Mathematics (2018) | Study Mode

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Question 51
A
0
B
1
C
26
D
46
Explanation

Since they are of different base, convert to base 10

243\(_{six}\) = (2 x 62) + (4 x 61) + (3 x 60)

= 72 + 24 + 3 = 99 base 10

243\(_{six}\) = 2 x 52 + 4 x 51 +3 x 50

50 + 20 + 3 = 73 base 10

Subtracting them, 99 - 73

= 26


Correct option: C
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Question 52
A
1.47
B
2.67
C
3.23
D
3.47
Explanation

\(\frac{5}{x}\) dx = 5 \(\frac{1}{x}\) = 5Inx

Since the integral of \(\frac{1}{x}\) is Inx

\(^2\) \(_1\) \(\frac{5}{x}\) dx = 5

dx = 5 (In<2 InIn1)

= 3.4657

= 3.47


Correct option: D
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Question 53
A
\(\frac{1}{16}\)
B
\(\frac{1}{6}\)
C
\(\frac{1}{4}\)
D
\(\frac{3}{8}\)
Explanation

P( tail on a coin) = \(\frac{1}{2}\)

Even numbers on a care 2, 4 and 6

P( even number on a die) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

P( tail on a coin and even number on a die) = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)


Correct option: C
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Question 54
A
210
B
1050
C
21400
D
25200
Explanation

To form words having 3 consonants and 2 vowels out of 7 consonants and 4 vowels, the number of such words is 7/3C x 4/2C = 35 x 6

= 210


Correct option: A
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Question 55
A
5km
B
12km
C
13km
D
17km
Explanation

Using Pythagoras theorem

PR\(^2\) = 5\(^2\) + 12\(^2\)

25 + 144 = 169

PR = (169)= 13km

Answer is C


Correct option: C
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Question 56
A
1110
B
10111
C
11101
D
111100
Explanation

First we convert the numbers to base ten

23\(_{five}\)= 2 x 51 + 3 x 50

= 10 + 3 = 13

101\(_{five}\) = (1 x 32) + (0 x 31) + (1 x 30)

= 9 + 0 + 1 = 10

So, y = 13 + 10 = 23

To convert 23 to base 2 (as in the diagram above)

Y = 23

= 10111\(_{five}\)

Answer is B


Correct option: B
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Question 57
A
\(\frac{2}{3}\)
B
\(\frac{1}{3}\)
C
\(\frac{1}{4}\)
D
\(\frac{1}{9}\)
Explanation

Total number of balls = 2 + 4 = 6

P(of picking a red ball) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

P(of picking a blue ball) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

With replacement,

P( picking two red balls) = \(\frac{1}{3}\) \(\frac{1}{3}\) = \(\frac{1}{9}\)


Correct option: D
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Question 58
Mathematics (2018) Q 58 image
A
MATHEMATICS A 9 image
B
{s}
MATHEMATICS B 9 image
C
{t, u}
MATHEMATICS C 9 image
D
{y, z}
MATHEMATICS D 9 image
Explanation

A1 = Elements in the universal set but not in A = {s, w, x, y, z}

B = {r, s. t, u}

C = {t, u, v, w, x}

A1 n B n C = elements common to the three sets = none = empty set =


Correct option: A
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Question 59
A
T = \(\frac{R + P^3}{15Q}\)
B
T = \(\frac{R - 15P^3}{Q}\)
C
T = \(\frac{R - 15P^3}{Q}\)
D
T = \(\frac{15R - Q}{P^3}\)
Explanation

Taking the cube of both sides of the equation give

P\(^3\) = \(\frac{Q(R - T)}{15}\)

Cross multiplying

15P\(^3\) = Q(R - T)

Divide both sides by Q

\(\frac{15P^3}{Q}\) = R - T

Rearranging gives

T = R - \(\frac{15P^3}{ Q}\)

= \(\frac{RQ - 15P^3}{Q}\)

Answer is C


Correct option: C
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Question 60
A
?1000
B
?2000
C
?3000
D
?4000
Explanation

Let the monthly expenditure angle for school fees is x, then that of housing will be 2x. Since the total angle in the circle is 360.

Using 90 as the angle for transport

So, x + 2x + 120 + 90 = 360

3x + 210 = 3605

3x = 360 - 210

= 150

x = \(\frac{150}{3}\) = 50

So the angle for housing is 2x = 2 50

= 100

Amount spent on housing = \(\frac{100}{360}\) 7200

= ?2000


Correct option: B

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