2017/2018 WAEC General Mathematics Answer

Here is the 2017/2018 WAEC General Mathematics Answers.


OBJ;
1-10:   CBBACCCBBA
11-20:  ADBBAABCDB
21-30:  BCADBCCAAB
31-40:  CDCACDBACB
41-50:  BDABCDDCAD


THEORY;
NOTE: Where ever you see
^ It means raise to power
/ It means Division
* It means Multiplication
X It means normal X



SECTION A Answer ALL

1a)
(y-1)log4^10= ylog16^10
log4^10 (y-1)= log16^y10
4^(y-1)=16y
4^y-1=4^2y
y-1=2y
-1=2y=y
-1=y
y= -y

1b)
let the actual time for 5km/hr be t
for 4km/hr=30mint + t
4km/hr=0.5 + t
distance = 4(0.5+t)
=2*4t
for 5km/hr, time= t
distance =5t
1+4t=5t
t=2hrs
actual distance = 5*2=10km


2a)
2/3(3x-5)-3/5(2x-3)=3/1
L C M =15
10(3x-5)-9(2x-3)=45
30x-50-18x+27=45
30x-18x=45+50-27
12x-23=45
12x=45+23
12x=68
x=68/12
x=34/6
x=17/3

2b)
U’aS=180-(n+88)
=180-n-88=92-n
also, u’TQ=18m
80degree + 92-n+180-m=180degree
80+92+180-n-m=180degree
352-n-m=180degree
-n-m=180-352
-n-m=-172
+(n+m)= +172
m+n=172dgree


3a)
Tan 23.6° = h/50
Cross multiply
Tan 23.6° x h/50
h = 50 tan 23.6°
= 21.844m


4a)
T6=37
T6=a+(6-1)d
T6=a+5d
a+5d=37 —–(eq1)
s6=147
sn=n/2(2a+(n-1)d)
147=3(2a+5d)
49=2a+5d
2a+5d=49 —-(eq2)
a+5d=37 —(eq1)
2a+5d=49 —(eq2)
a=12

4b)
S15=15/2(2(12)+14d)
S15=15/2(24+14d)
from(1)
a+5d=37
12+5d=37
5d=37-12
5d=25
d=5
S15 = 15/2(24+14(15)
S15= 15/2(24+70)
S15=15/2*94
S15=15*42
S15=630


5a)
draw
U=20
B= y-45
S= y-34
B=bag
S=shoe
let n(B)=y
n(S)=y+11
for bag only y-45
for shoe only y-11-45=y-34

5b)
y-45+45+y-34=120
2y-34=120
2y=154
y=154/2
y=77
number of customers who bought shoe = y+11
77+11=88

5c)
n(bag)=77customers
probability =77/120
=0.642



SECTION B Answer 5 QUESTIONS ONLY

10a).
Sin x = 5/13
Using pythagoras rule
M^2 = 13^2 – 5^2 (^ means Raise to power)
M^2 = 169 – 25
M ^2 = 144
M = √144
M = 12
Hence:
Cos x – 2sin x / 2tan x
12/13 – 2(5/13) / 2(5/12)
= 12/13 – 10/23 / 5/6
FIND LCM
= 12 – 10/13 / 5/6
= 12/65

10bi)
Considering < LMB /MB/^2. = 12^2 - 9.6^2 /MB/^2 = 51.84 /MB/ = √51.84 /MB/ = 7.2m From < AML /LA/^2 = 2.8^2 + 9.6^2 /LA/ ^2 = 100 /LA/ = √100 /LA/ = 10m 10bii) Let the angle be. θ


13ai)
Given
x(*)y=x+y/2
i)3(*)2/5=3+2/5/2
=(15+2/5)*1/2
=17/5*1/2
=17/10= 1,7/10

13aii)
8(*)y=8^1/4
=8+y/2 =33/4
32+4y=66
4y=66-32
4y=34
y=34/4
y=17/2
y=8^1/2

13b)
Given DABC
AB=(^-4/6) and AC =(3/^-8)
so AP =1/2(^-4/6)
AP=(^-2/3)
hence
CP = CA + AP
CP= -(3/^8)+(^-2/3)
CP = (^-5/11)


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