## 2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

COMPLETE MATHEMATICS OBJ & ESSAY QUESTIONS & ANSWERS NOW AVAILABLE. subscribe for the expo if you don’t want late answers

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MATHS OBJ:

1-10: CDAAEABAEC

11-20: ACDDCDCDAC

21-30: CEADEDCABC

31-40: CBEECCBDCC

41-50: DDCBCDDBBA

51-60: BCECCBBCEE

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(1a)

Log10(20x – 10) – log10(x+3) = log10^5

Log10(20x-10/x+3)= log10^5

20x – 10/x + 3 = 5

Cross multiplying

20x – 10 = 5(x + 3)

5(4x – 2) = 5(x + 3)

4x – 2 = X + 3

4x – x = 3+2

3x = 5

X = 5/3 OR 1 whole no 2/3

(1b)

Let actual amount be #X

15% of #x = #600

15x/100 = 600

X = (100/15)*600

X = 100*40

X = 4,000

Actual amount = #4,000

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(2a)

(X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5

= (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/Z^5

= X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5

=X^3/2+1 * Y^-9/4-4 * Z^3/4-5

=X^5/2 * Y^-25/4 * Z^-17/4

=X^10/4 * Y^-25/4 * Z^-17/4

=(X^10/Y^25 Z^17)^1/4

(2b)

√2/k + √2 = 1/k – √2

Multiply both sides by (k+√2)(k-√2)

√2(k-√2) = k+√2

√2k-√2 = k+√2

√2k-k = 2+√2

K(√2 -1) = 2+√2

K = 2+√2/√2-1

K = -(2+√2)/1-√2

Rationalizing

K = -(2+√2) * 1+√2/1-√2

K = -(2+√2)(1+√2)/1 – 2

K = (2+√2)(1+√2)

K = 2+2√2 + √2+2

K = 4+3√2

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(3)

V = Mg√1 – r²

Square both sides

V² = m²g²(1-r²)

V²/m²g² = 1-r²

r² = 1 – v²/m²g²

r = √1-(v/mg)²

If v = 15, m = 20, and g = 10

r = √1 – (15/20*10)²

r = √1 – (0.075)²

r= √(1.075)(0.925)

r = √0.994375

r = 0.9972

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

(4)

Draw the diagram

(i) Arc length = Tita/360*2πr

= 72/360*2*22/7*14

=1/5*44*2

=88/5

=17.6cm

(ii) Perimeter of Sector = arc length +2r

=17.6+2(14)

=17.6+28

=45.6cm

(iii) Area of sector = Tita/360*πr²

=72/360*22/7*14/1*14/1

=1/5*22*2*14

=616/5

=123.2cm2

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

(5a)

Mode = mass with highest frequency = 35kg

Median is the 18th mass

= 40kg.

(5b)

In a tabular form

Under Masses(x kg)

30,35,40,45,50,55

Under frequency(f)

5,9,7,6,4,4

Ef = 35

Under X-A

-10, -5, 0, 5, 10, 15

Under F(X-A)

-50, -45, 0, 30, 40, 60

Ef(X – A) = 35

Mean = A +

(Ef(X – A)/Ef)

= 40 + 35/35

= 40 + 1

= 41kg

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

(7ai)

T3=>a+2d=6(eqi)

T7=>a+6d=30(eqii)

Eqii minus eqi gives

6d-2d=30-6

4d=24

d=24/4

d=6

Common difference=6

(7aii)

Putting d=6 into eqi

a+2(6)=6

a+12=6

a=6-12

a=-6

(7aiii)

10th term T10=a+9d

=-6+9(6)

=-6+54

=48

(7bi)

T3=>ar²=9/2(eqi)

T6=>ar^5=243/16(eqii)

Dividing eqii by eqi

ar^5/ar²=243/16 divided by 9/2

r³=243/16*2/9

r³=27/8

r³=3³/2³

r=3/2

Putting this into eqi

a(3/2)²=9/2

a(9/4)=9/2

a=9/2*4/9

a=4/2=2

(7bii)

Common ratio r=3/2 as above

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(8a)

x=a+by(eqi)

when y=5 and x=19

19=a+5b(eqii)

when y=10 and x=34

34=a+10b(eqiii)

solving eqii and eqiii

a+10b=34

a+5b=19

=>5b=15

b=15/5=3

putting b=3 in eqii

19=a+5(3)

19=a+15

a=19-15

a=4

(8ai)

Putting a=4 and b=3 in eqi

x=4+3y

This is the relationship between xand y

(8aii)

When y=7

x=4+3(7)

x=4+21

x=25

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

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

(11a)

x+y/2 =11

x+y= 11*2

x+y= 22 —(1)

x-y= 4 —-(11)

x+y = 22—-(1)

–

x-y= 4—-(11)

____

2y = 18

y= 18/2

y=9

Substitute y=9 in equ 1

x+9=22

x=22-9

x=13

x=13, y=9

x+y= 13+9= 22

Sum of the two number

(11b)

(6x + 3) dx

(6x + 3)dx

(6x +3)^6 – (6x + 3)^1

(6 x + 3)^5

(7776x^5 + 243)

38,880x/6 + 243

6480 x^6 + 243x

9(720x^6 + 27x)

(11c)

y = x² + 5x – 3 (x = 2)

y = 2² + 5(2) – 3

y = 4 + 10 – 3

y = 14 – 3

y = 11

Gradient of the curve = 11

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

(12a)

Pr of Abu to pass = 3/7

Pr of Abu to fail = 1 – 3/7 = 7-3/7 = 4/7

Pr of kuranku to pass = 5/9

Pr of kuranku to fail = 1 – 5/9 = 9 – 5/9 = 4/9

Pr of musa to pass = 12/13

Pr of musa to fail = 1 – 12/13 = 13 – 12/13 = 1/13

Pr of only one of them passing is

=(3/7*4/9*1/13)+(5/9*4/7*1/13)+(12/13*4/7*4/9)

=12/819+ 20/819 + 192/819

=12+20+192/819 = 224/819

= 32/117

(12b)

10Red + 8green + 7blue = 25

(i)

pr of different colour is

Prof(RG)+(RB)+(GB)+(BG)+(BR) +(GR)

=(10/25*8/24)+(10/25*7/24)+(8/25*7/24)+(7/25*8/24)+(7/25*10/24)+(8/25*10/24)

=80/100 + 70/600 + 56/600 + 56/600 + 70/600 + 80/600

= 80+70+56+56+70+80/600

= 412/800 = 103/200

(ii)

pr of atleast one must be

=Pr[RB+BR+GB+BG+BB]

= (10/25*7/24)+(7/25*10/24)+(8/25*7/24)+ (7/25*8/24) + (7/25*7/24)

=70/600+70/600+56/600+56/600+49/600

=70+70+56+56+49

/600

=301/600

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2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

2018/2019 NECO RUNZ MATHEMATICS OBJ & ESSAY QUESTIONS AND ANSWERS

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