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Solution Centre for All Examinations both WAEC/WASSCE/SSCE, NECO, NABTEB, JAMB (UTME), Post-UTME, WAEC GCE, NECO GCE, NABTEB GCE etc

F/Maths OBJ:

1BCADDDDDBB

11DBABCCDDCC

21BCABBAABAB

31CBADBDDCAB

15a)

(2.2) + 80(0.6) = 300(1.2) + 100 (3)

2.2B + 48 = 360 + 300

2.B + 48 = 660

2.2B = 660 – 48

2.2B = 612

Reation at B = 278.18N

Taking moment about B

80(2.1) + 300(1) = 2.2A + 100 (0.8)

224 + 300 = 2.3A + 80

524 – 80 = 2.2A

A = 444/2.2

Raction a+ A = 201.82N

CLICK HERE FOR THE DIAGRAM

15b)

CLICK HERE FOR THE DIAGRAM

12)

TABULATE

No of head|x 0|12|3|45|6|7|8|9|10

Frequency|f|2|7|23|36|11|61|100|12|8|5|3

Fx| 0|7|46|108|44|305|600|84|64|45|30

Ef = 268 Efx = 1333

(a) mean no heads = Efx/Ef = 1333/268 = 4.97

(b) prob (even no heads) = 23/268 + 11/268 + 100/268 + 8/268 + 3/268

= 145/268

(c) prob (odd as heads) 7/268 + 26/268 + 61/268 + 12/268 + 5/268

= 121/268

14)

CLICK HERE FOR THE DIAGRAM

9)

Draw the pie chart diagram

radius is the same hence

(y – 2)² + (X – 3)² = (y + 4)² + (X – 5)² = (y + 1)² + (X + 2)²

Taking the first pair

Y² – 4y + 4 + x² – 6x + 9 = y² + 8y + 16 + x² – 10x + 25

– 6x – 4y + 13 = -10x+8y+41

4x – 12y = 28

X = 3y = 7——–(1)

Taking the second pair

Y² + 3y + 16+x² – 10x+25 = y²+ 2y + 1 + x²+ 4x + 4

-10x + 3y + 41 = 4x + 2y + 5

14x – 6y = 36

7x – 3y = 18———(2)

Eqn 2 minus Eqn 1

6x = 11

X = 11/6

Put this into Eqn (1)

11/6 – 3y = 7

3y = 11/6 – 7

3y = 11 – 42/6

3y = -31/6

y = -31/18

(a) coordinates of centre is (11/6, -31/18)

(b) Radius r = √(y-2)²+(x-3)²

r = √-31/18 -2)² + 11/6 – 3)²

r= √13.85 + 1.36

r = 3.9

(c) (X – 11/6)² + (y + 31/18)² = 3.9²

(X – 11/16)² + (y + 31/18)² = 15.21

2)

Log 3x – 3logx³+ 2 = 0

Log 3x – 3log3³/log3x + 2 = 0

Log3x – 3/log3x + 2 = 0

P – 3/p + 2 = 0 where P = log3x

P² – 3 + 2p = 0

P² + 2p – 3 = 0

(p + 3) ( p – 1) = 0

P = -3 OR p = 1

But log3x = p

when p = -3

Log3x = -3

X = 3-³ = 1/27

And when P = 1

Log3x = 1

X = 3¹ = 3

And X = 1/27 OR 3

3a)

U=x-2 hence x=u+2

therefore

x^2+5/(x-2)^4

=(u+2)^3+5/(u+2-2)^2

=(u+2)^3+5/u^4

3b)

(u+2)^3+5/u^4

=(u+2)^3/u^4+5/u^4

4)

sn=n/2[2a+(n-1)d]

s12=12/2[2a+11d]

s12=6(2a+11d]=168

2a+11d=28 eq—–(1)

multipply 1by2 to obtain

2a+4d=14 eq——(3)

subtract (3)from(2)

7d=14

d=2

sub for d in (1)

a+4=7

a+7-4-3

a=3

common difference d=2

first term a=3

5)

tabulate

X,y,d,d^2

A 8,6,2,4

B 5,3,2,4

C 1,4,-3,9

D 7,8,-1,1

E 2,5,-3,9

F 6,7,-1,1

G 3,1,2,4

H 4,2,2,4

Ed^2 =4+4+9+1+9+1+4+4

=36

r=1-6£d^2/n(n^2-1)

=1-6*36/8(8^2-1)=1-216/504

=1-0.43

=0.57

1)

X-3(12-2x+8)+4(30-5x-4)+3(-20-4)=24

X-3(20-2x)+4(26-5x)+3(-24)= -24

20x-2x^2+60+6z+104-20x-72=24

20x^2+6x-60+104-72+24=0

-2x^2+6x-4=0

Divide 2x by -2

x2^2-3x+2=0

x2^2-2x-x+2=0

x(x-2)-1(x-2)-0

(x-1)(x-2)=0

x-1 or x=2

6)

It follow that:

P(n) = 1/3 and p(k)¹ = 1 – 1/3 = 2/3

P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

Hence, probability that only one if the them will be solve the questions will be:

= (1/3 × 4/5) + (1/5 × 2/3)

= 4/15 + 2/15

= 6/15

= 2/5

8a)

m1 = 20kg

u1 = 8ms-1

m2 = 30kg

u2 = 50ms-1

(a) in same direction

m1u1 + m2u2 = (m1+m2)v

20 × 80 + 30 × 50 = (20+30)v

1600+1500 = 50v

3100/50 = 50/50

V = 62ms-1

8b)

M1=20kg

u1=80mls

m2=30kg

u2=50mls

v=?

M1u1+m2u2=v(m1+m2)

(20*80)+(30*(-50))=v(20+30)

1600-1500=50v

100=50v

v=100/50

v=2mls

7)

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

Therefore:

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

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