Mathematics 2010 | Study Mode

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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

Correct option: A
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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 \(\frac{19}{24}\)

= \(\frac{1}{36}\)


Correct option: C
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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%

Correct option: C
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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%

Correct option: B
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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

Correct option: A
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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

Correct option: B
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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

Correct option: D
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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

Correct option: B
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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

Correct option: A
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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

Correct option: A

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