WAEC GCE 2018 Jan/Feb Mathematics Obj & Essay/Theory Expo Answer

The Real and Verified WAEC GCE Mathematics OBJ, ESSAY/Theory Questions and Answer – Nov/Dec 2018 Expo Runz

Verified WAEC GCE 2018 Maths Expo Answer

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MATHS OBJ:
1-10: ADCAACCBBB
11-20: BACBBBCABB
21-30: ABDBBCBCCB
31-40: BAADBDDBAA
41-50: DABDDBBDAC

(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
1198/63
=19^1/63
(1b)
Sin 48 =x / 250
X =250 sin 48 degrees
X = 250 * 0 . 7431
X =185 . 7775 m
=186 m
=======================
(2a)
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
(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
======================
(3)
CIRCUMFERENCE OF TWO SEMI CIRCLES* =PIEd
22 / 7 X 120
= * 377 . 142 *
2 ( 377 . 142 + 60)
=874 . 29 Km
(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
=======================
(4a)
Rate = 2/100 * N0.02 per month
Rate per annum = 0.02 * 12 = 0.24 per annum
(4b) Draw the Diagram
<ZWY = <XWY = 180 degree(opp angles of cyclic quad. are supplimantary)
<ZWY = 100 = 180
<ZWY = 180-100
<ZWY = 100degree
==========°==°===================
(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)
====================
(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%
========================
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
9bii)(RV)²=(OR)²+(OV)²
==>32²=(10.6065)²+(OV)²
1024=112•4978+(OV)²
1024-112•4978=(OV)²
(OV)²=911•50215
OV= √911•50215
OV=height=30•1911≈
Height = 30•2cm
(ii)Volume=⅓×base area×height
=⅓×15×15×30•1911
=2,264•33≈
2,264cm³
===========================
(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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