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2017 CORE MATHS SOLUTIONS (Objective and THEORY ) Pass it on and help others.

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NEWS: 2017 CORE MATHS SOLUTIONS (Objective and THEORY ) Pass it on and help others. [New Education News] » Naijacrawl
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2017 CORE MATHS SOLUTIONS , this is the objective and theory we are posting it for free.
Read Also : Physics Objective answer for WAEC today 28/8/2020


Mathz- obj from Www.naijacrawl.com1 CBBACCCBBA
11 ADBBAABCDB
21 BCADBCCAAB
31 CDCACDBACB
41 BDABCDDCAD

Theor- Answers
+ ++ + ++ + ++ + + ++ + ++ + ++ + + ++ + SECTION (A ) ANSWER ALL
1 a)
(y - 1 )log 4 ^ 10 = ylog 16 ^ 10log 4^ 10 (y - 1 )= log 16 ^ y 10
4 ^ ( y - 1)= 16 y
4 ^ y - 1 = 4^ 2 y
y - 1 =2 y
- 1 = 2y = y
- 1 = y
y = - y
1 b)
let the actual time for 5km / hr be t
for 4km / hr=30 mint + t
4 km/ hr= 0. 5 + t
distance = 4( 0 . 5 +t )
= 2* 4 t
for 5km / hr, time = t
distance =5 t
1 +4 t= 5 t
t= 2 hrs
actual distance = 5 * 2= 10 km
================================
2 a)
2 /3 (3 x - 5 )- 3 / 5( 2x - 3)= 3 / 1
L C M = 15
10 ( 3x - 5 ) - 9 (2 x - 3 )= 45
30 x - 50 - 18 x + 27 = 45
30 x - 18 x = 45 +50 - 27
12 x - 23 =45
12 x = 45 + 23
12 x = 68
x =68 /12
x =34 /6
x =17 /3
2 b)
U ' aS= 180- (n +88 )
= 180- n - 88 = 92 - n
also , u' TQ = 18 m
80 degree + 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
================================
3 a)
Tan 23 . 6 = h/ 50
Cross multiply
Tan 23 . 6 x h/ 50
h = 50 tan 23 . 6
= 21 . 844m
3 b)
Area of
A = 1 /2 bh
45 = 1/ 2 x 10 x h
45 = 5h
h = 9 cm
Area of < QTUS = 1 / 2 ( QT + US )h
= 1 /2 ( 6 + 16 ) 9
= 99 cm ^ 2
================================
4 a)
T 6= 37
T 6= a +( 6 - 1 )d
T 6= a +5 d
a +5 d =37 - - - - - (eq 1)
s 6= 147
sn= n /2 (2 a +( n- 1) d )
147= 3( 2a + 5d )
49 = 2 a+ 5 d
2 a+ 5 d= 49 - - - - ( eq2 )
a +5 d =37 - - - ( eq1 )
2 a+ 5 d= 49 - - - (eq 2)
a =12
4 b)
S 15 = 15 / 2( 2( 12 ) +14 d)
S 15 = 15 / 2( 24 + 14 d )
from( 1 )
a +5 d =37
12 + 5 d= 37
5 d= 37 - 12
5 d= 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
================================
5 a)
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
5 b)
y - 45 + 45 + y - 34 =120
2 y - 34 = 120
2 y = 154
y =154/ 2
y =77
number of customers who bought shoe = y + 11
77 + 11 = 88
5 c )
n (bag)= 77 customers
probability = 77 /120
= 0. 642
================================
SECTION (B ) ANSWER 5 QUESTIONS ONLY
10 a )
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 - 2 sin x / 2 tan x
12 / 13 - 2 (5 /13 ) / 2 (5 / 12 )
= 12 / 13 - 10 /23 / 5/ 6
FIND LCM
= 12 - 10 /13 / 5/ 6
= 12 / 65
10 bi )
Considering < LMB
/MB / ^ 2 . = 12 ^ 2 - 9 . 6^ 2
/MB / ^ 2 = 51 . 84
/MB / = √ 51 . 84
/MB / = 7 . 2 m
From < AML
/LA /^ 2 = 2 . 8 ^ 2 + 9 . 6 ^ 2
/LA / ^ 2 = 100
/LA / = √ 100
/LA / = 10 m
10 bii )
Let the angle be. θ
From < AML
Tanθ = 9 . 6 /2 . 8
Tan θ = 3 . 4288
θ = Tan ^ - 1 ( 3. 4288)
= 73 . 74

(11a) for 8 students
2/3 of an hour = 12 tanks
Rate = 12/1 x 2/3
=18 tanks/hour
8 students = 18
4 students= x
x= 18 x 4/8
x=9 tanks/hour
Rate= Number of tanks/Time
Number of Tanks= rate x time
= 9x1/3
=3 tanks
(11bi) Tan Alpha = 10/12 = 0.8333
Alpha= Tan^-1(0.8333)
Alpha= 39.80 degrees
But Tita= 2xAlpha
Tita = 2 x 39.8= 79.6 degrees
(11bii) x^2 = 10^2 + 12^2
x^2= 100 + 144
x^2= 244
x= sqroot of 244
x= 15.62
Length of the chord
= Tita/360 x 2 x pie x r
= 79.6/360 x 2 x 22/7 x 15.62
= 21.71cm
The perimeter of the minor
Segment= 21.71 + 20
= 41.71cm




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