WAEC 2017 AGRIC OBJ AND THEORY ANSWERS – MAY/JUNE EXPO

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Maths OBJ:
1CBBACCCBBA
11ADBBAABCDB
21BCADBCCAAB
31CDCACDBACB
41BDABCDDCAD

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SECTION B ANS 5 QUESTIONS ONLY
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10)
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
10b)
Draw a triangle LACB
in triangle LCB
Hyp^2 = Opp^2 + Adj^2
12^2 = 9.6^2 + |CB|^2
144 = 92.16 = |CB|^2
144 – 92.16 = |CB|^2
51.84 = |CB|^2
therefore, |CB| = √51.84
|CB| = 7.2m
|AC| + |CB| =|AB|
|AC| + 7.2m = 10m
|AC| = 10m – 7.2m
|AC| = 2.8m
In triangle LCA
Hyp^2 = Opp^2 + Adj^2
|LA|^2 = |AC|^2 + |LC|^2
|LA|^2 = 2.8^2 + 9.6^2
|LA|^2 = 7.84 + 92.16
|LA|^2 =100
|LA| = √100
|LA| = 10m
10bii)
in triangle LCA
sinθ = Opp/Hyp
sinθ = |LC|/|LA|
sinθ = 9.6/10
sinθ = 0.96
θ = sin^-1 (0.96)
θ = 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)

12a)
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)=+sqr1/36
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*2y] [4m+3n, 4x+5y] = [2,3, 1,4] therefore
m+2n=2——(i)
4m+3n=3——(ii)
from ——(i)
m=2-2n
4(2-2n)+3n=3
8-8n+3n=3
8-5n=3
8-3=5n
5=5n
n=1
hence
m=2-2(1)
M=0
also
x+2y=1——(i)
4x+3y=4——(ii)
from ——(iii)
x=1-2y
4(1-2y)+3y=4
4-8y+3y=4
y=0
therefore x=1-2(0)
x=1
this N=[i i]

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 = 4x20x12/8×40
= 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

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4b )
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
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)=77 customers
probability = 77 /120
=0 . 642

SECTION B ANS 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 / 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

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 ° If you want to Get the Answers Direct To Your Phone As Sms, Kindly Send MTN Recharge Card worth of N1000

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Objective

1. Look at this series: 2, 1, 12, 14 ,… What number should come next?

A. 13
B. 18
C. 28
D. 116
E. 132

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

2. A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?
A. 25
B. 19
C. 18
D. 17

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3. If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X.

A. 12
X
B. X-14
C. X-13
D. X-12

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

4. Find the value of X if 2√x+2√=1x−2√
A. 3√2+4
B. 3√2-4
C. 3-2√2
D. 4+2√2

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

5. If (a2b−3c)34a−1b4c5=apbqcr What is the value of p+2q?
A. (5/2)
B. -(5/4)
C. -(25/4)
D. -10

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

6. If {(a2b-3c)3/4/a-1b4c5} = apbqcr; what is the value of p+2q?
A. (5/2)
B. -(5/4)
C. -(25/4)
D. -10

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

WAEC Mathematics Questions 2017 [Theory and Objective] |Check Maths Random Repeated Questions Here.

7. A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss.
A. 30% gain
B. 25% gain
C. 30% loss
D. 25% loss

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8. What is the answer when 24346 is divided by 426?
A. 236
B. 356
C. 526
D. 556

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

9. If 29 x (3Y)9 = 35 x (3Y)5, find the value of Y.
A. 4
B. 3
C. 2
D. 1

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

10. The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers
A. p/2
B. 3p/2
C. 5p/2
D. 3p

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

11. A binary operation * is defined by a*b = ab+a+b for any real number a and b. if the identity element is zero, find the inverse of 2 under this operation.
A. 2/3
B. 1/2
C. -1/2
D. -2/3

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

12. Factorize completely X2+2XY+Y2+3X+3Y-18
A. (x+y+6)(x+y-3)
B. (x-y-6)(x-y+3)
C. (x-y+6)(x-y-3)
D. (x+y-6)(x+y+3)

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

13. Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X.
A. 4≤X≤5
B. 5≤X≤8
C. 5≤X≤10
D. 8≤X≤10

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

14. Three consecutive positive integers k, l and m are such that l2 = 3(k+m). Find the value of m
A. 4
B. 5
C. 6
D. 7

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

15. Express 1X3−1 in partial fractions
A. 1/3{(1/x-1)-(x-2/x2-x+1)}
B. 1/3{(1/x-1)-(x+2/x2+x+1)}
C. 1/3{(1/x-1)-(x-2/x2+x+1)}
D. 1/3{(1/x-1)-(x-2/x2-x-1)}

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