Posted on: 03:44 Fri, 29 Nov 2019




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Exam Time:- Saturday 30th Nov. 2019
General Mathematics Paper III (Objective) – 1.00pm – 2.45pm
General Mathematics Paper II (Essay) – 3.00pm – 5.30pm
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MATHS OBJ:-
1-10: BCBEDDCCDC
11-20: DEBCBBDDEC
21-30: BCBCCBBEBD
31-40: EBABEDCDEB
41-50: DDAECACBAB
51-60: DDBCEACBBE
=•=•=•=•=•=•=•=•=•=•=

MATHS ANSWERS:
(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]

= [â…”(3) â…”(4) â…”(0)]
[â…”(2) â…”(0) â…”(3)]
[â…”(1) â…”(4) â…”(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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(7a)
Y= 2x²+7x-6
Dy/dx = 4x+7
Dy/dx = 4(3)+7
= 12+7
= 9
(7b)
x= price of 1kg
Y= price of 1kg of meat
7x+4y = 4240 .........(i) x5
3x+5y=4010............(ii) x4
35x+5y=4010...........(iii)
12x+20=18440/23x=2760
X = 2x0/23
X = #120
Substitute for x in equa (1)
7x + 4y = 4240
7(120)+4y=4240
840+4y=4240
4y= 4240-840
4y= 3400/4
Y= #850

(7bii)
10x+5y
10(120)+5(850)
1200+4250
= #5450