1

Which of the following correctly expresses 48 as a product of prime factors?

A

3 x 4 x 4

B

2 x 3 x 8

C

2 x 2 x 2 x 3 x 4

D

2 x 2 x 2 x 2 x 3

CORRECT OPTION:
d

48 = 2 x 2 x 2 x 2 x 3

2

Evaluate \((20_{three})^2 - (11_{three})^2\) in base three

A

101

B

121

C

202

D

2020

CORRECT OPTION:
c

\((20_{three})^2 - (11_{three})^2\)

= \((20_{3} - 11_{3})(20_{3} + 11_{3})\)

= \((2_{3})(101_{3})\)

= \(202_{3}\)

3

Evaluate \(log_{10}5 + log_{10}20\)

A

2

B

3

C

4

D

5

CORRECT OPTION:
a

\(log_{10}5 + log_{10}20\

log_{10}[5\times 20]\

log_{10}100\

log_{10}10^2\

2log_{10}10=2\)

log_{10}[5\times 20]\

log_{10}100\

log_{10}10^2\

2log_{10}10=2\)

4

In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects?

A

15

B

30

C

35

D

45

CORRECT OPTION:
c

Let c = no of students that offered both subjects

\(\therefore\) No of students offering Economics = 65 - c

No of students offering Geography = 50 - c

65 - c + c + 50 - c = 80

115 - c = 80

c = 35

35 students offer both Economics and Geography.

5

If N varies directly as M and N = 8 when M = 20 find M when N = 7

A

13

B

15

C

\(17\frac{1}{2}\)

D

\(18\frac{1}{2}\)

CORRECT OPTION:
c

N ∝ M

∴N = kM where k is a constant

N = 8 when M = 20

∴ 8 = k x 20

\(k = \frac{8}{20} = \frac{2}{5}\

∴ N = \frac{2}{5}M\

If N = 7\

∴ 7 = \frac{2}{5}M\

M = \frac{35}{2} = 17\frac{1}{2}\)

∴N = kM where k is a constant

N = 8 when M = 20

∴ 8 = k x 20

\(k = \frac{8}{20} = \frac{2}{5}\

∴ N = \frac{2}{5}M\

If N = 7\

∴ 7 = \frac{2}{5}M\

M = \frac{35}{2} = 17\frac{1}{2}\)

6

Express 0.0462 in standard form

A

0.460 x 10^{-1}

B

0.462 x 10^{-2}

C

4.62 x 10^{-1}

D

4.62 x 10^{-2}

CORRECT OPTION:
d

7

In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether?

A

81

B

72

C

54

D

45

CORRECT OPTION:
a

Ratio of good ones to bad ones is 5:4; If 36 is bad;

∴ the good ones = \(\frac{5\times 36}{4}=45\) oranges.

The total number of oranges is 36 + 45 = 81.

8

A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate

A

95%

B

47.5%

C

5.26%

D

5%

CORRECT OPTION:
d

The percentage error is \(\frac{error}{actual}\times \frac{100}{1}\%\

=\frac{200-190}{200}\times \frac{100}{1}\

=\frac{10}{200}\times \frac{100}{1}\% = 5\%\)

=\frac{200-190}{200}\times \frac{100}{1}\

=\frac{10}{200}\times \frac{100}{1}\% = 5\%\)

9

Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\)

A

\(\frac{8}{25}\)

B

\(\frac{12}{25}\)

C

\(3\frac{3}{5}\)

D

\(4\frac{1}{8}\)

CORRECT OPTION:
c

\(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 \left(1\frac{1}{2}\right)^{-1}\

\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5}

\)

\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5}

\)

10

The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\)

A

3rd

B

4th

C

5th

D

6th

CORRECT OPTION:
c

\(T_{n} = 2^{2n - 1}\)

\(2^{2n - 1} = 2^9\)

\(2n - 1 = 9 \implies 2n = 9 + 1\)

\(2n = 10 \implies n = 5\)

The 5th term = 2\(^9\)

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