PERFECT WAEC MATHEMATICS EXPO RUNZ 2018 – MATHS QUESTIONS AND ANSWER FREE

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PERFECT WAEC MATHEMATICS EXPO RUNZ 2018 – MATHS QUESTIONS AND ANSWER FREE

WAEC 2018 Mathematics Obj And Essay Answer – May/June Expo – PERFECT WAEC MATHEMATICS EXPO RUNZ 2018 – MATHS QUESTIONS AND ANSWER FREE – 2018 WAEC LIVE ANSWERS HOT WAEC 2018 Mathematics Obj And Essay Answer – Expo – This is random repeated questions and answers 2018 mathematics examinations. No doubt questions in this year exam will be repeated, therefore we urge you to be careful while making use of this site.

PERFECT WAEC MATHEMATICS EXPO RUNZ 2018 – MATHS QUESTIONS AND ANSWER FREE

WAEC GCE Mathematics Obj And Essay/Theory Solution Questions and Answer – 2018 Expo Runz.

 

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

COMPLETED

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 200 seconds 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

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.

GET ANSWERS 2018 MATHS

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

4b) Draw the Diagram

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)

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.

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

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%

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.

WAEC 2018/2019 FINAL ANSWERS CHECK HERE

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