`Real & Confirmed WAEC GCE Mathematics Questions and Expo Answer - Jan/Feb 2018`

`Verified WAEC GCE 2018 Jan/Feb Mathematics OBJ and Essay Answer and Solution to the questions.`

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**MATHS OBJ:1-10: ADCAACCBBB11-20: BACBBBCABB21-30: ABDBBCBCCB31-40: BAADBDDBAA41-50: 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`

=>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

`=========================`

**GOODLUCK!!!**

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