2017 - JAMB Mathematics Past Questions and Answers - page 1
1
Given T = { even numbers from 1 to 12 }
N = {common factors of 6, 8 and 12}
Find T ∩ N
A
{2, 3}
B
{2, 3, 4}
C
{3, 4, 6}
D
{2}
correct option: d
T = {evenn numbers from 1 to 12}
N = {common factors of 6,8 and 12}
Find T ∩ N
T = {2, 4, 6, 8, 10, 12}
N = {2}
T ∩ N = {2} i.e value common to T & N
2
What is the next number in the series 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)...
A
\(\frac{1}{3}\)
B
\(\frac{2}{8}\)
C
\(\frac{3}{7}\)
D
\(\frac{1}{8}\)
correct option: d
2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\).....
There are 4 terms in the series
Therefore the next number will be the 5th term
Tn = ar\(^{n − 1}\) (formular for geometric series)
a = first term = 2
r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\)
n = number of terms
T5 = 5th term = ?
T5 = ar\(^{5 - 1}\)
= ar\(^4\)
= 2 × (ar\(^{n − 1}\))4
= 2 × \(\frac{1}{16}\)
= \(\frac{1}{8}\)
Users' Answers & CommentsThere are 4 terms in the series
Therefore the next number will be the 5th term
Tn = ar\(^{n − 1}\) (formular for geometric series)
a = first term = 2
r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\)
n = number of terms
T5 = 5th term = ?
T5 = ar\(^{5 - 1}\)
= ar\(^4\)
= 2 × (ar\(^{n − 1}\))4
= 2 × \(\frac{1}{16}\)
= \(\frac{1}{8}\)
3
If U = {x : x is an integer and 1 ≤ x ≤ }
E1 = {x: x is a multiple of 3}
E2 = {x: x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E2
E1 = {x: x is a multiple of 3}
E2 = {x: x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E2
A
\(\frac{3}{4}\)
B
\(\frac{3}{10}\)
C
\(\frac{1}{4}\)
D
\(\frac{1}{20}\)
correct option: a
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
E1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
Probability of E2 = \(\frac{5}{20}\) i.e \(\frac{\text{Total number in}E_2}{\text{Entire number in set}}\)
Probability of set E2 = 1 − \(\frac{5}{20}\)
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
Users' Answers & CommentsE1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
Probability of E2 = \(\frac{5}{20}\) i.e \(\frac{\text{Total number in}E_2}{\text{Entire number in set}}\)
Probability of set E2 = 1 − \(\frac{5}{20}\)
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
4
The curved surface area of a cylinder 5cm high is 110cm2. Find the radius of its base
π = \(\frac{22}{7}\)
π = \(\frac{22}{7}\)
A
2.6cm
B
3.5cm
C
3.6cm
D
7.0cm
correct option: b
Curved surface area of cylinder = 2πrh
110 = 2 × \(\frac{22}{7}\) × r × 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
Users' Answers & Comments110 = 2 × \(\frac{22}{7}\) × r × 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
5
If two graphs Y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of p in terms of q
A
q − \(\frac{8}{7}\)
B
7 − \(\frac{q}{4}\)
C
8 − \(\frac{q}{2}\)
D
7 + \(\frac{q}{8}\)
correct option: b
Y = Px2 + q
Y = 2x2 - 1
Px2 + q = 2x2 - 1
Px2 = 2x2 - 1 - q
p = \(\frac{2x^2 - 1 - q}{x^2}\)
at x = 2
P = \(\frac{2(2)^2 - 1 - q}{2^2}\)
= \(\frac{2(4) - 1 -q}{4}\)
= \(\frac{8 - 1 - q}{4}\)
P = \(\frac{7 - q}{4}\)
Users' Answers & CommentsY = 2x2 - 1
Px2 + q = 2x2 - 1
Px2 = 2x2 - 1 - q
p = \(\frac{2x^2 - 1 - q}{x^2}\)
at x = 2
P = \(\frac{2(2)^2 - 1 - q}{2^2}\)
= \(\frac{2(4) - 1 -q}{4}\)
= \(\frac{8 - 1 - q}{4}\)
P = \(\frac{7 - q}{4}\)
6
A
\(\frac{\sqrt{3}}{2\sqrt{2}}\)
B
√3 − \(\frac{1}{2}\)
C
\(\frac{1}{2}\)√2
D
1 + \(\frac{\sqrt{2}}{2}\)
correct option: d
hypotenuse
sin = \(\frac{1}{2}\)
\(\sin45 = \frac{1}{\sqrt{2}}\)
= \(\frac{2}{2}\)
∴ (sin45 + sin30)
= \(\frac{1}{\sqrt{2}} + \frac{1}{2}\)
= \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\)
= \(\frac{\sqrt{2} + 1}{2}\)
= \(\frac{1 + \sqrt{2}}{2}\)
Users' Answers & Commentssin = \(\frac{1}{2}\)
\(\sin45 = \frac{1}{\sqrt{2}}\)
= \(\frac{2}{2}\)
∴ (sin45 + sin30)
= \(\frac{1}{\sqrt{2}} + \frac{1}{2}\)
= \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\)
= \(\frac{\sqrt{2} + 1}{2}\)
= \(\frac{1 + \sqrt{2}}{2}\)
7
If y = x Sin x, find \(\frac{dy}{dx}\) when x = \(\frac{\pi}{2}\)
A
\(\frac{- \pi}{2}\)
B
-1
C
1
D
\(\frac{ \pi}{2}\)
correct option: c
y = xsinx
\(\frac{dy}{dx}\) = \(1 \sin x + x \cos x\)
= \(\sin x + x \cos x\)
At x = \(\frac{\pi}{2}\)
= sin\(\frac{\pi}{r}\) + \(\frac{\pi}{2} \cos {\frac{\pi}{2}}\)
= 1 + \(\frac{\pi}{2}\) × 10
= 1
Users' Answers & Comments\(\frac{dy}{dx}\) = \(1 \sin x + x \cos x\)
= \(\sin x + x \cos x\)
At x = \(\frac{\pi}{2}\)
= sin\(\frac{\pi}{r}\) + \(\frac{\pi}{2} \cos {\frac{\pi}{2}}\)
= 1 + \(\frac{\pi}{2}\) × 10
= 1
8
If temperature t is directly proportional to heat h, and when t = 20oC, h = 50 J, find t when h = 60J
A
24oC
B
20oC
C
34oC
D
30oC
correct option: a
t ∝ h, t = 20, h
t = ? h = 60
t = kh where k is constant
20 = 50k
k = \(\frac{20}{50}\)
k = \(\frac{2}{5}\)
when h = 60, t = ?
t = \(\frac{2}{5}\) × 60
t = 24oC
Users' Answers & Commentst = ? h = 60
t = kh where k is constant
20 = 50k
k = \(\frac{20}{50}\)
k = \(\frac{2}{5}\)
when h = 60, t = ?
t = \(\frac{2}{5}\) × 60
t = 24oC
9
Evaluate 1 - (\(\frac{1}{5}\) x \(\frac{2}{3}\)) + ( 5 + \(\frac{2}{3}\))
A
4
B
3
C
2\(\frac{2}{3}\)
D
3\(\frac{2}{3}\)
correct option: d
Users' Answers & Comments10
Given m = N\(\frac{\sqrt{SL}}{T}\) make T the subject of the formula
A
\(\frac{\text{NSL}}{M}\)
B
\(\frac{N^2SL}{M^2}\)
C
\(\frac{N^2SL}{M}\)
D
\(\frac{NSL}{M^2}\)
correct option: b
M = \(\frac{\sqrt{SL}}{T}\),
make T subject of formula square both sides
M2 = \(\frac{N^2SL}{T}\)
TM2 = N2SL
T = \(\frac{N^2SL}{M^2}\)
Users' Answers & Commentsmake T subject of formula square both sides
M2 = \(\frac{N^2SL}{T}\)
TM2 = N2SL
T = \(\frac{N^2SL}{M^2}\)