Today, we shall discuss the following:

- Neco 2021
**Mathematics**Objective Questions - 2021 Neco
**General**Essay Questions And Answers.**Mathematics** - Instructions To Pass Neco 2021 Examination.

As usual, you will be given questions and options A to E to choose from. Normally, the number of **objective questions (OBJ)** you are to answer in Neco 2021 **Mathematics** Science is 50. Below are sample Neco Mathematics questions.

1. If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term

A. -45

B. -15

C. 15

D. 33

E. 45

A. -45

B. -15

C. 15

D. 33

E. 45

2. If sinθθ = K find tanθθ, 0o ≤≤ θθ ≤≤ 90o

A. 1-K

B. kk−1kk−1

C. k1−k2√k1−k2

D. k1−kk1−k

E. kk2−1√kk2−1

A. 1-K

B. kk−1kk−1

C. k1−k2√k1−k2

D. k1−kk1−k

E. kk2−1√kk2−1

3. Evaluate (101.5)2 – (100.5)2

A. 1

B. 2.02

C. 20.02

D. 202

E. 2020

A. 1

B. 2.02

C. 20.02

D. 202

E. 2020

4. Express the product of 0.06 and 0.09 in standard form

A. 5.4 * 10-1

B. 5.4*10-2

C. 5.4*10-3

D. 5.4*102

E. 5.4*103

A. 5.4 * 10-1

B. 5.4*10-2

C. 5.4*10-3

D. 5.4*102

E. 5.4*103

5. Simplify 361/2 x 64-1/3 x 50

A. o

B. 124

C. 2/3

D. 11/3

E. 71/2

A. o

B. 124

C. 2/3

D. 11/3

E. 71/2

6. Find the quadratic equation whose roots are x = -2 or x = 7

A. x2 + 2x – 7 = 0

B. x2 – 2x + 7 = 0

C. x2 + 5 +14 = 0

D. x2 – 5x – 14 = 0

E. x2 + 5x – 14 = 0

A. x2 + 2x – 7 = 0

B. x2 – 2x + 7 = 0

C. x2 + 5 +14 = 0

D. x2 – 5x – 14 = 0

E. x2 + 5x – 14 = 0

7. A sales girl gave a change of N1.15 to a customer instead of N1.25. Calculate her percentage error

A. 10%

B. 7%

C. 8.0%

D. 2.4%

E. 10%

A. 10%

B. 7%

C. 8.0%

D. 2.4%

E. 10%

8. What is the probability of having an odd number in a single toss of a fair die?

A. 1/6

B. 1/3

C. 1/2

D. 2/3

E. 5/6

A. 1/6

B. 1/3

C. 1/2

D. 2/3

E. 5/6

9. If the total surface area of a solid hemisphere is equal to its volume, find the radius

A. 3.0cm

B. 4.5cm

C. 5.0cm

D. 9.0cm

A. 3.0cm

B. 4.5cm

C. 5.0cm

D. 9.0cm

10. If 23x + 101x = 130x, find the value of x

A. 7

B. 6

C. 5

D. 4

A. 7

B. 6

C. 5

D. 4

11. Simplify: (34−2334−23) x 11515

A. 160160

B. 572572

C. 110110

D. 1710710

A. 160160

B. 572572

C. 110110

D. 1710710

12. Simplify:(103√5√−15‾‾‾√1035−15)2

A. 75.00

B. 15.00

C. 8.66

D. 3.87

A. 75.00

B. 15.00

C. 8.66

D. 3.87

13. The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?

A. 90cm

B. 135cm

C. 180cm

D. 225cm

A. 90cm

B. 135cm

C. 180cm

D. 225cm

14. Which of following is a valid conclusion from the premise. “Nigeria footballers are good footballers”?

A. Joseph plays football in Nigeria therefore he is a good footballer

B. Joseph is a good footballer therefore he is a Nigerian footballer

C. Joseph is a Nigerian footballer therefore he is a good footballer

D. Joseph plays good football therefore he is a Nigerian footballer

A. Joseph plays football in Nigeria therefore he is a good footballer

B. Joseph is a good footballer therefore he is a Nigerian footballer

C. Joseph is a Nigerian footballer therefore he is a good footballer

D. Joseph plays good football therefore he is a Nigerian footballer

15. On a map, 1cm represent 5km. Find the area on the map that represents 100km2.

A. 2cm2

B. 4cm2

C. 8cm2

D. 8cm2

A. 2cm2

B. 4cm2

C. 8cm2

D. 8cm2

16. Simplify; 3n−1×27n+181n3n−1×27n+181n

A. 32n

B. 9

C. 3n

D. 3n + 1

A. 32n

B. 9

C. 3n

D. 3n + 1

17. What sum of money will amount to D10,400 in 5 years at 6% simple interest?

A. D8,000.00

B. D10,000.00

C. D12,000.00

D. D16,000.00

A. D8,000.00

B. D10,000.00

C. D12,000.00

D. D16,000.00

18. The roots of a quadratic equation are 4343 and -3737. Find the equation

A. 21×2 – 19x – 12 = 0

B. 21×2 + 37x – 12 = 0

C. 21×2 – x + 12 = 0

D. 21×2 + 7x – 4 = 0

A. 21×2 – 19x – 12 = 0

B. 21×2 + 37x – 12 = 0

C. 21×2 – x + 12 = 0

D. 21×2 + 7x – 4 = 0

19. Find the values of y for which the expression y2−9y+18y2+4y−21y2−9y+18y2+4y−21 is undefined

A. 6, -7

B. 3, -6

C. 3, -7

D. -3, -7

A. 6, -7

B. 3, -6

C. 3, -7

D. -3, -7

20. Given that 2x + y = 7 and 3x – 2y = 3, by how much is 7x greater than 10?

A. 1

B. 3

C. 7

D. 17

A. 1

B. 3

C. 7

D. 17

21. Simplify; 21−x−1×21−x−1x

A. x+1x(1−x)x+1x(1−x)

B. 3x−1x(1−x)3x−1x(1−x)

C. 3x+1x(1−x)3x+1x(1−x)

D. x+1x(1−x)x+1x(1−x)

A. x+1x(1−x)x+1x(1−x)

B. 3x−1x(1−x)3x−1x(1−x)

C. 3x+1x(1−x)3x+1x(1−x)

D. x+1x(1−x)x+1x(1−x)

22. Make s the subject of the relation: P = S + sm2nrsm2nr

A. s = mrpnr+m2mrpnr+m2

B. s = nr+m2mrpnr+m2mrp

C. s = nrpmr+m2nrpmr+m2

D. s = nrpnr+m2nrpnr+m2

A. s = mrpnr+m2mrpnr+m2

B. s = nr+m2mrpnr+m2mrp

C. s = nrpmr+m2nrpmr+m2

D. s = nrpnr+m2nrpnr+m2

23. Factorize; (2x + 3y)2 – (x – 4y)2

A. (3x – y)(x + 7y)

B. (3x + y)(2x – 7y)

C. (3x + y)(x – 7y)

D. (3x – y)(2x + 7y)

A. (3x – y)(x + 7y)

B. (3x + y)(2x – 7y)

C. (3x + y)(x – 7y)

D. (3x – y)(2x + 7y)

24. The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take π=227π=227]
A. 2.6cm

B. 3.5cm

C. 3.6cm

D. 7.0cm

B. 3.5cm

C. 3.6cm

D. 7.0cm

25. The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x

A. 3

B. 5

C. 6

D. 8

A. 3

B. 5

C. 6

D. 8

26. Calculate the gradient of the line PQ

A. 3535

B. 2323

C. 3232

D. 5353

A. 3535

B. 2323

C. 3232

D. 5353

27. A straight line passes through the point P(1,2) and Q

(5,8). Calculate the length PQ

A. 411‾‾‾√411

B. 410‾‾‾√410

C. 217‾‾‾√217

D. 213‾‾‾√213

(5,8). Calculate the length PQ

A. 411‾‾‾√411

B. 410‾‾‾√410

C. 217‾‾‾√217

D. 213‾‾‾√213

28. If cos θθ = x and sin 60o = x + 0.5 0o < θθ < 90o, find, correct to the nearest degree, the value of θθ

A. 32o

B. 40o

C. 60o

D. 69o

A. 32o

B. 40o

C. 60o

D. 69o

29. Age(years)Frequency13101424158165173Age(years)1314151617Frequency1024853

The table shows the ages of students in a club. How many students are in the club?

A. 50

B. 55

C. 60

D. 65

A. 50

B. 55

C. 60

D. 65

30. The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range

A. 16

B. 14

C. 13

D. 11

A. 16

B. 14

C. 13

D. 11

31. If log2(3x – 1) = 5, find x.

A. 2.00

B. 3.67

C. 8.67

D. 11

A. 2.00

B. 3.67

C. 8.67

D. 11

32. A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r

A. 2323

B. 2

C. 3

D. 6

A. 2323

B. 2

C. 3

D. 6

33. Express 1975 correct to 2 significant figures

A. 20

B. 1,900

C. 1,980

D. 2,000

A. 20

B. 1,900

C. 1,980

D. 2,000

34. A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue

A. 2929

B. 518518

C. 20812081

D. 5959

A. 2929

B. 518518

C. 20812081

D. 5959

35. The relation y = x2 + 2x + k passes through the point (2,0). Find the value of k

A. – 8

B. – 4

C. 4

D. 8

A. – 8

B. – 4

C. 4

D. 8

36. Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8…

A. 13, 19, 23

B. 9, 11, 13

C. 11, 15, 19

D. 13, 21, 34

A. 13, 19, 23

B. 9, 11, 13

C. 11, 15, 19

D. 13, 21, 34

37. If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x

A. 2

B. 5

C. 15

D. 18

A. 2

B. 5

C. 15

D. 18

38. The sum 110112, 11112 and 10m10n02. Find the value of m and n.

A. m = 0, n = 0

B. m = 1, n = 0

C. m = 0, n = 1

D. m = 1, n = 1

A. m = 0, n = 0

B. m = 1, n = 0

C. m = 0, n = 1

D. m = 1, n = 1

39. A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.070 to N1.00, how much did the engine cost in Niara?

A. N250,000.00

B. N200,000.00

C. N150,000.00

D. N100,000.00

A. N250,000.00

B. N200,000.00

C. N150,000.00

D. N100,000.00

40. If 27x×31−x92x=127x×31−x92x=1, find the value of x.

A. 1

B. 1212

C. -1212

D. -1

A. 1

B. 1212

C. -1212

D. -1

The following are the kind of questions you should expect in Neco 2021 **Mathematics** Theory or Essay. They are hot cake questions:

- Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles
- The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q?
- n what modulus is it true that 9 + 8 = 5?
- In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years?
- In parallelogram PQRS, QR is produced to M such that |QR| = |RM|. What fraction of the area of PQMS is the area of PRMS?

- Do not open your question paper until you are told to do so
- USE HB pencil throughout in the OBJ Section
- You are free to use Biro in the theory part
- You are allowed to use calculator to solve
- Make sure that you fill your name, Subject, paper, paper code and other examination details where necessary.
- Ensure that the texts in your question papers are boldly printed
- Behave yourself.
- Don’t let invigilators catch you using expo

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That’s all for now… I shall update you when more real live questions and answers come up. However, I advice that you are hardworking so as to pass your Neco once and for all.

Read Also: How to read and pass Neco in one day

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The National Examinations Council is an examination body in Nigeria that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and December/January respectively.

1a)

(2x+1)/(3-4x)=2/3

3(2x+1)=2(3-4x)

6x+3=6-8x

6x+8x=6-3

14x/14=3/14

x=3/14

1bi)

E=MV^2/2

2E/M =MV^2/M

V^2=2E/M

V=sqr2E/M

1bii)

Vsqr2E/M

Vsqr2*64/2

Vsqr64

V=8

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

2 a )

number of sides =12

radius of circle =10 cm

area =?

n ×© 2= 360

12 © 2= 360

© 2 = 360 / 12 =30 °

© 1 + © 2= 180 – 30

© 1 = 150

When © 1 and © 2 are interior and exterior angle

of a polygon A sector has are .

Area of sector =© / 360 × rot 8 ^2

= 150 / 360 ×22 / 7× 100 / 1

A =130 – 95 cm^2

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

2 b )

1 /2 ( 2x + 1 ) – 2 /5 ( x – 2 )= 3

2 x +1/ 3 – 2 x – 4/ 3= 3

10 x +5 – 6x + 12 / 15 =3 / 1

Cross multiple

4 x +17 =45

4 x /4 =28 / 4

x = 7.

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

3 )

Apply 5m rule to find C P

t / sin T = P /sin P

t / sin 110 = 6/ sin 40

t =6 * 0. 9396 / 0. 6428

= 56376 / 6. 6428

= 87704

= 877 km

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

4 )

Total Fruit = 80 + 60 = 140

( a)

( i ) Pr one of each fruit is picked

( 79 / 140 * 60 / 139 ) + ( 59 / 140 * 80 / 139 )

= 4740 / 19, 460 + 4720 / 19 , 460

= 9460 / 19, 460 = 0.486

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

4 aii )

Pr one type of fruit is picked

( 79 / 140 * 78 / 139 ) + ( 39 / 140 * 5 p/ 139 )

= 6162 / 19, 460 + 3422 / 19 , 460

= 9584 / 19 , 460 = 0.492

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

4 b )

5 X / 8 – 1 / 6 ≤ X / 3 + 7 /24

Multiply through by 24 i : e

15 X – 4≤ 8X + 7

15 X – 8X ≤ 7 + 4

7 X = 11

X ≤ 11 / 7 ===> X ≤ 1 4 / 9

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

5 a )

3 /X + 2 – 6 /3 X – 1

3 ( 3 X – 1) – 6 ( X + 2) / ( X + 2) ( 3X – 1)

9 X – 3 – 6X – 12 /( X + 2) ( 3X – 1)

3 X – 15 / ( X + 2) ( 3X – 1)

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

5 b )

C .I= P [1 +r / 100 ]^

= 25000 [ 1+ 12/ 100 ]^ 3

= 25000 [ 1+ 0. 12 ]^ 3

= 25000 * 1 . 4049

= 35122 . 50

= N 35 ,122 . 50

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

6 a )

X +- 3/ 2

X =2/ 3 or X =2

( X + 3/ 2)^ 2 or ( X – 2 )

( X + 3/ 2) ( X – 2 )

X ( X – 2) + 3 / 2 ( X – 2)

X ^2 – 2X + 3 X / 2 – 3

2 X ^2 – 4X + 3X – 6

2 X ^2 – X – 6

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

6 b )

h /h +8 = 6 / 10

10 h = 6 h + 48

h = 12

H = h + 8

H = 12 + 8

H = 20

Volume = 1 /3 A . h

= 1 / 3 ( 10 * 10 ) * 20 – 432 / 3

= 200 / 3 – 432 /3

= 1568 / 3

= 522 . 67 cm3

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

7 a )

titan / 360 ×2 pie r cos t

d = 55 / 360 ×2 ×22 / 7×640 cos 4

d = 55 × 44 × 6400 cos 4/ 2520

d = 55 × 44 × 6400 ×0496 / 2520

d = 15 , 449 . 28 / 2520

d = 6130. 67

d ~ 6130 km .

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

ii ) distance along gent circle

D = tita / 360 ×2 pie r

D = 55 / 360 × 2/ 7× 22 /7 × 6400/ 1

D = 55 × 44 ×6400 / 2520

D = 15 , 488 ×6400 / 2520

D = 6144 .03

D =~ 6146 km .

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

7 b )

Length of sector tita/ 360 × 2 pie r

L= 120 / 360 ×2 /1 × 22 / 7× 42/ 1

L= 120 ×44 ×42 / 2520

L= 221760 / 2520

L= 88cm

L= 2pie r

Where r is the radius of circumference

88 =2 × 22 / 7× r

88 ×7 = 44 r

R =88 ×7 /44

R =616 / 44

R =14 cm.

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

Curved surface area

= pie rc

A =22 / 7 ×14 × 42 / 1

A =22 ×14 ×4 ^2 /7

A =12936 /7

A =1848 cm^ 2.

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

8 a )

X =60 / t — – – – – – – – – > ( i )

Y = 180 / t – – – – – — – – – > ( ii )

T 1=60 / X

T 2=100 /Y

T 1+T 2= 5

60 / X + 180 / Y = 300 – — – – – – – – – > ( i )

180 / X + 60 / Y = 260 – – – – – – – – — > ( ii )

Let P =1 / X

2 = 1 / Y

60 p + 180 Q = 300

180 p + 6Q = 200

P + 3 Q =5

9 P + 3Q = 13

Substract ( i ) from ( ii )

8 p = 8

P = 8 / 8 ÷ P = 1

Subtract P into ( i )

P + 3 Q =5

1 + 3 Q = 5

3 Q =5 – 1

3 Q =4

Q = 4/ 3

P = 1 ÷ 1 = 1/ X ÷ X = 1

4 /3 = 1/ Y ÷ Y = 3/ 4

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

8 b )

2001 – – – – – — – – 25 , 700

2002 – – – – – — – – 15 / 100 X 25, 700 + 25,

700 = 29 , 555

Amount of tax in 2002

= 29 , 555 * 12 .5 /100

= N 3694 .375

= N 3690

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

8 c )

Log 25

Log 16 25 / 100

Log 16 2/ 4

Log 4 6 – 1

– 1/ 2 Log 4^4

– 1/ 2

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

9 ai)

W= K +C / 2

24 = k + C / 16

384 = 16 K + C – — – – – – ( i )

18 = K + C / 4

72 + 4K + C – – – – – – – — ( ii )

16 K + C = 384

4 K + C = 72

Substact ( ii ) from ( i )

12 k / 12 = 312 /12

K = 26

Substract K into ( i )

16 k + C = 384

C = 384 – 416

C = – 32

W= k +C / t 2

W= 26- 32 / t 2

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

( 9aii )

When W= – 46, t =?

– 46 = 26 – 32 / t 2

t 2 = – 32 / -72

t = Sqrt 16 / 36 = 4 /6

= 2 / 3

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

9 b )

V = Pie r 2 . d = 14 , r = 7cm

1232 = 22 / 7 * 7 ^2 * h

h = 7 * 1232/ 22* 49

h = 8624 /1074

h = 8 cm

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

11 a)

y ^ 1 =x ^2 ( 3x +1 )^ 2

v = ( 2x + 1 )^ 2

v = m ^2

dm / dx =2

dv / dm =2 m

dy /dx = dv / dm ×dm / dx

= 2 m× 2

= 4 m

dy /dx = 4( 2x + 1)

dy /dx = udv /dx +v whole no . dy / dx

= x ^2 4( xx + 1)^ 2, × 3x

= 4 x ^2 ( 2x +1) + 2x ( 2x + 1)^ 2.

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

11 b)

[ 3 3 -1 ] [ 1 0 2] [ 3 – 2 3] + 2 [0 – ( – 4) – 3[ 63 ] + – 1 ( – 2)

8 + 9+ 2

= 19 .

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

11 c )

m= y 2- y 1 /x 2- x 1

y – y 1 = m ( x – x 2)

m= 4- 3/ – 1- 2

m= – 1/ 3

y 1 -y 2=- 1 /3 ( x – x 2)

y – 3= 1/ 3 ( x – 2)

y – 3= – 4/ 3 + 2/ 3

3 y =- x + 1[truncated by WhatsApp]

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

MATHEMATICS OBJ 100% VERIFIED :

1-10: ADADDDCEBE

11-20: CCDCAACECA

21-30: DCDAECAEED

31-40: EAADBEEADC

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

8a)

|AD|^2=13^2-5^2

|AD|^2=169-25

|AD|^2=144

AD=sqr144

AD=12CM

|AD|=12-r

r^2=(12-r)^2 – 5^2

r^2=(12-r)(12-r)+25

r^2=144-24r+25

r^2=169-24r

r^2+24r-169=0

r^2+24r=169

r^2+24r+14^2=169+14^2

(r+14)^2=169+196

(r+14)^2=365

(r+14=sqr365

r+14=19.105

r=19.105-14

r=5.105

r=5.1cm

8aii)

circumfrenece of a circle=2pie R

C=2×22/2*(5.1)^2

C=1144.44/7

C=163.4914cm

C=163.5cm

8b)

y2-y1/x2-x1=y-y1/x-x1

6-2/2-(-1)=y-2/x-(-1)

4/2+1 = y-2/x+1

4/3=y-2/x+1

3(y-2)=4(x+1)

3y-6=4x+4

3y-4x=4+6

3y-4x=10

y=4x/3+10/3

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

9a)

let Xy represent the two digit number

x-y=5 —–(i)

3xy – (10x +y)=14

3xy – 10x – y =14 —-(ii)

from eqn (i)

x=5+y

3y(5+y)-10(5+y)-y=14

15y+3y^2 – 50 – 10y – y =14

3y^2 + 4y -50 = 14

3y^2 + 4y -50 – 14 =0

3y^2 + 4y – 64 =0

3y^2 + 12y + 16y – 64 =0

(3y^2 – 12y) (+16y – 64)=0

by

(y-4)+16(y-4)=0

(y-4)=0

9aii)

(3y+16)(y-4)=0

3y+16=0 or y-4=0

3y=-16 or y=4

y=-16/3 or y=4

when y=4

x=5+y

x=5+4

x=9

the no is 94

9b)u

3-2x/4 + 2x-3/3

=3(3-2x)+4(2x-3)/12

=9-6x+8x-12/12

=2x-2/12