MATHEMATICS WAEC 2017

MATHEMATICS WAEC ANSWER LOADIND...

Mathz- obj
1 CBBACCCBBA
11 ADBBAABCDB
21 BCADBCCAAB
31 CDCACDBACB
41 BDABCDDCAD
Theor- Answers
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SECTION (A ) ANSWER ALL
1 a)
(y - 1 )log 4 ^ 10 = ylog 16 ^ 10
log 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
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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
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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
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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
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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
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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 °
==================
SECTION B ANS 5 QUESTIONS ONLY
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=-=
(8)
Tabulate:
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
(8b)
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
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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
10bii)
Let the angle be. θ
From <AML
Tanθ = 9.6/2.8
Tan θ = 3.4288
θ = Tan^-1 ( 3.4288)
= 73.74°
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=-=
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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