2011 - JAMB Mathematics Past Questions and Answers - page 3
21
If | 2 3 | = | 4 1 |. find the value of y. 7
A
-6
B
6
C
-12
D
12
correct option: a
\(\begin{vmatrix} 2 & 3 \ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \ 3 & 2x \end{vmatrix}\)
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = \(\frac{12}{-2}\)
= -6
Users' Answers & Comments(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = \(\frac{12}{-2}\)
= -6
22
Evaluate \(\begin{vmatrix} 4 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 \end{vmatrix}\)
A
25
B
45
C
15
D
55
correct option: a
\(\begin{vmatrix} 4 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 \end{vmatrix}\)
4 \(\begin{vmatrix} 3 & -1 \ 1 & 3 \end{vmatrix}\) -2 \(\begin{vmatrix} 2 & -1 \ -1 & 3\end{vmatrix}\) -1 \(\begin{vmatrix} 2 & 3 \ -1 & 1 \end{vmatrix}\)
4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]
= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]
= 4(10) - 2(5) - 1(5)
= 40 - 10 - 5
= 25
Users' Answers & Comments4 \(\begin{vmatrix} 3 & -1 \ 1 & 3 \end{vmatrix}\) -2 \(\begin{vmatrix} 2 & -1 \ -1 & 3\end{vmatrix}\) -1 \(\begin{vmatrix} 2 & 3 \ -1 & 1 \end{vmatrix}\)
4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]
= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]
= 4(10) - 2(5) - 1(5)
= 40 - 10 - 5
= 25
23
The inverse of matrix N = \(\begin{vmatrix} 2 & 3 \\
1 & 4\end{vmatrix}\) is
A
\(\frac{1}{5}\) \(\begin{vmatrix} 2 & 1 \ 3 & 4\end{vmatrix}\)
B
\(\frac{1}{5}\) \(\begin{vmatrix} 4 & -3 \ -1 & 2\end{vmatrix}\)
C
\(\frac{1}{5}\) \(\begin{vmatrix} 2 & -1 \ -3 & 4\end{vmatrix}\)
D
\(\frac{1}{5}\) \(\begin{vmatrix} 4 & 3 \ 1 & 2\end{vmatrix}\)
correct option: b
N = [2 3]
N-1 = \(\frac{adj N}{|N|}\)
adj N = \(\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix}\)
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix}\)
Users' Answers & CommentsN-1 = \(\frac{adj N}{|N|}\)
adj N = \(\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix}\)
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix}\)
24
What is the size of each interior angle of a 12-sided regular polygon?
A
120o
B
150o
C
30o
D
180o
correct option: b
Interior angle = (n - 2)180
but, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = \(\frac{(n - 2)180}{n}\)
x = \(\frac{1800}{12}\)
= 150o
Users' Answers & Commentsbut, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = \(\frac{(n - 2)180}{n}\)
x = \(\frac{1800}{12}\)
= 150o
25
A circle of perimeter 28cm is opened to form a square. What is the maximum possible area of the square?
A
56cm2
B
49cm2
C
98cm2
D
28cm2
correct option: b
Perimeter of circle = Perimeter of square
28cm = 4L
L = \(\frac{28}{4}\) = 7cm
Area of square = L2
= 72
= 49cm2
Users' Answers & Comments28cm = 4L
L = \(\frac{28}{4}\) = 7cm
Area of square = L2
= 72
= 49cm2
26
A chord of circle of radius 7cm is 5cm from the centre of the maximum possible area of the square?
A
4√6 cm
B
3√6 cm
C
6√6 cm
D
2√6 cm
correct option: a
From Pythagoras theorem
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = \(\sqrt {24}\)
= \(\sqrt {4 \times 6}\)
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm
Users' Answers & Comments|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = \(\sqrt {24}\)
= \(\sqrt {4 \times 6}\)
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm
27
A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold
A
48 cm3
B
33 cm3
C
60 cm3
D
27 cm3
correct option: b
Volume of cube = L3
33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3
Users' Answers & Comments33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3
28
The perpendicular bisector of a line XY is the locus of a point B. whose distance from Y is always twice its distance from X. C
A
whose distance from X is always twice its distance from Y
B
whose distance from Y is always twice its distance from X.
C
which moves on the line XY
D
which is equidistant from the points X and y
correct option: d
Users' Answers & Comments29
The midpoint of P(x, y) and Q(8, 6). Find x and y. midpoint = (5, 8)
A
(2, 10)
B
(2, 8)
C
(2, 12)
D
(2, 6)
correct option: a
P(x, y) Q(8, 6)
midpoint = (5, 8)
x + 8 = 5
\(\frac{y + 6}{2}\) = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)
Users' Answers & Commentsmidpoint = (5, 8)
x + 8 = 5
\(\frac{y + 6}{2}\) = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)
30
Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2).
A
5y - 2x -18 = 0
B
5y + 2x - 18 = 0
C
5y - 2x + 18 = 0
D
5y + 2x - 2 = 0
correct option: b
2y = 5x + 4 (4, 2)
y = \(\frac{5x}{2}\) + 4 comparing with
y = mx + e
m = \(\frac{5}{2}\)
Since they are perpendicular
m1m2 = -1
m2 = \(\frac{-1}{m_1}\) = -1
\(\frac{5}{2}\) = -1 x \(\frac{2}{5}\)
The equator of the line is thus
y = mn + c (4, 2)
2 = -\(\frac{2}{5}\)(4) + c
\(\frac{2}{1}\) + \(\frac{8}{5}\) = c
c = \(\frac{18}{5}\)
\(\frac{10 + 5}{5}\) = c
y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\)
5y = -2x + 18
or 5y + 2x - 18 = 0
Users' Answers & Commentsy = \(\frac{5x}{2}\) + 4 comparing with
y = mx + e
m = \(\frac{5}{2}\)
Since they are perpendicular
m1m2 = -1
m2 = \(\frac{-1}{m_1}\) = -1
\(\frac{5}{2}\) = -1 x \(\frac{2}{5}\)
The equator of the line is thus
y = mn + c (4, 2)
2 = -\(\frac{2}{5}\)(4) + c
\(\frac{2}{1}\) + \(\frac{8}{5}\) = c
c = \(\frac{18}{5}\)
\(\frac{10 + 5}{5}\) = c
y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\)
5y = -2x + 18
or 5y + 2x - 18 = 0