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WAEC GCE 2018 1ST SERIES - MATHS ANSWER

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WAEC GCE 2018 1ST SERIES - MATHS

Saturday 3rd February, 2018
General Mathematics [Core] 2 [Essay] 9:30am – 12:00pm
General Mathematics [Core] 1 [Objectives] 2:00pm – 3:30pm
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MATHS OBJ:
1-10: ADCAACCBBB
11-20: BACBBBCABB
21-30: ABDBBCBCCB
31-40: BAADBDDBAA
41-50: DABDDBBDAC..

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VIEW IMAGE ANSWERS HERE WHILE WE SOLVE MORE AND TYPE
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VIEW IMAGE ANSWERS HERE WHILE WE SOLVE MORE AND TYPE
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VIEW IMAGE ANSWERS HERE WHILE WE SOLVE MORE AND TYPE
===================================
VIEW IMAGE ANSWERS HERE WHILE WE SOLVE MORE AND TYPE
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(1a) 
[1/4 × 64/7 + 2/5(8 + 9/12)] + (8 - 5/20) 
[1/4 × 64/7 + 2/5(17/12)] + 3/20
[16/7 + 17/30] + 3/20
( 480 + 119/210) + 3/20
599/210 + 3/20
1198+63/420 = 1261/420

(1b)
Sin 48 =x / 250
X =250 sin 48 degrees
X = 250 * 0 . 7431
X =185 . 7775 m
=186 m

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(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

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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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(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 supplimentary)
< ZWY = 100 = 180
< ZWY = 180-100
< ZWY = 100degree

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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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(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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(9a)
Let the distance which the donckey can reach = x
Hence 17² - 15² = x²
289 - 225 = x²
64 = x²
√64 = √x²
8 = x
But since the donkey can move both sides, the lenght of the fence = 8*2 = 16m

(9bi)
Draw the diagram

(9bii)
(I)
(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

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(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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(13a)
1/3 [3] + [6] + [P] = [3]
-6 -2 q 4

[1] + [6] + [P] = [3]
-2 -2 q 4

1+6+p = 3
7 + p = 3
P = -4

-2 -2 + q = 4
-4 + q = 4
q = 4+4 = 8
T = [-4]
[ 8]


(13b)
2[3] + n[1] = [8]
m 2 4

[6] + [n] = [8]
2m 2 4

Hence 6 +n = 8 => n = 8-6
n = 2
Also 
2m + 2n = 4
2m + 2(2) = 4
2m = 4 - 4
2m = 0
m = 0/2 = 0
Hence m = 0 , n = 2


(13c) 
(Y+4)*y = 12
(Y+4+1)*y = 12
(Y+5)*y = 12
Y+5+Y+1 = 12
2y+6 = 12
2y = 12-6
2y = 6
Y = 6/2
Y = 3

















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