2008 - JAMB Mathematics Past Questions and Answers - page 6
51
The cost of a kerosene per litre increases from N60 to N85. What is the percentage rate of increase?
A
42%
B
41%
C
40%
D
25%
correct option: a
Users' Answers & Comments52
Simplify 16\(\frac{1}{2}\) x 4\(\frac{1}{2}\) x 27\(\frac{1}{3}\)
A
\(\frac{3}{8}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{4}\)
D
\(\frac{3}{2}\)
correct option: a
Simplify 16\(\frac{1}{2}\) x 4\(\frac{1}{2}\) x 27\(\frac{1}{3}\)
= 16\(\frac{1}{2}\) x 4\(\frac{1}{2}\) x 27\(\frac{1}{3}\)
= \(\frac{1}{\sqrt{16}^1} \times \frac{1}{\sqrt{4}^1}\) x (3\(\sqrt{27}\))1
= \(\frac{1}{4}\) x \(\frac{1}{2}\) x \(\frac{3}{1}\)
= \(\frac{3}{8}\)
Users' Answers & Comments= 16\(\frac{1}{2}\) x 4\(\frac{1}{2}\) x 27\(\frac{1}{3}\)
= \(\frac{1}{\sqrt{16}^1} \times \frac{1}{\sqrt{4}^1}\) x (3\(\sqrt{27}\))1
= \(\frac{1}{4}\) x \(\frac{1}{2}\) x \(\frac{3}{1}\)
= \(\frac{3}{8}\)
53
If \(logX^{\frac{1}{2}}\) 64 = 3, find the value of x.
A
4
B
16
C
32
D
64
correct option: b
\(logX^{\frac{1}{2}}\) 64 = 3, find the value of X
recall that X = 10p
log10X = P
64 = \(X^{\frac{1}{2}}\)
43 = \(X^{\frac{1}{2}}\)
\(4^{3 \times \frac{1}{2}}\) = \(X^{\frac{1}{2} \times \frac{1}{3}}\)
4 = \(X^{\frac{1}{2}}\)
(4)2 = \(X^{\frac{1}{2} \times 2}\)
x = 16
Users' Answers & Commentsrecall that X = 10p
log10X = P
64 = \(X^{\frac{1}{2}}\)
43 = \(X^{\frac{1}{2}}\)
\(4^{3 \times \frac{1}{2}}\) = \(X^{\frac{1}{2} \times \frac{1}{3}}\)
4 = \(X^{\frac{1}{2}}\)
(4)2 = \(X^{\frac{1}{2} \times 2}\)
x = 16
54
If \(\frac{1 + \sqrt{2}}{1 - \sqrt{2}}\) is expressed in the form x + y\(\sqrt{2}\). Find the values of x and y
A
(-3, -2)
B
(-2, 3)
C
(3, 2)
D
(2, -3)
correct option: a
\(\frac{1 + \sqrt{2}}{1 - \sqrt{2}}\) x \(\frac{1 + \sqrt{2}}{1 + \sqrt{2}}\)
= \(\frac{(1 + \sqrt{2}) (1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}\)
= \(\frac{1 + \sqrt{2} + \sqrt{2} + (\sqrt{2})^2}{1 + \sqrt{2} - \sqrt{2} - (\sqrt{2})^2}\)
= \(\frac{1 + 2\sqrt{2} + \sqrt{2}}{1^2 - 2}\)
= \(\frac{3 + 2\sqrt{2}}{-1}\)
-3 - 2\(\sqrt{2}\) = x + y\(\sqrt{2}\)
x = -3, y = -2
= (-3, -2)
Users' Answers & Comments= \(\frac{(1 + \sqrt{2}) (1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}\)
= \(\frac{1 + \sqrt{2} + \sqrt{2} + (\sqrt{2})^2}{1 + \sqrt{2} - \sqrt{2} - (\sqrt{2})^2}\)
= \(\frac{1 + 2\sqrt{2} + \sqrt{2}}{1^2 - 2}\)
= \(\frac{3 + 2\sqrt{2}}{-1}\)
-3 - 2\(\sqrt{2}\) = x + y\(\sqrt{2}\)
x = -3, y = -2
= (-3, -2)
55
if X = {n2 + 1 : n = 0, 2, 3} and y = {n + 1 : n = 2, 3, 4, 5}, find x ∩ y
A
{1, 3}
B
{5,10}
C
\(\Phi\)
D
(4, 6)
correct option: c
X = {n2 + 1 : n = 0, 2, 3}
y = {n + 1 : n = 2, 3, 4, 5}
find x ∩ y
x = (n2 + 1)(22 + 1)(32 + 1)
y = (2 + 1) (3 + 1)(5 + 1)
x = {1, 5, 10}
y = {3, 4, 6}
x ∩ y = \(\Phi\)
Users' Answers & Commentsy = {n + 1 : n = 2, 3, 4, 5}
find x ∩ y
x = (n2 + 1)(22 + 1)(32 + 1)
y = (2 + 1) (3 + 1)(5 + 1)
x = {1, 5, 10}
y = {3, 4, 6}
x ∩ y = \(\Phi\)
56
Make Q the subject of the formula when L = \(\frac{4}{3}\) m\(\sqrt{PQ}\)
A
\(\frac{9L^2}{16m^2P}\)
B
\(\frac{3L}{4m \sqrt{P}}\)
C
\(\frac {\sqrt{3L}}{4mP}\)
D
\(\frac{3L^2}{16m^2P}\)
correct option: a
L = \(\frac {4}{3}\) m \(\sqrt{PQ}\)
make Q the subject
\(\frac {L}{1}\) = \(\frac {4}{3}\) m \(\sqrt{PQ}\)
3L = 4m \(\sqrt{PQ}\)
Taking the square of both sides
(3L)2 = (4m \(\sqrt{PQ}\))
\(\frac {9L^2}{16m^2P}\) = \(\frac {16m^2PQ}{16m^2P}\)
Q = \(\frac {9L^2}{16m^2P}\)
Users' Answers & Commentsmake Q the subject
\(\frac {L}{1}\) = \(\frac {4}{3}\) m \(\sqrt{PQ}\)
3L = 4m \(\sqrt{PQ}\)
Taking the square of both sides
(3L)2 = (4m \(\sqrt{PQ}\))
\(\frac {9L^2}{16m^2P}\) = \(\frac {16m^2PQ}{16m^2P}\)
Q = \(\frac {9L^2}{16m^2P}\)
57
If 2x2 - kx - 12 is divisible by x - 4, find the value of k
A
4
B
5
C
6
D
7
correct option: b
Users' Answers & Comments58
Factorize completely (4x + 3y)2 - (3x - 2y)2
A
(x + 5y)(7x + y)
B
(x + 5y)(7x - y)
C
(x - 5y)(7x + y)
D
(x - 5y)(7x -y)
correct option: a
(4x + 3y)2 - (3x - 2y)2
= 16x2 + 24xy + 9y - (9x2 - 12xy + 4y2)
= 7x2 + 36xy + 5y
= 7x2 + xy + 35xy + 5y2
= (7x2 + xy) + (35xy + 5y2)
= x(7x + y) + 5y(7x + y)
= (7x + y)(x + 5y)
Users' Answers & Comments= 16x2 + 24xy + 9y - (9x2 - 12xy + 4y2)
= 7x2 + 36xy + 5y
= 7x2 + xy + 35xy + 5y2
= (7x2 + xy) + (35xy + 5y2)
= x(7x + y) + 5y(7x + y)
= (7x + y)(x + 5y)
59
If x - 3 is directly proportional to the square of y and x = 5 when y = 2, find x when y = 6
A
30
B
21
C
16
D
12
correct option: b
x - 3 x y2
x - 3 = ky2
when x = 5, y = 2
5 - 3 = k(2)2
2 = 4k
k = \(\frac{2}{4}\)
k = \(\frac{1}{2}\)
Find x when y = 6
from x - 3 = ky2
x - 3 = \(\frac{1}{2}\)ky2
x - 3 = \(\frac{1}{2}\) x 36
x - 3 = 18
x = 18 + 3
x = 21
Users' Answers & Commentsx - 3 = ky2
when x = 5, y = 2
5 - 3 = k(2)2
2 = 4k
k = \(\frac{2}{4}\)
k = \(\frac{1}{2}\)
Find x when y = 6
from x - 3 = ky2
x - 3 = \(\frac{1}{2}\)ky2
x - 3 = \(\frac{1}{2}\) x 36
x - 3 = 18
x = 18 + 3
x = 21
60
If p varies inverse as the square of q and p = 8 when q =4, find q when p = 32
A
\(\pm\)16
B
\(\pm\)8
C
\(\pm\)4
D
\(\pm\)2
correct option: d
P \(\alpha\) \(\frac{1}{q^2}\)
p = \(\frac{k}{q^2}\)
k = pq2
k = 8 x 42
k = 8 x 16 = 128
find k = pq2
q2 = \(\frac{k}{p}\)
q = \(\frac{k}{p}\)
q = \(\frac{128}{32}\)
q = √4
q = \(\pm\)2
Users' Answers & Commentsp = \(\frac{k}{q^2}\)
k = pq2
k = 8 x 42
k = 8 x 16 = 128
find k = pq2
q2 = \(\frac{k}{p}\)
q = \(\frac{k}{p}\)
q = \(\frac{128}{32}\)
q = √4
q = \(\pm\)2