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Maths Objective

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NECO SUBSCRIPTION IS GOING ON STILL #2000

ANSWER DROP A DAY TO THE EXAM

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Answer only 10 questions

NO1) 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

[4/18, 10:08] ‪+234 810 856 7860‬: NO2) Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)²  – p² (3) – 14

10 = 18p – 3p²  – 14

3p²  – 18p + 24 = 0

p²  – 6p + 8 = 0

using factor method,

p²  – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2



‬: 🇳🇬🇳🇬🇳🇬🇳🇬👇🏽NIGERIA



*MATHEMATICS  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


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

(2)
 Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)²  – p² (3) – 14

10 = 18p – 3p²  – 14

3p²  – 18p + 24 = 0

p²  – 6p + 8 = 0

using factor method,

p²  – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2


2b)
The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4)  3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)
[4/18, 10:19] ‪+234 810 856 7860‬: (3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80

(3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hypH
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1

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

[4/18, 10:33] OSHY....: (4ai) x + 90 degrees = 3x+15(sum of two inter <s= exterior <)

Collecting like terms

90-15= 3x -x

75= 2x

x= 75/2

x= 37.5

(4aii) x^degrees + 90+<RSQ = 180deggrees(sum of <s in a Triangle)

37.5 + 90+ <RSQ= 180

127.5 + <RSQ= 180

<RSQ= 180 - 127.5

<RSQ= 52.5 degrees

(4b) If 2N4 seven = 15N nine.  Find N

Converting all to base 10

2x7^2+Nx7^1+4x7^0= 1x9^2+5x9^1+Nx9^0

2x49+7N+4x1= 1x81+5x9+Nx1

98+7N+4= 81+45+N

Collecting like terms

7N-N= 81+45-98-4

6N= 24

N=24/6

N=4

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

NO 5

5a)
m + n + s + p + q / 5 = 12
m + n + s + p + q =  60  ----> equ 1

Now;

(m+4) + (m-3) + (5+6) + (p-2) + (q+8)  / 5
= (m + n + s + p + q) + (4 - 3 + 6 - 2 + 8) / 5
= 60 + 13 / 5
= 73 / 5
= 14.6

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

[4/18, 10:33] OSHY....: (3a)Using Pythagoras theorem let the side of the rhombus = x

x^2= (10.2/2)^2 + (9.3/2)^2

x^2= 5.1^2 + 4.65^2

x^2= 26.01 + 21.62

x^2= 47.63

x= sqroot of 47.63

x= 6.9

Sides= 6.9cm

But perimeter of rhombus = 4x

Perimeter = 4x6.9

Perimeter = 27.6cm

(3b) Sin x= 3/5

Opp= 3, Hyp= 5

Let adjacent = x

From Pythagoras theoren

5^2= 3^2+x^2

25= 9+x^2

x^2= 25-9

x^2= 16

x= sqroot of 16

x= 4

5cosx-4tanx

=4 - 3= 1

5cosx-4tanx= 1

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

[4/18, 10:33] OSHY....: (10a) Using Pythagoras theorem from SPQ

|SQ|^2 = 12^2 + 5^2

= 144+25

=169

SQ= sqroot of 169

= 13cm

Sin tita= 5/13 = 0.3846

Tita= Sin^-1(0.3846)

= 22.6degrees

From PRQ

Sin tita= |PR|/12

Sin 22.6 = PR/12

Sin 22.6= PR/12

PR= 12xsin 22.6

PR= 12x0.3843

PR= 4.61cm

(10bii)Let the height at which m touches the wall= y

Cos x^degrees= 8/10= 0.8

x^degrees= Cos^-1(0.8)

= 36.87degrees

Sin x^degrees = y/12

Sin 36.87= y/12

y= 12xsin36.87

y= 12x0.60000

y= 7.2m

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

(6a)

Draw the Venn diagram

Let the number of cars with faults in brakes only be x

(6b) Number that passed = 60% × 240 = 144

Number that failed =

240 - 144 = 96

Therefore; 28+2x+x+14+6+6-x+8 = 96

2x + 62 = 96

2x = 96 - 62

2x = 34

X = 34/2

X = 17

(i) faulty brakes cars = 8+6+x+6-x

= 8+6+6

=20

(ii) only one fault = 28+x+2x

=28+3x

=28+3(19)

=28+51

= 79

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

[4/18, 11:17] ‪+234 816 198 6400‬: (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
(b)
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
[4/18, 11:18] ‪+234 902 983 6226‬: (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

========Mr Tim sure runz========

(7a) Given the points (2,5) and (-4,-7)

Gradient (m)= -7-5/-4-2= -12/-6 = 2

Let (x, y)  be a point on the line and (2,5) on the line

2/1 = y-5/x-2

y-5= 2(x-2)=2x-4

y=2x-4+5

y= 2x+1

(7bi) (QR) ^2= 8^2 + 5^2

= 64+25

= 89

QR= sqroot of 89

= 9.43km

(7bii) 1. Distance between Q and R

9.43/sin90 = 5/sinR

5sin90= 9.43SinR

Sin R= 5Sin90/9.43

Sin R= 5/9.43

= 0.5302

R= Sin^-1 0.5302

=32

2. Bearing of R from Q

= 32+90

=122degrees

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

(8a)
Cost price for Lami= #300.00
  Profit made by lami = x%
Ie selling price for lami=(100+x/100)×#300
=#3(100+x)
=#(300+3x)

Bola's cost price = #3(100+x)
Profit made by bola =x%
Selling price for bola =(100+x/100)×#3(100+x)
=#3/100(100+x)²

James cost price =#3/100(100+x)²=300+(6x+3/4)
expanding;
3/100(10000+200+x²) = 300+3/4+6x
3(10000+200x+x²)=30000+75+600x
30000+600x+3x²=30000+75+600x
3x²=75
X² = 75/3
X² = 25
X = square root 25
X = 5

(8b)
3x-2<10+x<2+5x
3x-2<10+x & 10+x<2+5x
3x-x<10+2 & 10-2<5x-x
2x<12              8<4x
X<12/2            4x>8
X<6                   x>8/4
                          X>2

Also; 3x-2<2+5x
          -4<2x
          2x > -4
          X > -2
Therefore; Range is -2<X<6

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

NO9) Using cosine rule,

|TQ|ˆ² = 4ˆ² + 6 ˆ² – 2(4)(6) cos30°

|TQ|ˆ² = 16 + 36 – 48(0.8660)

|TQ|ˆ² = 52 – 41.568

|TQ|ˆ² = 10.432

TQ = √10.432

TQ = 3.23CM

From similar triangles;

|PT|/|TQ| = |PS|/|SR|

4/3.23 = 10/|SR|

4|SR| = 32.3

|SR| = 32.3/4

|SR| = 8CM (nearest whole number)



=======!

(10a) Using Pythagoras theorem from SPQ
|SQ|^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^-1(0.3846)
= 22.6degrees
From PRQ
Sin tita= |PR|/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12x0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^-1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12x0.60000
y= 7.2m

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