TIE

OAT

Semaphores

for a good quality work.

You can get a uneducated person for 1/4 to 1/10 of salary
why don't you hire him. Answer is quality, understanding,
future value addition etc.

2520.

0 is the smallet digit 0 is divisible

1024 which is nothing but 10!(factorial)

1050. I multiplied all the prime numbers from 1 - 10 and 
then found its(210) smallest 4 digit multiple which is 1050

sorry It's 2520 only... I multiplied wrongly...

10*9*7*4

prime number multiplicastion is not useful...

1050 is not divisible by 8,9

1080 is the answer

1080 is not divisible by 7...

1*2*3*5*7*4*3=2520

Yes 2520 is right answer let me elaborate bit more on how to
get to this answer. 

required number = 9 * 8 * 5* 7

5 and 7 are in list since they are prime number and they
have no other multiple in 1 to 10. 3 is also a prime number
but is not used in this multiplication since we already have
used 3 (9 is divisible by 3).

Since its an even number and divisible by 5 we can neglect 10.

Since its even and divisible by 3 we can neglect 6 too.

Similarly divisible by 8 means divisible by 4 and 2.


Let me ask another question. What is the smallest number
which is divisible by all number from 1 to 9 (note I left
out 10)

Let me ask another question. What is the smallest number
which is divisible by all number from 1 to 9 (note I left
out 10)

Answer would be the same!!
2520

question has some doubt. number of flash will be same. it
will not be change. Both will flash together after 40sec

6 minutes

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.

577500 Sqm

32986.72

10500rupees

33000

Rate should be Rs 150 per meter (not sqm).

(Rs 150 / m) * circumference
= (Rs 150 / m) * pi * diameter
= (Rs 150 / m) * (22/7) * 70 m
= Rs 150 * 22 * 10
= Rs 33000

33000

3297.00

IT SHOULD BE 
RS.32970
=3.14*70*150
=32970

Just look into the problem rate is per Squar meter...so

Area of circle is pie*radius ki power 2
R is 35m
area=3.14*35*35

cost=150*22/7*35*35
 =calculate it

Seems like a trick in the question.
Rate is 150 per sq meters. Which can be interpreted as "150
Rs to fence a square whose area is 1 sq meter". Perimeter of
a square whose area is 1 sq meter is 4 meters (each side of
one meters).

So 4 meters => 150 Rs
Cost per meter => 75/2
Circumference of field => 2*22/7*35
Total cost for fencing the field => 2*(22/7)*35*(75/2) 

= Rs 8250

26500

=30*20/100
=6
24 apple * 1 = 24
06 apple * 2 = 12
             ------
             = 36
             ------
Amount(capital) = 30
profit 20%      = 06
 
He should sell in 1 Rs = 24 apples

25
explanation:
let s.p of 'x' apples=100p
20%profit
also,SP = 1.20 OF CP
THEREFORE,CP of 'x' apples = 100/1.20=83.3333p
COST OF 1 APPLE=3.33p(30 apples in 100p)
no. of apples bought in 83.33p=(83.33/3.33)=25

25

25

24 apples.

Number of apples for 20% profit= Total number of apples-20%
of total apples=30-(20/100*30)=30-6=24.

24.(logical xpln)
assume 100 rs & 30 apples.i.e 3.33/ apple.
for 20 % profit i.e 20 rs, each apple 2 b sold at 4 rs.
but he has to sell it in rs 100 therefore no of apples shud
b reduced.
to earn 20 rs =3.33*(6) approx.
therefore 6 less apples thus 24 apples.

25 Apples.

Explanation :
he got 30 Apples for Rs 1 or 100 paisa.
Profit is 25% on CP, so SP is 120 paisa.

so now, he will sell 30 Apples for 120 Paisa means SP of 1 
Apple is 120/30 = 4 paisa.

But he has to sell foe only 100 paisa, hence he has to sell
100/4= 25 Apples.

30 apples = 100paise
he want 20% profit from 100 paise
20% profit in 100 paise is 20 paise
so now he planned to sell 30 apples for 100+20=120 paise
now the rate of 1 apple =120/30=4paise
for 100 paise =100/4=25.

23apples

24 apples

25 apples

30apls =100p
1apls =100/30p
x apls =100p
1apls =100/xp
profit=(100/x-100/30)p
profit%=profit*100/puchase price
20=(100/x-100/30)*100/(100/30)

we will get x=25.
so he will sell 25 apls in 1 Rs.

cost of one apple is 1\30 i.e.0.033
now he wants to earns 20% profit
             cost of one apple now .033*20\100
                   =.033 ( cost)+.0066(profit per apple ) 
=.0396
selling cost of one apple = .0396
 
no. of apples sold = 1\.0396 = 25.25

so the ans. is 25