Mathematics (Essay & Objective) Waec 2017 Free Expo Answers - GidiRocks.com (mp3)Mathematics (Essay & Objective) Waec 2017 Free Expo Answers - GidiRocks.com
Mathematics (Essay & Objective) Waec 2017 Free Expo Answers - GidiRocks.comMathematics (Essay & Objective) Waec 2017 Free Expo Answers - GidiRocks.com

Mathematics (Essay & Objective) Waec 2017 Free Expo Answers

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Maths (Essay & Objective) Waec 2017 Free Expo Answers

============================
Thursday, 20th April, 2017
General Mathematics 2 (Essay) 09.30am – 12.00pm
General Mathematics 1 (Objective) 2.00pm – 3.30pm
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MATHS OBJ:
1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB

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West African Examination Council, WAEC 2017 Mathematics Question and Answer for May/June Examination EXPO had been posted here for free.

waec 2017 mathematics expo
waec 2017 mathematics expo

Note
Where ever u c
^ it means raise to power
/ division
* multiplication
X normal X

Waec 2017 Mathematics Question And Answer Free Exam Expo

Waec 2017 Mathematics Question And Answer Free Exam Expo

Waec 2017 Mathematics Question And Answer Free Exam Expo

Waec 2017 Mathematics Question And Answer Free Exam ExpoWaec 2017 Mathematics Question And Answer Free Exam Expo

Waec 2017 Mathematics Question And Answer Free Exam Expo

Note
Where ever u c
^ it means raise to power
/ division
* multiplication
X normal X

 

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

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8a )
In Table Form / Tabular form
X = 1, 2, 3, 4, 5
F = m +2, m – 1, 2m- 3, m +5, 3 m- 4 = 8m – 1
Fx = m+2 , 2m – 2, 6 m- 9, 4m +20, 15m – 20 = 28 m – 9
But x̄ ( this symbol ( x̄) means X bar )
= 75 /23
ΣFx / Σf = 75 /23 = 28m – 9/8 m- 1
75/23 = 28m – 9/8m – 1
Cross multiply
75( 8m- 1) = 23 (28 m- 9)
600m – 75 = 644m – 207
– 75 + 207 = 644m – 600 m
132 = 44 m
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
Q 1 = ( N +1 /4) = ( 23+1/4 )
= 6
Q 3 = ( 3N + 1 /4) = ( 3*23+1 /4)
= 18
Inter quarter range = Q 3 – Q 1
=. 18 – 6
= 12
8bii)
Pr . (at least 4 mark )
= 8 +3+2+ 5/23
= 18/23
================================

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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

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10bii)
Let the angle be. θ
From <AML
Tanθ = 9.6/2.8
Tan θ = 3.4288
θ = Tan^-1 ( 3.4288)
= 73.74°
=============

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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