Core Banking is normally defined as the business conducted 
by a banking institution with its retail and small business 
customers. Many banks treat the retail customers as their 
core banking customers, and have a separate line of 
business to manage small businesses. Larger businesses are 
managed via the Corporate Banking division of the 
institution.Core banking basically is depositing and 
lending of money.

modulation means analog to digital signal
demodulation means digital to analog signal

CAN ANYBODY ANSWER THIS QUESTION ...

It is also a A.C SIGNAL so output goes to a square pulse

if the freequency is high we will get same square wave with 
modified magnitude but if the freqcency is low impulse will 
appiare

if in case u are applying dc signal o/p is triangular wave

IN SOME CASES THE TRANSFORMER IS DESIGNED TO APPLY DC 
PULSE( NOT CONTINUOUS DC SIGNAL ) TO ITS PRIMARY WINDING 
AND THE TRANSFORMER IS CALLED AS "PULSE TRANSFORMER". 
  IF DC VOLTAGE IS GIVEN TO WDG; AS WDS IS AS GOOD AS 
INDUCTOR HENCE CURRENT I=(1/L)*INTEGRAL(Vdc)dt, WHICH WILL 
CONTINUOUSLY INCREASE THE CURRENT FLOWING THROUGH THE 
WINDING. HENCE THE WIDTH OF THE PULSE SHOULD BE SMALL 
OTHERWISE THE TRANSFORMER WINDING WILL BURN. 
  IN THE PULSE TRANSFORMERS THE RESISTANCE OF THE WINDING 
SHOULD BE KEPT MINIMUM TO LINIT THE I^2*R LOSSES AND LIMIT 
THE TEMP OF THE WINDING BEFORE INSULATION FAILURE.

In Transformer secondary the emf induced depends on the 
rate of change of flux,If square pulse or pulsating DC is 
given thn output will be triangular wave(due to inductnce 
of secondary)

what will be width of pulse and min voltage needed for
transformation.

It depends on the width of pulse,otherwise transformer will
go in saturation.also at secondary the pulse will not be
simple pulse but it will be rather impulse.

Induced voltage=4.44*f*N*A*B = 4*form factor*f*N*A*B
f=freq
N=no of primary turns
A=area of wdg
B=max flux density

For sinusoidal input, form factor= 1.1 , its a peaky wave 
form

But for square ac input signal, form factor is less than 
1.1. (as its non peaky)

So other factors remaining same, induced voltage decreases 
for square input ac signal.

yes

9

YES

Nine

i will write the kvb exam .so i have a model paper

i've written the kvb clerical exam.so iwant a model paper.

I want KVB aptitude questions. plz send to this mail

hi i apply for po in kvb bank.. please send model question
paper for kvb for post of po.. this is my email id..
dhinathayalan@gmail.com

    kindly forward kvb bank question paper...

hi i hav applied for po post in kvb plz send me the model 
paper for tat

i need po model question papper

hi i have apply for kvb po exam,so i'm not having 
materials,so if any previous exam papers having send this 
mail:
 kamalaprakash@yahoo.co.in

thans&regards

Dear Friends, Brothers & sisters, Plz send KVB PO Model 
questions to my mail id. 
VERY VERY URGENT.

answer is 
(2*2)+1=5
(4*4)+1=17
(6*6)+1=37
(8*8)+1=65
(10*10)+1=101
(11*11)+1=122
(12*12)+1=145

If we consider the set of odd numbers
1,3,5,7,9,11... as x1,x2,x3,x4,x5,x6....

and the expression(Xi*Xi+1)+2

(1*3)+2=5
(3*5)+2=17
(5*7)+2=37
(7*9)+2=65
(9*11)+2=101

If the question was like this -> 5,17,37,65,101.... i.e. 65 
instead of 55 then the number next in the series would be 
(11*13)+2 = 143

125

ans is 145

6 races

k...if timing devices ver not yet invented :P


1st hold 5 prelim races.....

then


1.hold a race among all the 1st place horses[of the prelims]
 (let it be race x)......this will determine the fastest horse..

2.hold a race among all the 2nd place horses[of the prelims]
(let it be race y)...... 

3.hold a race among all the 3rd place horses[of the prelims]
(let it be race z)........

4.then hold a race among the 2nd & 3rd placed horses of race
x,1st & 2nd of race y  and 1st place of race z..........this
will determine the 2nd&3rd fastest horses.....

  .
.   .   5+4=9..........9 



v need a min of 9 races 2 determine the top 3
horses(according 2 me)

u c ..........v humans have invented sumtin called a stop
watch...... 


conduct 5 races......note the timings of the horses....&
select the top 3 :P 


ps : it is no mentioned anywhere in the Q tat v r not
allowed 2 use timing devices

I m sorry for the wrong answer I posted earlier. So the 
answer is definitely not 11. At the same time Anjali, it is 
not even 6. It would be wrong to assume that the horses 
which came second and third in the first set of five races 
cannot come second and third overall. So let me provide the 
correct answer now. 

We conduct the 5 races. Choose the top 3 from all the 5 
races. Now conduct a race of the top horses of all the 
races. Choose the top 3 of the last race. They could 
probably be the top 3 of the horses. We discard the horses 
which came 4th and 5th along with all the horses of their 
group. Let me explain it pictorially.
The top represents the standing of the 6th race. The race 
among the top horses of each group. Now we can discard all 
the horses which came 4th and 5th in their respective 
groups. So we are left with the top 3 horses of each 
group.Now consider the horses which came 4th and 5th in the 
6th race. They cannot fall in the top 3. Nor can any of 
their group members. So we get rid of all the remaining 
horses of group 4 and group 5. So we are left with 9 horses 
to choose from.
Now consider the horse which came 3rd in the 6th race and 
its group memebers. We call it group 3. Now the 2nd and 3rd 
horses of group 3 cannot come even 3rd overall. Hence they 
can be discarded. SO we consider only the horse which came 
first(1 selected for the final race). Consider group 2. We 
can discard the horse which  came 3rd in the group. And we 
select the horses which came 1st and 2nd( 1+2 selected for 
the final race). Similarly we consider all the top three 
horses from group 1. But we can safely ignore the horse 
which came first overall.( 1+2+2 selected for the final 
race). So we have 5 horses to choose the best two from. We 
already have the horse which came 1st overall. No need to 
put that in the race. So the minimum number of races 
required is 7 and 11 or 6.   

1 2 3 4 5

1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5

MINIMUM 6 RACES REQUIRED

conduct 5 races, in each race 5 horeses

out of 5 races to select 1 horse from each race (total 5 
horses selected)

conduct 6 race with selected 5 horeses
out of that select top 3 horses

6

Could you please explain me how you got 6 as your answer. I 
could do it in a minimum of 11 races. There has to be 
atleast 5 races to select 15 horses. The 15 horses then 
compete among themselves and best 9 horses are chosen. The 
9 horses then run in groups of 5 and 4. From the group of 
5, three horses are selected. The horse which comes third 
is made to run in the next group as well. We again choose 
three horses from the group. So that makes it a group of 5 
fastest horses. Then run the last race by which we can 
determine the three fastest horses.

If you want best 3, then we need 11 races as Alok Chandra 
said. 
But ideally we no need to arrang horses for races. Race any 
5 horses in each race and select first 3. Repeat this 11 
times, you will get best 3.
This is for the worst case.

We will only need 7 races to get the best three out of 25.

Let us Name horses as H1, H2, H3...H25. Now,

Race 1 - H1 - H5
Race 2 - H6 - H10
Race 3 - H11 - H15
Race 4 - H16 - H20
Race 5 - H21 - H21

Let us now assume 

First Rank of Race 1  - W11 
Second Rank of Race 1 - W12
.
.
First Rank of Race 2  - W21
.
Third Rank of Race 2  - W23
.
.
Fifth Rank Of Race 5  - W55 

Race 6 - W11 , W21 , W31, W41, W51 ( toppers from first 
five races)

Race 7 - 2nd from Race 6, 3rd from Race 6, X, Y , Z where

X= 2nd winner from the team of winner of race 6
Y= 3rd winner from the team of winner of race 6
Z= 2nd winner from the team of 2nd winner of race 6

So the three best horses will be , Winner of Race 6 and 
Winner and Second Winner of Race 7.

13 races in total.

It might also happen that all best 3 horses are running in 
one of the 5 races all together. Therefore we will be 
selecting top 3 for every race.

After 5 races-->we will have top 15 horses
+ After 3 races-->we will have top 9 horses
+ after 2 races--> we will have top 6 horses
+ after 2 races--> we will have top 4 horses
+ after 1 race--> top 3 horses
Total races = 13

if we note the time its only 5 race

if we don t have the timer its obviously six race

atleast 7 races are required.

5 races will select top 5 horses.6th race will help out in
finding 4 best horses. 7th race will give us 3 best horses.

a re scientist logo bus six races would be required......

using stop watch 5race is enough
(they hav not abandoned use of count down timer...dont
assume anything untill stated )

without timer

13  surabh has ansered it already

Hi everybody!

Correct answer is 7.

First conduct 5 races, top horse from each group selected.
Now conduct 1 race among these five, select top three from
these. Now discard the groups of the horses who stood 3rd,
4th and 5th in last race because no one of them can make it
to top three. Now take 2nd and 3rd horse from first group
(the group to which the first ranker belongs) and 2nd horse
from second group (the group to which 2nd ranker belongs).
Now we have total of six horses but first rank is already
decided, so we are left with five horses from these, we have
to select 2. Now conduct 1 race of these five horses and
select top 2.

Thanks for reading this long answer!

Hello Everyone!

My answer is 6

Race1: Conduct race for 5 horse(H1 to H5), The Winner is W1(say)

Race2: Conduct race for 5 horses(incuding Winner from Race1)
i.e race between W1 and H6:H9
The Winner of the Race 2  is W2 (Say)

Race 3: Conduct race for 5 horses(including winner from
Race2) i.e. Race between W2 and H10 to H13
The Winner of the Race 3  is W3 (Say)

Race 4: Conduct race for 5 horses(including winner from
Race3) i.e. Race between W3 and H14 to H17
The Winner of the Race 4  is W4 (Say)

Race 5: Conduct race for 5 horses(including winner from
Race4) i.e. Race between W4 and H18 to H21
The Winner of the Race 5  is W5 (Say)

Race 6: Conduct race for 5 horses(including winner from
Race5) i.e. Race between W5 and H22 to H25

The first coming 3 horses will be considered as best.

I think the above answer is correct... If you think so.
Please drop a mail to dinesh_ram7@hotmail.com

I calculated 8 races. Here is the logic

1. 5 races, R1 to R5 s.t. each race has 5 horses and every
horse appears in exactly one of R1 to R5. Discard 4th and
5th positioned horses. Note that 3rd positioned horses
cannot be discarded as 3rd horse of Ri may be faster than
toppers of other 4 races.

2. R6 between 1st ranked horses of races R1 to R5. Topmost
ranked horse is the fastest horse.

3. R7 between
     - 2nd and 3rd toppers of R6
     - 2nd  position holders of those three races of R1 to
R5 whose topmost horses were ranked in top 3 in R6.
   Winner of R7 is the 2nd fastest horse.

4. R8 between
     - All horses except winner from R7
     - 3rd position holder of the race out of R1 to R5 whose
topmost horse won R6.
     Fastest horse will be 3rd fastest horse.

It should be 13
(5+3+2+2+1)- please see descriptions.
25= 5 5 5 5 5 - 5 races
select top three from each as it may possible three of any
group are best three.
3+3+3+3+3=15
now
15= 5 5 5 - 3 races
again select top three from each
3+3+3=9
9= 5 4 - 2 races
6= 5 1 - 2 races
4= 4   - 1 race
that's y (5+3+2+2+1)

An explanation as to why the answer is 7, assuming no method
to time the races:

Divide the 25 horses into groups of 5 labeled A to E. Hold 5
races, one for each group and label the horses A1, A2, etc.,
where the number designates what place the horse finished.
Hold a 6th race out of the five winners of each group, and
arbitrarily pick A1 as the winner, B1 as second place, etc.
After the 6th race, you'll have the following tree for
fastest horse:

  A1
  | \
  A2 B1 
  |  | \
  A3 B2 C1
  |  |  | \
  A4 B3 C2 D1
       .
       .
       .    

As you can see, A1 is the fastest horse and is faster than
all the horses in the other groups since he beat their
winners, and likewise, B1 is faster than the horses in
groups C to E since he beat their winners. But we don't know
if B1 is faster than the rest of the horses in group A,
since they never raced against each other, and likewise, we
don't know if C1 is faster than the remaining horses in
group B, since they didn't race each other, etc. To
determine the 2nd fastest horse, let's hold a hypothetical
7th race between A2 and B1. There are two possible outcomes
that result in the following trees:

     A1		    A1
     |		    |
     B1		    A2
    / | \          / |
   A2 B2 C1       A3 B1
   |  |  |        |  | \
   A3 B3 C2       A4 B2 C1

In the first outcome (left tree), B1 has won and is the
second fastest horse. Now we need to find out which horse is
3rd fastest in an 8th race between A2, B2, and C1, since
they have never raced each other.

In the second outcome (right tree), A2 has won and is the
second fastest. Therefore, an 8th race needs to happen
between A3 and B1 to determine the 3rd fastest horse.

If you consider which horses run in the hypothetical 7th and
8th races (horses A2, A3, B1, B2, and C1), you'll notice
that only 5 horses need to race, so all 5 horses can
actually run in the 7th race to determine the 2nd and 3rd
fastest horses.

So, is it possible to find the top 3 fastest horses in only
6 races? Yes, there are certain situations where this is
possible. Take the scenario where the winner always gets to
run in the next race: the first race has 5 horses (Group F)
that haven't raced; the winner goes on to race 4 more horses
that haven't raced (Group E), and that winner gets to race 4
more (Group D), etc. until the 6th and last race (Group A).

As long as the winner from the 5th race does not finish in
1st or 2nd, then you can determine the 3 fastest horses;
otherwise, you won't be able to determine the 2nd and/or 3rd
fastest horses, since the horses in group A have not raced
any of the other horses.

Race	1(F) 	2(E)	3(D)	4(C)	5(B)	6(A)
----------------------------------------------------
1st	F1-.	E1----->E1-._	C1.	B1-.	A1
2nd	F2  \	E2	D1   `->E1 \	B2  \	A2
3rd	F3   `->F1	D2	C2  \	B3   `->B1
.	F4	E3	D3	C3   `->C1	A3
.	F5	E4	D4	C4   	B4	A4

Thus, to guarantee you can determine the fastest 3 horses,
you need 7 races.

It is seven.....
It is already given by some one .....
But I will try to make it more clear.....

First conduct 5 races with the given 25 horses......
so we will get :
   Races  :      1  2  3  4  5
   First  :      1  1  1  1  1 
   Second :      2  2  2  2  2
   Third  :      3  3  3  3  3
   Fourth :      4  4  4  4  4
   Fifth  :      5  5  5  5  5

So we are sure that the fastest would be from the horses 
which stood first in each race........
that is from First: 1  1  1  1  1

So next we must conduct the race among all the horses which 
stood first in each race,i.e. race no 6:

Than we will get the fastest......

Next we have to get the second fastest and third fastest..

Now we have the options like this:

Now the second might be the second from the 6th race or the 
second from the earlier conducted race where the fastest 
was the first.....

Third might be the third from the 6th race or the third 
from the earlier conducted race where the fastest was the 
first or the second from the earlier conducted race where 
the second fastest from the 6th race was the first.....

So we have to find out the second and third from this group 
by conducting the race between:

1)Second from the 6th race
2)Second from the earlier conducted race where the fastest  
was the first
3)Third from the 6th race
4)Third from the earlier conducted race where the fastest 
was the first
5)Second from the earlier conducted race where the second 
fastest from the 6th race was the first

So after conducting this 7th race we will get the second 
fastest and third fastest....

Hence the result is 7 races.......

Hope u will understand this.....
if not u can mail me

My answer is 6.
Let we divide 25 horses into 5 groups,each having 5 horses 
each. If we conduct race for each group, we ll get 5 
leading horses. Let we conduct another race among them, so 
we can get best three horses..........

6 Races

First Five Races

3+3+3+3+3=15 (first three winners from each race)

6th, 7th & 8th Races are 1+1+1=3 (first winner from each 
race)

One confusion if 6 then five races for top five and then 
one to find top three but how could u decide that 
first ,second third of first race is not the top three as 
it may happen third of first race can fatser than first of 
rest races we have to consider top three from each race.

correct answer is 7... why do yiou people keep writing 
anything... atleast look at the top explainations....
answer#4 given by alok chandra is perfectly correct..
atleast have a look at it and then write something

After 5 races-->we will have top 15 horses
+ After 3 races-->we will have top 9 horses
+ after 2 races--> we will have top 6 horses

Conduct 1 race for any 5 horses and select the best three.

make the remaining one horse run with those three 
(after 1 race for 4 Horses --> top 3 horses)


Total races = 12

6
organise 5 races consisting of 5 horses each 
select the fastest horse of each race
then organise a race for the 5 selected horses
the winner amongst them will be the best one

My ans is 12. I will explain with a diagram

5 5 5 5 5 -> now 5 races each with 5 horses
| | | | |
3 3 3 3 3-> selected 3 toppers from each race  so tot 15  
            horses remaining
5 5 5-> now again 3 more races ie tot 8 races yet
| | |
3 3 3->    selected 3 toppers from each race  so tot 9  
            horses remaining 
5 4-> now again 2 more races ie tot 10 races yet
| |
3 3  selected 3 toppers from each race  so tot 6 
            horses remaining 
5  _> one more race tot 11 races
|
3 + 1

4 _> one more race tot 12 races
|
3 

So tot no of races 12.

But here we assume tht two horses never reach the
destination at the same time :-) if we consider that then no
of races will be much more.

my answer is 12

first best 15 horses from 5races
next bes 9 horses from 3 races
next best 6 horses from 2 races
here evry one made a mistake one race
is held because there cant be a race with one horse
and final race btn final 4

5+3+2+1+1=12

Hi All,
The Answer is 11.
25 horses can be diveded into 5 batch and we can select 
best 15 horses. 

Here we are looking for best 3 from 25. So in each batch 
the horse placed 4th and 5th place cannot beat the 1st 
three placed horses in their batch. So we can dis qualify 
them.

Note: race count=5

Now the 15 horses can be divided into 3 batches and can 
select best 9 among them.

Note: race count = 5 + 3 = 8

These 9 can be divided into two batches. 5 in 1st batch and 
4 in 2nd batch.

1st batch 5 horses result                
        1(Qualify for final)
        2(Qualify for final)
        3(include in 2nd batch for next race)
        4(disqualify)
        5(disqualify)

Note: race count = 5+3+1 = 9

2nd batch 4 horses + 1st batch 3rd place horse result.
        1(Qualify for final)
        2(Qualify for final)
        3(Qualify for final)
        4(disqualify)
        5(disqualify)

Note: race count = 5+3+1+1 = 10

Now there are 5 horses qualified for final.
Among these 5 we can select best 3 horses.

Note: Total race count = 5+3+1+1+1 = 11.

If any doubt in my explanation please reply..

6
IS THE RIGHT ANSWER BECOZ FIRST HELD 5 RECES and among them 
choose 5 horces which are top of the each races.and after 
that among that top 5 horces held 1 races.and choose top 
three.so total 5+1=6 races

6 races

7 required.....
none others right...

first five races,
    5   5   5   5   5
    |   |   |   |   |
    3   3   3   3   3     total horses remaining=15

take the top one from five races 
   now you can select the best horse by keeping one 
race.take second and third position horse from that race 
and neglect forth and fifth.
    keep two races for other two batches and select the top 
two horses.
[total race: 5+1+2, remaining horses: 2+2+2 and the best is 
selected]
     now aim is to select second and third.
     keep race for 5 horses and select top 2 and make it to 
race with the one remaining.now you can select the second 
and third best.
 thus total race=5+1(selecting the best)+[(2+2)->for 
selecting second and third] 
     TOTAL RACE: 10
(if you want to convey anything,you can mail me)

Smartwork is better than Hardwork where Smartwork only 
comes in if you have talent within you.

Hard work is the most importent for me.

I think if you are hard working person, then talent 
automatically comes in you. As you work hard for doing 
something nthere is no way that talent can't be achieved by 
you.

Hardworking donkey is better than the smart working horse.

so talent is more improtant

Smart work is that which is done by your brain and hard work
is done by your mind,so both work is important my life to
achieve sucess

i think both is important cause suppose if u r a hard 
worker u should have a talent otherwise if u have talent u 
must have a hardworker

in my views talent is mandatory as if we have talent in us 
then we can work hard to acheive our gaol....

Talented people achieve better than hard workers.A talented 
person completes any thing very sooner than the hard worker

talent

Talented person should work hard... without hard work your
talent won't come out.... so if you worked hard surely your
talent will make a print in this universe... this is my opinion

i think some people may not understand for the first 
attempt.
so if they work hard, obveosly they become  talent.
this is my opinion & experiance

talent is most important for me

working hard with immense talents will surely make our life 
colorful and there is no doubt tat the two things will 
bring laurels to our life

hard work is more important for me!...because one's talent is simply a waste without it! a very general example could be of a talented singer,who might not be that good at singing without hard work as he actually is.

annie frank