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Re: University Of Ibadan 2015/16 Applicants by mikechibuzor(m): 7:58pm On May 31, 2015
horpeyemmi66:
*PHYSICS
1. Two Submarines The knight and The cascade were submerged gradually into a river at a depth of 120m and 180m respectively below the water surface. If the atmospheric pressure is 15m of Water, find the ratio of the pressure of The Cascade to The knight.

2. A pilot records the atmospheric pressure outside his aircraft as 83cmHg while a ground observer has a reading of 94cmHg for the atmospheric pressure on the ground. If the density of the atmosphere is constant, calculate the height of the aircraft above the ground[Relative density of Mercury=15 and Air=0.0015]

3. If the normal atmospheric pressure in a chemistry Laboratory supports a column of Mercury 0.85m high and the relative density of Mercury is 18.5, then the height of the water column which the atmospheric pressure will support in the same Chemistry laboratory at the same time is?

4. A man walks 8km north and then 5km in a direction 60° east of north. Find the distance from his starting point.

5. A boy travels 12Km eastwards to a point B and then 5km southwards to another point C. Cakculate the difference between the magnitude of the displacement of the boy and the distance travelled by him.

6. The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resultant graph is linear.
The slope of this graph is a measure of

A. Initial displacement

B. Initial velocity

C. Acceleration

D. Half the acceleration

7. Which of the following is a set of Vectors?

A. Force, mass and moment

B. Acceleration, velocity and moment

C. Mass, weight and density

D. Mass, Volume and density

8. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m/s and rebounds with the same speed. Determine the impulse experienced by the wall.

A. 20.0kgm/s

B. 10.0kgm/s

C. 5.0kgm/s

D. 1.3kgm/s

9. Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y, the ratio of the time taken by X to that taken by Y is?

A. 1/2

B. 2

C. 1/√2

D. √2

10. The driver in a motor car of which the total mass is 800kg and which is travelling at 20m/s, suddenly observes a stationary dog in his path 50m ahead. If the car brakes exert a force of 2000N, what will most likely happen?

A. The car will be able to stop immediately the driver notices the dog
B. The car will stop 30m after hitting the dog
C. The car will stop 20m in front of the dog
D. The driver will quickly reverse the car
E. The driver will have to find a banking angle to ensure traction.


cc Polymath, Pr0ton, Dr. Host, Dr. Sage, Mikechibuzor, thaotech, AleXis0r, francistony, MedwhiteO, Bhuumhite, Funkysilver etc


*******UNA GOOD MORNING*******

4. Let the distance from his starting point be x.
So sin 30/5= sin 120/x
x=5sin 120/sin 30.
X= 8.7km.
5. Using pythagoras rule
x^2= 12^2 + 5^2
x=13km. Which is the displacement. The total distance travelled is the sum of the distance he travelled in both directions which is 17km (12 + 5).
So difference is equal to -4km.
6. D
7. B. Vectors have both magnitue and direction and every watchamacallit in option b does.
8. Impulse = momentum change.
Ft= mv - mu.
I believe the answer is 20kgm/s but using the formula gives something diff. Please kindly do extensive explanation here.
9. Is definitely A.
10. Tough but i think i'm tougher.. Hehe
, F= ma right? No owo e so ke to ba n doubt. No hands up seems i can go on.
2000= 800 x a
a= 2.5m/s/s.
After the force was applied, the car moved with 2.5m/s/s before it completely stopped.
The distance it took the car to finally stop should be
uhm.. V^2= u^2 + 2as
20 x 20 =0 + 5s
S = 80m.
Apparently, option B is the correct answer.
RIP to the dog.

I can't remember the last time i read but the way things are(your questions pushing me), i'm sure it won't be long for me to re-marry my books.
For the record,we're divorced.
Re: University Of Ibadan 2015/16 Applicants by thaotech: 9:08pm On May 31, 2015
horpeyemmi66:
*PHYSICS
1. Two Submarines The knight and The cascade were submerged gradually into a river at a depth of 120m and 180m respectively below the water surface. If the atmospheric pressure is 15m of Water, find the ratio of the pressure of The Cascade to The knight.

2. A pilot records the atmospheric pressure outside his aircraft as 83cmHg while a ground observer has a reading of 94cmHg for the atmospheric pressure on the ground. If the density of the atmosphere is constant, calculate the height of the aircraft above the ground[Relative density of Mercury=15 and Air=0.0015]

3. If the normal atmospheric pressure in a chemistry Laboratory supports a column of Mercury 0.85m high and the relative density of Mercury is 18.5, then the height of the water column which the atmospheric pressure will support in the same Chemistry laboratory at the same time is?

4. A man walks 8km north and then 5km in a direction 60° east of north. Find the distance from his starting point.

5. A boy travels 12Km eastwards to a point B and then 5km southwards to another point C. Cakculate the difference between the magnitude of the displacement of the boy and the distance travelled by him.

6. The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resultant graph is linear.
The slope of this graph is a measure of

A. Initial displacement

B. Initial velocity

C. Acceleration

D. Half the acceleration

7. Which of the following is a set of Vectors?

A. Force, mass and moment

B. Acceleration, velocity and moment

C. Mass, weight and density

D. Mass, Volume and density

8. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m/s and rebounds with the same speed. Determine the impulse experienced by the wall.

A. 20.0kgm/s

B. 10.0kgm/s

C. 5.0kgm/s

D. 1.3kgm/s

9. Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y, the ratio of the time taken by X to that taken by Y is?

A. 1/2

B. 2

C. 1/√2

D. √2

10. The driver in a motor car of which the total mass is 800kg and which is travelling at 20m/s, suddenly observes a stationary dog in his path 50m ahead. If the car brakes exert a force of 2000N, what will most likely happen?

A. The car will be able to stop immediately the driver notices the dog
B. The car will stop 30m after hitting the dog
C. The car will stop 20m in front of the dog
D. The driver will quickly reverse the car
E. The driver will have to find a banking angle to ensure traction.


cc Polymath, Pr0ton, Dr. Host, Dr. Sage, Mikechibuzor, thaotech, AleXis0r, francistony, MedwhiteO, Bhuumhite, Funkysilver etc


*******UNA GOOD MORNING*******

1.104000cm
2. 2:3
3. 15.75
4. 13KM
5. 4km
6. C
7. B
8. A
9. A
10. C



Good evening everyone!
Re: University Of Ibadan 2015/16 Applicants by thaotech: 9:15pm On May 31, 2015
mikechibuzor:


4. Let the distance from his starting point be x.
So sin 30/5= sin 120/x
x=5sin 120/sin 30.
X= 8.7km.
5. Using pythagoras rule
x^2= 12^2 + 5^2
x=13km. Which is the displacement. The total distance travelled is the sum of the distance he travelled in both directions which is 17km (12 + 5).
So difference is equal to -4km.
6. D
7. B. Vectors have both magnitue and direction and every watchamacallit in option b does.
8. Impulse = momentum change.
Ft= mv - mu.
I believe the answer is 20kgm/s but using the formula gives something diff. Please kindly do extensive explanation here.
9. Is definitely A.
10. Tough but i think i'm tougher.. Hehe
, F= ma right? No owo e so ke to ba n doubt. No hands up seems i can go on.
2000= 800 x a
a= 2.5m/s/s.
After the force was applied, the car moved with 2.5m/s/s before it completely stopped.
The distance it took the car to finally stop should be
uhm.. V^2= u^2 + 2as
20 x 20 =0 + 5s
S = 80m.
Apparently, option B is the correct answer.
RIP to the dog.

I can't remember the last time i read but the way things are(your questions pushing me), i'm sure it won't be long for me to re-marry my books.
For the record,we're divorced.

For the impulse

Ft=mv-mu
after hitting the wall the ball rebounded and moved in the opposite direction
Ft=mv-(-mu) the - sign was introduced cos of the opposite direction

10+10= 20

horpeyemmi60 over to you

1 Like

Re: University Of Ibadan 2015/16 Applicants by mikechibuzor(m): 9:23pm On May 31, 2015
thaotech:

For the impulse

Ft=mv-mu
after hitting the wall the ball rebounded and moved in the opposite direction
Ft=mv-(-mu) the - sign was introduced cos of the opposite direction

10+10= 20

horpeyemmi60 over to you
i see.thanks for putting me through.
Re: University Of Ibadan 2015/16 Applicants by enry54(m): 9:58pm On May 31, 2015
horpeyemmi66:
*PHYSICS
1. Two Submarines The knight and The cascade were submerged gradually into a river at a depth of 120m and 180m respectively below the water surface. If the atmospheric pressure is 15m of Water, find the ratio of the pressure of The Cascade to The knight.

2. A pilot records the atmospheric pressure outside his aircraft as 83cmHg while a ground observer has a reading of 94cmHg for the atmospheric pressure on the ground. If the density of the atmosphere is constant, calculate the height of the aircraft above the ground[Relative density of Mercury=15 and Air=0.0015]

3. If the normal atmospheric pressure in a chemistry Laboratory supports a column of Mercury 0.85m high and the relative density of Mercury is 18.5, then the height of the water column which the atmospheric pressure will support in the same Chemistry laboratory at the same time is?

4. A man walks 8km north and then 5km in a direction 60° east of north. Find the distance from his starting point.

5. A boy travels 12Km eastwards to a point B and then 5km southwards to another point C. Cakculate the difference between the magnitude of the displacement of the boy and the distance travelled by him.

6. The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resultant graph is linear.
The slope of this graph is a measure of

A. Initial displacement

B. Initial velocity

C. Acceleration

D. Half the acceleration

7. Which of the following is a set of Vectors?

A. Force, mass and moment

B. Acceleration, velocity and moment

C. Mass, weight and density

D. Mass, Volume and density

8. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m/s and rebounds with the same speed. Determine the impulse experienced by the wall.

A. 20.0kgm/s

B. 10.0kgm/s

C. 5.0kgm/s

D. 1.3kgm/s

9. Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y, the ratio of the time taken by X to that taken by Y is?

A. 1/2

B. 2

C. 1/√2

D. √2

10. The driver in a motor car of which the total mass is 800kg and which is travelling at 20m/s, suddenly observes a stationary dog in his path 50m ahead. If the car brakes exert a force of 2000N, what will most likely happen?

A. The car will be able to stop immediately the driver notices the dog
B. The car will stop 30m after hitting the dog
C. The car will stop 20m in front of the dog
D. The driver will quickly reverse the car
E. The driver will have to find a banking angle to ensure traction.


cc Polymath, Pr0ton, Dr. Host, Dr. Sage, Mikechibuzor, thaotech, AleXis0r, francistony, MedwhiteO, Bhuumhite, Funkysilver etc


*******UNA GOOD MORNING*******
1• P = A•P + Pwater
Pk = 15 + 120 = 135m of Water
Pc = 15 + 180 = 195m of Water
Thus: Pc/Pk = 1.44

2• A•P Read By Pilot = 83cm of Hg= 0.83m of Hg
A•P = 0.83 x 15000 (Conv. R•D of mecury i.e 15 to Density in Kg/m3) x 10= 124500N/m2

Thus: Distance between the Top Atmospheric Level and Aircraft
1.5(R•D of Air in Kg/m3) x 10 x h=124500

H = 8300m


A•P Read by Ground Observer = 94cm of Hg = 0.94m of Hg
A•P = 0.94 x 15000 x 10 = 141000N/m2

Thus: Distance between the Top Atmospheric Level and Ground Observer
1.5(R•D of Air in Kg/m3) x 10 x h=141000

H= 9400m

Height of The Aircraft Above Ground Level = (9400 - 8300)m
=1100m

3• A•P = 0.85m of Hg = 0.85 x 18.5 x 1000 ( R•D of Hg in Kg/m3) x 10
= 157250N/m2

Thus: Height of Water Column =
1000 x 10 x h = 157250

H=15.725m

4• If you really meant Distance = 13km but if it is displacement

D^2=25+64-2(5)(8 )Cos120
=25+64-2(5)(8 )(-Cos60)
=25+64+40
=128
D = 11.3 km

5• Magnitude Of Displacement = 13km
Distance Travelled = 17km
Their Diff. = 4km

6• C
7• B
8•A. 9• C
10• B
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 5:39am On Jun 01, 2015
Well, well, what can I possibly say, I have been so impressed by you guys from Mikechibuzor to thaotech to enry54...your attempts are so worthy of an applause.

Basically, you chewed those questions...I want us to go over them again, so you can mark yourself.

1. There is a formula

→Pabs= Ap + h
For the first Submarine, The knight
→Pabs= 15+120= 135

For the second Submarine, The cascade
→Pabs= 15+180= 195

The ratio of the cascade to the Knight is given as Pc/Pk

Therefore Pc/Pk= 195/135
→Pc/Pk= 1.44

2. enry54 had it solved well, but I'd have to go through another route.

The said aircraft(Aneroid barometer) is obviously in the air and the observer is with a barometer(Mecury) on the ground.

One can infer from that above that

AIR= hρg
AIR= h x 0.0015 x 10
AIR= 0.015h

ON THE GROUND= (h2-h1)ρg
ON THE GROUND=( 94-83) x15 x10
ON THE GROUND= 1650

Therefore,
→0.015h=1650
→h= 110,000cm
→h=110m

3. h1ρ1g=h2ρ2g
obviosly, we are to look for h2

→ h2=h1ρ1g/ρ2g
when g cancels out

→h2=h1 x ρ1/ ρ2.......

recall from question that relative density/ Specific gravity of Mecury was given as 18.5

→ρ1/ρ2=18.5

Invariably,
→h2= 0.85 x 18.85
→h2=15.73m

4. I want to draw an attention here,
when using the cosine formula, one thing is considered, the Sign.

The formula goes with a minus when Θ>90°, and with a plus when Θ<90°.

I am not solving that question, you were asked to find the displacement although the ans is 11.36km when done correctly it's not the total distance they are asking after.

I guess you should be able to do justice now.

5. No need to touch that...you all smashed it

6. D, cross check with S=ut+ 1/2gt^2

7. B

8. Thanks for the explanation at thaotech, that's just it. But the examiners could be silly and say the ball bounced 3times, 4times etc.

The thing to do is, solve normally and multiply your final answer by the number of times the ball bounced.

9. C, why C abi....come along.

X and Y cover the same distance,

So,

Sx=Sy

but it was from rest then

Uxt + 1/2axtx^2= Uyty + 1/2ayty^2

0+1/2axtx^2=0+1/2ayty^2

also, "the acceleration of X is twice that of Y"

ax=2ay
Substituting the relationship,

1/2• 2ay•tx^2=1/2•ay•ty^2

when 1/2 cancel out and ay follow suit then,

the ratio becomes,

2tx^2=ty^2

tx^2/ty^2=1/2

tx/ty= 1/√2

10. I can see a couple of ways in which this question was dealt with...you can also go about it this way

Workdone= Change in K.E

F x D= 1/2mu^2-1/2mv^2

F x D= 1/2m(U^2-V^2)

D= [1/2m(U^2-V^2)]/F

when done correctly, D=80m(the vehicle actually did hit the poor dog which was 50m from the car's initial position).

After hitting the dog, the driver stopped 80m-50m=30m after hitting the dog.

It's not something terrible afterwards, the dog is now in Dog Heaven...lol!

And btw, banking angle is something else oh, banking angle is that angle which a car, bicycle, motorcycle, turns through when negotiating a bend on a sloppy road or surface...it has a formula, but not needed in O-levels.

Thanks for staying tuned!
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 7:39am On Jun 01, 2015
horpeyemmi66:
Pr0ton(I guess I got it now), thaotech, MedwhiteO and Tholuwaniey(nice meeting you!)

You guys had almost thesame scores...MedwhiteO took the lead nevertheless!

I want you to confirm the following numbers well, numbers 2,8 and 9 that was where you guys made mistakes.

Anyways I salute you all...kudos!





Bless you all...

PS: @Pr0ton, I referred to you as the baba Isale group(god father of the group), cause during my debut in here, you first caught my attention; Very amiable.

Oh oh-kay. I missed your Physics questions though.
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 7:42am On Jun 01, 2015
Pr0ton:


Oh oh-kay. I missed your Physics questions though.
You can still make do with what you met...I have got to grab bread with Jam spread.
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 7:58am On Jun 01, 2015
horpeyemmi66:

You can still make do with what you met

Yh, I will

....I have got to grab bread with Jam spread.

That alone? How about some coffee?

Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 8:04am On Jun 01, 2015
Pr0ton:


Yh, I will



That alone? How about some coffee?


I am not a fan of Caffeine...I'd stick with Horlicks
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 8:08am On Jun 01, 2015
horpeyemmi66:

I am not a fan of Caffeine...I'd stick with Horlicks

Horlicks undecided
Ok.
Re: University Of Ibadan 2015/16 Applicants by Greatwonders(m): 11:34am On Jun 01, 2015
Pls is dere any tin yet abt d post utme[color=#006600][/color]
Re: University Of Ibadan 2015/16 Applicants by Charlesschools: 1:42pm On Jun 01, 2015
Want to gain admission into University of Ibadan and your score is low for the course you applied for and also you need help processing your admission and accomodation to stay during post utme examination, contact Mr Charles for assistance: 08160433274 or 08153049342...
Re: University Of Ibadan 2015/16 Applicants by Nobody: 8:27pm On Jun 01, 2015
Fembleez1:

I wish I am - as the hype! cry


If I am able to
Re: University Of Ibadan 2015/16 Applicants by Fembleez1(m): 10:22pm On Jun 01, 2015
shoyemiayodeji:




If I am able to

Able to what?
Re: University Of Ibadan 2015/16 Applicants by tundeks: 1:26am On Jun 02, 2015
P0lyMath:
To my questions, u guys got two out of the three.

No2. Let them meet at 'x' measured from the top of the tower. Let O' be the ball from the top and O'' be the ball from the foot of the tower.
Distance travelled by O'=x
Distance travelled by O"=50-x
X=gt²/2=4.9t².........…………1
50-x=ut-gt²/2
50-x=25t-4.9t²
X=50-25t+4.9t²…………………2
Equating 1 and 2
50-25t+4.9t²=4.9t²
25t=50
t=2s
But x=4.9t²
X=4.9(2²)
X=19.6m

Therefore, their time of crossing each other is after 2 seconds and they crossed each other at 19.6m measured from the tower top.


FIGHTING
is this how UI used to set post ume question cos this ur questions can make candidates to run away frm UI
Re: University Of Ibadan 2015/16 Applicants by horpeyemmi66(m): 3:30am On Jun 02, 2015
tundeks:
is this how UI used to set post ume question cos this ur questions can make candidates to run away frm UI
All these struggles you see is all a contingency plan...If the questions come soft, bless God! And if not the Contingency plan will help us contend favourably...

We are preparing to face anything...hash tag saynotomediocrity!
Re: University Of Ibadan 2015/16 Applicants by Luukasz(m): 5:08am On Jun 02, 2015
Guys good Morning.Pls i want to know the mode of U.I's Post-Utme,will it be CBT or PAPER TYPE & again will Current Affairs be among the list of Subjects.
Re: University Of Ibadan 2015/16 Applicants by Nobody: 8:27am On Jun 02, 2015
Fembleez1:

Able to what?
Quote miss ROAD.
Re: University Of Ibadan 2015/16 Applicants by Aktripple7(m): 8:52am On Jun 02, 2015
Gd Mrning here...... Abeg make the whatsapp admin add me on this number: 09024190321. Thanks.
Re: University Of Ibadan 2015/16 Applicants by John1305(m): 1:43pm On Jun 02, 2015
Keep it up guys, I'm following this thread for someone who also intends to get into UI, I must say from the level of preparedness I see here that, the Majority of you must make it, as it were I plead that any information about the procedure for the PUTME should not be kept.
This is funny, I'm am actually a student of UI, but I also know that your wires are higher than ours when it comes to getting info, when in ui, u'll understand. (Books books n books). All we pride in is the experience we have gotten.
With that said, I pray the Lord helps us. All.

Are there DE applicants here? I have some experiences that might answer ur questions.
Re: University Of Ibadan 2015/16 Applicants by niffiegee: 6:35pm On Jun 02, 2015
Hi I'm ola, i'd like to join the Whatsapp group.my number 08133072404
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:31pm On Jun 02, 2015
If all of una wey dey cry to join that rubbish whatsapp group no shutup i go hide all of una phone numbers with bans inclusive.


Even with earlier warning una no wan hear word shebi?
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:35pm On Jun 02, 2015
horpeyemmi66:

Sey you miss road??angry
Dat guy will either serve his ban or create a new moniker.


Ogunsinamayowa, Kingjay18, T33jay18 i'm back o make we run things here.
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:38pm On Jun 02, 2015
shoyemiayodeji:


Quote miss ROAD.
I dey hail sir! Saw you on last years thread, are re-writting or just passing by?
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:41pm On Jun 02, 2015
Luukasz:
Guys good Morning.Pls i want to know the mode of U.I's Post-Utme,will it be CBT or PAPER TYPE & again will Current Affairs be among the list of Subjects.
The subjects you filled and wrote in jamb. What course and jambscore?
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:46pm On Jun 02, 2015
Horpeyemmi66 you are a very wierd person, are you a nightwalker or daywalker or you don't sleep at all?


DrHost=I sight you!
Re: University Of Ibadan 2015/16 Applicants by Nobody: 10:47pm On Jun 02, 2015
cassyrooy:
I dey hail sir! Saw you on last years thread, are re-writting or just passing by?


Just passing by Bro......A uite already wink
Re: University Of Ibadan 2015/16 Applicants by DrHost: 10:51pm On Jun 02, 2015
John1305:
Keep it up guys, I'm following this thread for someone who also intends to get into UI, I must say from the level of preparedness I see here that, the Majority of you must make it, as it were I plead that any information about the procedure for the PUTME should not be kept.
This is funny, I'm am actually a student of UI, but I also know that your wires are higher than ours when it comes to getting info, when in ui, u'll understand. (Books books n books). All we pride in is the experience we have gotten.
With that said, I pray the Lord helps us. All.

Are there DE applicants here? I have some experiences that might answer ur questions.
Pls what the chances of someone applying with poly ibadan nd upper class in civil engineering
Re: University Of Ibadan 2015/16 Applicants by cassyrooy(m): 10:53pm On Jun 02, 2015
shoyemiayodeji:



Just passing by Bro......A uite already wink
What course bro, cos i'm kinda sorting for clarity on literature. If you did lastyear what were the topics and text you read last year?


But if you didn't= geedot helpout.

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