2025/2025 WAEC/NECO Answers: Neco Gce 2025. Mathematics answers



+234 8106996452 +234 8106996452

Notice
Chat us on WhatsApp now …. 2025 WAEC Subscription is on. 10K if you pay today or tomorrow... Subscribe now.. daily subscription should be done at least a day before exam... Full WhatsApp Group Subscribers get the answers much earlier before other people. Food and nutrition midnight answers is available at examplaza.com. Also note that Our 2025 WAEC Midnight Answers Subscription ends on Sat 3rd May 2025. Note: we don't accept calls. Please, we beg you in the name of God, do all your daily subscriptions AT LEAST a day before the exam, we may not have time to attend to you for daily subscriptions on the exam day. This is examplaza.com, we source other websites.
ID: 26
To get 2025 WAEC Answers midnight before the exam, WhatsApp this

Refresh this page in every 5 minutes starting from 3 hours to the exam. 




OBJ

1-10: DABBABBBAD

11-20: CDBDCCCBBA

21-30: CDABDACCCC

31-40: BCBCABABBC

41-50: DACCCCDBDB

This is actually the Neco Gce 2018. Mathematics answers Answers as questions sent to our subscribers on WHATSAPP, if you are looking for 2025 WAEC/NECO Answers midnight before the exam, try contacting us.



(10)

Draw the triangle 


Hence we have A+B+C = 180

50+80+C = 180

130+C = 180

C = 180 - 130

C = 50


(i)

The bearing of B from C = 90 - 50 = 40°


(ii) 

Bearing of A from B

90+50(alternate angle to A) 

= 140°


(iii) 

Distance between B and C

Using sine rule 

C/sinC = a/SinA

40/sin50 = a/sin50

Cross multiply 

asin50 = 40sin50

a = 40km

Hence distance between B and C = 40km


(iv) 

Using cosine rule 

b²= a²+c² - 2acCosB

where a = 40, c= 40 and B=80

b² = 40²+40² -2*40*40cos80

b² = 1600+1600 - 3200(0.1736)

b² = 3200 - 555.52

b² = 2644.48

b = √2644.48

b = 51.42km


(v)

Height of ΔABC

Draw the triangle 


Area of the triangle 

=1/2acSinB

=1/2*40*40sin80

=800(0.9848)

=787.84km²

Hence 1/2bh=787.84

bh=2*787.84

h=2*787.84/b

h=2*787.84/51.42

h=30.64

=31






(12)

TABULATE:

Marks | frq | c.f | boundary

11-20 | 8 | 8 | 10.5 – 20.5

21-30 | 6 | 14 | 20.5 – 30. 5

31-40 | 10 | 24 | 30.5 – 40.5

41-50 | 12 | 36 | 40.5 – 50.5

51-60 | 8 | 44 | 50.5-60.5

61-70 | 6 | 50 | 60.5 – 70.5


(12b)

DRAW THE CUMULATIVE FREQUENCE CURVE


(12ci)

Medium mark = N + ½

= 50 + 1/2 = 51/2

= 25. 5

= 48


(12cii)

Lower quartile Q1

= ¼ x 50

= 12.5

= 28


(12ciii)

Upper quartile Q2

= ¾ x 50

= 37.5

=53


(12civ)

Both percentile 

70/100 x 50 

= 35

= 50

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

(1) 

TABULATE:

No| Log

3081| 3.4887

0.775 | 1.8893

0.456 | 1. 6589

| UNDER Log

| 3.4887

| 1. 5482

| 3.9405

| 0.9851

Square root of 4/3081/0.775.0456

Antilog = 9663

= 9.663

= 2.9






(2a)

1101â‚‚ = 2x + 1

1 x 2² + 0 x 2¹+ 1 x 2º = 2x + 1

8 + 4 + 1 = 2x +1

13 – 2x

X = 12/2 = 6


(2b)

Sin x = 12/13 = 0.9231

Z = sin-¹ 0.9231 = 67.4º

Therefore 3 sin x + ½ cos x

= 3 sin 67.4 + ½ cos 67.4

= 3 x 0.9232 + ½ x 0.3843

= 7.7696 + 0.1921






(3a)

PQ x 90º  [angle in a semicircle]

QPO = 90º - 62 [ angle in a triangle]

= 28º

Therefore POZ = 28º [alternative angles]

(ii)

PXZ = ½ x 28º = 14º [angle at center is twice angle at cirumfeence]


(3b)

6/5 + 3/x+3 – 9/5(x+3)

= 6 (x+3) + 3 (5) – 9/5(x+3)

= 6x + 18 + 15 – 9/5 (x+3)

= 6x + 24/5(x+3) 

= 6(x+4)/5(x+3)






(4a)

1/3(y-1)+2>1/2(2y-1)+1

2y-2+12>6y-3+1

2y-6y>-2-10

-4y>-12

y<3


(4bi)

M=y2-y1/X2-X1

=-1-2/2-3=-3/-1=3


(4bii)

2X+1=X+3

2X-X=3-1

X=2





(6)COMPLETED 

(6)

126y = 86

1 x y² 2 x y¹ + 6 x yº = 86

y² + 2y + 6 = 86

y² + 2y + 6 – 86 = 0

y² + 2y – 80 = 0

y + 10y – 8y – 80 = 0

y (y+10)-8y (y+10) = 0 

y – 8 = 0 or y + 10 = 0

the positive value of y


(6b) Area of triangle = ½ x b x h

Let the Acheal Area = x

Area = ½ x b x h

Base = x – 9x/100 = 93/100

Height = 9x/100 + x = 107x/100

Area = ½ x 93x/100 x 107x/100

=963x/2000

% error = actual Area – wrong/actual area x 100

= x – 963x/2000 /x 100

= 20000x – 963x/20000 x 100

= 19037/20000 x 100

= 95.185%


(6c)

p/100 + 2p + 7 = 11.02 x 100

p + 200p + 700 = 1102

201p = 1102 – 700

201p = 402

P = 402/201

P = N2

P = 200K




(8)

S²â‚ (3x – 1) (x+2) dx

By expansion’

S²â‚ 3x² + 6x – x – 2 dx

S²â‚ 3x² + 5x – 2 dx

By integrating using d formula

Xn+1/n+1

Therefore 3x ²+¹/2+1 + 5x¹+¹/1+1

= 2xº+¹/0+1 + c

Therefore 3x³/3 + 5x²/2 + 2x/1 + c

Therefore x³ + 5x²/2 + 2x

But x = 2 at higher and are at lower by substituting 2 in x than of value of 1

(2)³ + 5/2(2)² + 2(2) – ((1)³ + 3/2(1) + 2 (1)

8 + 5/2 x 4 + 4 – (1 + 5/2 + 2)

8 + 10 + 4 – (3 + 5/2)

22 – 11/2 = = 22-11/2

= 44- 11/2 

= 33/2


(8bi)

T = 2Ï€ square l/g

Dividing both side by 2Ï€

T/2Ï€ = square root of L/g

By squaring both side

(T/2 Ï€)² = (L/g)²

T²/4 Ï€² = l/g

Cross multiplication

gT²/T² = 4Ï€²L/T² 

g = 4Ï€²L/T²


(8bii)

T = (0.4)1/2 = square root of 0.4

L = 0.04   T1 = 3.14

G = 4Ï€²L/T2 

= 4 x (3.14)² x 0.04/(0.4)²

= 4 x 9.8596 x 0.04/0.4 

= 1.578/0.4

g = 3.94




(9a)

2p-q=10.......(1)

3p+q^2=22........(2)

Eq(1) x3 and eq(2) x2

6p-3q=30

6p+2q=44

Subtracting eq(1) from eq(2)

-3q-2q= -14

2q^2+3q= 14

2q^2+3q-14=0

2q^2+7q-4q-14=0

q(2q+7)-2(2q+7)=0

(q-2)(2q+7)=0

q-2=0 or 2q+7=0

q=2 or q= -7/2

Substitute q into eq(1)

2p-q=10

2p-2=10

2p=10+2

2p=12

p=12/2=6

When q= -7/2

2p-q=10

2p-(-7/2)=10

2p+7/2=10

4p+7=10

4p=10-7

4p=3

p=3/4


(9b)

z^2 -25/z^2-9z+20

If Z is undefined 

Z^2-9z+20=0

Using factorization method 

-4z and -5z

Z^2 -4z-5z+20=0

Z(z-4) - 5(z-4)=0

(z-4)(z-5)=0

Z-4=0 or Z-5=0

Z=4 or Z=5

Z=4 or 5

Z is undefined when it is equal to  4 or 5



Answers loading.... 



How do I getNeco Gce 2025. Mathematics answers BEFORE THE EXAM?
. You can get 2025/2025 Exam answers directly from examplaza.com, visit HERE NOW

Name: examplaza.com

Founded: 2010 (15 years)

Founder: Mr. Onuwa

Headquarters: Borno, Nigeria

Official Website: https://examplaza.com/

Official Contact: +2348106996452


Tags:

: Related Posts


WAEC expo: This is 2025 COMPLETE WAEC QUESTIONS AND ANSWERS

waec: Ace your upcoming WAEC 2025 Science exam withplaz's reliable and verified WAEC science answers. Discover how you can get ready for the exam at examplaza.com.

waec: Get ahead in your WAEC exams with the help of verified answers from Examplaza. Explore details Midnight WAEC Expo and how to take advantage of this service in this blog post.

waec: Obtaining WAEC Geography Answers for the 2025 exam is attainable. Read through on how to get it.

waec: Looking for 2025 WAEC Physics Runs? Learn about the most reliable sources and how to obtain WAEC Physics Expo for free in this article.

waec: Using our insider's knowledge, Examplaza can provide you with 100% correct WAEC midnight expo a day or hours before the start of the exam. Subscribe now to be ahead.

waec: Don't allow fear over WAEC Runs to cloud your sense of self-worth. Your future success doesn't depend on your exam performance alone. Read more.

waec: Maximize your success in the 2025 WAEC exams by following this insightful guide. Learn about the top runz subscription site, Examplaza, and discover unbeatable deals on WAEC Expo.

WAEC TIMETABLE: Verified 2025 Complete and Updated WAEC Timetable for School Candidates which states that 2025 WAEC will start on Thursday, 24th April 2025 and end on Friday, June 20th April 2025

jamb: In this article, we've uncovered the ultimate blueprint for UTME exam success using JAMB Runz 2025. From detailed steps to obtain JAMB UTME Runz to tips for exam preparation, you'll find everything you need to excel in 2025.