THE OFFCICAL CAT2008: DI - lOGICAL REASOING

[federaloshima]

Originally Posted by sareet

Question:
You are in a completely dark room with two tables. One of the tables (call it A) has a large number of coins lying on it, exactly 100 of which are showing heads. The second table (call it B) has nothing on it. As you cannot see, you can perform only two operations: you can shift a coin from A to B without flipping the coin over OR you can you can shift a coin from A to B after flipping it over. How many minimum operations you need to perform in order to ensure that there are equal numbers of heads on both the tables?

dude its 100.

[deep@k]

You are in a completely dark room with two tables. One of the tables (call it A) has a large number of coins lying on it, exactly 100 of which are showing heads. The second table (call it B) has nothing on it. As you cannot see, you can perform only two operations: you can shift a coin from A to B without flipping the coin over OR you can you can shift a coin from A to B after flipping it over. How many minimum operations you need to perform in order to ensure that there are equal numbers of heads on both the tables?



is the answer is 100....

[vishrock]

Originally Posted by deep@k

You are in a completely dark room with two tables. One of the tables (call it A) has a large number of coins lying on it, exactly 100 of which are showing heads. The second table (call it B) has nothing on it. As you cannot see, you can perform only two operations: you can shift a coin from A to B without flipping the coin over OR you can you can shift a coin from A to B after flipping it over. How many minimum operations you need to perform in order to ensure that there are equal numbers of heads on both the tables?



is the answer is 100....

Can u people explain the approach?

[r_bajla]

plz...somebody give the approach

[mahi101987]

i am joining this thread today only

i hope to solve the problem posted by the PUYS

thanks a ton for this thread

[the_hate]

Originally Posted by deep@k

You are in a completely dark room with two tables. One of the tables (call it A) has a large number of coins lying on it, exactly 100 of which are showing heads. The second table (call it B) has nothing on it. As you cannot see, you can perform only two operations: you can shift a coin from A to B without flipping the coin over OR you can you can shift a coin from A to B after flipping it over. How many minimum operations you need to perform in order to ensure that there are equal numbers of heads on both the tables?



is the answer is 100....

i think ans is 50....tk ny 50 coins...flip all of them...and keep on table B

approach->
suppose 50 heads(h) and 0 tail(t) on A

now when u tk all 50 nd flip...result wud be...
0 H on A nd 0 H on B(since all 50 will become T on flipping)

now lets say thr r 1000 T and 50 H on A
if we tk 50 coins from A thr can be many possiblities->
if all 50 are H-> same as above...
if all 50 are T-> A will have 50 H and B too will have 50 H
if say 20T and 30 H-> 20 H left on A and on flipping 20 H will be on B

u can solve this using x T on A and y out of those selcted in total 50...
i like using observation...

i hope u get my point
---------------------
editops! i thought A has 50 h on it!! yeah the ans is 100....

[abhinav321]

Hi all,
The answer to this prob is 100
my approach....
suppose there were x tails on table A, along with 100 Heads

now on taking each coin from table A to table B flip the coin....
so now suppose I take a coins from table A to table B out of which b were heads on table A.
so a-b were tails on table A--->these coins will be heads on table B as i am flipping every coin.
so now equating heads on both tables 100 - b(no of heads taken from table A to B) = a-b
therefore a=100

[mohitjha]

the answer should be 100.
pick 100 random coins from A, flip it and put it on B.
case1. u picked all heads frm A then each table has 0 heads.
case2. u picked all tails frm A then each table has 100 heads
case3. u picked x heads and 100-x tails from A then each table has 100 - x heads.

[Madrock]

dude, as far as i see this, it seems u r wrong.
may b m doing something wrong but u gotta atleast support ur answer with explanation.
lez see q.78 fr eg.

max no. of eggs tht can be delivered by B & C, without disturbing the condition of A & D's total to be a multiple of 4 and 6 factors is 64. which can be delivered in 6 ways. by B & C.
and remaining 12 can be delivered in 4 ways by A & D. so total ways 6X4 = 24.
so option 4 is right.
may be i am rong in my approach. plz help me out in d way u have reached the conclusion.

Originally Posted by implex

78 is wrong ...
it should be 2
79 is wrong as well
its 1
80 is wrong again
it will be 1

and sadly 81 is wrong again
it will be 4 ..

Originally Posted by implex

here is a new question then :

[Madrock]

There can be another alternate key which i found out, m i right ?

Backstroke: Adam, Carl, Doug, Eric,Brad
Breaststroke: Carl, Brad, Adam, Eric, Doug
Butterfly: Brad, Adam, Carl, Doug, Eric
Freestyle: Eric, Brad, Adam, Doug, Carl

Originally Posted by naresh007

Hey…Puy’s, i’m sorry for delay….as I was busy I can’t post answers for the Questions 4 & 5..
In fact, I hardly seen any answers reply for them!…ya, I understand they are bit hard…

Anyway this is Answer key for the Question.No: 4


Backstroke: Adam, Carl, Doug, Brad, Eric
Breaststroke: Doug, Brad, Adam, Eric, Carl
Butterfly: Eric, Adam, Carl, Brad, Doug
Freestyle: Eric, Adam, Doug, Carl, Brad

Hint: Here clues 5,6 & 14 are very imp please observe it carefully..you will get..clincher..

Very soon…I post the key for the encrypted kind of logic walaaa...i.e. Q.No:5.. (Omg!... its tough)…, till then keep Solving….people..

Quote:
Originally Posted by naresh007 (THE OFFCICAL CAT2008: DI - lOGICAL REASOING)
Hello..Puy's..

Question No- 4 : Order
Five swimmers (Adam, Brad, Carl, Doug, and Eric) have been preparing for the Olympics. It is now time for the swimming time trials. The five swimmers each compete in the four different strokes (backstroke, breaststroke, butterfly, and freestyle). The top three finishers in each event will qualify for the Olympic swim team in that stroke. Using the following clues, determine the order of finish in each of the four strokes.

1) Only one contestant qualified in all four strokes.
2) No contestant finished last in more than one event.
3) Adam finished better in the backstroke than he did in the butterfly.
4) Brad finished better than Doug in the butterfly.
5) Adam finished just behind Brad and just ahead of Eric in the breaststroke.
6) Doug finished just ahead of Carl in the freestyle.
7) Neither Brad nor Eric finished third in any event.
Eric's finish in the backstroke was the same as Doug's in the butterfly.
9) Doug only finished in the same position in the backstroke and the freestyle.
10) Carl finished in a different position in each event.
11) Brad finished only two events in the same position.
12) The contestant who finished second in the butterfly beat Doug in the freestyle.
13) The contestant who finished first in the freestyle did not qualify in the backstroke.
14) The contestant who finished fifth in the backstroke did not finish third in the butterfly.
15) No contestant finished in the same position in both the breaststroke and the butterfly.