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2018 WAEC GENERAL MATHS THEORY AND OBJ ANSWERS AVAILABLE

 

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GENERAL MATHS ANSWERS.

GENERAL MATHS OBJ

1-10:::: abbcddbbaa
11-20: cdbabcccdc
21-30: dadbabcdad
31-40: badadbcacb
41-50: adddbdadca


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

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

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 opeople

(-3, 5)


NO3) DRAW THE DIAGRAM.

Using pythagoras theorem,

lˆ² = 5.1ˆ² + 4.65ˆ²

lˆ² = 26.01 + 21.6225

lˆ² = 47.6325

l = √47.6325

l = 6.9cm (1d.p)

perimeter of phombus = 41

4 * 6.9

= 27.6cm

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


6a) Draw the diagram

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

6bi) faulty brakes cars= 8+6+x+6-x

= 8 + 6 + 6

= 20

bii) Only one fault = 28 + x + 2x

= 28 + 3x

= 28 + 3(17)

= 28 + 51

= 79


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)

9b) Now: |PT|/|PQ| = |PS|/|PR|

4/6 = 10/|PR|

4|PR| = 60

|PR| = 60/4

= 15CM

Therefore, |QR| = 15 – 6 = 9cm

Area of TQRS = Area of ΔPRS – Area of ΔPQT

=1/2(PS)(PR)sin30° – 1/2(PT)(PQ)sin30°

= 1/2sin30°(|PS| (PR) – |PT| |PQ|)

=1/2 * 1/2((10)(15) – (4)(6))

=1/4(150 – 24)

=1/4 * 126

=31.5 apx 32cm²


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


5a) m+n+s+p+q/5 = 12

m+n+s+p+q =60

Now; (m+4) + (n-3) + (s+6) + (p-2) + (q+8)

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

= 60+13/5

= 73/5 = 14.6

5b) 75% of 500 = 375 people

Number of people above 65 yrs = 500 – 375 =125

25% of 500 = 125

Number of people below 15 yrs = 125

Number between 15yrs and 65yrs

= 500 – (125 + 125)

= 500 – 250

= 250 people


(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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