WAEC GCE 2018 Mathematics Expo Answer Obj And Essay Answer – Free Maths waec gce expo 2018 - Jan/Feb Expo

WAEC GCE 2018 Mathematics Expo Answer Obj And Essay Answer – Free Maths waec gce expo 2018 - Jan/Feb Expo

WAEC GCE 2018 Mathematics Expo Answer Obj And Essay Answer – Free Maths waec gce expo 2018 - Jan/Feb Expo 

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MATHS OBJ:
1-10=ACCAACDBBB
11-20=BCABABCABB
21-30=DCCCBCBCCA
31-40=BDADBBDBAA
41-50=DABCDBADAC

COMPLETED

11a)
Loga(y + 2) = 1 + LogaX
=> Log^y a + Log^2 a = Log^a a + Log^x a

Loga^(y + 2) = Loga^(ax)

Y + 2 = ax
Hence y+2/a = ax/a
X = y+2/a

11bi)
Bibiani = 600
Amenji = 700
Oda = 1800
Wawso = 1500
Sankose=2400
Total = 7200

Bibiani = 600/7200 × 360/1 = 30°
Amenji = 700/7200 × 360/1 = 45°
Oda = 1800/7200× 360/1 = 90°
Wawso = 1500/7200× 360 = 75°
Sankose = 2400/7200× 360/1 = 120°
Total = 30°+45°+90°+75°+120° = 360°

11bii)
% of timber produced from Amenji = 900/7200 × 100/1 = 12.5%

11biii)
Revenue received by Bibiani = 600×$560 = $336,000
Revenue received by Oda = 1800×560 = $1,008,000
Oda will receive(1,008,000 - 336000) = $672,000 more than Bibiani

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4a)
Rate = 2/100 * N0.02 per month
Rate per annum = 0.02 * 12 = 0.24 per annum

4b) Draw the Diagram

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3a)
[Diagram]
Distance covered by an athlete = Perimeter of A + Perimeter of rectangle CDEF + perimeter of B
Perimeter of A = 2πr/2 = π =22/7, r = d/2 = 120/2 = 60m
= 22/7 × 60 = 1320/7 = 188.57m

Perimeter of B = perimeter of A = 188.57m
Perimeter of rectangle CDEF= 2(L + B)
L = 120m; B = 60m
Perimeter = 2(120+60) = 2(180)
=360m
Distance covered by an athlete = 188.57 + 360 + 188.57
=737.14m
If the athlete runs the track two times = 2 × 737.14
= 1474.28m

3b)
If the athlete spends 200seconds for the race
Speed = distance/time
Distance = 1474.28m
Time = 200second
Distance = 1474.28m = 1.47428km
Time= 200seconds = 3.3333hrs
Speed = 1.47428/3.3333 = 0.44kmhr-1

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6a)
Tanx = 5/12
Using the diagram
Sinx = 5/13
Cosx = 12/13
Sinx/(sinx)² + cosx = 5/13/(5/13)² + 12/13
= 5/13all over 25/169 + 12/13
= 5/13/25+156/169
=5/13/181/169
= 5/13 × 169/181 = 65/18

6b)

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9b
(PR)²=(PS)²+(SR)²
(PR)²=15²+15²
(PR)²=225+225
(PR)²=450
PR=sqr root 225×2
PR=15root2cm
But OR=PR÷2 = 15root 2÷2

=7.5×1.4142
=10.6065

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7a)
Reduction in the first sales = 40%
Reduction in the second sales = 30%
Price sold Ghc 3500 = 70% ie (100 - 30)%
GHc y = 100% second reduction sale
35 × 100 = 70y
35 × 100/70 = 70/70
Y = 350/7 = 50
Hence price after first sale = GHc50
But GHc50 = 60% ie (100-40)%
Therefore GHcx = 100% first reduction sale
100 × 50/60 = 60x/60
X=> 500/60 = GHc83.33
=>GHc83.3
Hence price before the first sales = GHc83.33

7b)
Initil price of article = GHc = 180.00
In the first sales, reduction = 40%
i.e 100% - GHc 18.00
40% - GHc x
100x/100 = 40*180/100
.:. x = 4*18 = GHc 72.00
Since reduction in the first sale is GHc 72.00
Then reduction in the second = 30%
100% = GHc 108
30% = y
100y/100 = 30*108/100 = 324/10 = GHc 32.4
(i) Hence reduction in the price due to the two sales = (72+32.4)GHc = GHc 104.4
(ii) % reduction = Reduction/Original price * 100/1
=104.4/180 * 100/1 = 58%

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1)
1/4 * 9 1/7 + 2/5 [2/3 + 3/4] / (2/5 - 1/4)
(1/4 * 64/7 + 2/5)[17/12)] /8-20/20]
16/7 +2/5(17/2) *[20/3
(16/7 +1/5 *17/6)*20/3
(16/7+17/30)*20/3
(16*30+17*7 /210)*20/3
(480+119/210)*20/3 599/210 *20/3
599*2/63
1198/69
=19^1/63

1b)
Sin 48=x/250
X=250 sin 48 degrees
X= 250 * 0.7431
X=185.7775m
=186m

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2)
Let musa's age=x.
Manya's age=y.
x-y=3---------(1)
Also x=3+y------(2)

7years ago
Musa's age=x-7
Manya's age=y-7
x-7=2(y-7)
x-7=2y-14
x-2y=-14+7
x-2y=-7-------eqn(3)

Put eqn(2) into eqn(3)
3+y-2y=-7
-y=-7-3
-y=-10
Y=10

But x=3+10=====>x=13
Also therefore Musa's age is x =13,
And Manya's age is y=10

2b)
Let the time be y
( x + y) + (x + 3 + y) = 45
(10 + y) + (10 + 3 +y) = 45
10+10+3+2y = 45
23+2y = 45
2y = 45-23
2y = 22
Y = 22/2
Y = 11years
The sum of their ages will be 45 after 11 years

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