2007 - JAMB Mathematics Past Questions and Answers - page 2
11
\(\begin{array}{c|c} Marks & 3 & 4 & 5 & 6 & 7 & 8\\
\hline Frequency & 5 & y - 1 & y & 9 & 4 & 1\end{array}\)
The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, the value of y
The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, the value of y
A
11
B
15
C
9
D
13
correct option: a
(3 x 5) + (4(y - 1)) + (5y) + (6 x 9) + (7 x 4) + 8 = 200
15 + 4y - 4 + 5y + 54 + 28 + 8 = 200
9y + 105 - 4 = 200
9y + 101 = 200
9y = 200 - 101
9y = 99
y = 11
Users' Answers & Comments15 + 4y - 4 + 5y + 54 + 28 + 8 = 200
9y + 105 - 4 = 200
9y + 101 = 200
9y = 200 - 101
9y = 99
y = 11
12
What is the mean deviation 0f 3, 5, 8, 11, 12 and 21?
A
37
B
60
C
10
D
4.7
correct option: d
mean(\(\bar{x}\)) = \(\frac{\sum x}{n}\)
= \(\frac{7 + 5 + 2 + 1 + 2 + 11}{6}\) = 10
M.D = \(\frac{7 + 5 + 2 + 1 + 2 + 11}{6}\)
= \(\frac{28}{6}\)
= 4.7
Users' Answers & Comments= \(\frac{7 + 5 + 2 + 1 + 2 + 11}{6}\) = 10
M.D = \(\frac{7 + 5 + 2 + 1 + 2 + 11}{6}\)
= \(\frac{28}{6}\)
= 4.7
13
Find the sum to infinity of the series 2 + \(\frac{2}{3}\) + \(\frac{9}{8}\) + \(\frac{27}{32}\) + ....
A
1
B
2
C
4
D
8
correct option: d
a = 2, r = \(\frac{{\frac{3}{2}}}{3}\)
= \(\frac{3}{4}\)
S\(\infty\) = \(\frac{a}{1 - r}\)
\(\frac{2}{1 - \frac{3}{4}}\)
= \(\frac{2}{\frac{1}{4}}\)
= 8
Users' Answers & Comments= \(\frac{3}{4}\)
S\(\infty\) = \(\frac{a}{1 - r}\)
\(\frac{2}{1 - \frac{3}{4}}\)
= \(\frac{2}{\frac{1}{4}}\)
= 8
14
W \(\alpha\) L2 and W = 6 when L = 4. If L = \(\sqrt{17}\), find W.
A
6\(\frac{3}{8}\)
B
\(\frac{7}{8}\)
C
\(\frac{1}{8}\)
D
\(\frac{5}{8}\)
correct option: a
W \(\alpha\) L2, \(\frac{W}{L^2}\) = k(constant)
\(\frac{6}{42}\) = k
\(\frac{W}{(\sqrt{17})^2}\)
W = \(\frac{17 \times 6}{16}\)
= \(\frac{51}{8}\)
= 6\(\frac{3}{8}\)
Users' Answers & Comments\(\frac{6}{42}\) = k
\(\frac{W}{(\sqrt{17})^2}\)
W = \(\frac{17 \times 6}{16}\)
= \(\frac{51}{8}\)
= 6\(\frac{3}{8}\)
15
A binary operation \(\Delta\) is defined is defined by a \(\Delta\) b = a + b + 1 for any real numbers a and b. Find the inverse of the real number 7 under the operation \(\Delta\), if the identify element is -1
A
-1
B
5
C
-7
D
-9
correct option: d
a \(\Delta\) b = a + b + 1
a \(\Delta\) a-1 = e
7 \(\Delta\) 7-1 = 7 + 7-1
therefore 7-1 = -1 - 8
= -9
the inverse of 7 under the operation \(\Delta\) is -9
Users' Answers & Commentsa \(\Delta\) a-1 = e
7 \(\Delta\) 7-1 = 7 + 7-1
therefore 7-1 = -1 - 8
= -9
the inverse of 7 under the operation \(\Delta\) is -9
16
Solve the inequality -3(x - 2) < -2(x + 3)
A
x > -12
B
x > 12
C
x < -12
D
x < 12
correct option: b
-3 (x - 2) < -2(x + 3), -3x + 6 < -2x - 6
6 + 6 < 3x - 2x, 12 < x or x > 12
Users' Answers & Comments6 + 6 < 3x - 2x, 12 < x or x > 12
17
If f (x) = 3x - 2, P = \(\begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix}\) and I is 2 X 2 identity matrix, evaluate f(P)
A
\(\begin{pmatrix} 6 & 3 \ -3 & 0 \end{pmatrix}\)
B
\(\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\)
C
\(\begin{pmatrix} 8 & 3 \ -3 & 2 \end{pmatrix}\)
D
\(\begin{pmatrix} 4 & 3 \ -3 & -2 \end{pmatrix}\)
correct option: d
f(x) = 3x - 2, f(P) = 3p - 2I
= 3\(\begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix}\) - 2\(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) = \(\begin{pmatrix} 6 & 3 \ -3 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 6 & -2 \ -3 & 0 \end{pmatrix}\)\(\begin{pmatrix} 3 & -0 \ 0 & -2 \end{pmatrix}\) = \(\begin{pmatrix} 4 & 3 \ -3 & -2 \end{pmatrix}\)
Users' Answers & Comments= 3\(\begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix}\) - 2\(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) = \(\begin{pmatrix} 6 & 3 \ -3 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 6 & -2 \ -3 & 0 \end{pmatrix}\)\(\begin{pmatrix} 3 & -0 \ 0 & -2 \end{pmatrix}\) = \(\begin{pmatrix} 4 & 3 \ -3 & -2 \end{pmatrix}\)
18
Factorize 2t2 + t - 15
A
(t + 3)(2t - 5)
B
(2t + 3)(t - 5)
C
(2t - 3)(t + 5)
D
(t + 3)(t - 5)
correct option: a
2t2 + t - 15 = 2t2 - 5t + 6t - 15
= t(2t - 5) + 3(2t - 5) = (t + 3)(2t - 5)
Users' Answers & Comments= t(2t - 5) + 3(2t - 5) = (t + 3)(2t - 5)
19
A binary operation \(\oplus\) on real numbers is defined by x \(\oplus\) y = xy + x + y for any two real numbers x and y. The value of (-\(\frac{3}{4}\)) \(\oplus\) 6 is
A
-\(\frac{3}{4}\)
B
\(\frac{45}{4}\)
C
-\(\frac{4}{3}\)
D
\(\frac{3}{4}\)
correct option: d
x \(\oplus\) y = xy + x + y
(-\(\frac{3}{4}\)) \(\oplus\) 6 = -\(\frac{3}{4.6}\) - \(\frac{3}{4}\) + 6
= -\(\frac{9}{2}\) - \(\frac{3}{4}\) + 6
= -\(\frac{3}{4}\)
Users' Answers & Comments(-\(\frac{3}{4}\)) \(\oplus\) 6 = -\(\frac{3}{4.6}\) - \(\frac{3}{4}\) + 6
= -\(\frac{9}{2}\) - \(\frac{3}{4}\) + 6
= -\(\frac{3}{4}\)
20
The solution of the quadratic inequality (x2 + x - 12) \(\geq\) 0 is
A
x \(\geq\) 3 or x \(\geq\) -4
B
x \(\leq\) 3 or x \(\leq\) -4
C
x \(\geq\) 3 or x \(\leq\) -4
D
x \(\geq\) -3 or x \(\leq\) 4
correct option: c
(x2 + x - 12) \(\geq\) 0 , (x - 3)(x + 4) \(\geq\) 0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 \(\geq\) 0 and x + 4 \(\geq\) 0
or x - 3\(\leq\) 0 and x + 4 \(\leq\) 0
when x \(\geq\) 3, the condition is satisfied
when x \(\geq\) -4, the condition is not satisfied.
when x \(\leq\) 3, the condition is not satisfied
when x \(\leq\) -4 , the condition is not satisfied. Thus, the solution of the inequality is x \(\geq\) 3 or x \(\leq\) -4 ,
Users' Answers & CommentsFor the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 \(\geq\) 0 and x + 4 \(\geq\) 0
or x - 3\(\leq\) 0 and x + 4 \(\leq\) 0
when x \(\geq\) 3, the condition is satisfied
when x \(\geq\) -4, the condition is not satisfied.
when x \(\leq\) 3, the condition is not satisfied
when x \(\leq\) -4 , the condition is not satisfied. Thus, the solution of the inequality is x \(\geq\) 3 or x \(\leq\) -4 ,