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.
31. 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)
Correct Option: Answer is D
Solution:
factorize m3 – m2 + 2m – 2
do the grouping m3 – m2 + 2m – 2 =( m3 – m2 )+(2m-2)
factor out m2 in the first and 2 in the second group
m2(m-1)+2(m-1) => factor out common term m-1
= (m2+2) (m-1)
=(m2 + 2)(m – 1)
32. 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
Correct Option: Answer is C
Solution:
Sum of all 4 angles of a quadrilateral = 360°
(5x-30) + (3x + 60) + (60-x) + (4x+ 50) = 360°
(12x – x) + ( 170 – 30) = 360°
11x + 140 = 360°
11x = 360 – 140 = 220
x = 220/11 = 20°
Each angles is :
5x – 30 = 100 – 30 = 70°
3x+ 60 = 60 + 60 = 120°
60 – x = 60 – 20 = 40°
4x + 50 = 80 + 50 = 130°
Smallest of these angles is 40°
33. What is 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
Correct Option: Answer is C
Solution:
Given that 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
34. 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
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. Factorize completely 81a4– 16b4
- A. (3a + 2b)(2a – 3b)(9a2 + 4b2)
- B. (3a – 2b)(2a – 3b)(4a2 – 9b2)
- C. (3a – 2b)(3a + 2b)(9a2 + 4b2)
- D. (6a – 2b)(8a – 3b)(4a3 – 9b2)
Correct Option: Answer is C
Solution:
81a4 – 16b4 = (9a2)2 – (4b2)2
= (9a2 + 4b2) (9a2 – 4b2)
(9a2 – 4b2)= (3a – 2b)(3a + 2b)
36. Find x if log9x = 1.5
- A. 27
- B. 15
- C. 3.5
- D. 32
Correct Option: Answer is A
Solution:
if log9x = 1.5 = 91.5 = x
93/2= x =>(√9)3 = 3
X= 27
37. 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
Correct Option: Answer is D
Solution:
-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
As 3 and 4
38. 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°
Correct Option: Answer is A
Solution:
Sinθ = 6/12
Sinθ = 1/2
θ = Sin0.5
θ = 30°
Bearing of Z from X, (270 – 30)° = 240°
39. 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
Correct Option: Answer is 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
Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.
40. If (x + 2) and (x – 1) are factors of the expression Lx+2kx2, 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
Correct Option: Answer is 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 = −72/12
k = -6
where L = -24 – 2k
L = -24 – 2(-6)