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2018/19 WASSCE GENERAL MATHEMATICS THEORY AND OBJECTIVE ANSWER AVAILABLE HERE

By | On April 17, 2018 (7 months ago)

NOTE: OUR SUBSCRIBERS WILL GET THERE ANSWERS 2HRS BEFORE WE POST FREE ANSWER HERE

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🇳🇬🇳🇬🇳🇬🇳🇬🇳🇬🇳🇬👇🏽NIGERIA

*MATHEMATICS ANSWERS

MATHS OBJECTIVES

1-10 ACBCDDCBAA

11-20 CDBABCCCDC

21-30 DADBABCDAD

31-40 BADADBCACB

41-50 ADDDBDADCA

(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)

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

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/hyp
= 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
========================

4ai)
sum of angle in a D =180degree
xdegree + 90degree + 180degree – (3x+15)=180degree
xdegree + 90degree + 180degree – 3x+15=180degree
-2x=180degree – 255
+2x/2=+75/2
x=37.5

===================================
4aii)
<RsQ =180 – (3x+15)
<RsQ =180-(3*37.5+15)
=180-(112.5 + 15)
=180 – 127.5
<RsQ= 52.5degree
=========================

4b)
2N4seven =15Nnine
2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree
9*49+N*7+4*1=1*81+5*9+N*1
98+7N+4=81+45+N
7N+102=126+N
7N-N=126-102
6N/6 =24/6
N=4
========================
No 6

240 cars
60% passed
40% passed

40/100 * 240

=96

sketch the diagram

6b

2x+x+2x+6+6+8+12=96

28+6+6+8+12+x+2x =96
60+3x=96
3x=96-60
3x=36
x=36/6
x=12

i. forty braked

=12+8+6+12
=38

ii. only one fault

=28+12+(2 * 12)
=28+12+24
=64

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

(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

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

(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= 12×0.3843
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= 12×0.60000
y= 7.2m

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