# Mathematics (2015) | Study Mode

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

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

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

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)

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

Question 46
A

2 log5y + 5log5 y2-3

B

log5 y2 + 5log5 x + 3

C

25logy 5 + 3

D

2log5y + 5log5x - log5b - 3

##### Explanation

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

= $$6x^2 + 3x + 1$$

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

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

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

= $$x^2/2 - 4x - 1/x^2 + C$$

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

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)

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