# Waec Gce 2018 Mathematics Obj And Essay Answer – Jan/Feb Expo

## WAEC GCE Mathematics Obj And Essay/Theory Solution Questions and Answer – Jan/Feb 2018 Expo Runz.

**MATHS OBJ:**

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

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

4a)

Rate = 2/100 * N0.02 per month

Rate per annum = 0.02 * 12 = 0.24 per annum

4b) Draw the Diagram

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)

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

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%

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

1198/69

=19^1/63

1b)

Sin 48=x/250

X=250 sin 48 degrees

X= 250 * 0.7431

X=185.7775m

=186m

2)

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

Y=10

But x=3+10=====>x=13

Also therefore Musa's age is x =13,

And Manya's age is 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

**MATHS OBJ:**

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

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

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

4a)

Rate = 2/100 * N0.02 per month

Rate per annum = 0.02 * 12 = 0.24 per annum

4b) Draw the Diagram

4a)

Rate = 2/100 * N0.02 per month

Rate per annum = 0.02 * 12 = 0.24 per annum

4b) Draw the Diagram

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

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)

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)

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

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

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%

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%

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

1198/69

=19^1/63

1b)

Sin 48=x/250

X=250 sin 48 degrees

X= 250 * 0.7431

X=185.7775m

=186m

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

1198/69

=19^1/63

1b)

Sin 48=x/250

X=250 sin 48 degrees

X= 250 * 0.7431

X=185.7775m

=186m

2)

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

Y=10

But x=3+10=====>x=13

Also therefore Musa's age is x =13,

And Manya's age is 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

2)

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

Y=10

But x=3+10=====>x=13

Also therefore Musa's age is x =13,

And Manya's age is 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

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

*042tvseries.Com send all exam expo answer earlier than others*============================

*WAEC GCE ssce 2018 Mathematics expo answers, Ssce 2018 Mathematics objective and essay Expo answers now available, nov/dec Mathematics theory Answers for 2018, Mathematics obj and theory Questions WAEC GCE Ssce Expo Answers Here, Free wassce Mathematics objective, essay & theory correct Expo 2018 answers Runz WAEC GCE Mathematics, WAEC GCE Ssce 2018 Free Mathematics essay, obj and theory Questions and Answers Expo for WAEC GCE Mathematics Paper 1 and Paper 2 objective and essay Questions and Answers Expo Now Posted, Free Mathematics objective ans for internal ssce, see WAEC GCE ssce expo Mathematics obj and theory, essay questions for free, loadedexpo correct wassce Economics ans, sure WAEC GCE ssce Mathematics objective answers runz runs expo free real and correct WAEC GCE ssce 2018 nov/dec Mathematics free expo runz runs ans answers online for free see free answers online, 2018 Verified WAEC GCE Free Ssce Mathematics Obj and theory Answers have been posted REAL WAEC GCE SSCE Mathematics QUESTIONS & ANSWERS Direct To Phone Number AS Text message, RE: 2018 WASSCE Mathematics (Expo) WAEC GCE 2018 certified runs Mathematics questions and answers WAEC GCE SSCE November/December 2018 Mathematics THEORY / ESSAY / OBJ QUESTION AND EXPO I need WAEC GCE ssce free expo site, how can I get WAEC GCE Mathematics answer for free, WAEC GCE 2018 Mathematics essay, Objective And Theory Question and Answer Now Posted, Real 2018/2019 nov/dec wassce WAEC GCE Mathematics objective, essay and theory 100% correct expo questions and answers runz chokes runs, verified 2018 WAEC GCE Mathematics theory, essay and obj WAEC GCE Mathematics expo website, WAEC GCE ssce Mathematics live cheats, sure November/December WAEC GCE SSCE 2018 Mathematics original objective, theory and essay questions and answer now available here, 100% WAEC GCE ssce Mathematics expo, 2018 wassce Mathematics obj, essay and theory runz now posted, correct WAEC GCE ssce Mathematics objective, essay (theory) answer now ready, free 2018 wassce Mathematics obj, theory and essay best expo site WAEC GCE ssce Mathematics questions and answers, password for certified WAEC GCE ssce Mathematics essay theory obj expo, how to get/ where to get wassce Mathematics theory, essay and objective expo answer direct to my phone as text message, trusted WAEC GCE ssce Mathematics questions and answer, i need WAEC GCE Ssce nov/dec Mathematics essay theory objective obj Expo 2018 answers on whatsapp, gidifans Mathematics answer, solutionclass Mathematics answer, 042tvseries Mathematics answer, codedexam WAEC GCE Mathematics answer, wapbaze Mathematics answer, wapextra Mathematics expo, Martinlibrary Mathematics expo, examgists WAEC GCE Mathematics expo, WAEC GCE Mathematics objectives answer, naija WAEC GCE Mathematics expo free answer*

Real and 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.

**GOODLUCK!!!**

## No comments