2020 WAEC MATHEMATICS QUESTIONS AND ANSWERS
MATHEMATICS OBJ
MATHS-OBJ
1 -10 CBCDACDCCD
11-20 AADBDACBBC
21-30 BDDABDADAD
31-40 CDACCCCCDA
41-50 BBBCDCACDB
Completed.
MATHEMATICS THEORY
WAEC MATHS ANSWERS
(1a)
Given A={2,4,6,8,…}
B={3,6,9,12,…}
C={1,2,3,6}
U= {1,2,3,4,5,6,7,8,9,10}
A’ = {1,3,5,7,9}
B’ = {1,2,4,5,7,8,10}
C’ = {4,5,7,8,9,10}
A’nB’nC’ = {5, 7}
(1b)
Cost of each premiere ticket = $18.50
At bulk purchase, cost of each = $80.00/50 = $16.00
Amount saved = $18.50 – $16.00
=$2.50
2ai).
p = (rk/Q -ms)2/3
p3/2 = (rk/Q -ms)2/8 * 3/2
p3/2 = rk/Q = ms
p3/2 + ms = rk/Q
q(p3/2 + ms ) = rk
Q =rk/p3/2 + ms
2aii)
Where :
r = 10, k = 4, s = 0.2
p = 3, m = 15
Q = 10 * 4/3^3/2 + (15 * 0.2)
Q = 40/5.2 + 3
Q = 40/8.2
Q = 4.88
(4a)
Total Surface Area = 224πcm²
r:l = 2:5
r/l = 2/5
Cross multiply
2l/2 = 5r/2
L = 5r / 2
Total surface = πrl + πr²
= πr (l + r)
24π/π = πr (5r/2 + r )/ π
224 = 5r²/2 + r²/1
L.c.m = 2
448 = 5r² + 2r²
448 / 7= 7r²/7
r² = 64
r = √64 = 8cm
L = 5*8/2 = 20cm
(4b)
Volume = 1/2πr²h
= 1/3 * 22/7 * 8 * 8 * 18.33
= 1228.98cm³
L² = h² + r ²
20² = h² + 8²
400 – 64 = h²
h² = 336
h = √ 336
h = 18.33cm
(5a)
Total income = 32+m+25+40+28+45
=170+m
PR(²)=m/170+m = 0.15/1
M=0.15(170+m)
M=25.5+0.15m
0.85m/0.85=25.5/0.85
M=30
(5b)
Total outcome = 170 + 30 = 200
(5c)
PR(even numbers) = 30+40+50/200
=115/200 = 23/40
Number: 8
Let ms Maureen income = x
1/4 x = shopping mall
1/3 x = at an open market
Hence shopping mall and open market = 1/4x. + 1/3x
=3x+4x /12. =. 7/12x
Hence de remaining amount
=X-7x/12.
=12x-7x/12. = 5x/12
Then 2/5(5x/12) = mechanic workshop
=2x/12 *x/6
Amount left = #225,000
Total expenses
7/12 * x/6 *225,000. =#x
7x+2x+2,700,000/12 = #x
9x+2,700,000 /12 = #x
9x+2,700,000= 12x
2,700,000= 12x-9x
2,700,00 =3x
X= 2,700,000/3
=#900,000
8aii:) amount spent on open market
=1/3x=1/3*900,000
=#300,000
(9a)
Draw the triangle
(9b)
(i)Using cosine formulae
q² = x² + y² – 2xycosQ
q² = 9² + 5² – 2×9×5cos90°
q² = 81 + 25 – 90 × 0
q² = 106
q = square root 106
q = 10.30 = 10km/h
Distance = 10 × 2 = 20km
(ii)
Using sine formula
y/sin Y = q/sin Q
5/sin Y = 10.30/sin 90°
Sin Y = 5 × sin90°/10.30
Sin Y = 5 × 1/10.30
Sin Y = 0.4854
Y = sin‐¹(0.4854), Y = 29.04
Bearing of cyclist X from y
= 90° + 19.96°
= 109.96° = 110°
(9c)
Speed = 20/4, average speed = 5km/h
(10)
10 and graph complete
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11b) 8y = – 4x + 24
Y = – 4/8 x + 24/8
Y = – 1/2 x + 3
Gradient = m = – 1/2
:. Y- y1 = m(x-x)
Y- 12 = 1/2 (x + 8)
2(y-12) = 1 (x+8)
2y – 24 = -x-8
X + 2y – 24 + 8 = 0
X + 2y – 24 + 8 = 0
X + 2y – 16 = 0
Is the line above
12
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