/16
- D. 7/8
Correct Option: Answer is A
- 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) = 2/16 = 1/8
26. Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe?
- A. N10.80
- B. N10.67
- C. N2.80
- D. N2.67
Correct Option: Answer is C
S.I = PRT/100
=> 10 x 2 x 4 / 100
=> 4/5 = 0.8
Total amount = N10.80
He pays N8.00
Remainder = 10.80 – 8.00 = N2.80
27. Simplify 2log2/5 – log72/125 + log 9
- A. 1 – 4 log3
- B. -1 + 2 log 3
- C. -1 + 5 log2
- D. 1 – 2log 2
Correct Option: Answer is D
28. A car travels from calabar to Enugu, a distance of P km with an average speed of U km per hour and continues to benin, a distance of Q km, with an average speed of Wkm per hour. Find its average speed from Calabar to Benin
- A. (\frac{(p + q)}{pw + qu}\)
- B. \(\frac{uw(p + q)}{pw + qu}\)
- C. \(\frac{uw(p + q)}{pw}\)
- D. \(\frac{uw}{pw + qu}\)
Correct Option: Answer is D
Average speed = Total Distance / Total Time
Calabar to Enugu in time t1, hence
t1 = P/U also from Enugu to Benin
t2 = q/w
Averagw speed = p + q / t1 + t2
= \(\frac{p + q}{\frac{p}{u} + \frac{q}{w}\)
= p + q x \(\frac{uw}{pw + qu}\)
= \(\frac{uw(p + q)}{pw + qu}\)
29. If w varies inversely as uv / v+u and is equal to 8 when
u = 2 and v = 6, find a relationship between u, v, w.
- A. uvw = 16(u + v)
- B. 16uv = 3w(u + v)
- C. uvw = 12(u + v)
- D. 12uvw = u + v
Correct Option: Answer is D
30. If g(x) = x2+ 3x find g(x + 1) – g(x)
- A. (x + 2)
- B. 2(x + 2)
- C. (2x + 1)
- D. (x2 + 4)
Correct Option: Answer is B
g(x) = x2 + 3x
When g(x + 1) = (x + 1)2 + 3(x + 1)
= x2 + 1 + 2x + 3x + 3
= x2 + 5x + 4
g(x + 1) – g(x) = x2 + 5x + 8 – (x2 + 3x)
= x2 + 5x + 4 – x2 -3x
= 2x + 4 or 2(x + 4)
= 2(x + 2)