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

=====================

(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)**
=====================

(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

=====================

(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

=====================

(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

=====================

(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

=====================

(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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