# 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 m^{3}– m^{2}+ 2m – 2

**A.**(m^{2}+ 1)(m – 2)**B.**(m – 1)(m + 1)(m + 2)**C.**(m – 2)(m + 1)(m – 1)**D.**(m^{2}+ 2)(m – 1)

**Correct Option: Answer is D**

Solution:

factorize m^{3} – m^{2} + 2m – 2

do the grouping m^{3} – m^{2} + 2m – 2 =( m^{3} – m^{2} )+(2m-2)

factor out m^{2 }in the first and 2 in the second group

m^{2}(m-1)+2(m-1) => factor out common term m-1

= (m^{2}+2) (m-1)

=(m^{2} + 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(3^{n – 1})**C.**n^{2}+ n**D.**n^{2}+ 3n + 2

**Correct Option: Answer is C**

Solution:

Given that 2, 6, 12, 20…?

the nth term = n^{2} + 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 = 3/30×360^{o}=36^{o}

**35. **Factorize completely 81a^{4}– 16b^{4}

**A.**(3a + 2b)(2a – 3b)(9a^{2}+ 4b^{2})**B.**(3a – 2b)(2a – 3b)(4a^{2}– 9b^{2})**C.**(3a – 2b)(3a + 2b)(9a^{2}+ 4b^{2})**D.**(6a – 2b)(8a – 3b)(4a^{3}– 9b^{2})

**Correct Option: Answer is C**

Solution:

81a^{4} – 16b^{4 } = (9a^{2})^{2} – (4b^{2})^{2}

= (9a^{2} + 4b^{2}) (9a^{2} – 4b^{2})

(9a^{2} – 4b^{2})= (3a – 2b)(3a + 2b)

**36. **Find x if log_{9}x = 1.5

**A.**27**B.**15**C.**3.5**D.**32

**Correct Option: Answer is A**

Solution:

if log_{9}x = 1.5 = 9^{1.5 }= x

9^{3/2}= 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+2kx^{2}, 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)