2017 Waec Mathematics Answers Essay/Obj Free Expo

Thursday, 20th April, 2017
General Mathematics (Core) 2 (Essay) 09.30am – 12.00pm
General Mathematics (Core) 1 (Objective) 3.00pm – 4.30pm

MATHS OBJ

1-10: CBBCCACBBC

11-20: CACBAAAADB

21-30: BBADBCCACB

31-40: BBCACBDADA

41-50: DDACCCDDCB

SECTION A ANS 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



3b)

Area of A = 1/2bh
45 = 1/2 x 10 x h
45 = 5h
h = 9cm
Area of < QTUS = 1/2 ( QT + US)h
= 1/2 ( 6 + 16)9
= 99cm^2


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







8a)

In Table Form / Tabular form
X = 1,2,3,4,5
F = m+2, m-1, 2m-3, m+5, 3m-4 = 8m - 1
Fx = m+2, 2m-2, 6m-9, 4m+20, 15m-20 = 28m - 9
But x̄ ( this symbol (x̄) means X bar)
= 75/23
ΣFx / Σf = 75/23 = 28m - 9/8m-1
75/23 = 28m - 9/8m - 1
Cross multiply
75(8m-1) = 23(28m-9)
600m - 75 = 644m - 207
-75 + 207 = 644m - 600m
132 = 44m
M = 3



8bi)

In tabular form
X = 1,2,3,4,5
F = 5,2,3,8,5
Cum Freq= 5,7,10,18,23
Q1 = (N+1/4) = (23+1/4)
= 6
Q3 = (3N + 1/4) = (3*23+1/4)
= 18
Inter quarter range = Q3 - Q1
=. 18-6
= 12


8bii)

Pr. (at least 4 mark)
= 8+3+2+5/23
= 18/23





10)






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
From Tanθ = 9.6/2.8
Tan θ = 3.4288
θ = Tan^-1 ( 3.4288)
= 73.74°






11a)

8 students finished 
12 tanks in 2/3 (60) mins
= 40 mins
4 student wil finish
X tanks in 1/3 (60)min
= 20mins
X = 4x20x12/8x40
= 3tanks



11b)

L(AB) = 200m |ON| = 12cm
r2 = (AN)2 + (ON )2
r2 = (10)2 +(12)2
r2 = 100 + 144
r2 = 244
r = Sqr 244
r = 15.6CM 



11bii)

L(AB) = 2r sin 0/2
20 = 2 (15.6) sin 0/2
20 = 31.2 sin 0/2
sin0/2 = 20/31.2
sin0/2 = 0.6410
0/2 = sin -1 (0.6410)
0/2 = 39.87
0 = 2 (39.87)
0 = 79.74
= 79.7' (1 d.p )



11bii)

p 2r + 0/360 x 2TTr
= 2 (15.6 ) + 79.7/360 x 2x 3 x42x15.6
=31.2 + 21.7
= 52.9 cm



12)




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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