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1 Find the value of 1101112 + 101002
A 11010112 B 10010012 C 10010112 D 10011112

2 A woman bought a grinder for N60,000. She sold it at a loss of 15%. How much did she sell it? A N53,000 B N52,000 C N51,000 D N50,000

3 Express the product of 0.00043 and 2000 in standard form. A 8.6 x 10-3 B 8.3 x 10–2 C 8.6 x 10-1 D 8.6 x 10

4 A man donates 10% of his monthly net earnings to his church. If it amounts to N4,500, what is his net monthly income? A N40,500 B N45,000 C N52,500 D N62,000

5 If log7.5 = 0.8751, evaluate 2 log75 + log750
A 6.6252 B 6.6253 C 66.252 D 66.253

6 Solve for x in 8x-2 = 2/25
A 4 B 6 C 8 D 10

7 Simplify
A 3 B 3 C 3 D 3

8 Evaluate Log28 + Log216 – Log24
A 3 B 4 C 5 D 6

9 If P = {1,2,3,4,5} and P  Q = {1,2,3,4,5,6,7}, list the elements in Q A {6} B {7} C {6,7} D {5,6}

10 If gt2 – k – w = 0, make g the subject of the formula

A k+w t2
B k−w t2
C k+w t
D k−w t

11 Factorize 2y2 – 15xy + 18×2
A (2y – 3x) (y + 6x) B (2y – 3x) (y – 6x) C (2y + 3x) (y – 6x) D (3y + 2x) (y – 6x)

12 Find the value of k if y – 1 is a factor of y3 + 4y2 + ky – 6 A -6 B -4 C O
D 1

13 y varies directly as w2. When y = 8, w = 2. Find y when w = 3 A 18 B 12 C 9 D 6

14 P varies directly as Q and inversely as R. When Q = 36 and R = 16, P = 27. Find the relation between P, Q and R.

15 What is the solution of ?
A -3 < x < 1 B x < -3 or x > 1 C -3 < x < 5 D x < –3 or x > 5

17 The 4th term of an A.P. is 13 while the 10th term is 31. Find the 24th term. A 89 B 75 C 73 D 69

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18 What is the common ratio of the G.P. ?

19 A binary operation * is defined by x * y = xy. If x * 2 = 12 – x, find the possible values of x A 3,4 B 3,-4 C -3,4 D -3,-4

23 How many sides has a regular polygon whose interior angle is 135o

24 A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if the radius of its base is 28cm? A 8.0m B 7.5m C 5.0m D 2.5m

25 The locus of a dog tethered to a pole with a rope of 4m is a A circle with diameter 4m B circle with radius 4m C semi-circle with diameter 4m D semi-circle with radius 4m

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26 Find the mid point of S(-5, 4) and T(-3, -2)
A -4, 2 B 4, -2 C -4, 1 D 4, -1

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27 The gradient of a line joining (x,4) and (1,2) is . Find the value of x
A 5 B 3 C -3 D -5

28 Calculate the mid point of the line segment y 4x + 3 = 0, which lies between the x-axis and y-axis. A B C D

29 Find the equation of the straight line through (-2, 3) and perpendicular to 4x + 3y – 5 = 0 A 3x – 4y + 18 = 0 B 3x + 2y – 18 = 0 C 4x + 5y + 3 = 0
( ) 3 −3 8 2
( ) 3 3 8 2
( ) −2 2 2 2
( ) −2 3 3 2
D 5x – 2y – 11 = 0

…
36 The mean of 2 – 4, 4 + t, 3 – 2t and t – 1 is
A t B -t C 2 D -2

38 Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6 A 6 B 5 C 4 D 3

45 From the figure above, what is the value of p? A 135o B 90o C 60o D 450

46 Find the value of x in the figure above A 20 cm B 10 cm 3 –√ 3 –√
C 5 cm D 4 cm

47 in the figure above, what is the equation of the line that passes the y-axis at (0,5) and passes the x-axis at (5,0)? A y = x + 5 B y = -x + 5 C y = x – 5 D y = -x – 5

48 The pie chart above shows the monthly distribution of a man’s salary on food items. If he spent N8,000 on rice, how much did he spent on yam? A N42,000 B N18,000 C N16,000 D N12,000

Answers: JAMB Past Questions: JAMB » Mathematics

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1. C

2. C

3. C

4. B

5. B

6. D
7. A
8. C

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9. C  Q = (P  Q) – P {6,7}

10. A  gt2 – k – w = 0  gt2 = k + w
11. B  2y2 – 15xy + 18×2  2y2 – 12xy – 3xy + 18×2
2y(y – 6x) – 3x(y – 6x)
= 5lo 2 g2
= 5 × 1
= 5 ∪
g = k+w t2
(2y – 3x) (y – 6x)

12. D  if y – 1 is a factor of y3 + 4y2 + ky – 6, then  f(1) = (1)3 + 4(1)2 + k(1) – 6 = 0 (factor theorem)
1 + 4 + k – 6 = 0
5 – 6 + k = 0
-1 + k = 0
k = 1

13. A
y = kw2  8 = k(2)2
8 = k(4)
k = 8/4
y ∝ w2
k = 2  Thus y = 2w2 When w = 3, y = 2(3)2
y = 2 x 9 = 18

14. B
When Q = 36, R = 16, P = 27
Then substitute into the equation
So the equation connecting P, Q and R is
P ∝ Q R
P = K Q R
27 = K 36 16
K = 27×16 36
K = 12

15. A  Consider the range -3 < x < -1
= { -2, -1, 0}, for instance
When x = -2,
When x = -1,
= -3 < -1
When x = 0,
P = 12Q R
< −1− 2−5 −2+3
< −1− 7 1
< −1− 1−5 −1+3
< −1− 6 2
< −10 −5 0+3

Hence -3 < x < 1

16. A
6x + 9  10x – 7
6x – 10x  – 7 – 9
-4x  -16
-4x/-4  -16/-4
x  4 17. C  a + 3d = 13 ………. (1) a + 9d = 31 ………. (2)
(2) – (1): 6d = 18
d = 18/6 = 3
< −1− 5 3
+ ≤ −x 2 3 4 5x 67 12
12 + 12 ≤ 12 − 12 x 2 3 4 5x 6
7 12
≤
From (1), a + 3(3) = 13
a + 9 = 13
a = 13 – 9 = 4
Hence, T24 = a + 23d T24 = 4 + 23(3) T24 = 4 + 69 T24 = 73

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18. A  Common ratio r of the G.P is
r = = +1T n Tn
T2 T1
r = +2 10√ 5 √ +10√ 5 √
r = × +210√ 5 √ +10√ 5 √
−10√ 5 √ −10√ 5 √
= ( )( )+( )(− )+(2 )( )+(2 )(− ) 10√ 10 √ 10 √ 5 √ 5 √ 10 √ 5 √ 5 √ ( −( 10√ )2 5 √ )2
10− +2 −10 50√ 50 √ 10−5

19. B  x * y = xy x * 2 = 12 – x
Thus by comparison,
x = x, y = 2
But x * y = x * 2  xy = 12 – x  x2 = 12 – x  x2 + x – 12 = 0  x2 + 4x – 3x – 12 = 0
50√ 5
25×2√ 5
5 2 √ 5
2 –√
x(x + 4) – 3(x + 4) = 0
(x – 3)(x + 4) = 0
x – 3 = 0 or x + 4 = 0
So x = 3 or x = -4

20. C
By matrices multiplication;
5x – 6y = 7 ……..(1) 2x – 7y = -11 ……(2) 2 x (1): 10x – 12y = 14 …….(3) 5 x (2): 10x – 35y = -55 ……(4)
(3) – (4): 23y = 69
y = 69/23 = 3

21. D
( )( )=( ) 5 2 −6 −7 5 2 7 −11

-4x – (-1)12 = -12
-4x + 12 = -12
-4x = -12 – 12
-4x = – 24
x = 6

22. D
= 0(28 – 40) – 3(4 – 0) + 2(5 – 0)
= 0(-12) – 3(4) + 2(5)
= 0 – 12 + 10
= -2

23. D
= −12
∣ ∣ ∣
−x −1
12 4
∣ ∣ ∣
0 − 3 + 2 ∣ ∣ ∣ 7 5 8 4 ∣ ∣ ∣ ∣ ∣ ∣ 1 0 8 4 ∣ ∣ ∣
∣ ∣ ∣
1 0
7 5
∣ ∣ ∣
If each interior angle of the polygon is 135o, then each exterior angle is 180o – 135o = 45o. Hence, number of sides =
= 8

24. D  Using
6160 = 22/7 x 28 x 28 x h

25. B

26. C  Mid point of S(-5, 4) and T(-3, -2) is
360o one exterior angle
360o 45o
V = π h r2
h = 6160 22×4×28
h = 2.5m
[ (−5 + −3), (4 + 2)] 1 2 1 2

27. A
1 – x = 2(2 – 4)
1 – x = 4 – 8
1 – x = -4
-x = -4 – 1
x = 5

28. A  y – 4x + 3 = 0
When y = 0, 0 – 4x + 3 = 0
[ ( + ), ( + )] 1 2 x1 x2 1 2 y1 y2
[ (−8), (2)] 1 2 1 2
[−4,1]
Gradient m = − y2 y1 −x 2 x1
=1 2
2−4 1−x
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y – 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y – 4x + 3 = 0, which lies between the x-axis and the y-axis is;

29. A
[ ( + ), ( + )] 1 2 x1 x2 1 2 y1 y2
[ ( + 0), (0 + −3)] 1 2 3 4 1 2
[ ( ), (−3)] 1 2 3 4 1 2
[ , ] 3 8 −3 2
4x + 3y – 5 = 0 (given)
The equation of the line perpendicular to the given line takes the form 3x – 4y = k
Thus, substitution x = -2 and y = 3 in 3x – 4y = k gives;
3(-2) – 4(3) = k
-6 – 12 = k
k = -18
Hence the required equation is 3x – 4y = –18
3x – 4y + 18 = 0

30. B

31. D  If y = 4×3 – 2×2 + x, then;   = 3(4×2) – 2(2x) + 1 δ y δx
= 12×2 – 4x + 1

32. D  y = cos 3x
Let u = 3x so that y = cos u
Now, ,
By the chain rule,

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33. D  y = x2 – 2x – 3,
= 3δ y δx
= −sinuδ y δx
= ×δ y δx δy δu
δu δx
= (−sinu)(3)δ y δx
= −3sinuδ y δx
= −3sin3xδ y δx
Then
But at minimum point, ,
Which means 2x – 2 = 0
2x = 2
x = 1.  Hence the minimum value of y = x2 – 2x – 3 is;  ymin = (1)2 – 2(1) – 3  ymin = 1 – 2 – 3  ymin = -4

34. C

35. C
= 2x − 2δ y δx
= 0δ y δx
∫ sin2xdx = (−cos2x) + k 1 2
− cos2x + k 1 2

let u = 2x + 3,
Now

36. C
Mean x =
= [(2 – t) + (4 + t) + (3 – 2t) + (2 + t) + (t – 1) ] 5
= [11 – 1 + 3t – 3t]  5
= 10  5
∫(2x + 3 δx ) 1 2
= 2δ y δx
δx = δu 2
∫(2x + 3 δx = ∫ . ) 1 2 u 1 2
δx 2
= ∫ δu 1 2 u 1 2
= × + k 1 2 u 3 2 2 3
= + k 1 3 u 3 2
= (2x + 3 + k 1 3 ) 3 2
∑ x n
= 2

37. D

38. B  First arrange the numbers in order of magnitude; 1,2,3,3,4,5,5,5,5,6,7,8,9,9,10
Hence the median = 5

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39. A
Mean x =
= 3
∑ x n
= 5+4+3+2+1 5
= 15 5

Hence, standard deviation;

40. C  A team of 2 girls can be selected from 7 girls in
x 5 4 3 2 1
d = x − 3 2 1 0 −1 −2
d2 4 1 0 1 4 ∑ + 10 d2
= = ∑ d2 n − − −−√ 10 5 −−√
= 2 –√
7C3
= 7! (7−3)!3!
= ways 7! 4!3!

41. C  Let E demote the event of obtaining at least a 4 Then n(E) = 16 + 10 + 14 = 40
Hence, prob (E) =

42. C  Sample space S = {10, 11, 12, … 30}
Let E denote the event of choosing a number divisible by 3
Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7
Prob (E) =
Prob (E) =
Prob (E) =

43. A
n(E) n(S)
= 40 100
= 2 5
n(E) n(E)
7 21
1 3

44. C  In the diagram above,  = 54o(alternate angles; KL||MN) < KNM = 2 (LN is bisector of < KNM) = 108o  35o + < KMN + 108o = 180o(sum of angles of )  < KMN + 143o = 180o  < KMN = 180o – 143o  = 37o

45. B  In the figure above, qo = 30o (vertically opposite angles)  (P + 2q)o + 30o = 180o(angles on a straight line)  p + 2 x 30o + 30o = 180o  p + 60o + 30o = 180o  p + 90o = 180o  α α △
p = 180o – 90o  = 90o

46. B  In the figure above,  (Sine rule)
x =
= 10 x
= 10 x
= 10 cm

47. B  (x1, y1) = (0,5)  (x2, y2) = (5, 0)
Using
=xsin 60o
10 sin 30o
10 sin 60o sin 30o
×3√ 2
1 2
×3√ 2
2 1
3 –√
=y −y1 −y 1 y1
x−x1 −x 1 x1
=y −5 0−5
x−0 5−0

5(y – 5) = -5x
y – 5 = -x
x + y = 5
y = -x + 5

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48. C  Angle of sector subtended by yam   = 360o – (70 + 80 + 50)o  = 360o – 200o  = 160o
But  x T = 8000
T =
= N36,000
Hence the amount spent on yam =
=y −5 −5
x 5
80o 360o
8000×360o 80o
× N36,000160 o 260
= N16,000

 

 

 

 

 

 

 

 

 

 

 

 

 

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