CRK-OBj
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1DABDBABCDA
11DDCADBCCDD
21ACADBBCBDC
31BDDCBDBBDB
41BCCBDDCACB
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(1)
110â‚“ = 40â‚…
change all to base ten.
110â‚“ = 40â‚…
1x²+1x ¹+0xâ° = 4×5¹+0×5â°
x² + x + 0 = 20
x² + x – 20 = 0
Solve quadratically
x² + 5x – 4x – 20 = 0
x(x+5) – 4(x+5) = 0
(x-4)(x+5) = 0
Either x-4 = 0 or x + 5 = 0
x = -5 or 4
therefore, x= 4
1b)
15/√75 + √108 + √432
= 15/√253 + √363 + √144×3
= 15/5√3 + 6√3 + 12√3
= 3/√3 + 18√3
= 3√3/3 = 18√3
= √3 + 18√3
=19√3
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(2)
(2a)
The equation of the line through the points
A(-2,7) and B(2,-3)
Using the equation Y=mx+b
Where m=slope of the gradient, b=the intercept at the vertical axis
Hence slope=Change in Y/Change in X
=>Y2-Y1/X2-X1=Y-y1/X-x1
=(-3-7)/(2--2)=Y-7/X+2
=-10/4=(Y-7)/(X+2)
=4(y-2)=-10(x+2)
4y-28=-10x-20
4y=-10x+8
5x+2y=4
(2b)
(5b-a)/(8b+3a)=1/5
5(5b-a)=1(8b+3a)
25b-5a=8b+3a
25b-8b=3a+5a
17b=8a
Therefore a/b=17/8
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(3a)
Total ratio =3+5+8=16
Ali's share =3/16 * 420,000/1
=210,000
Son's of Ali and Yusuf's share = N78,750+ N210,000
= N288,750
(3b)
2(1/8)x=32x-1
21(2-3)x=25(x-1)
21* 2-3x=25(x-1)
2-3x+1=25(x-1)
-3x+1=5(x-1)
-3x+1=5x-5
-5x-3x= -5-1
-8x/-8= -6/-8
x=3/4
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(4)
Since
|PR|² = |PQ|² + |QR|²
|PR|² = 3² + 4²
|PR|² = 9 + 16
|PR|² = 25 PR = √25
|PR| = 5cm
Considering PRS
|PS|² = |PR|²+|SR|²
13² = 5² + |SR|²
169 = 25 + |SR|²
|SR|² = 169 - 25
|SR|² = 144
|SR| = √144 = 12cm
Hence the area of the quadrilateral = Area of triangle PQR + area of PRS
= 1/2bh + 1/2bh
= 1/2×4×3 + 1/2×12×5
= 6+30 = 36cm
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(5a)
No of red balls = 3
No of green balls = 5
No of blue balls = x
Prob.(red ball) = no of total outcome/no of possible outcome
Pr(red) = 3/3+5+x = 1/6
3/8+x = 1/6
6(3) = 1(8+x)
18 = 8 + x
X = 18 - 8 = 10
Therefore the no of blue ball = 10
(5b)
Probability of picking a green ball
P(g) = no of green balls/no of possible outcome
P(g) = 5/3+5+10 = 5/18
=5/18
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(6ai)
F α M1M2/d²
F = KM1M2/d²
Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m
20 = k(25)(10)/5²
250k = 500
k = 500/250 = 2
Expression is
F = 2M1M2/d²
(6aii)
Making d subject
d = √2M1M2/F
d = √2 ×7.5×4/30
d = √60/30 = √2
d = √2m or 1.41m
(6b)
Draw the diagram
X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)
5x + 200 = 540
5x = 540 - 200
5x = 340
X = 340/5
X = 68
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(8a)
1/3x - 1/4(x+2)>_ 3x -1â…“
1/3x - 1/4(x+2)>_3x - 4/3
Multiply through by the L. C. M(12), we have
4x - 3(x + 2)>_36x - 16
4x - 3x - 6 >_ 36x - 16
-6+16 >_36x + 3x - 4x
10 >_ 35x
35x _< 10
X = 10/35
X = 2/7
(8bi)
Draw the triangle
|AB|/66 = sin35
|AB| = 66sin35 = 66×0.5736 = 37.8576
Draw the right angled triangle
|AD|/|AB| = Tan52
|AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m
Height of tower = 48.45m
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(10)
130kg of tomatoes for #52,000
Half of the tomatoes
130/2 = 65kg sold at 30%
Profit = #52,000/2 = 26,000
#26,000 = 100%
X = 130%
X = 26000 × 130/100
= #33,800
Then 65kg was then sold at reduction of 12% per kg
Recall that the initial cost price = 52000/130
=400kg
65kg sold at = 33,000/65
=#520/kg
Then for 12% reduction
520 × 88/100 = 457.6/kg
(a)
The new selling price per kg = #457.6/kg
(b) 65kg - 5kg = 60kg
(60kg×457.6kg)+33,800
= #61,256.00
#profit = selling price /cost price × 1000/1
=61256/52000×100/1= 117.8
= 17.8%
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(11ai)
ar² = 1/4 ......(1)
ar^5= 1/32 .....(2)
Divide eqn (2) by eqn(1)
ar^5/ar² = 1/32÷1/4
r³ = 1/32 × 4/1
r³= 1/8
r³ = 2-³
r = 2-¹
r = 1/2
Common ratio = 1/2
Put this into eqn (1)
a(1/2)² = 1/4
a(1/4) = 1/4
a = (1/4)/(1/4) = 1
First term, a = 1
(11aii)
Seventh term, T7 = ar^6
=(1)(1/2)^6
=1/64
(11b)
Given : X = 2 and X = -3
(X - 2)(X + 3) = 0
X² + 3x - 2x - 6 , 0
X² + x - 6 = 0
Comparing with ax²+bx+c = 0
a = 1
b = 1
C = -6
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(12a)
Given : siny = 8/17
Draw the right angle
From Pythagorean triple, third side is 15
Draw the right angle triangle
tan y = 8/15
tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15
tany /1+2tan y = 8/31
(12b)
Amount shared = #300,000
Otobo's share = #60,000
Ade's share = 5/12 × #(300,000-60,000)
= 5/12 × #240,000
=#100,000
Adeobi's share = #300,000 - (#60,000 + #100,000)
= 300,000 - 160,000
=#140,000
Ratio : Otobo : Ade : Adeola
60,000 : 100,000 : 140,000
60 : 100 : 140
6 : 10 : 14
3 : 5 : 7
You are required to get at least a credit pass (50%) in your core subjects. Under the new WAEC grading system, A1 is Excellent, B2 Very Good, B3 Good, C4, C5 and C6 are interpreted as Credit, D7 and D8 are interpreted as Pass, while F9 is Fail.
A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.
A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.
Under the new WAEC grading system, A1 is Excellent, B2 Very Good, B3 Good, C4, C5 and C6 are interpreted as Credit, D7 and D8 are interpreted as Pass, while F9 is Fail. A1 and B2 in the WASSCE means Excellent, B3 is B (Very Good), C4 is C (Good), C5 and C6 are D (Credit), D7 and E8 are E (Pass) and F9 is F (Fail).