Mathematics 2016 JAMB Past Questions

1. Integrate

\frac{2 x^{3} + 2 x}{x}

with respect to x

A. $\frac{2 x^{3}}{3}$

- 2x + k
B. x³ + 2x + k
C. $\frac{2 x^{3}}{3}$

+ 2x + k
D. x³ - 2x + k

Correct Answer: Option C

Explanation

\int

\frac{2 x^{3} + 2 x}{x}

\int

\frac{2 x^{3}}{x}

\frac{2 x}{x}

\int

(2x² + 2) dx

\frac{2 x^{2} + 1}{2 + 1}

+ 2x + k

=

\frac{2 x^{3}}{3}

+ 2x + k

2. If the mean of 4, y, 8 and 10 is 7. Find Y?

A. 6
B. 10
C. 7
D. 9

Correct Answer: Option A

Explanation

\frac{4 + y + 8 + 10}{4}

= 7

22 + y = 4 x 7
22 + y = 28
y = 28 - 22 = 6

3. Find the derivative of y = (

\frac{1}{3}

X + 6)

A. $\frac{2 x^{3}}{3}$ - 2x + k
B. $\frac{2}{3}$ ( $\frac{1}{3}$ X + 6)
C. $\frac{2}{3}$ ( $\frac{1}{3}$ X + 6)²
D. $\frac{1}{3}$ ( $\frac{1}{3}$ X + 6)²

Correct Answer: Option B

Explanation

Y = (

\frac{1}{3}

X + 6)²

\frac{d y}{d x}

= 2(

\frac{1}{3}

X + 6)

\frac{1}{3}

\frac{2}{3}

(

\frac{1}{3}

X + 6)

4. In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?

A. 45
B. 10
C. 35
D. 30

Correct Answer: Option B

Explanation

80 - x + x + 60 - x + 20 = 150
160 - x = 150
x = 160 - 150 = 10

5. If the 2nd term of a G.P is

\frac{8}{9}

and the 6th term is 4

\frac{1}{2}

. Find the common ratio.

A. 2
B. $\frac{3}{2}$

C. $\frac{2}{3}$
D. 3

Correct Answer: Option B

Explanation

T₂ =

, T₆ = 4

ar =

ar⁵ =

r =

From the diagram above, find the value of < OTQ

A. 230^o
B. 55^o
C. 115^o
D. 65^o

Correct Answer: Option D

Explanation

< OTQ =

\frac{130}{2}

= 65^o

7. The sum of the interior angles of a polygon is a given as 1080^o. Find the number of the sides of the polygon.

A. 5
B. 7
C. 6
D. 8

Correct Answer: Option D

Explanation

$\frac{(n - 2) 180}{180}$ = $\frac{1080}{180}$

(n - 2) = 6
n = 6 + 2 = 8

In the diagram above, l₁ is parallel to l₂, Find the value of < PMT

A. 82^o
B. 36^o
C. 72^o
D. 118^o

Correct Answer: Option A

Explanation

< MPT = 180^o - 118^o = 62^o

< PML = 62^o ( Alternative angle)

y + 2y + 10^o + 62^o = 180^o (Angles on a straight line)

3y = 180 - 72

\frac{3 y}{3}

\frac{108}{3}

y = 36^o

< PMT = 2y + 10 = 2(36) + 10 = 82^o

From the diagram above, Find the value of < ROP

A. 110^o
B. 70^o
C. 95^o
D. 85^o

Correct Answer: Option C

Explanation

< ROP = 95^o (Exterior angle of cyclic quadrilateral)

The venn diagram shows a class of 50 students with the games they play. How many students play only two games?

A. 15
B. 16
C. 20
D. 18

Correct Answer: Option A

Explanation

Number of students that plays only two games = 4 + 8 + 3 + = 15

11. If line p = 5x + 3 is parallel to line p = wx + 5. Find the value of w.

A. 7
B. 3
C. 6
D. 5

Correct Answer: Option D

Explanation

P = 5x + 3, P = wx + 5
M₁ = 5, M₂ = w
M₁ = M₂ (parallel lines)
w = 5

12. Evaluate

\int

(cos4x + sin3x)dx

A. sin4x - cos3x + k
B. sin4x + cos3x + k
C. $\frac{1}{4}$ sin4x - $\frac{1}{3}$ cos3x + k
D. $\frac{1}{4}$ sin4x + $\frac{1}{3}$ cos3x + k

Correct Answer: Option C

13. If x₁₀ = 23₅. Find x

A. 15
B. 12
C. 14
D. 13

Correct Answer: Option D

Explanation

x₁₀ = 23₅
x₁₀ = 2 x 5¹ + 3 x 5⁰
= 10 + 3
= x₁₀ = 13

The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class. What is the size of the angle of the sector in economics?

A. 24^o
B. 39^o
C. 32^o
D. 36^o

Correct Answer: Option C

Explanation

x^o + 36^o + 54^o + 28^o + 90^o + 120^o = 360^o

x + 328^o = 360^o
x = 360^o - 328^o
x = 32^o

15. Calculate the range of 20, -6, 25, 30, 21, 28, 32, 33, 34, 5, 3, 2, and 1.

A. 32
B. 36
C. 33
D. 40

Correct Answer: Option D

Explanation

Range = 32- (-6) = 34 + 6 = 40

The bar chart above shows the number of visitors received in a week. How many visitors were received on Friday, Tuesday and Sunday?

A. 17
B. 22
C. 20
D. 16

Correct Answer: Option D

Explanation

No of visitors = 10 + 5 + 1 = 16

17. Factorize k² - 2kp + p².

A. (k +p)²
B. (k - p)²
C. k² + p²
D. k² - p²

Correct Answer: Option B

Explanation

k² - 2kp + p² = (k - p)²

18. Calculate the perimeter of a sector of a circle of raduis 12cm and angle 60^o.

A. (12 + 4 $π$

B. (24 + 4

π

)cm

C. (12 + 6

π

)cm

D. (24 + 6

π

)cm

Correct Answer: Option B

Explanation

Perimeter = 2r +

\frac{0}{360}

x 2

π

r
= 2(12) +

\frac{60}{360}

x 2 x

π

x 12
= (24 + 4

π

)cm

\begin{array}{cc} M a r k s & 2 & 3 & 4 \\ F r e q u e n c y & 4 & 4 & y \end{array}

The table above shows the frequency distribution of marks obtained by a group of students. If the total mark is 48, find the value of y.

A. 6
B. 8
C. 7
D. 5

Correct Answer: Option C

Explanation

u = {1, 2, 3, 4, 5......14}
p = {2, 4, 6, 8, 10, 12, 14}
p¹ = {1, 3, 5, 7, 9, 11,13}

20. Given U = {xx is a positive integer less than 15} and P = {xx is even number from 1 to 14}. Find the compliment

A. {1, 3, 5, 7, 9, 11, 13, 15}
B. {2, 3, 5, 7, 9, 11, 13}
C. {1, 3, 5, 7, 9, 11, 13}
D. {2, 3, 5, 7, 11, 15}

Correct Answer: Option C

Explanation

u = {1, 2, 3, 4, 5......14}
p = {2, 4, 6, 8, 10, 12, 14}
p¹ = {1, 3, 5, 7, 9, 11,13}

21. Simplify

\frac{0.026 \times 0.36}{0.69}

. Leave your answer in standard form

A. 1.36 x 10^-4
B. 1.36 x 10^-3
C. 1.36 x 10^-1
D. 1.36 x 10^-2

Correct Answer: Option D

Explanation

= 0.01356 = 1.36 x 10^-2

22. A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2:3:5 respectively. If Bisi got 5, how many were shared out?

A. 15
B. 25
C. 30
D. 50

Correct Answer: Option B

Explanation

Let x represent total number of pencils shared
B : S : T
2 : 3 : 5

2+3+5 = 10

\frac{2}{10}

x x = 5

\frac{2 x}{10}

= 5
2x = 50
x = 25

23. Calculate the perimeter of a sector of a circle of raduis 9cm and angle 36^o.

A. 18cm
B. (18 + $\frac{9 π}{5}$ )cm
C. (9 + $\frac{9 π}{5}$ )cm
D. $\frac{9 π}{5}$ cm

Correct Answer: Option B

Explanation

Perimeter = 2r +

\frac{θ}{360}

x 2

π

r
2(9) +

\frac{36}{360}

x 2 x

π

x 9 = 18 +

\frac{9 π}{5}

24. Evaluate

\frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2}

A. $\frac{23}{5}$

B. - $\frac{23}{5}$
C. $\frac{1}{8}$
D. $\frac{21}{7}$

Correct Answer: Option C

Explanation

\frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2}

\frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2}

\frac{3 \sqrt{27} - 3 {\sqrt{8}}^{2}}{\sqrt{16} \times 2}

\frac{3 - 4}{4 \times 2}

\frac{- 1}{8}

\begin{array}{cc} S c o r e s & 3 & 6 & 5 & 2 \\ F r e q u e n c y & 2 & 3 & 4 & 6 \end{array}

From the table above, find the median

A. 3
B. 5
C. 4
D. 6

26. Find

\frac{d y}{d x}

if y =

\frac{2}{3}

X³ -

\frac{4}{x}

A. 3x² - $\frac{4}{x}$

B. 2x² +

\frac{4}{x^{2}}

C. 3x² +

\frac{4}{x^{2}}

D. 2x² -

\frac{4}{x}

Correct Answer: Option B

Explanation

y =

\frac{2}{3}

x³ -

\frac{4}{x}

\frac{d y}{d x}

= 2x² - Lx^-2 = 2x² +

\frac{4}{x^{2}}

27. Evaluate

\frac{0.8 \times 0.43 \times 0.031}{0.05 \times 0.72 \times 0.021}

. Correct to four significant figures.

A. 14.09
B. 14.1
C. 14.12
D. 14.11

Correct Answer: Option B

Explanation

Length or arc =

\frac{θ}{360}

x 2

π

r

=

\frac{30}{360}

x 2 x

\frac{22}{7}

x 12

=

\frac{1}{12}

x 2 x

π

x 2 = 2

π

28. An arc subtends an angle of 30^o at the centre of a circle radius 12cm. Calculate the length of the arc.

A. 6 $π$

B. 2

π

C. 3

π

D. 9

π

Correct Answer: Option B

Explanation

Length or arc =

\frac{θ}{360}

x 2

π

r

=

\frac{30}{360}

x 2 x

\frac{22}{7}

x 12

=

\frac{1}{12}

x 2 x

π

x 2 = 2

π

cm

29. Evaluate

\int_{0}^{\frac{π}{2}}

sin xdx

A. 2
B. -1
C. 1
D. -2

Correct Answer: Option C

Explanation

\int_{0}^{\frac{π}{2}}

sin xdx = -cosx|

\frac{π}{2}

= -(cos

\frac{π}{2}

- cos0) = -(0-1) = 1

30. An arc of the length 16

π

cm subtends an angle of 80^o at the centre of the circle. Find the radius of the circle.

A. 24cm
B. 28cm
C. 36cm
D. 32cm

Correct Answer: Option C

Explanation

L =

\frac{θ}{360}

x 2

π

r
16

π

\frac{80}{360}

x 2

π

\frac{16 π \times 360}{80 \times 2 π}

= r

r = 36cm

31.The mean of 2-t, 4+t, 3-2t, 2+t and t-1 is

A. 2
B. t
C. -t
D. -2

Correct Answer: Option A

Explanation

Mean =

\frac{2 - t + 4 + t + 3 - 2 t + 2 + t + t - 1}{5}

\frac{10}{5}

= 2

32. Evaluate

\frac{12.02 \times 20.06}{26.04 \times 60.06}

, correct to three significant figures.

A. 0.157
B. 0.154
C. 0.155
D. 0.158

Correct Answer: Option B

Explanation
$\frac{12.02 \times 20.06}{26.04 \times 60.06}$ = $\frac{241.1212}{15698.96}$ = 0.154

33. If y = 2x³ + 6x² + 6x + 1, Find

\frac{d y}{d x}

A. 6x² + 12x + 1
B. 6x² + 6x + 1
C. 6x² + 6x + 6
D. 6x² + 12x + 6

Correct Answer: Option D

Explanation
y = 2x³ + 6x² + 6x + 1

$\frac{d y}{d x}$ = 6x² + 12x + 6

34.The nth term of the sequence 3, 9, 27, 81.....is

A. 3 x 3^n-2
B. 3 x 3^n-1
C. 3 x 3ⁿ⁺²
D. 3 x 3ⁿ⁺¹

Correct Answer: Option B

Explanation

3,9,27,81,.....
r =

\frac{9}{8}

= 3
T_n = ar^n-1

T_n = 3(3)^n-1 =

\frac{3 (3^{2})}{3}

35. Simplify 1 - (

\frac{1}{7}

x 3

\frac{1}{2}

)

\div

\frac{3}{4}

A. 2
B. $\frac{1}{3}$

C. 1
D. $\frac{2}{3}$

Correct Answer: Option B

Explanation

1 - (

\frac{1}{7}

x 3

\frac{1}{2}

)

\div

\frac{3}{4}

1 - (

\frac{1}{7}

\frac{7}{2}

) x

\frac{4}{3}

1 -

\frac{1}{2}

\frac{4}{3}

1 -

\frac{2}{3}

\frac{1}{3}

36. If a car travels 120km on 45 litres of petrol, how much petrol is needed for a journey of 600km?
- A. 720 litres
- B. 160 litres
- C. 225 litres
- D. 960 litres
Correct Answer: Option C

Explanation
120km $\to$ 45litres
1km $\to$ $\frac{45}{120}$ litres
600km $\to$ $\frac{45}{120}$ x 600 = 225litres

37. Find $\frac{d y}{d x}$

. If y = 3x³ + 2x² + 3x + 1
- A. 9x² + 4x + 3
- B. 9x² - 4x - 3
- C. 9x² + 4x - 3
- D. 9x² - 4x + 3
Correct Answer: Option A

Explanation
y = 3x³ + 2x² + 3x + 1
= 9x² + 4x + 3
38. Evaluate $\int$ (sinx - 5x²)dx
- A. -cosx - 10x + k
- B. cosx - $\frac{5 x^{3}}{3}$ + k
- C. -cosx - $\frac{5 x^{3}}{3}$ + k
- D. cosx - 10x + k
Correct Answer: Option C

Explanation
$\int$ ^{(sin x - 5x^2)} = -cosx - $\frac{5 x^{3}}{3}$ + k

39. If N = $\frac{p}{2}$ ( $\frac{T_{1} - T_{2}}{T_{1}}$ ). Find P when N = 12, T₁ = 27 and T₂ = 24. $\frac{T_{1} - T_{2}}{T_{1}}$
- A. 48
- B. 108
- C. 54
- D. 216
Correct Answer: Option D

Explanation
N = ()
12 = ()
24 = P(
P = 24 x 9 = 216
40, Solve for x and y respectively
3x - 5y = 9
6x - 4y = 12
A. $\frac{3}{4}$ , 1

B. $\frac{4}{3}$ , 1

C. $\frac{3}{4}$ , -1

D. $\frac{4}{3}$ , -1
Correct Answer: Option D

Explanation
3x - 5y = 9 ------x2
6x - 4y = 12 -----x1
6x - 10y = 18
-6x - 4y = 12
____________
=

y = -1
in eq (1) 3x - 5y = 9
3x - 5(-1) = 9
3x + 5 = 9
3x = 4
x =

41. If Q is a factor of 18 and T is prime numbers between 2 and 18. What is Q

T?
- A. (2,3)
- B. (2,3,18)
- C. (2,3,9)
- D. (2,3,6)
Correct Answer: Option A

Explanation
Q = {1,2,3,6,9,18}, T = {2,3,5,7,11,13,17}
Q $\cap$
T = {2,3}

42
From the cyclic quadrilateral above, find < TVS
- A. 60^o
- B. 80^o
- C. 70^o
- D. 50^o
Correct Answer: Option B

Explanation
< TVS = 180^o - (80 + 20)
= 180 - 100 = 80^o

43.Find the mean of 10, 8, 5, 11, 12, 9, 6, 3, 15, and 23.
A. 11.2

B. 10

C. 10.2

D. 11.3

Correct Answer: Option C

Explanation
$\frac{10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23}{10}$
= $\frac{102}{10}$ = 10.2
44
In the cyclic quadrilateral above . Find < PRO
- A. 70^o
- B. 20^o
- C. 50^o
- D. 30^o
Correct Answer: Option D

Explanation
< PRO = 180^o - (100 + 50)
= 180 - 150 = 30^o

45.OGIVE is constructed using

A. Third quartile range

B. Semi-quartile range

C. Cummulative frequency table

D. Inter-quartile table

Correct Answer: Option C

46. Rationalize

\frac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}}

A. 5 + 2

\sqrt{6}

B. 5 - 4 $\sqrt{6}$
C. 5 + 4 $\sqrt{6}$
D. 5 - 2 $\sqrt{6}$

Correct Answer: Option D

Explanation

\frac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}}

\frac{6 - 2 \sqrt{6} - 2 \sqrt{6} + 4}{6 - 4}

\frac{10 - 4 \sqrt{6}}{2}

= 5 - 2

\sqrt{6}

The bar chart above shows the marks obtained by students in a mathematics test.
How many students in all took the test?

A. 40
B. 30
C. 10
D. 20

Correct Answer: Option A

Explanation

3 + 2 + 5 + 5 + + 5 + 10 + 4 + 3 + 1 + 2 = 40

48

Determine the mean score of the student that took the mathematics test?

A. 4.5
B. 4.3
C. 4.2
D. 4.6

Correct Answer: Option C

Explanation

(3 x 0) + (1 x 2) + (5 x 2) + (5 x 3) + (5 x 4) +
(10 x 5) + (4 x 6) + (3 x 7) + (1 x 8) + (2 x 9)

\div

40

=

\frac{168}{40}

= 4.2

49

From the diagram above. Find the fraction of the shaded position?

A. $\frac{1}{3}$

B. $\frac{1}{5}$
C. $\frac{1}{4}$
D. $\frac{1}{6}$

Correct Answer: Option D

Explanation

θ

= 180^o -(90 + 60)

θ

= 180^o - 150^o = 30^o

Fraction of shaded position =

\frac{30}{360}

\frac{30}{360}

\frac{1}{12}

\frac{1}{12}

\frac{1}{6}

From the diagram above, find the required roots/region of x.

A. x

\leq

-1, x

\geq

B. x

\leq

C. -1

\leq

\leq

D. x

\leq

-1, o

\leq

\leq

Correct Answer: Option C

Explanation

The required roots/region of x is -1

\leq

\leq