1

Express 302.10495 correct to five significant figures

A

302.10

B

302.11

C

302.105

D

302.1049

CORRECT OPTION:
a

2

Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)

A

\(\sqrt{2}\)

B

\(\sqrt{5}\)

C

2\(\sqrt{2}\)

D

2\(\sqrt{5}\)

CORRECT OPTION:
d

\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)

= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\)

= \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\)

= 2\(\sqrt{5}\)

= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\)

= \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\)

= 2\(\sqrt{5}\)

3

In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y

A

50:11

B

15:17

C

13:55

D

12:11

CORRECT OPTION:
b

Divide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = \(\frac{1050}{70}\) = 15

and y = \(\frac{1190}{70}\) = 17

ratio of enrollment of x to y = 15:17

and y = \(\frac{1190}{70}\) = 17

ratio of enrollment of x to y = 15:17

4

Convert 35_{10} to number in base 2

A

1011

B

10011

C

100011

D

11001

CORRECT OPTION:
c

5

The n^{th} term of a sequence is T_{n} = 5 + (n - 1)^{2}. Evaluate T_{4} - T_{6}

A

30

B

16

C

-16

D

-30

CORRECT OPTION:
c

T_{4} = 5 + (4 - 1)^{2}; where n = 4

= 5 + (3)^{2} = 5 + 9

= 14

T = 5 + (6 - 1)^{2}

where n = 6

= 5 + (5)^{2}

= 5 + 25 = 30

T^{4} + T^{6} = 14 - 30

= -16

= 5 + (3)

= 14

T = 5 + (6 - 1)

where n = 6

= 5 + (5)

= 5 + 25 = 30

T

= -16

6

Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?

A

25m/s

B

150m/s

C

250m/s

D

500m/s

CORRECT OPTION:
a

\(\frac{720 \times 1000}{8 \times 60 \times 60}\) = \(\frac{20 \times 10}{8}\)

= \(\frac{200}{8}\)

= 25 m/s

= \(\frac{200}{8}\)

= 25 m/s

7

If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged

A

5%

B

7\(\frac{1}{2}\)%

C

8%

D

10%

CORRECT OPTION:
d

A = 3,500, P = N2,500

A = P + I

But I = N3,500 - N2,500

I = N1,000

I = \(\frac{PRT}{100}\)

N1,000 = \(\frac{2.500 \times R \times 4}{100}\)

1000 = 100R

R = \(\frac{1000}{100}\)

= 10%

A = P + I

But I = N3,500 - N2,500

I = N1,000

I = \(\frac{PRT}{100}\)

N1,000 = \(\frac{2.500 \times R \times 4}{100}\)

1000 = 100R

R = \(\frac{1000}{100}\)

= 10%

8

Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)

A

5

B

4

C

3

D

1

CORRECT OPTION:
a

\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)

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

\(\frac{5}{3x} = \frac{1}{3}\)

3x = 15

x = \(\frac{15}{3}\)

= 5

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

\(\frac{5}{3x} = \frac{1}{3}\)

3x = 15

x = \(\frac{15}{3}\)

= 5

9

Simplify: \(\frac{54k^2 - 6}{3k + 1}\)

A

6(1 - 3k^{2})

B

6(3k^{2} - 1)

C

6(3k - 1)

D

6(1 - 3k)

CORRECT OPTION:
c

\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\)

= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\)

= 6(3k - 1)

= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\)

= 6(3k - 1)

10

Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)

A

p = \(\frac{2q - rs}{6}\)

B

p = 2qr - sr - 3

C

p = \(\frac{2qr - s}{6}\)

D

p = \(\frac{2qr - rs}{6}\)

CORRECT OPTION:
d

q = \(\frac{3p}{r} + \frac{s}{2}\)

q = \(\frac{6p + rs}{2r}\)

6p + rs = 2qr

6p = 2qr - rs

p = \(\frac{2qr - rs}{6}\)

q = \(\frac{6p + rs}{2r}\)

6p + rs = 2qr

6p = 2qr - rs

p = \(\frac{2qr - rs}{6}\)

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