Mathematics (2015) | Study Mode

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Question 41
A

1458

B

1485

C

1345

D

1258

Explanation

1st G.P. = a =2

2nd G.P. = ar - 1 = 54

2(r) = 54

r = 54/2 = 27

r = 27

3rd term = ar2 = (2) (27)2

2 \(\times\) 27 \(\times\) 27

= 1458


Correct option: A
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Question 42
A

-3 < x \(\leq\)3

B

-2 < x \(\leq\) 5

C

2< x \(\geq\) - 5

D

-1\(\leq\) x \(\geq\) 2

Explanation

S = {0, 1, 2, 3, 4, 5}

T = {- 1, 0, 1, 2}

S \(\cup\) T = {- 1, 0, 1, 2, 3, 4, 5 }

- - 2 < x\(\leq\) 5


Correct option: B
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Question 43
A

N14,950.50K

B

N25,150.30K

C

N15,000.00K

D

N38,888.90K

Explanation

[Return Investment] as a ratio ;

i.e The Ratio is Return : Investmen

[(Return1 Investment1 ) = (Return2 Investment2)]

R1 = N 25000

R2 =?

I1 = N450,000,

I1 = N 700000

(25000 450000) = (R2 700000)

R2 = [(25000 \(\times\) 700000 ) 450000]

= N38,888.90K

The return on a investment of Y = N38888.90K


Correct option: D
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Question 44
A
1.35
B
1.353
C
1.455
D
0.455
Explanation
log717

= [log 17 log7]

= [1.2304 0.8451]

[100.0899 101.9270]

= 1.455(antilog)

Correct option: C
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Question 45
A

11112

B

110112

C

101112

D

110012

Explanation

Convert the binary to base 10 and they convert back to base two

100112 + xxxxx2 + 111002 + 1012 = 10011112


(1 \(\times\) 24 + 0 \(\times\) 23 + 1 \(\times\) 22 + 1 \(\times\) 21 + 1 \(\times\) 20) + xxxxx2 +(1 \(\times\) 24 + 1 \(\times\) 23 + 1 \(\times\) 22 + 0 \(\times\) 21 + 0 \(\times\) 20) + (1 \(\times\) 22 + 0 \(\times\) 21 + 1 \(\times\) 20)

= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)

=(64 + 0 + 0 + 8 + 4 + 2 + 1)

19 + xxxxx2 + 33 = 79

xxxxx2 + 52 = 79

xxxxx2 = 79 - 52

xxxxx2 = 2710

\( \begin{array}{c|c}
2 & 27 \\
\hline
2 & 13 \text{ rem 1}\\
2 & 6 \text{ rem 1}\\
2 & 3 \text{ rem 0}\\
2 & 1 \text{ rem 1}\\
& 0 \text{ rem 1}\\
\end{array}\uparrow \)

2710 = 110112

Therefore xxxxx2 = 2710 = 110112


Correct option: B
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Question 46
A

2 log5y + 5log5 y2-3

B

log5 y2 + 5log5 x + 3

C

25logy 5 + 3

D

2log5y + 5log5x - log5b - 3

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: D
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Question 47
A

\( (3x^2 + 8x + 7) (3x + 4)^2 \)

B

\( x^2 + 2x + (1 (x + 1)^2) \)

C

\( 2x^2 + 4x + (1 (2x + 4)^2) \)

D

\( x^2 + 3x + (2 (x + 3)^2) \)

Explanation

dy/dx \( (4x^3 + 3x^2 + 2x + 1) \)

= \( 12x^2 + 6x + 2 \)

= \( 12x^2 + 6 + 2 \)

Divide both sides by 2

\( \frac{12x^2}{2} +\frac{6x}2 +\frac{2}{2} \)

= \( 6x^2 + 3x + 1 \)

dy/dx

= \( 6x^2 + 3x + 1 \)


Correct option: A
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Question 48
A

\( \frac{5x^4}{4} + \frac{7x^3}{3} + 2x + C \)

B

\( \frac{5x}{4} + \frac{7x^3}{3} - x^2 + 5x + C \)

C

\( \frac{5x^3}{3} + \frac{7x^2}{x} - x + C \)

D

\( \frac{2x^2}{3} + \frac{x}{5} - C \)

Explanation

\(- [(x^2 + 4x^2 + 1 ) x^2]dx = - (x^3/x^2)dx + -(1/x^2)dx \)

-\( - xdx + - 4^{-1} or + - (1/x^2)dx\)

= \( x^2/2 - 4x - 1/x^2 + C \)


Correct option: B
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Question 49
A

17.1cm2

B

27.2cm2

C

47.1cm2

D

37.3cm2

Explanation

Find the slant height

\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)

\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)

\( l^2 = 25 \)

Squaring both sides

l = 5cm

The area of curved surface (s) =(3)(5)

15= 15 \(\times\) 3.14

= 47.1cm2


Correct option: C
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Question 50
A

\(\frac{2x}{(x + 1)(x-3)} \)

B

\(\frac{2}{(x + 1)(x-1)} \)

C

\(\frac{2x}{(x + 1)2}\)

D

2x(x+1)2

Explanation

[1 (x+1)] + [1 (x - 1)]

= ((x - 1) + [(x + 1)) (x+1)(x - 1)]

Using the L.C.M.

= (x - 1 + x + 1) (x + 1)(x - 1)

= (x + 2 - 1 + 1) (x + 1)(x - 1)

= 2x (x + 1)(x - 1) =2x (x + 1)(x - 1)


Correct option: B

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