# Mathematics 2012 | Study Mode

Question 31
A
1
B
2
C
3
D
4

Question 32
A
126
B
180
C
216
D
224
##### Explanation
$$\frac{x}{7} = \frac{96}{1}$$ ==> $$\frac{672 + x}{8} = 112$$

Therefore x = 224

Question 33
A
9
B
8
C
7
D
4
##### Explanation
Arrange all the values in ascending order,

2,2,3,3,3,4,4,4,4,5,5,5,7,8,9,9

Question 34
A
11
B
9
C
8
D
4
##### Explanation
Range = Highest Number - Lowest Number

Range = 11 - 2 = 9

Question 35
A
3.9
B
4.9
C
5.9
D
6.9
##### Explanation
$$\begin{array}{c|c} x & (x - \varkappa) & (x - \varkappa)^2 \\ \hline 2 & -5 & 25 \\ \hline 3 & -4 & 16 \\ \hline 8 & 1 & 1 \\ \hline 10 & 3 & 9 \\ \hline 12 & 5 & 25 \\ \hline & & 76 \end{array}$$

S.D = $$\sqrt{\frac{(x - \varkappa)^2}{n}}$$

S.D = $$\sqrt{\frac{76}{5}}$$

S.D = 3.9

Question 36
A
$$\sqrt{13}$$
B
$$3\sqrt{2}$$
C
$$\sqrt{26}$$
D
$$10\sqrt{5}$$
##### Explanation
P1 (4, 3), P2 (x, y)

y = 2x + 4 .....(1)

y = 7 - x .....(2)

Substitute (2) in (1)

7 - x = 2x + 4

7 - 4 = 2x + x

3 = 3x

x = 1

Substitute in eqn (2)

y = 7 - x

y = 7 - 1

y = 6

P2 (1, 6)

Distance between 2 points is given as

D = $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

D = $$\sqrt{(1 - 4)^2 + (6 - 3)^2}$$

D = $$\sqrt{(-3)^2 + (3)^2}$$

D = $$\sqrt{9 + 9}$$

D = $$\sqrt{18}$$

D = $$\sqrt{9 \times 2}$$

D = $$3\sqrt{2}$$

Question 37
A
3630
B
3360
C
1120
D
560
##### Explanation
$$\frac{n + 1 (n - 2)}{(n + 1)!}$$

$$\frac{(n + 1) + (n - 2)!(n - 2)!}{(n + 1)!}$$

$$\frac{(n + 1)(n + 1 -1)(n+1-2)(n+1-3)!}{3!(n-2)!}$$

$$\frac{(n + 1)(n)(n-1)(n-2)!}{3!(n-2)!}$$

$$\frac{(n + 1)(n)(n-1)}{3!}$$

Since n = 15

$$\frac{(15 + 1)(15)(15-1)}{3!}$$

$$\frac{16 \times 15 \times 14}{3 \times 2 \times 1}$$

= 560

Question 38
A
6720
B
6270
C
6207
D
6027
##### Explanation
$$\frac{8!}{3!}$$

$$\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2\times 1}$$

==> 8 x 7 x 6 x 5 x 4 = 6720

Question 39
A
$$\frac{2}{27}$$
B
$$\frac{3}{27}$$
C
$$\frac{4}{27}$$
D
$$\frac{5}{3}$$
##### Explanation
Pass(P) = $$\frac{2}{3}$$, Fail(F) = $$\frac{1}{3}$$

T = P.P.F ==> $$\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}$$ = $$\frac{4}{27}$$

Question 40
A
$$\frac{2}{15}$$
B
$$\frac{3}{15}$$
C
$$\frac{7}{15}$$
D
$$\frac{13}{15}$$
##### Explanation
Man lives = $$\frac{2}{3}$$ not live = $$\frac{1}{3}$$

Wife lives = $$\frac{3}{5}$$ not live = $$\frac{2}{5}$$

$$P(\frac{2}{3} \times \frac{2}{5}) + (\frac{2}{5} \times \frac{1}{3}) + (\frac{2}{3} \times \frac{3}{5})$$

= $$\frac{4}{15} + \frac{3}{15} + \frac{6}{15}$$

= $$\frac{13}{15}$$

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