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 ACCAACDBBB

11BCABABCABB

21DCCCBCBCCA

31BDADBBDBAA

41DABCDBADAC

completed

*2018 WAEC GCE JAN/FEB MATHEMATICS THEORY ANSWERS*

(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

=======================

(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

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