## Mathematics 2011 Study Mode

Try this quiz in CBT Mode

Question 1
A
B
C
D
Explanation.
2q35 = 778

2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80

2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1

50 + 5q + 3 = 56 + 7

5q = 63 - 53

q = $$\frac{10}{5}$$

q = 2

Question 2
A
$$5\frac{2}{3}$$
B
30
C
$$4\frac{1}{3}$$
D
50
Explanation.
$$\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}$$

$$\frac{\frac{11}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}$$

$$\frac{110}{54} \div \frac{66}{1620}$$

50

Question 3
A
N150
B
N220
C
N130
D
N250
Explanation.
S.I. = $$\frac{P \times R \times T}{100}$$

If T = 9 months, it is equivalent to $$\frac{9}{12}$$ years

S.I. = $$\frac{5000 \times 4 \times 9}{100 \times 12}$$

S.I. = N150

Question 4
A
B
C
D
Explanation.
M:N:Q == 5:4:3

i.e M = 5, N = 4, Q = 3

Substituting values into equation, we have...

$$\frac{2N - Q}{M}$$

= $$\frac{2(4) - 3}{5}$$

= $$\frac{8 - 3}{5}$$

= $$\frac{5}{5}$$

= 1

Question 5
A
$$\frac{2}{3}$$
B
$$\frac{1}{2}$$
C
$$\frac{8}{9}$$
D
$$\frac{1}{3}$$
Explanation.
$$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}$$

$$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}$$

$$(\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}$$

$$\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}$$

$$\frac{2}{3} \div \frac{4}{3}$$

$$\frac{2}{3} \times \frac{3}{4}$$

$$\frac{2}{4}$$

$$\frac{1}{2}$$

Question 6
A
B
C
D
Explanation.
log$$_{3}^{18}$$ + log$$_{3}^{3}$$ - log$$_{3}^{x}$$ = 3

log$$_{3}^{18}$$ + log$$_{3}^{3}$$ - log$$_{3}^{x}$$ = 3log33

log$$_{3}^{18}$$ + log$$_{3}^{3}$$ - log$$_{3}^{x}$$ = log333

log3($$\frac{18 \times 3}{X}$$) = log333

$$\frac{18 \times 3}{X}$$ = 33

18 x 3 = 27 x X

x = $$\frac{18 \times 3}{27}$$

= 2

Question 7
A
$$\frac{1 - \sqrt5}{2}$$
B
$$\frac{1 - \sqrt5}{4}$$
C
$$\frac{ \sqrt5 - 1}{2}$$
D
$$\frac{1 + \sqrt5}{4}$$
Explanation.
$$\frac{2 - \sqrt5}{3 - \sqrt5}$$ x $$\frac{3 + \sqrt5}{3 + \sqrt5}$$

$$\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}$$ = $$\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}$$

= $$\frac{6 - \sqrt5 - 5}{9 - 5}$$

= $$\frac{1 - \sqrt5}{4}$$

Question 8
A
$$\frac{7}{3}$$
B
$$\frac{5}{3}$$
C
$$\frac{5}{2}$$
D
$$\frac{3}{2}$$
Explanation.
($$\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}$$)

$$\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}$$

= 2 - $$\frac {1}{3}$$

= $$\frac {16 - 1}{3}$$

= $$\frac{5}{3}$$

Question 9
A
49
B
170
C
21
D
210
Explanation.
The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.

= 7 x 6 x 5

= 210 ways

or $$\frac{7}{P_3}$$ = $$\frac{7!}{(7 - 3)!}$$ = $$\frac{7!}{4!}$$

= $$\frac{7 \times 6 \times 5 \times 4!}{4!}$$

= 210 ways

Question 10
A
$$\sqrt\frac{3T - K}{M}$$
B
$$\sqrt\frac{3T - M}{K}$$
C
$$\sqrt\frac{3T + K}{M}$$
D
$$\sqrt\frac{3T - K}{M}$$
Explanation.
T = $$\frac{KR^2 + M}{3}$$

3T = KR2 + M

KR2 = 3T - M

R2 = $$\frac{3T - M}{K}$$

R = $$\sqrt\frac{3T - M}{K}$$