2025 NECO MATHEMATICS: 2025 NECO MATHEMATICS ANSWERS (9942)

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Tuesday 10th Nov
Paper III: Objective – General Mathematics
10:00am – 11:45am
Paper II: Essay – General Mathematics
12:00noon – 2:30pm
===============

MATHS OBJ
1-10: ACBAECABAE
11-20: CCABBCAAEC
21-30: BBEBBBACBD
31-40: CCADBAEABD
41-50: DEBBBCCBBD
51-60: DACEADAEEA

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No. 8 corrected

making some correction to No. 8

No. 8 added

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Welcome to official 2025 Mathematics NECO answer page. We provide 2025 Mathematics NECO Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct NECO Maths Expo. NECO Mathematics Questions and Answers 2025. NECO Maths Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2025 NECO MATHEMATICS ANSWERS. 2025 NECO EXAM Mathematics Questions and Answers

(1a)
x¹ y¹. x¹ y¹
A (6,5). B ( -2 ,7)

The equation of a straight line
y² - y¹/x² - x¹ = y-y/x¹-x¹

7-5/-2-6 = y-5/x-6
2/-8 = y-5/x-6
-1/4 = y-5/x-6
4(y-5)= -1(x-6)
4y-20= -x+6
4y+x=6+20
4y+x=26

(1b)
∫² (2x + 9)dx
-²
2x²/2+9x + c|²_¹
x² + 9x + c|²_¹
(2² + 9(2) + C) - ((-1)² + 9 (-1)+ c)
(4+18+c) - (1-9+C)
(22 + C) - (-8 + C)
22+C + 8 - C
=30.
===================

(2a)
To calculate u; 18+5+12+u=52
35+u=52
u=52-32=17
u=17

To calculate v
6+5+12+v=35
23+v=35
v=35-23=12
v=12

To calculate w
w+18+17+6+5+12+12
=100
w+70=100
w=100+70=30
w=30

(2b)
at least two means =18+6+5+12=41
; 41 tourist travelled by at least two means of transportation.
==============

(3a)
(i)median =6+6/2=12/2=6
(ii)median + range = 6+7=13
Range = highest value - lowest value = 10-3=7

(3b)
Pr(both pass)
=2/5*3/4=6/20=3/10.
=====================

(4a)
x²+3x-28=0
x²+7x-4x-28=0
x(x+7)-4(x+7)=0
(x-4)(x+7)=0
x-4=0 or x+7=0
x=0+4 or x=0-7
x=4 or -7

(4b)
8x/9 - 3x/2 =5/6 - x/1
L.c.m=54
6*8x-27*3x=9*5-54*x
48x-81x=45-54x
48x-81x+54x=45
21x/21=45/21
x=45/21=15/7=2⅐

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

(5)

(5a)
d/dx(4x³ - 2x + 4)⁵
= 5(4x³-2x+4)⁴ × (12x² - 2)
= 10(6x² - 1)(4x³ - 2x + 4)⁴

(5b)
5/3(2-x) - (1-x)/(2-x) = 2/3
Multiply through with 3(2-x)
3(2-x)[5/3(2-x)] -3(2-x)[1-x/2-x] = 3(2-x)(2/3)
5 - 3(1-x) = (2 - x)(2)
5 - 3 + 3x = 4 - 2x
3x + 2x = 4 + 3 - 5
5x = 2
X = 2/5


=================
(6)

(6ai)
Volume of sphere = 9 ⅓ ×(its surface area)
4/3πr³ = 28/3×4πr²
r = 28 units

Surface area = 4πr²
= 4x22/7×28×28
= 9856 units squared.

(6aii)
Volume = 9 ⅓ × 9856
= 28/3 × 9856
= 91989.33
=91989 cubic units

(6b)
log10(3x - 5)² - log10(4x -3)² = log10 25
Log10(3x - 5/4x - 3)² = log10 25
(3x - 5/4x - 3)² = 25
Square root of both sides gives
3x-5/4x-3 = ±5
3x-5 = 5(4x-3) OR 3x-5 = 5(4x-3)
3x-5 = 20x-15 OR 3x-5 = 20x+15
3x-20x = 5-15 OR 3x+20x = 15+5
-17x = -10 OR 23x = 20
X = 10/17 OR x = 20/23


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

(7)

(7a)
Given: 3x + 5y = 10
5y = -3x + 10 OR y = -3/5x+ 2
Gradient of the straight line = -1 ÷ (-2/5)
= 5/3
Equation of line:y-2/x-3 = 5/3
3y-6 = 5x - 15
3y = 5x - 15 + 6
3y = 5x - 9
y = 5/3x - 3
Intercept of line = -3

(7b)
Amount = P(1+R/100)³
=8000(1+5/100)³
=8000(1.05)³ OR 8000(1.05/100)³
=8000×1.157625
=#9261

Compound interest = Amount - principal
= 9261 - 8000
= ₦1,261

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

(8a)
Ta; a+8d=50 --------(1)
T12; a+11d=65---------(2)
Subtract equation (1) from (2)
a+11d-(a+8d)=65-50
a+11d-a-8d=15
11d-8d=15
3d/3=15/3
d=5

Substitute for d=5 in each equation (1)
a+8d=50
a+8(5)=50
a+40=50
a=50-40=10
Sn=n/2(2a+(n-1)d)
S70=70/2(2*10+(70-1)5)
=35(20+69*5)
=35*365=12,775


===========

(10a)
Let the woman's age be W
Let the daughter's age be d
Given: w = 4d-----(1)
Given:(w+5)²=(d+5)²+120--(2)
Put eqn(1)into(2)
(4d+5)² = (d+5)² + 120
(4d+5)² - (d+5)² = 120
[4d+5+d+5][4d+5-d-5] = 120
[5d+10][3d] = 120
5(d+2)(3d) = 120
15(d+2)(d) = 120
d(d+2) = 8
d² + 2d - 8 = 0
d² + 4d - 2d - 8 = 0
d(d+4) -2(d+4) = 0
(d - 2)(d + 4) = 0
d - 2 = 0 (only)
d = 2
Daughter is 2 years old

(10b)
t = w + wy²/PZ
Multiply through by PZ
Pat = PWZ + wy²
Wy² = ptz - pwz
y² = Pz(t - w)/w
y = ±√PZ(t - w)/w
If P = 5, Z = 10, t = 9, w=3
y = ±√(5)(10) (9-3)/3
y = ±√(5)(10)(6)/3
y = ±√100
y = ±10


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

(11)

(11)
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|Frequency|3|10-T|3T|8|2T+2|T+2|

(a)Ʃ+=50
3+10-T+3T+8+2T+2+T+2=50

3+10+8+2+2-T+3T+2T+T=50

25+5T=50
5T=50-20
5T/5=25/5
T=5

(11b)
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|Frequency|3|5|15|8|12|7|

The frequency of the modal class is 15.

(11c)
Tabulate.
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|F|3|5|1|5|8|1|2|7|50

|X|7|12|17|22|27|32|

|FX|21|60|255|176|324|224|1060

|X-x̅|-14.2|-9.2|-4.2|0.8|5.8|10.8|

|(X-x̅)²|201.64|84.64|17.64|0.64|33.64|116.64|

F|(X-x̅)|604.92|423.2|264.6|5.12|403.68|816.48|2518

Mean (x̅) =Ʃfx/Ʃf=1060/50=21.2

Variance =Ʃf(x-x̅)²/Ʃf
=2518/50
=50.36

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

(12a)
y=²-2x-3
Tabulate
|x|-2|-1|0|1|2|3|
|y|5|0|-3|-4|-3|0|

(12b)
Graph.


(12c)
y=1-3x
|y|-2|-1|0|1|2|3|
|x|7|4|1|-2|-5|-8|

(12d)
(i)the root of the equation x² - 2x - 3= 1-3x are -3 and 1.6

(ii) the minimum value of y is - 4 and the corresponding value of x is 1

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