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Mathematics (Essay & Objective) Waec 2017 Free Expo Answers
By: On: April 20, 2017

.Maths-Obj
1-10 : CBBDCADBBC
11- 20: CACBAAACAB
21- 30: BBACBCCAAB
31- 40: CBCACBDADB
41- 50: DBDCCDBDAB
===========
SCROLL DOWN . . . NO 12 ADDED
2017 CORE MATHS SOLUTIONS
Note
Where ever u see
^ it means raise to power
/ division
* multiplication
X normal X
================================
SECTION A ANS ALL
================================
1a)
( y -1 ) log4^10= ylog 16 ^10
log4^10 ( y – 1) = log 16^y 10
4^( y – 1) =16y
4^y – 1=4^2y
y -1 =2y
-1= 2y =y
-1= y
y = -1
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=180 degree
352- n-m =180 degree
-n- m= 180- 352
-n- m= -172
+( n+m ) = +172
m+ n=172 dgree
================================
3a)
Tan 23 . 6° = h /50
Cross multiply
Tan 23 . 6° x h /50
h = 50 tan 23. 6 °
= 21 . 844 m
3b)
Area of < TRU = 45cm ^2 ( Note : This ^ means
Raise to power )
A = 1/2 bh
45 = 1/2 x 10 x h
45 = 5h
h = 9cm
Area of < QTUS = 1/2 ( QT + US ) h
= 1 /2 ( 6 + 16) 9
= 99 cm ^2
================================
4a)
T 6=37
T 6=a+( 6 -1) d
T 6=a+5 d
a+5d =37 — — ( eq1)
s 6=147
sn =n /2( 2a+( n -1) d)
147= 3( 2a+5 d)
49=2 a+5d
2a+5 d=49 — –( eq 2)
a+5d =37 — ( eq1)
2a+5 d=49 — -( eq2 )
a=12
4b)
S 15=15 /2( 2( 12) + 14d)
S 15=15 /2( 24+14 d)
from ( 1)
a+5d =37
12+5 d=37
5d=37 -12
5d=25
d=5
S 15 = 15/2 ( 24+14 ( 15)
S 15= 15/2( 24 +70)
S 15=15 /2*94
S 15=15 *42
S 15=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 ANS 5 QUESTIONS ONLY
================================
8a)
In Table Form / Tabular form
X = 1, 2, 3, 4, 5
F = m+2 , m- 1, 2m -3 , m+ 5, 3m -4 = 8 m – 1
Fx = m +2, 2 m- 2, 6m -9, 4m+ 20, 15 m- 20 = 28m –
9
But x̄ ( this symbol ( x̄ ) means X bar )
= 75 /23
ΣFx / Σf = 75 /23 = 28 m – 9 /8m- 1
75/23 = 28m – 9/8m – 1
Cross multiply
75( 8m -1 ) = 23( 28m -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
Q1 = ( N +1/4) = ( 23 +1/4)
= 6
Q3 = ( 3 N + 1/4) = ( 3 *23 +1/4)
= 18
Inter quarter range = Q3 – Q 1
=. 18 -6
= 12
8bii )
Pr . ( at least 4 mark )
= 8 +3+2+ 5/23
= 18/23
================================
10a)
Sin x = 5/13
Using pythagoras
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
From < AML
Tanθ = 9. 6/2. 8
Tan θ = 3. 4288
θ = Tan ^- 1 ( 3. 4288 )
= 73 . 74 °
================================
12a) Using completing the square method
3y ^2- 5y +2 =0
y ^2 – 5 /3y + 2/3= 0
y ^2- 5/3y =- 2/3
y ^2- 5/3y +(^ – 5/6) ^2= ( -^5/6) ^2 -2/3
( y -5 /6) ^2=25 /36-2 /3
( y -5 /6) ^2=25 /-24/36
( y -5 /6) ^2=1/36
( y -5 /6) =+sqr 1/36
y =5/6 +1/6
y =5+ 1/6 or 5- 1/6
y =6/6 or 2/3
y =1 or 2 /3
12b)
given
M N = [ 2, 3 1, 4]
hence
[1, 4 2, 3] * [m , n x , y ] =[ 2, 3 1 ,4 ]
[m +2n, x *2 y ]
[4m +3n , 4x +5 y ] = [ 2, 3, 1, 4]
therefore
m+ 2n=2- — — ( i )
4m +3n=3 — — -( ii )
from – — — ( i )
m= 2-2n
4( 2- 2n) +3n =3
8-8 n+3n= 3
8-5 n=3
8-3 =5n
5=5n
n=1
hence
m= 2-2( 1 )
M= 0
also
x +2y =1– — –( i )
4x +3 y =4- — — ( ii )
from – — — ( iii )
x =1- 2y
4( 1- 2y ) +3y =4
4-8 y +3y = 4
y =0
therefore x =1-2 ( 0)
x =1
this N=[ i i ]
================================
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+4 y =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)
Enjoy

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