1

If 23_{x} + 101_{x} = 130_{x}, find the value of x

A

7

B

6

C

5

D

4

CORRECT OPTION:
d

23_{x} + 101_{x} = 130_{x}

2 x X^{1} + 3 x X^{o} + 1 x X^{2} + 0 x X^{1} + 1 x X^{o}

= 1 x X^{o} = 1 x X^{2} + 3 x X^{1} + 0 x X^{o}

= X^{2} + 3x + 0

2x + 3 = x^{2} + 0 + 1 + x^{2} + 3x

2x - 3x + x^{2} - x^{2} = -3 - 1

- x = -4

x = 4

2 x X

= 1 x X

= X

2x + 3 = x

2x - 3x + x

- x = -4

x = 4

2

Simplify: (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

A

\(\frac{1}{60}\)

B

\(\frac{5}{72}\)

C

\(\frac{1}{10}\)

D

1\(\frac{7}{10}\)

CORRECT OPTION:
c

(\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

= (\(\frac{9 - 8}{12} \times \frac{6}{5}\))

= \(\frac{1}{12} \times \frac{6}{5}\)

= \(\frac{1}{10}\)

= (\(\frac{9 - 8}{12} \times \frac{6}{5}\))

= \(\frac{1}{12} \times \frac{6}{5}\)

= \(\frac{1}{10}\)

3

Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))^{2}

A

75.00

B

15.00

C

8.66

D

3.87

CORRECT OPTION:
b

Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\)

= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\)

hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))^{2} = (\(2\sqrt{15} - \sqrt{15}\))^{2}

= (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\))

= 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\)

= 4 x 15 - 2 x 15 - 2 x 15 + 15

= 60 - 30 - 30 + 15

= 15

= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\)

hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))

= (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\))

= 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\)

= 4 x 15 - 2 x 15 - 2 x 15 + 15

= 60 - 30 - 30 + 15

= 15

4

The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?

A

90cm

B

135cm

C

180cm

D

225cm

CORRECT OPTION:
c

d \(\alpha\) t^{2}

d = t^{2} k

where k is a constant. d = 45cm, when t = 3s; thus 45 = 3^{2} x t

k = \(\frac{45}{9}\) = 5

thus equation connecting d and t is d = 5t^{2}

when t = 6s, d = 5 x 6^{2}

= 5 x 36

= 180cm

d = t

where k is a constant. d = 45cm, when t = 3s; thus 45 = 3

k = \(\frac{45}{9}\) = 5

thus equation connecting d and t is d = 5t

when t = 6s, d = 5 x 6

= 5 x 36

= 180cm

5

Which of following is a valid conclusion from the premise. "Nigeria footballers are good footballers"?

A

Joseph plays football in Nigeria therefore he is a good footballer

B

Joseph is a good footballer therefore he is a Nigerian footballer

C

Joseph is a Nigerian footballer therefore he is a good footballer

D

Joseph plays good football therefore he is a Nigerian footballer

CORRECT OPTION:
c

From the venn diagram, Nigeria footballers from a subset of good footballers.

6

On a map, 1cm represent 5km. Find the area on the map that represents 100km^{2}.

A

2cm^{2}

B

4cm^{2}

C

8cm^{2}

D

8cm^{2}

CORRECT OPTION:
b

On a map, 1cm represents 5km. Then it follows that 1cm^{2} represents 25km^{2}. Acm^{2} represents 100km^{2}. By apparent cross-multiplication, 1cm^{2} x 100km^{2} = Acm^{2}x 25km^{2}

therefore A = \(\frac{100}{25}\) = 4cm^{2}

therefore A = \(\frac{100}{25}\) = 4cm

7

Simplify; \(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)

A

3^{2n}

B

9

C

3^{n}

D

3^{n + 1}

CORRECT OPTION:
b

\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)

= \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\)

= 3\(^{n - 1 + 3n + 3 - 4n}\)

= 3\(^{4n - 4n - 1 + 3}\)

= 3^{2}

= 9

= \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\)

= 3\(^{n - 1 + 3n + 3 - 4n}\)

= 3\(^{4n - 4n - 1 + 3}\)

= 3

= 9

8

What sum of money will amount to D10,400 in 5 years at 6% simple interest?

A

D8,000.00

B

D10,000.00

C

D12,000.00

D

D16,000.00

CORRECT OPTION:
a

A = P + 1

I = A - P

= 10,400 - P

Now using I = \(\frac{P \times T \times R}{100}\)

i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\)

= 100(10,400 - P) = 30P

10(10,400 - P) = 3P

104,000 - 10P = 3P

104,000 - 10P = 3P

104,000 = 3P + 10P

= 104,000 = 13P

P = \(\frac{104,000}{100}\)

P = D8,000

I = A - P

= 10,400 - P

Now using I = \(\frac{P \times T \times R}{100}\)

i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\)

= 100(10,400 - P) = 30P

10(10,400 - P) = 3P

104,000 - 10P = 3P

104,000 - 10P = 3P

104,000 = 3P + 10P

= 104,000 = 13P

P = \(\frac{104,000}{100}\)

P = D8,000

9

The roots of a quadratic equation are \(\frac{4}{3}\) and -\(\frac{3}{7}\). Find the equation

A

21x^{2} - 19x - 12 = 0

B

21x^{2} + 37x - 12 = 0

C

21x^{2} - x + 12 = 0

D

21x^{2} + 7x - 4 = 0

CORRECT OPTION:
a

Let x = \(\frac{4}{3}\), x = -\(\frac{3}{7}\)

Then 3x = 4, 7x = -3

3x - 4 = 0, 7x + 3 = 0

(3x - 4)(7x + 3) = 0

21x^{2} + 9x - 28x - 12 = 0

21x^{2} - 19x - 12 = 0

Then 3x = 4, 7x = -3

3x - 4 = 0, 7x + 3 = 0

(3x - 4)(7x + 3) = 0

21x

21x

10

Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

A

6, -7

B

3, -6

C

3, -7

D

-3, -7

CORRECT OPTION:
c

\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\)

Factorize the denominator;

Y^{2} + 7y - 3y - 21

= y(y + 7) -3 (y + 7)

= (y - 3)(y + 7)

Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

when y^{2} + 4y - 21 = 0

ie. y = 3 or -7

Factorize the denominator;

Y

= y(y + 7) -3 (y + 7)

= (y - 3)(y + 7)

Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

when y

ie. y = 3 or -7

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