Mathematics Objective and Theory Questions And Answers.

MATHS OBJECTIVE
MATHS OBJ:

1-10: ACBCDDCBAA

11-20: CDCABCCCAC

21-30: DADBABCDAD

31-40: BADADBCACB

41-50: ADDDBDADCA


MATHS THEORY ANSWERS

(1)
On February 28th 2012, value = (100-30/100) * #900,00.00
= 70/100 * #900,00
= #630,000.00
On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
= #299,000
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(2a)
Given y-2px^2-p^2x-14
At (3,10)
10=2p(3)^2-p^2(3)-14
3p^2-18p 24=0
p^2-6p 8==0
Using factor method
P^2-2p-4p 8=0
P(p-2)-4(p-2)=0
(p-4)(p-2)=0
p=4 or p=2
(2b)
The lines must be solved simultaneously
3y-2x=21--(i)
4y 5x=5--(ii)
Using elimination method
=>12y-8x=84--(iii)
=>12y 15x=15--(iv)
eq(iv) minus (eqiii)
23x=-69
x=-69/23
x=-3
Put x into eq(i)
3y-2(-3)=21
3y 6=21
3y=21-6
3y=15
y=15/3
Coordiantes of Q is (-3,5)
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(3a)
Using pythagoras theorem
L^2=5.1^2 4.65^2
L^2=26.01 21.6225
L^2=47.6325
L=sqroot(47.6325)
L=6.9cm(1 d.p)
Perimeter of rhombus=4
=4*6.9
=27.6cm
(3b)
Sin x=3/5
DRAW THE TRIANGLE
Using pythagoras tripple the third side=4
therefore cosx=4/5
tanx=3/4
therefore 5cosx-4tanx
=5(4/5)-4(3/4)
=4-3
=1
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(4ai)
Draw the diagram
aiX 90 = 3x 15
90 = 3x - X 15
90 = 2x 15
2x 15 = 90
2x = 90 - 15
X = 75/2
X = 37.5•
(4aii)
(4b)
2N4seven = 15Nnine
Converting both to base 10
2×7 N×7¹ 4×7 = 1×9² 5×9¹ N×9
98 7N 4 = 81 45 N
7N 102 = 126 N
7N - Ń = 126 - 102
6N = 24
Ń = 24/6
Ń = 4
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(5a)
m n s p q/5=12
m n s p q=60......(1)
Now;
(m 4) (n-3) (5 6) p-2) (q 8)/5
=(m n s p q) (4-3 3 6-2 8)/5
=60 13/5
=73/5
=14.6
(5b)
75% of 500 = 375 people
Number of people above 65 years = 500-375
=125
25% of 500 = 125
Number of people below 15 years = 125
Number between 15 years and 65 years
=500-(125 125)
=500-250
=250 people
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(6)
Total number of cars on road worthiness = 240
60% passed ie 60/100×240/1 = 144cars.
Number that failed = 240-144 = 96cars
(6a)
Draw the Venn diagram
C = clutch
B = brakes
S = steering
(6b)
From the diagram above
E = 28 12 8 6 x 6 2x 96=60 3x
96-60=3x
36/3 = 3x/3
Therefore X = 12
(i) The no of cars that had faulty brakes
=12 8 6 x (Since X = 12)
=12 8 6 12 = 38
(ii) Only one fault = (no of clutch only) (no of brakes only) (no of steering only)
=28 x 12x = 28 12 24
=64cars
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(7a)
(y-y1)/(x-x1)=(y2-y1)/(x2-x1)
(y-5)/(x-2)=(-7-5)/(-4-2)
(y-5)/(x-2)=-12/-6
(y-5)/(x-2)=2
Cross multiply
y-5=2(x-2)
y-5=2x-4
2x-y-4 5=0
2x-y 1=0
(7bi)
DRAW THE DIAGRAM
(7bii)
(I)
p^2=q r^2-2qrcosP
p^2=8^2 5^2-2*8*5*cos90
p^2=64 25-0
p^2=89
p=sqroot(89)
p=9.4339km
Therefore: |QR|=9.43km(3 sf)
(II)
q/sinQ=p/sinP
8/sinQ=9.4339/sin90
sinQ=(8*sin90/9.4339
sinq=(8*1)/9.4339 =0.8480
Q=sin^1(0.8480)=57.99 degrees
but Q=30 A
A=Q-30
=57.99-30
A=27.99 degrees
The bearing of R from Q
=180-A
180-27.99
=155.01
=>152 degrees
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