Mathematics 2011 | Study Mode

Question 21
A
-6
B
6
C
-12
D
12
Explanation
$$\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}$$ = $$\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}$$

(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

Question 22
A
25
B
45
C
15
D
55
Explanation
$$\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}$$

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

Question 23
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
$$\frac{1}{5}$$ $$\begin{vmatrix} 4 & 3 \\ 1 & 2\end{vmatrix}$$
Explanation
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 = $$\frac {1}{5}$$ $$\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}$$

Question 24
A
120o
B
150o
C
30o
D
180o
Explanation
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

Question 25
A
56cm2
B
49cm2
C
98cm2
D
28cm2
Explanation
Perimeter of circle = Perimeter of square

28cm = 4L

L = $$\frac{28}{4}$$ = 7cm

Area of square = L2

= 72

= 49cm2

Question 26
A
46 cm
B
36 cm
C
66 cm
D
26 cm
Explanation
From Pythagoras theorem

|OA|2 = |AN|2 + |ON|2

72 = |AN|2 + (5)2

49 = |AN|2 + 25

|AN|2 = 49 - 25 = 24

|AN| = $$\sqrt {24}$$

= $$\sqrt {4 \times 6}$$

= 26 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|

26 + 26 = |AB|

|AB| = 46 cm

Question 27
A
48 cm3
B
33 cm3
C
60 cm3
D
27 cm3
Explanation
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

Question 28
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

Question 29
A
(2, 10)
B
(2, 8)
C
(2, 12)
D
(2, 6)
Explanation
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)

Question 30
A
5y - 2x -18 = 0
B
5y + 2x - 18 = 0
C
5y - 2x + 18 = 0
D
5y + 2x - 2 = 0
Explanation
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

Try this quiz in in E-test/CBT Mode
switch to

Question Map