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MATHS OBJ
1-10: CBCDACDCCD
31-40: CBACCCCCDA

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(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

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(2ai)
P = (rk/Q - ms)â…”
P^3/2 = rk/Q - ms
rk/Q = P^3/2 + ms
Q= rk/P^3/2 + ms

(2aii)
When P =3, m=15, s=0.2, k=4 and r=10
Q = rk/p^3/2 + ms = 10(4)/(3)^3/2 + (15)(0.2)
= 40/8.196 = 4.88(1dp)

(2b)
x + 2y/5 = x - 2y
Divide both sides by y
X/y + 2/5 = x/y - 2
Cross multiply
5(x/y) - 10 = x/y + 2
5(x/y) - x/y = 2 + 10
4x/y = 12
X/y = 3
X : y = 3 : 1

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(3a)
Draw the diagram

CBD = CDB(Base angles of an issoceles triangle)
BCD + CBD + CDB = 180Â°(sum of angles in a triangle)
2CDB + BCD = 180Â°
2CDB + 108Â° = 180Â°
2CDB = 180Â° - 108Â° =72Â°
CDB = 72/2 = 36Â°
BDE = 90Â°(angle in a semi-circle)
CDE = CDB + BDE
= 36Â° + 90Â°
= 126Â°

(3b)
(CosX)Â² - SinX/(SinX)Â²+ CosX
Using Pythagoras theorem, third side of triangle
yÂ² = 1Â² + |3Â²
yÂ² = 1 + 3 = 4
y = square root e = 2
Sin X = root 3/2(opp/hyp)
(CosX)Â² - SinX/(SinX)Â² + CosX
= (1/2)Â² - root3/2 / (root3/2)Â² + 1/2
= 1/4 - root3/2 / 3/4 + 1/2
= 1 - 2root3/4 / 3+2/4
= 1-2root3/5

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(4a)
Given: r : l = 2 : 5 (ie l = 5/2r)
Total surface area of cone =Ï€rÂ² + url
224Ï€ = Ï€(rÂ² + r(5/2r))
224 = rÂ² + 5/2rÂ²
224 = 7/2rÂ²
7rÂ² = 448
rÂ² = 448/7 = 64
r = root 64 = 8.0cm

(4b)
L = 5/2r = 5/2 Ã— 8 = 20cm
Using Pythagoras theorem
LÂ² = rÂ² + hÂ²
hÂ² = lÂ² - rÂ²
hÂ² = 20Â² - 8Â²
hÂ² = (20 + 8)(20 - 8)
hÂ² = 28 Ã— 12
h = root28Ã—12
h = 18.33cm

Volume of cone = 1/3Ï€rÂ²h
= 1/3 Ã— 22 Ã— 7 Ã— 8Â² Ã— 18.33
=1229cmÂ³

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(5a)
Prob(2) = no of 2s/Total outcomes
0.15 = m/32+m+25+40+28+45
0.15 = m/m + 170
m = 0.15m + 25.5
m - 0.15m = 25.5
0.85m = 25.5
m = 25.5/0.85 = 30

(5b)
Number of times dice was rolled = m + 170
= 30 + 70
= 200

(5c)
Prob(even number) = no of even numbers/Total outcome
= m+40+45/200
=30+40+45/200
=115/200
= 23/40 = 0.575

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(7a)
Total surface area = url + 2Ï€rÂ²
=Ï€r(l + 2r)

Draw the diagram
From pythagoras theorem
LÂ² = 14Â² + 48Â²
LÂ² = 196 + 2304
LÂ² = 2500
L = /2500 = 50m

=Ï€r(L + 2r)
= 22/7 Ã—14(50 + 2(14))
= 44(50 + 28)
= 3432mÂ²
Total surface area = 3432mÂ²
~3430mÂ²(to 3s.f)

(7b)
Five years ago,
Let Musa's age = x
Let Sesay's age = y
X - 5 = 2(Y - 5)
X - 5 = 2y - 10
X - 2y = 5 - 10
X - 2y = -5 ..... (1)
-X + y = 100 ..... (2)
-3y = -105
Subtracting eqn 2 from 1
-3y/3 = -105/-3
y = 35
Sesay's present age = 35 years

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(8a)
Let Ms Maureen's Income = Nx
1/4x = shopping mall
1/3x = at an open market

Hence shopping mall and open market = 1/4x + 1/3x
= 3x + 4x/12 = 7/12x

Hence the remaining amount
= X-7/12x = 12x-7x/12 =5x/12

Then 2/5(5x/12) = mechanic workshop
= 2x/12 = x/6
Amount left = N225,000
Total expenses
= 7/12x + X/6 + 225000
= Nx

7x+2x+2,700,000/12 =Nx
9x + 2,700,000 = 12x
2,700,000 = 12x - 9x
2,700,000/3 = 3x/3
X = N900,000

(ii) Amount spent on open market = 1/3X
= 1/3 Ã— 900,000
= N300,000

(8b)
T3 = a + 2d = 4m - 2n
T9 = a + 8d = 2m - 8n
-6d = 4m - 2m - 2n + 8n
-6d = 2m + 6n
-6d/-6 = 2m+6n/-6
d = -m/3 - n
d = -1/3m - n

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(9a)
Draw the triangle

(9b)
(i)Using cosine formulae
qÂ² = xÂ² + yÂ² - 2xycosQ
qÂ² = 9Â² + 5Â² - 2Ã—9Ã—5cos90Â°
qÂ² = 81 + 25 - 90 Ã— 0
qÂ² = 106
q = square root 106
q = 10.30 = 10km/h
Distance = 10 Ã— 2 = 20km

(ii)
Using sine formula
y/sin Y = q/sin Q
5/sin Y = 10.30/sin 90Â°
Sin Y = 5 Ã— sin90Â°/10.30
Sin Y = 5 Ã— 1/10.30
Sin Y = 0.4854
Y = sinâ€Â¹(0.4854), Y = 29.04

Bearing of cyclist X from y
= 90Â° + 19.96Â°
= 109.96Â° = 110Â°

(9c)
Speed = 20/4, average speed = 5km/h

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(11a)

(11b)
Given 8y+4x=24
8y=-4x + 24
y=4/8x + 24/8
y=-1/2x +3
Using m = y-y/x-xÂ¹ and given (xÂ¹=-8) (yÂ¹=12)
-1/2=y-12/x+8
2(y-12)=-x-8
2y-24=-x-8
2y+x=24-8
2y+x=16

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(12a)
BCD=ABC=40Â°(alternate D)

DDE=2*BCD(
DDE = 2*40 = 80Â°
OD3=OED(base < of I sealed D ODE)
ODE + OED + DOE= 180Â°(sum of < is in D)
2ODE+DOE=180Â°
2ODE+80Â°=180
2ODE+180=180
2ODE+100Â°
ODE+100/2=50Â°

(12bi)
Digram

(12bii)
Area of parallelogram = absin
=5*7*sin125Â°
=35*sin55Â°
=35*0.8192
=28.67
=28.7cmÂ²(1dp)

(12c)
Given x=1/2(1-âˆš2)
2xÂ²-2x=2[1/2(1-âˆš2]Â²-2(1/2(1-âˆš2)}
=2[1-2âˆš2+2/4]-(1-âˆš2)
=(3-2âˆš2/2)-(1-âˆš2)
=3-2âˆš2-2+2âˆš2/2=1/2

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PLS TAKE NOTE OF THESE SYMBOLS BELOW

(i) / means division or divide
Examples:
2/2 means 2/2 ie 2 over 2.
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(ii) * means multiplication
Example: 2*2 means 2x2
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(iii) RtP means Raise to power
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(iv) Sqr root means means square root sign ie _/

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