JAMB Mathematics Previous Year Questions: 2021 | Mathematics for JAMB PDF Download

Q1: Solve the following equation:
(a)
(b)
(c)
(d) (2, 1)
Ans: (b)
Given:


L.C.M. ⇒ (2r - 1) (r + 2)

Q2: In how many ways can 2 students be selected from a group of 5 students for a debating competition?
(a) 20 ways
(b) 10 ways
(c) 15 ways
(d) 25 ways
Ans: (b)

Q3: Find the rate of change of volume V of a hemisphere with respect to its radius r, when r = 2
(a) 8π
(b) 2π
(c) 16π
(d) 4π
Ans: (a)

Q4: Find the maximum value of y = 3x² + 5x - 3
(a) 6
(b) 0
(c) 2
(d) No correct option
Ans: (a)
y = 3x² + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = -5/6
At maximum:

App the L.C.M.: 36

Q5: A trapezium has two parallel sides of lengths 5cm and 9cm. If the area is 91cm², what is the distance between the parallel sides?
(a) 13 cm
(b) 12 cm
(c) 8 cm
(d) 9 cm
Ans: (a)
Area of Trapezium = 1/2(sum of parallel sides) x h
Hence,
91 = 1/2 (5 + 9)h
91 = 7h
h = 91/7
h = 13 cm

Q6: Find the value of p if the line which passes through (-1, -p) and (-2, 2) is parallel to the line 2y + 8x - 17 = 0
(a) -2/7
(b) 7/6
(c) -6/7
(d) 2
Ans: (d)
Given  2y + 8x - 17 = 0
Equation of the line: y = mx + c
2y = -8x + 17
y = -4x  + 17/2
The slope, m₁ = 4
For parallel lines, m₁. m₂ = -4
where slope ( -4) =  at points (-1, -p) and (-2, 2)
-4(x₂ - x₁) = y₂ - y₁
-4 ( -2 - (-1)) = (2 - (-p))
p = 4 - 2 = 2

Q7: The ratio of the length of two similar rectangular blocks is 2:3. If the volume of the larger block is 351cm³, find the volume of the other block.
(a) 234.00 cm³
(b) 166.00 cm³
(c) 526.50 cm³
(d) 687cm³
Ans: (a)
Let x = total volume, 2 : 3 = 2 + 3 = 5
3/5x = 351
x = 351 x 5 / 3
= 585
Volume of the smaller block = 2/5 x 585
= 234.00 cm³

Q8: Find the derivative of the function y = 2x²(2x - 1) at the point x = -1
(a) -4
(b) 16
(c) 18
(d) -8
Ans: (b)
y = 2x²(2x - 1)
y = 4x³ - 2x²
dy/dx= 12x² - 4x
at x = -1,
dy/dx = 12(-1)² - 4(-1)
= 12 + 4 = 16

Q9: Correct 241.34(3 x 10^-3)² to 4 significant figures.
(a) 0.0014
(b) 0.001448
(c) 0.0022
(d) 0.002172
Ans: (d)
(3 x 10^-3)²
= 3²x²

x² = 1/x³
= 24.34 x 3² x 1/10⁶
= 2172.06 / 10^6.
= 0.00217206
= 0.002172

Q10: Find the mean deviation of 1, 2, 3 and 4
(a) 1.0
(b) 1.5
(c) 2.0
(d) 2.5
Ans: (a)
Mean deviation formula:  (?|x - x|)/n
x = 2.5
= (|1 - 2.5| + |2 - 2.5| + |3 - 2.5| + |4 - 2.5|)/4
= 4/4
= 1

Q11: At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50?
(a) 1/2 %
(b)
(c) 1.5%
(d) 25%
Ans: (c)
Interest,
Hence, 

= 750/500
= 3/2
= 1.5%

Q12: Three children shared a basket of mangoes in such a way that the first child took 1/4 of the mangoes and the second 3/4 of the remainder. What fraction of the mangoes did the third child take?
(a) 3/16
(b) 7/16
(c) 9/16
(d) 13/16
Ans: (a)
When the first child takes 1/4 it will remain 1 - 1/4 = 3/4
Next, when the second child takes 3/4 of the remainder
⇒ 3/4 i.e. find 3/4 of 3/4
= 3/4 x 3/4
= 9/16
Fraction remaining = 3/4 - 9/16
= 12-9 / 16
= 3/16

Q13: Simplify and express in standard form
(a) 8.8 x 10^-1
(b) 8.8 x 10^-2
(c) 8.8 x 10^-3
(d) 8.8 x 10³
Ans: (c)

88 x 10^-4 = 88 x 10^-1 x 10^-4
= 8.8 x 10^-3

Q14: Simplify
(a) 4√3
(b) 4/√3
(c) 3√3
(d) √3/4
Ans: (a)

Q15: Three brothers in a business deal share the profit at the end of a contract. The first received 1/3 of the profit and the second 2/3 of the remainder. If the third received the remaining N12000.00 how much profit did they share?
(a) N60 000.00
(b) N54 000.00
(c) N48 000.00
(d) N42 000.00
Ans: (b)
Let T = total profit.
The first brother receives 1/3T
balance, 1 - 1/3
= 2/3T
The seconds brother receives the remnant: 2/3
2/3 x 2/3 x 4/9
The third brother receives the left over: 2/3T - 4/9T 

=
= 2/9T
The third brother receives 2/9T ⇒ N12000 
If 2/9T = N12, 000

= N54, 000

Q16: 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
Ans: (a)
PQ² = (x2 - x1)² + (y2 - y1)²
= 12² + 5²
= 144 + 25
= 169
PQ = √169
PQ = 13
PQ = Diameter = 2 x Radius, r
r = PQ/2
= 13/2
= 6.5

Q17: The angle of a sector of a circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector
(a) 8.8 cm
(b) 25.4 cm
(c) 25.6 cm
(d) 29.8 cm
Ans: (d)
Length of Arc AB =

Perimeter of sector = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm

Q18: Find the length of a side of a rhombus whose diagonals are 6cm and 8cm
(a) 8 cm
(b) 5 cm
(c) 4 cm
(d) 3 cm
Ans: (b)
For a rhombus,
The diagonals bisect each other at right angles.
The diagonals divide into four congruent right-angled triangles.
Hence,  6cm splits into 3cm each and 8cm to 4cm each
Hypotenuse(hyp) = Adjacent(adj) + Opposite(opp)
hyp² = adj² + opp²
hyp² = 3² + 4²
hyp² = 25
hyp = 5
Length (L) = 5cm since a rhombus is a quadrilateral with 4 equal lengths

Q19: 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
Ans: (a)
Interior angles of a regular polygon =

140n = 180n - 360
40n = 360
n = 9 sides

Q20: A cylinder pipe, made of metal is 3cm thick. If the internal radius of the pipe is 10cm, find the volume of metal used in making 3m of the pipe.
(a) 53πcm³
(b) 207πcm³
(c) 15 300πcm³
(d) 20 700πcm³
Ans: (d)
Volume of a cylinder = πr²h
Convert 3m to cm by multiplying by 100
Volume of exterior = c x 13²  x 300
Volume of interior= π x 10²  x 300
Difference: volume of the external cylinder - volume of the internal cylinder
Total volume (v) = π(169 - 100) x 300
V = π x 69 x 300
V = 20 700πcm³

Q21: 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
Ans: (d)
line parallel to the two lines

Q22: 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) 48^o
(b) 84^o
(c) 92^o
(d) 276^{o
Ans: (d}
Sum of 4, 16, 30, 20, 10, 14 and 26
= 120
numbers ≥  16 are 16 + 30 + 20 + 26 = 92
sum of angles =
= 276^o

Q23: 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
Ans: (c)
Given that the mean of 10 positive numbers = 16
Sum of numbers = 16 x 10 = 160
Mean of 11 numbers = 18
Total sum of numbers = 11 x 18
= 198
The 11^th number = 198 - 160
= 38

Q24: 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) 1/2
(c) 1/3
(d) 1/4
Ans: (b)
Sample space = 16
Sum of numbers removed: (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even numbers count: 8
Pr(even sum) = 8/16
= 1/2

Q25: 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
Ans: (c)

= 4/5
= 0.8
Total amount = N10.80
Musa repays N8.00
Remainder = 10.80 - 8.00
= N2.80

Q26: Find the probability that a number selected at random from 41 to 56 is a multiple of 9
(a) 1/8
(b) 2/15
(c) 3/16
(d) 7/8
Ans: (a)
Given the range of number 41 to 56, we have:
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56
The numbers that are multiple of 9 are: 45, 54
P(multiple of 9) = 2/16
= 1/8

Q27: Simplify
(a) 1 - 4 log3
(b) -1 + 2 log 3
(c) -1 + 5 log2
(d) -1 + 5 log2
Ans: (d)

= log 10 - log 4
= log10  - log2²
= 1 - 2 log2

Q28: 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)
(b)
(c)
(d)
Ans: (b)
Average speed = total distance / total time
Calabar to Enugu in time t1,
t₁ = P/U
Enugu to Benin:
t₂ = q/w
Average Speed  =

Q29: If w varies inversely as  and w = 8 when u = 2 and v = 6, find a relationship between u, v, w.
(a) uvw = 16(u + v)
(b) uvw = 16(u + v)
(c) uvw = 12(u + v)
(d) 12uvw = u + v
Ans: (c)

w = 8, u = 2 and v = 6

k = 12
12 ( u + v) = uwv

Q30: If g(x) = x² + 3x  find g(x + 1) - g(x)
(a) (x + 2)
(b) 2(x + 2)
(c) (2x + 1)
(d) (x² + 4)
Ans: (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)

Q31: Factorize m³ - m² + 2m - 2
(a) (m² + 1)(m - 2)
(b) (m - 1)(m + 1)(m + 2)
(c) (m - 2)(m + 1)(m - 1)
(d) (m² + 2)(m - 1)
And: (d)
(m² + 2) (m - 1)

 
Q32: The angles of a quadrilateral are 5x - 30, 4x + 60, 60 - x and 3x + 61.find the smallest of these angles?
(a) 5x - 30
(b) 4x + 60
(c) 60 - x
(d) 3x + 61
Ans: (c)
Sum of the 4 angles of a quadrilateral = 360°
(5x-30) + (3x + 61) + (60-x) + (4x+ 60) = 360°
11x + 151 = 360°
11x = 360 - 151 = 209
x = 209/11 = 19°
Each angle:
5x - 30 =  65°
4x+ 60 = 136°
60 - x =41°
3x + 61 = 118°
The smallest of the angles is 41° 

Q33: Find the n-th term of the sequence 2, 6, 12, 20...
(a) 4n - 2
(b) 2(3n - 1)
(c) n² + n
(d) n² + 3n + 2
Ans: (c)
Given the sequence 2, 6, 12, 20...
The nth term = n² + n
Check: n = 1, u₁ = 2
n = 2, u₂ = 4 + 2 = 6
n = 3, u₃ = 9 + 3 = 12
n = 4, u₄ = 16 + 4 = 20

Q34: If the binary operation * is defined by m * n = mn + m + n for any real number m and n, find the identity of the elements under this operation
(a) e = 1
(b) e = -1
(c) e = -2
(d) e = 0
Ans: (d)
Identity(e) : a * e = a
m * e = m...(i)
m * e = me + m + e
m * e = m
m = me + m + e
m - m = e(m + 1)
e = 0 / m + 1
e = 0

Q35: Factorize completely 81a⁴ - 16b⁴
(a) (3a + 2b)(2a - 3b)(9a² + 4b²)
(b) (3a + 2b)(2a - 3b)(9a² + 4b²)
(c) (3a - 2b)(3a + 2b)(9a² + 4b²)
(d) (6a - 2b)(8a - 3b)(4a³ - 9b²)
Ans: (c)
81a⁴ - 16b⁴ = (9a²)² - (4b²)²
= (9a² + 4b²)(9a² - 4b²)
9a² - 4b² = (3a - 2b) (3a + 2b)

Q36: Find x if log₉x = 1.5
(a) 27
(b) 15
(c) 3.5
(d) 32
Ans: (a)
Given log₉x = 1.5
9^1.5 = x
9^3/2 = x
(√ 9)³ = 3
x = 27

Q37: List all integers satisfying the inequality in -2 < 2x-6 < 4
(a) 2, 3, 4 and 5
(b) 2, 3
(c) 2, 5
(d) 3, 4
Ans: (d)
-2 < 2x - 6   AND  2x - 6 < 4
-2 + 6 <2x  AND  2x < 4 + 6
4 <2x  AND  2x < 10 : 2 <x  AND x <5
2 < x < 5
Hence,
3, 4

Q38: Give that X is due east point of Y on a coast. Z is another point on the coast but 6.0km due south of Y. If the distance ZX is 12km, calculate the bearing of Z from X
(a) 240°
(b) 150°
(c) 60°
(d) 270°
Ans: (a)
Sinθ = 6/12Sinθ = 1/2
θ = Sin⁰.5
θ = 30°
Bearing of Z from X, (270 - 30)° = 240°

Q39: A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?
(a) 25
(b) 19
(c) 18
(d) 17
Ans: (a)
Let the three items be M, Y and P
n{M ∩ Y} only = 4-3 = 1
n{M ∩ P) only = 5-3 = 2
n{ Y ∩ P} only = 2
n{M} only = 12-(1+3+2) = 6
n{Y} only = 10-(1+2+3) = 4
n{P} only = 14-(2+3+2) = 7
n{M∩P∩Y} = 3
The number of women in the group = 6 + 4 + 7 + (1 + 2 + 2 + 3)
= 25

Q40: If (x + 2) and (x - 1) are factors of the expression Lx + 2kx² + 24, find the values of L and k.
(a) l = -12, k = -6
(b) l = -2 , k = 1
(c) l = -2 , k = -1
(d) l = 0, k = 1
Ans: (a)
Given (x + 2) and (x - 1), i.e. x = -2 or +1
when x = -2
L(-2) + 2k(-2)² + 24 = 0
f(-2) = -2L + 8k = -24...(i)
And x = 1
L(1) + 2k(1) + 24 = 0
f(1):L + 2k = -24...(ii)
Subst, L = -24 - 2k in eqn (i)
-2(-24 - 2k) + 8k = -24
+48 + 4k + 8k = -24
12k = -24 - 48 = -72
k = -6
Where L = -24 - 2k
L = -24 - 2(-6)
L = -24 + 12
L = -12
K = -6, L = -12