WAEC GCE 2018 Mathematics Expo Answer Obj And Essay Answer – Free Maths waec gce expo 2018 - Jan/Feb Expo | 2019 JAMB Runs/Runz | 2019 JAMB Expo | 2019 JAMB CBT Expo / Runs / Runz/ Answers / Sites / Free BEST JAMB Expo Runs Website Online

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, Waec Gce Maths Expo Answer 2018, 2018 waec gce Mathematics answer expo, 2018 waec gce Mathematics Questions, 2018/2019 waec gce maths Expo Runs, 2018 waec Gce Mathematics Obj and Theory Expo, 2018 waec Gce Mathematics Question and answer..

WAEC GCE 2018 Mathematics Expo Answer, WAEC GCE 2018 Maths Expo Answer, waec GCE 2018 Mathematics Answer, waec GCE Mathematics expo 2018, FREE waec GCE Mathematics Answer 2018, FREE EXPO ON Mathematics waec GCE, waec GCE 2018 EXPO FOR FREE, FREE Mathematics waec GCE ANSWERS, Mathematics QUESTIONS FOR GCE, 2018 GCE ANSWERS, waec GCE 2018 Mathematics EXPO ANSWERS,Mathematics waec GCE RUNS, FREE waec GCE Mathematics ANSWERS, 2018 jan/feb GCE Mathematics RUNS, 2018 jan/feb waec GCE Mathematics ANS, 2018 jan/feb waec GCE Mathematics EXPO, 2018 jan/feb GCE Mathematics QUESTION, 2018 jan/feb GCE Mathematics QUESTIONS, waec GCE 2018 Mathematics EXPO ANSWERS, jan/feb GCE 2018 Mathematics EXPO, jan/feb GCE 2018 Mathematics ANSWERS, jan/feb waec GCE 2018 Mathematics ANS, jan/feb GCE 2018 Mathematics ANSWER, jan/feb waec GCE 2018 Mathematics RUNZ, GCE 2018 Mathematics ANSWERS, OBJECTIVE Mathematics EXPO GCE, GCE OBJECTIVE ANSWERS

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

(1630+177 /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

Adv: JAMB CBT EXPO 2018 JAMB RUNS 2018

2018 JAMB CBT EXPO / RUNS / ANSWERS

SOURCE: WAEC GCE 2018 Mathematics Expo Answer Obj And Essay Answer – Free Maths waec gce expo 2018 - Jan/Feb Expo JAMB Runs 2019 JAMB Runs Website | 2019 WAEC Expo | 2019 JAMB CBT Runs | TEELINK Jamb JAMB Expo site 2019 JAMB Runs | JAMB Expo Website| JAMB Expo 2019 JAMB / WAEC Expo 2019 | 2019 JAMB Expo | WAEC Expo | JAMB Expo 2019 2019 WAEC Expo | JAMB CBT Runs JAMB Runs WhatsApp Group Link 2019 JAMB CBT Expo | JAMB Runs JAMB Expo sites JAMB Expo Runs 2019/2020 2019 JAMB Runs | JAMB Expo Site Best jamb expo website JAMB Runs JAMB Runs Website| 2019/2020 JAMB CBT Expo 2019 JAMB Expo 2019 JAMB Runs 2019/2020 JAMB Runs | Site Map 1 2019 JAMB Expo 2019 JAMB CBT Expo JAMB Runs Online JAMB Expo 2019 JAMB CBT Expo
2019 JAMB CBT Expo | JAMB Runs 2019
WAEC GCE Expo
JAMB Expo Runs Site - WAEC GCE Expo - JAMB Runs websites Online
Examclass Jamb DAILY JAMB Expo DAILY JAMB Expo CodedExam Jamb Examsort Jamb 042Tvseries Jamb ExamLoaded Jamb GidiFans Jamb TODAY JAMB Expo Expocrew Jamb Donprosper Jamb ExamClass Jamb ExamCrown Jamb BEST JAMB Expo Webaite DAILY JAMB Expo JAMB Runs Webaite JAMB Runs Webaite JAMB Expo Site JAMB Expo Site 2019 JAMB Runs WhatsApp Group link 2019 JAMB Expo Runs WhatsApp Group link JAMB Runs 2019 2019 WAEC Expo DAILY JAMB Expo | WAEC Expo 2019 BEST JAMB Runs Webaite | WAEC GCE Expo || Site Map 2019 JAMB Expo 2019 JAMB Runs 2019 JAMB Expo Runs WhatsApp Group link 2019 JAMB CBT Expo JAMB Runs 2019 2019 JAMB CBT Runs JAMB Runs JAMB Expo JAMB Expo Runs 2019 JAMB CBT Runz JAMB Expo 2019

Don't Be Stingy, Share With Your Friends

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

2018 WAEC GCE JAN/FEB MATHEMATICS THEORY ANSWERS

(16/7 +1/5 17/6)20/3

(1630+177 /210)*20/3

(480+119/210)20/3 599/210 20/3