Waec Gce Mathematics (obj/Essay)Answers-Jan-Feb Expo -Examwaves

*Maths - Obj:*

1 ACCAACDBBB

11BCABABCABB

21DCCCBCBCCA

31BDADBBDBAA

41DABCDBADAC

completed!

*2018 WAEC GCE JAN/FEB MATHEMATICS THEORY ANSWERS*

*MR SHANE WILLIAMS*


(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


*MR SHANE*
=======================

(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
====== *08094186531* ===============
(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
=======================
(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

_compiled by:
*mr shane*
*08094186531*

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