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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**(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**

(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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WAEC English Questions and Answers 2022. WAEC Eng Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2022 NECO MATHEMATICS ANSWERS. 2022 WAEC EXAM English Questions and Answers

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