2010 - JAMB Mathematics Past Questions and Answers - page 1
1
If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.
A
2y + 3x
B
2y - 3x
C
3x + 2y
D
3x - 2y
correct option: d
27x3 - 8y3 = (3x - 2y)3
But 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y
Users' Answers & CommentsBut 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y
2
Factorize completely \(\frac{x^3 + 3x^2 - 10x}{2x^2 - 8}\)
A
\(\frac{x(x - 5)}{2(x + 2)}\)
B
\(\frac{x(x + 5)}{2(x + 2)}\)
C
\(\frac{x(x - 5)}{2(x - 2)}\)
D
\(\frac{x ^2 + 5}{2x + 4}\)
correct option: b
\(\frac{x^3 + 3x^2 - 10x}{2x^2 - 8}\) = \(\frac {x(x^2 + 3x - 10)}{2(x^2 - 4)}\)
= \(\frac {x(x^2 + 5x - 2x - 10)}{2(x + 2)(x - 2)}\)
= \(\frac {x(x - 2)(x + 5)}{2(x + 2)(x - 2)}\)
= \(\frac {x(x + 5)}{2(x + 2)}\)
Users' Answers & Comments= \(\frac {x(x^2 + 5x - 2x - 10)}{2(x + 2)(x - 2)}\)
= \(\frac {x(x - 2)(x + 5)}{2(x + 2)(x - 2)}\)
= \(\frac {x(x + 5)}{2(x + 2)}\)
3
Solve for x and y if x - y = 2 and x2 - y2 = 8
A
(-1, 3)
B
(3, 1)
C
(-3, 1)
D
(1, 3)
correct option: b
x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = \(\frac{12}{4}\)
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1
Users' Answers & Commentsx2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = \(\frac{12}{4}\)
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1
4
If y varies directly as the square root of x and y =3 when x = 16, calculate y when x = 64
A
6
B
12
C
3
D
5
correct option: a
y = \(\alpha\)√x ........(1)
y = k√x ........(2)
When y = 3, x = 16,
(2) becomes 3 = k√16 or 3 = k x 4
giving k = \(\frac{3}{4}\)
from (2), y = \(\frac{3}{4}\)√x
When x = 64, y = \(\frac{3}{4}\)√64
y = \(\frac{3}{4}\) x 8
= 6
Users' Answers & Commentsy = k√x ........(2)
When y = 3, x = 16,
(2) becomes 3 = k√16 or 3 = k x 4
giving k = \(\frac{3}{4}\)
from (2), y = \(\frac{3}{4}\)√x
When x = 64, y = \(\frac{3}{4}\)√64
y = \(\frac{3}{4}\) x 8
= 6
5
If x is inversely proportional to y and x = 2\(\frac{1}{2}\) when y = 2, find x if y = 4
A
4
B
5
C
1\(\frac{1}{4}\)
D
2\(\frac{1}{4}\)
correct option: c
x \(\alpha\) \(\frac{1}{y}\) .........(1)
x = k x \(\frac{1}{y}\) .........(2)
When x = 2\(\frac{1}{2}\)
= \(\frac{5}{2}\), y = 2
(2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\)
giving k = 5
from (2), x = \(\frac{5}{y}\)
so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)
Users' Answers & Commentsx = k x \(\frac{1}{y}\) .........(2)
When x = 2\(\frac{1}{2}\)
= \(\frac{5}{2}\), y = 2
(2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\)
giving k = 5
from (2), x = \(\frac{5}{y}\)
so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)
6
For what range of values of x is \(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)?
A
x < \(\frac{3}{2}\)
B
x > \(\frac{3}{2}\)
C
x < -\(\frac{3}{2}\)
D
x > -\(\frac{3}{2}\)
correct option: b
\(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)
Multiply through by through by the LCM of 2, 3 and 4
12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\)
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
\(\frac{2x}{2}\) > \(\frac{3}{2}\)
x > \(\frac{3}{2}\)
Users' Answers & CommentsMultiply through by through by the LCM of 2, 3 and 4
12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\)
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
\(\frac{2x}{2}\) > \(\frac{3}{2}\)
x > \(\frac{3}{2}\)
7
Solve the inequalities -6 \(\leq\) 4 - 2x < 5 - x
A
-1 < x < 5
B
-1 < x \(\leq\) 5
C
-1 \(\leq\) x \(\leq\) 6
D
-1 \(\leq\) x < 6
correct option: b
-6 \(\leq\) 4 - 2x < 5 - x
split inequalities into two and solve each part as follows:
-6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x
-10 \(\leq\) -2x
\(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\)
giving 5 \(\geq\) x or x \(\leq\) 5
4 - 2x < 5 - x
-2x + x < 5 - 4
-x < 1
\(\frac{-x}{-1}\) > \(\frac{1}{-1}\)
giving x > -1 or -1 < x
Combining the two results, gives -1 < x \(\leq\) 5
Users' Answers & Commentssplit inequalities into two and solve each part as follows:
-6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x
-10 \(\leq\) -2x
\(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\)
giving 5 \(\geq\) x or x \(\leq\) 5
4 - 2x < 5 - x
-2x + x < 5 - 4
-x < 1
\(\frac{-x}{-1}\) > \(\frac{1}{-1}\)
giving x > -1 or -1 < x
Combining the two results, gives -1 < x \(\leq\) 5
8
Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + .....
A
\(\frac{5}{8}\)
B
\(\frac{5}{7}\)
C
\(\frac{5}{11}\)
D
\(\frac{5}{9}\)
correct option: d
Using S\(\infty\) = \(\frac{a}{1 - r}\)
r = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\)
S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\)
= \(\frac{0.5}{({\frac{9}{10}})}\)
= \(\frac{0.5 \times 10}{9}\)
= \(\frac{5}{9}\)
Users' Answers & Commentsr = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\)
S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\)
= \(\frac{0.5}{({\frac{9}{10}})}\)
= \(\frac{0.5 \times 10}{9}\)
= \(\frac{5}{9}\)
9
If \(\begin{vmatrix} x & 3 \ 2 & 7 \end{vmatrix}\) = 15, find the value of x
A
4
B
5
C
2
D
3
correct option: d
If \(\begin{vmatrix} x & 3 \ 2 & 7 \end{vmatrix}\) = 15
7x - 2 x 3 = 15
7x - 6 = 15
7x = 15 + 6 = 21
therefore x = \(\frac{21}{7}\)
= 3
Users' Answers & Comments7x - 2 x 3 = 15
7x - 6 = 15
7x = 15 + 6 = 21
therefore x = \(\frac{21}{7}\)
= 3
10
Evaluate \(\begin{vmatrix} 2 & 0 & 5 \ 4 & 6 & 3 \ 8 & 9 & 1 \end{vmatrix}\)
A
5y - 2x -18 = 0
B
102
C
-102
D
-42
correct option: c
\(\begin{vmatrix} 2 & 0 & 5 \ 4 & 6 & 3 \ 8 & 9 & 1 \end{vmatrix}\)
= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102
Users' Answers & Comments= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102