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**Maths OBJ:**

1-10: BCBEDDCCDC

11-20: DEBCBBDDEC

21-30: BCBCCBBEBD

31-40: EBABEDCDEB

41-50: DACECDCBAB

51-60: DECCEABBBE

(1)

√((0.0024)×35000)/0.0105

**No | Log**

0.024 |_3.3802

35000 |4.5441

– |=1.9243

0.0105|_2.0212 –

|3.9031 ÷2

|1.9516

**√((0.0024)×35000)/0.0105**

**Antilog of .9516 =89.45**

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(2a)

Given that the roots of the equation are

X = -2/3 and X = -3/2

3x = -2 and 2x = -3

3x + 2 = 0 and 2x + 3 =0

(3x+2)(2x+3) = 0

6x² + 9x + 4x + 6 = 0

6x² + 13x + 6 = 0

**(2b)**

R = [3 4 0 ]
[2 0 3 ]
[1 2 2]

**(2ci)**

2/3R

**=2/3[3 4 0]
[2 0 3]
[1 2 2]**

**= [⅔(3) ⅔(4) ⅔(0)]
[⅔(2) ⅔(0) ⅔(3)]
[⅔(1) ⅔(4) ⅔(2)]**

**= [2 8/3 0 ]
[4/3 0 2 ]
[2/3 4/3 4/3]**

**(2cii)**

|R|

**= |3 4 0|**

|2 0 3|

|1 2 2|

**=3|0 3| -4|2 3| +0|2 0|**

|2 2| |1 2| |1 2|

**=3(2×0-3×2)-4(2×2-3×1)**

+0(2×2-0×1)

=3(0 – 6)-4(4 – 3) +0(4 – 0)

=3(-6) -4(1) +0(4)

= -18 – 4 + 0 = -22

**(2ciii)**

The transpose of R

= [3 2 1]
[4 0 2]
[0 3 2]
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(3a)

Given : R(3,5) and S(-2, -6)

equation to line through them is :

y-5/x-3 = -6-5/-2-3

y-5/x-3 = -11/-5

y-5/x-3 = 11/5

5(y-5)= 11(x-3)

5y-25 = 11x-33

5y-11x = 25-33

5y-11x = -8 or 11x-5y = 8

**(3b)**

RS= √(X1X2)² (y1y2)²

√(-2-3)² + (-6-5)²

√(-5)² + (-11)²

√25 + 121

√146

= 12.08

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(4a)

Given: curve; y x² – 3x

gradient ; dy/dx = 2x – 3

At (2-2), gradient = 2(2) – 3

4-3 = 1

**(4b)**

Given; y= 1+x²/1-x²

dy/dx = (1-x²)(2x) – (1+x²)(-2x)/(1-x²)²

= (1-x²)(2x) – (1+x²)(2x)/(1-x²)²

= 2x(1-x² + 1+x²)/(1-x²)²

= 2x(2)/(1-x²)²

= 4x/(1-x²)²

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(5)

No of blue balls = 6

No of red balls = 10

**(i)**

Prob (2 balls of some colour)

= BB or RR

Total no of balls = 6+10 = 16

BB or RR

(6/16 × 5/15) + (10/16 × 9/15)

=30/240 + 90/240

=120/240 = 1/2

**(ii)**

Prob (2 balls of different colours)

= BR or RB

= (6/16 × 10/15) or (10/16 × 6/15)

= 60/240 + 60/240 = 120/240

=1/2

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(6a)

27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)

= 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)

=3^6x+3-x = 3^4x-8-2x-4

5x+3 = 2x – 12

5x – 2x = -12-3

3x = -15

3x/3 = -15/3

X = -5

**(6b)**

X/x+101 = 11/1000

Since all the members are in binary, convert all to denary (base 10)

Xbase2 = Xbase10

101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10

11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10

**1000base2 = 1×2^3 = 8base10**

X/X+101 = 11/1000 –> X/X+5 = 3/8

3(x+5) = 8(x)

3x+15 = 8x

15 = 8x – 3x

15 = 5x

15/5 = 5/5

X = 3

Convert X=3 to base 10 to base 2

2|3

2|1R1

|0R1

.:. x = 11

**(6c)**

Given that log5 base 10 = 0.699

**and log3 base 10 = 0.477**

10

Log75 base 10 = log(3×5×5) base 10

**=log(3×5²) base 10**

**=log3 base 10 + 2log5 base 10**

**=0.477 + 2(0.699)**

= 0.477 + 1.398

= 1.875

log75 base 10 = 1.875

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(6b)

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(6c)

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(7a)

Given that Y=2×2+7x-6

To find the gradient of the curve at the point x=3

dy/dx=4x+7 (at x=3)

dy/dx=4(3)+7

dy/dx=19

**(7bi)**

Ade bought 7kg of maize + 4kg of meat=N4240

Kemi bought 3kg of maize + 5kg meat =N4610

Let x = 1kg of maize and y=1kg of meat

Therefore 7x+4y=4240—–(eq1)

3x+5y=4610———(eq2)

Substituting simultaneously

Multiply eq1 by 3 and eq2 by 7

21x+12y=12720—eq3

21x+35y=32270—eq4

Substract eq3 from eq4

23y=19550

23y/23=19550/23=850

y=850

Substitute y=850 in1

7x+4y=4240

7x+4(850)=4240

7x=4240-3400

x=840/7

=120

Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00

**(7bii)**

Total cost of 10kg of maize and 5kg of meat

=10x+5y

=10(120)+5(850)

1200+4250

=N5450

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(8)

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(8b)

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(9a)

Draw the diagram

Q(34.5°N, 22.3°W)

P(34.5°N, 38.7°E)

R(35.4°S, 38.7°E)

Distance between P and Q along the parallel of latitude

D = Φ/360 × 2πr

Where r = RcosΦ

Φ = 38.7° + 22.3° = 61°

Q = 34.6°

R = 6400km

D = 61/360 × 22/7 ×6400cos34.6

D =1717760(0.8231)/252

D = 5,610.66768

= 5,610km

**(9b)**

|PR|=(34.6+35.4)/360 * 2 *22/7*6400

=70/360 * 44/7 *6400

=19712000/2520

=7820km(3 sf)

Therefore the shortest distance between P and R = 7820km(3 sf)

**(9c)**

Circumference of the circle of latitude through R = 2πr where r = RcosΦ

= 2πRcosΦ

= 2 × 22/7 × 6400cos35.4

=281600/7(0.8151)

= 229,532.16/7

= 32,790.3

= 32,800km (3 s.f)

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(9c)

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(10)

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(12)

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OBJ QUESTIONS: ?

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ESSAY QUESTIONS: ?

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