## Get Free Live 2020 NECO June/July General Mathematics (MATHS) OBJ & THEORY Questions and Answers Free of Charge | NECO June/July Free Mathematics (Objectives and Theory) Questions and Answers EXPO Room (10th November, 2020).

**NECO MAY/JUNE 2020 FREE GENERAL MATHEMATICS (MATHS) QUESTION AND ANSWER ROOM**

Tuesday 10th November, 2020

Paper III: Objective – General Mathematics

12:00noon – 2:30pm

**2020 NECO MATHEMATICS (ESSAY) ANSWERS:**

**2020 NECO MATHEMATICS (ESSAY) ANSWERS:**

(1a)

Equation:

y – y1/x – x1=y2 – y1/x2 – x1

y – 5/x – 6 = 7 – 5/-2 – 6

y – 5/x – 6 = 2/-8 = 1/-4

y – 5 = 1/4(x – 6)

y – 5 = -1/4x + 6/4

y = -1/4x + 6/4 + 5

y = -1/4x + 3/2 + 5

y = -1/4x + 13/2

OR 4y = -x+26

X + 4y = 26

(1b)

2

S(2x + 9)dx

-1

= 2X¹+¹/1+1 + 9x]2

-1

= x²+9x]2

-1

=[2²+9(2)] – [(-1)² + 9(-1)]

=[4 + 18] – [1 – 9]

= 22 + 8

= 30

(2a)

No that traveled by bus

= u+18+5+12 = 52

=u+35 = 52

u = 52 – 35

u = 17

No that traveled by train

= 6+5+12+v = 35

V + 23 = 35

V = 35 – 23

V = 12

Total no of tourist

=u+v+w+18+12+6+5 = 100

17+12+w+18+12+6+5 = 100

W + 70 = 100

W = 100 – 70

W = 30

(2b)

No who travelled by at least two means of transportation

= 18 + 6 + 12 + 5

= 41

(3ai)

EF = 50

Median = 25th + 26th/2

= 6+6/2 = 6

(3aii)

Range = 10 – 3 = 7

Median + Range = 6 + 7 = 13

(3b)

Prob(Olu passes) = 2/5

Prob(Tony passes) = 3/4

Prob(Olu fails) = 3/5

Prob(Tony fails) = 1/4

Prob(both passes) =

2/5 × 3/4 = 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 = 4 OR x = -7

(4b)

8x/9 – 3x/2 = 5/6 – x

Multiply by 18

18(8x/9) -18(3x/2) = 18(5/6) -18x

16x – 27x = 15 – 18x

18x + 16x – 27x = 15

7x = 15

X = 15/7

X = 2 whole 1/7

(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

(6ai)

Volume of sphere = 9 ⅓ ×(its surface area)

4/3πr³ = 28/3×4πr²

r = 28 units

Surface area = 4πr²

= 4×22/7×28×28

= 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

(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

= #1261

(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

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**2020 NECO MATHEMATICS (OBJECTIVE) ANSWERS:**

**2020 NECO MATHEMATICS (OBJECTIVE) ANSWERS:**

1-10: ACBAECABAE

11-20: CCABBCAAEC

21-30: BBEBBBACBD

31-40: CCADBAAABD

41-50: DEBBBCCBBD

51-60: DACEADAEEA

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