# Mathematics 2010 | Study Mode

Question 21
A
3
B
2
C
6
D
5
##### Explanation
6r78 = 5119

6 x 82 + r x 81 + 7 x 8o = 5 x 92 + 1 x 91 + 1 x 9o

6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1

384 + 8r + 7 = 405 + 9 + 1

391 + 8r = 24

r = $$\frac{24}{8}$$

= 3

Question 22
A
$$\frac{1}{5}$$
B
$$\frac{1}{4}$$
C
$$\frac{1}{36}$$
D
$$\frac{1}{25}$$
##### Explanation

($$\frac{3}{4}$$ of $$\frac{4}{9}$$ $$\div$$ 9$$\frac{1}{2}$$) $$\div$$ 1$$\frac{5}{19}$$

Applying the rule of BODMAS, we have:

($$\frac{3}{4}$$ x $$\frac{4}{9}$$ $$\div$$$$\frac{19}{2}$$) $$\div$$$$\frac{24}{19}$$

($$\frac{3}{9}$$ $$\div$$ $$\frac{19}{2}$$) $$\div$$$$\frac{24}{19}$$ = ($$\frac{3}{9}$$ x $$\frac{2}{19}$$) $$\div$$$$\frac{24}{19}$$

($$\frac{1}{3}$$ x $$\frac{2}{19}$$) $$\div$$$$\frac{24}{19}$$

$$\frac{1}{3}$$ x $$\frac{2}{19}$$ x $$\frac{19}{24}$$

= $$\frac{1}{36}$$

Question 23
A
0.25%
B
0.01%
C
0.80%
D
0.40%
##### Explanation
Actual length of rope = 1.25m

Measured length of rope = 1.26m

error = (1.26 - 1.25)m - 0.01m

Percentage error = $$\frac{error}{\text{actual length}}$$ x 100%

= $$\frac{0.01}{1.25}$$ x 100%

= 0.08%

Question 24
A
3%
B
2%
C
5%
D
4%
##### Explanation
Using simple interest = $$\frac{P \times T \times R}{100}$$,

where: P denoted principal = N400
T denotes time = 3 years
R denotes interest rate = ?

24 = $$\frac{400 \times 3 \times R}{100}$$

24 x 100 = 400 x 3 x R

R = $$\frac{24 \times 100}{400 \times3}$$

= 2%

Question 25
A
12 : 15 : 10
B
12 : 15 : 16
C
10 : 15 : 24
D
9 : 10 : 15
##### Explanation
If p : q = $$\frac{2}{3}$$ : $$\frac{5}{6}$$, then the sum S1 of ratio = $$\frac{2}{3}$$ + $$\frac{5}{6}$$ = $$\frac{9}{6}$$

If q : r = $$\frac{3}{4}$$ : $$\frac{1}{2}$$, then the sum S2 of ratio = $$\frac{3}{4}$$ + $$\frac{1}{2}$$ = $$\frac{5}{4}$$

Let p + q = T1, then

q = ($$\frac{5}{6} \div \frac{9}{6}$$)T1 = ($$\frac{5}{6} \times \frac{6}{9}$$)T1 = $$\frac{5}{9}$$T1

Again, let q + r = T2, then

q = ($$\frac{3}{4} \div \frac{5}{4}$$)T2 = ($$\frac{3}{4} \times \frac{4}{5}$$)T2 = $$\frac{3}{5}$$T2

Using q = q

$$\frac{5}{9}$$T1 = $$\frac{3}{5}$$T2

5 x 5T1 = 9 x 3T2

$$\frac{T_1}{T_2}$$ = $$\frac{9 \times 3}{5 x 5}$$ = $$\frac{27}{5}$$

Giving that, T1 = 27 and T2 = 25

P = ($$\frac{2}{3} \div S_1$$)T1 = ($$\frac{2}{3} \div \frac{9}{6}$$)T1

= ($$\frac{2}{3} \times \frac{6}{9}$$)27 = 12

q = ($$\frac{5}{6} \div S_1$$)T1 = ($$\frac{5}{6} \div \frac{9}{6}$$)T1

= ($$\frac{5}{6} \times \frac{6}{9}$$)27 = 15

and r = ($$\frac{1}{2} \div S_2$$)T2 = ($$\frac{1}{2} \div \frac{5}{4}$$)T2

= ($$\frac{1}{2} \times \frac{4}{5}$$)25 = 10

Hence p : q : r = 12: 15 : 10

Question 26
A
2.1461
B
2.0491
C
3.1461
D
2.5441
##### Explanation
log 112 = log(16 x 7)

= log 16 + log7

= log 24 + log7

= 4log2 + log7

= 4(0.3010) + 0.8451

= 1.204 + 0.8451

= 2.0491

Question 27
A
25
B
11
C
55
D
36
##### Explanation
Given that,

X $$\ast$$ y = X + y2

(2 $$\ast$$ 3) $$\ast$$ 5 = (2 + 32)$$\ast$$5

= (2 + 9)$$\ast$$5 = 11 $$\ast$$5

Hence 11 $$\ast$$ 5 = 11 + 52

= 11 + 25 = 36

Question 28
A
p 1
B
p = 18
C
q 1
D
q = 18
##### Explanation
18(p+q) = (18+p)q
18p + 18q = 18q + pq
By comparison
18p = 18q or p = q
Again, 18q = pq or 18 = p which is required

Question 29
A
-44
B
-165
C
165
D
44
##### Explanation
3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44

Question 30
A
-cosx + 2x + k
B
cosx + 2x + k
C
-cosx + x2 + k
D
cosx + x2 + k
##### Explanation
$$\int$$(Sin x + 2)dx = -cos x + 2x + k

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