# WAEC GCE 2018 1ST SERIES - MATHS

Saturday 3rd February, 2018General 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

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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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