## CRK-OBj

This is actually the WAEC 2019 - MATHS ANSWERSAnswers as questions sent to our subscribers on WHATSAPP, if you are looking for 2021 WAEC/NECO Answers midnight before the exam, try contacting us.

1DABDBABCDA

11DDCADBCCDD

21ACADBBCBDC

31BDDCBDBBDB

41BCCBDDCACB

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

(1)

110â‚“ = 40â‚…

change all to base ten.

110â‚“ = 40â‚…

1x²+1x ¹+0xâ° = 4Ã—5¹+0Ã—5â°

x² + x + 0 = 20

x² + x â€“ 20 = 0

Solve quadratically

x² + 5x â€“ 4x â€“ 20 = 0

x(x+5) â€“ 4(x+5) = 0

(x-4)(x+5) = 0

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

1b)

15/âˆš75 + âˆš108 + âˆš432

= 15/âˆš253 + âˆš363 + âˆš144Ã—3

= 15/5âˆš3 + 6âˆš3 + 12âˆš3

= 3/âˆš3 + 18âˆš3

= 3âˆš3/3 = 18âˆš3

= âˆš3 + 18âˆš3

=19âˆš3

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

(2)

(2a)

The equation of the line through the points

A(-2,7) and B(2,-3)

Using the equation Y=mx+b

Where m=slope of the gradient, b=the intercept at the vertical axis

Hence slope=Change in Y/Change in X

=>Y2-Y1/X2-X1=Y-y1/X-x1

=(-3-7)/(2--2)=Y-7/X+2

=-10/4=(Y-7)/(X+2)

=4(y-2)=-10(x+2)

4y-28=-10x-20

4y=-10x+8

5x+2y=4

(2b)

(5b-a)/(8b+3a)=1/5

5(5b-a)=1(8b+3a)

25b-5a=8b+3a

25b-8b=3a+5a

17b=8a

Therefore a/b=17/8

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

(3a)

Total ratio =3+5+8=16

Ali's share =3/16 * 420,000/1

=210,000

Son's of Ali and Yusuf's share = N78,750+ N210,000

= N288,750

(3b)

2(1/8)^{x}=32^{x}-1

2^{1}(2^{-3})^{x}=2^{5}(x-1)

2^{1}* 2^{-3}x=2^{5}(x-1)

2^{-3}x+1=2^{5}(x-1)

-3x+1=5(x-1)

-3x+1=5x-5

-5x-3x= -5-1

-8x/-8= -6/-8

x=3/4

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

(4)

Since

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = âˆš25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 - 25

|SR|² = 144

|SR| = âˆš144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2Ã—4Ã—3 + 1/2Ã—12Ã—5

= 6+30 = 36cm

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

(5a)

No of red balls = 3

No of green balls = 5

No of blue balls = x

Prob.(red ball) = no of total outcome/no of possible outcome

Pr(red) = 3/3+5+x = 1/6

3/8+x = 1/6

6(3) = 1(8+x)

18 = 8 + x

X = 18 - 8 = 10

Therefore the no of blue ball = 10

(5b)

Probability of picking a green ball

P(g) = no of green balls/no of possible outcome

P(g) = 5/3+5+10 = 5/18

=5/18

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

(6ai)

F Î± M1M2/d²

F = KM1M2/d²

Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m

20 = k(25)(10)/5²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/d²

(6aii)

Making d subject

d = âˆš2M1M2/F

d = âˆš2 Ã—7.5Ã—4/30

d = âˆš60/30 = âˆš2

d = âˆš2m or 1.41m

(6b)

Draw the diagram

X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)

5x + 200 = 540

5x = 540 - 200

5x = 340

X = 340/5

X = 68

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

(8a)

1/3x - 1/4(x+2)>_ 3x -1â…“

1/3x - 1/4(x+2)>_3x - 4/3

Multiply through by the L. C. M(12), we have

4x - 3(x + 2)>_36x - 16

4x - 3x - 6 >_ 36x - 16

-6+16 >_36x + 3x - 4x

10 >_ 35x

35x _< 10

X = 10/35

X = 2/7

(8bi)

Draw the triangle

|AB|/66 = sin35

|AB| = 66sin35 = 66Ã—0.5736 = 37.8576

Draw the right angled triangle

|AD|/|AB| = Tan52

|AD| = 37.8576 Ã— Tan52° = 37.8576 Ã— 1.2799 = 48.45m

Height of tower = 48.45m

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

(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 Ã— 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 Ã— 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg - 5kg = 60kg

(60kgÃ—457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price Ã— 1000/1

=61256/52000Ã—100/1= 117.8

= 17.8%

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

(11ai)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32Ã·1/4

r³ = 1/32 Ã— 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

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

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo's share = #60,000

Ade's share = 5/12 Ã— #(300,000-60,000)

= 5/12 Ã— #240,000

=#100,000

Adeobi's share = #300,000 - (#60,000 + #100,000)

= 300,000 - 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

. You can get 2020/2021 Exam answers directly from examplaza.com, visit HERE NOW

2021 Neco Specimen | 2021 Neco 2021 NECO SPECIMEN | 2021 NECO SPECIMENS PRACTICAL ON PHYSICS

2021 NECO SCIENCE SPECIMENS

Get all your questions answered about the 0ne-time-only NECO exams for free and get a schedule of events. We will also help you with mock runs, tips on how best to prepare for the exam and even what to wear during the whole process

BEST NECO EXPO 2021 QUESTIONS AND ANSWERS | Neco Expo

Are you attempting the 2021 National Examinations Council (NECO) Senior Secondary School Examination (SSCE)? If yes, you can download or copy the timetable for the exam from this page.

2021 NECO OFFICIAL TIMETABLE

2021 NECO Science Specimens: 2021 Complete NECO Specimens. 2021 NECO Specimen | 2021 SPECIMEN for NECO

2021 NECO Specimen | 2021 SPECIMEN for NECO

Best Waec Expo runz site 2021, ARE YOU LOOKING FOR A PERFECT SOURCE FOR YOUR EXAM SUCCESS IN waec?, we provides real waec questions and answers.

2021 WAEC: BEST WAEC EXPO 2021 QUESTIONS AND ANSWERS

2021 Waec Science Specimens: 2021 Complete WAEC Specimens. 2021 Waec Specimen | 2021 SPECIMEN for WAEC

2021 Waec Specimen | 2021 SPECIMEN for WAEC

2021 NABTEB: NABTEB TIMETABLE FOR 2021/2021 ACADEMIC YEAR PDF DOWNLOAD. 2021 NABTEB TIMETABLE: 2021/2022 NABTEB TIMETABLE Free Download

2021 NABTEB: NABTEB TIMETABLE FOR 2021/2021 ACADEMIC YEAR PDF DOWNLOAD

Joint Admissions and Matriculation Board (JAMB) Form 2021 › See Registration Instructions and Guidelines. JAMB begins 2021 UTME registration, sets June 5 for examination

JAMB begins 2021 UTME registration, sets June 5 for examination - Joint Admissions and Matriculation Board (JAMB) Form 2021 › See Registration Instructions and Guidelines

This is 2021 WAEC Practical Specimens. Are you looking for WAEC specimen 2021? this page is all about 2021 WAEC Practical Specimens: 2021 WAEC Specimen - 2021 WAEC Science Specimens .

2021 WAEC Specimen | 2021 WAEC Science Specimens - WAEC Practical Specimens