WAEC 2017 MATHEMATICS THEORY AND OBJ 100% VERIFIED
MATHEMATICS OBJECTIVE
1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB
MATHEMATICS THEORY
1a. log4^y-1 = log 16^y
log4^y-1 = log4^2y
y-1
collect like terms
2y - y = -1
y = -1
1b. speed = dstance/ tme
distance = speed * time
30 minutes = 0.5 hour
4^-1 = o.5 + t
4(0.5 + t)
2 + 4t
for 5km/hr, time =t
therefore = 5t
time (t) = 2 hours
actual distance = 5(2) = 10 km
2a. 2/3(3x-5) - 3/5( 2x - 3) = 3
multiply through by 15
10(3x - 5) - 9 (2x-3) = 45
30x - 50 - 18x - 27 = 45
collect like terms
12x - 23 = 45
12x = 45 + 23
12x = 68
x = 5.6
2b. 180 - (n + 88)
(sum of angles on a straight line)
= 92 -n
180 - (m - 80)
(sum of angles in a triangle)
80+92+180-n-m = 180 degree
352 -n-m = 180 degree
-n-m = 180 - 352
-n-m = -172
insert negative sign to both sides
-(-n-m) = -(-1720
n+m = 172 degree
3a. tan 23.6degree = h / 50 (opp/adj)
0.4369 = h / 50
h = 0.4369 * 50
h = 21.8 m approx. 22 m
3b. area of triangle = 0.5 * b*h
area of the triangle = 45 cm ^2
45 = 0.5 * 10 * h
45 = 5h
h = 9 cm
area of trapezium = 0.5 ( a+ b) * h
= 0.5 (6+16)* 9
= 99 cm^2
4a. T6 = 37
T6 = a + 5d = 37 --- (i)
S6 = 3(2a+5d) = 147 ---(ii)
a + 5d = 37 ---(i) *3
6a +15d + 147 ---(ii) *1
3a + 15d = 111 ---(iii)
6a + 15d = 147 ---(iv)
3a = 36
a= 12
first term = 12
4b. S15= 15/2(2(12) + 14d)
to get d ; substitute 12 for a in equation (i)
12 + 5d = 37
5d = 37 - 12
5d = 25
d = 5
common difference = 5
S15= 7.5(24+ 70)
S15= 705
5a. Draw the venn diagram
5b. y-45+45+y= 120
2y- 34 = 120
2y = 120 + 34
2y = 154
y = 77
no. of the student who bought shoe= y + 11= 77+11= 88
5c. n(bags) = 77
probality = 77/120 = 0.64
WAEC 2017 MATHEMATICS THEORY AND OBJ ..... more update oncoming
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