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Let's Bell The CAT Together
Hi Guys,
Since only 2 months are left so we all need more and more practice. Here we get only a single question for the whole day. So I request you all to post some good questions covering every section under the above mentioned heading everyday. So come on guys let's bell the CAT together........
enjoy studying
(Amresh Singh)
Here is a question for you,
In an election for the president, 261 valid votes are cast, for the 5 contestants. Find the least number of votes a candidate requires to receive to win the election. No candidate gets less than 50 votes.
1) 53
2) 54
3) 55
4) 56
5) 57
edited by aaamresh on 9/15/2008 edited by aaamresh on 10/3/2008 Solid sometimes Liquid but never Gaseous. 

Intern
Let''s Bell The CAT Together
Originally Posted by aaamresh Hi Guys,
Since only 2 months are left so we all need more and more practice. Here we get only a single question for the whole day. So I request you all to post some good questions covering every section under the above mentioned heading everyday (don't forget to change the date). So come on guys let's bell the CAT together........
enjoy studying
(Amresh Singh)
Here is a question for you,
In an election for the president, 261 valid votes are cast, for the 5 contestants. Find the least number of votes a candidate requires to receive to win the election.
1) 53
2) 54
3) 55
4) 56
5) 57
Answer 53.
For minimum number all should be very close to each other. 261/5 = 52.2
So, winner gets 53 votes and the other 4 get 52 votes each.


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option (a) 53
as 261 votes are there and 5 candidates...so divide by 5 ..260/5 = 52...so now all 5 candidates have 52 votes each and 1 vote is left so that will goto one of the candidate and he/she will win..so least no. of votes required = 53


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Let''s Bell The CAT Together
hello amit & sumit,
i have updated the question by adding a line " no candidate gets less than 50 votes"
if your answer is still the same then,
Is getting 53 votes ensures a win?
there may be a tie as contestant gets 53,53,52,52,51 votes.
Solid sometimes Liquid but never Gaseous. 

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Originally Posted by aaamresh hello amit & sumit,
i have updated the question by adding a line " no candidate gets less than 50 votes"
if your answer is still the same then,
Is getting 53 votes ensures a win?
there may be a tie as contestant gets 53,53,52,52,51 votes. Hi aamresh
if u change the question such that no candidate gets less than 50 votes then the ans canl be option (4) 56
as in that case ..as each candidate gets atleast 50 votes...so total 250 votes have been cast..with each having 50 votes
now 11 votes are left..we can have two persons with 55 55 votes and still one vote left..so give that to one person and hsi count becomes 56...so votes required is 56
One Query
here you have asked least no. of votes required to win the election , that will be 53 only as that is the least no. of votes with other conditions favourable for a candidate to win the exam
if you are asking no. of votes necessarily needed to win...then the answer will be 56 as stated above
so i will say regardless of the extra condition least no. of votes = 53 only
what say


Let''s Bell The CAT Together
Originally Posted by aaamresh
Here is a question for you,
In an election for the president, 261 valid votes are cast, for the 5 contestants. Find the least number of votes a candidate requires to receive to win the election. No candidate gets less than 50 votes.
The answer is 53....
Sure... I did it by trial and error method....


Let''s Bell The CAT Together
As sumit said if the coundtion is 'least no. of votes required to win the election ' answer is 53 for sure.


Invisible
Let''s Bell The CAT Together
yes question is asking "least no. of votes required to win the election"
it is not asking "least no. of votes so that a candidate may win the election"
so i will go with the option (4).
There was a question in CAT 2007, which i think will clear your doubt.
here it is.......
In a football match, at the half time, M&M club was trailing by 3 goals. Did it win the match?
A) In the second half M&M club scored 4 goals.
B) The opponent scored 4 goals in the match.
Mark (1) if the question can be answered by using statement A alone but not by using statement B alone
Mark (2) if the question can be answered by using statement B alone but not by using statement A alone
Mark (3) if the question can be answered by using either of the statements alone
Mark (4) if the question can be answered by using both the statements together
Mark (5) if the question can't be answered on the basis of the two statements
It is clear that the question can't be answered by using either statements alone. Let's consider both the statements together. Under this condition 2 cases are possible.
Let opponent team is A
CASE 1 :
A M&M
first half 3 0
second half 1 4
total goals 4 4
RESULT= A TIE
CASE 2:
A M&M
first half 4 1
second half 0 4
total goals 4 5
RESULT= M&M WON
So what should be the answer 4 or 5
Obviously it should be 5.
This question is quite similar to what i asked.
So Sumit what should be the answer 53 or 56?
Solid sometimes Liquid but never Gaseous. 

Invisible
Let''s Bell The CAT Together
Here is another question in advance........
Find the square root of ( 32  ?1020 ).
1) ?15 + ?17
2) ?13  ?11
3) ?17  ?15
4) ?13 + ?11
5) ?18  ?14
you people are not posting the questions. it's not fair.
ok guys goodnight.........
Solid sometimes Liquid but never Gaseous. 

Virtuoso
Let''s Bell The CAT Together
answer is 3.
as 1020/4 = 255 = 15*17 and 15+17 = 32
so it is (?15 )^2 + (?17)^2  2[ ?15 *?17]
which is equal to square of ans 3...
so the ans is 3


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Invisible
Let''s Bell The CAT Together
Here is one more....
In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact center of the squares and only one ball is to be placed in one square) shown in the figure given below such that they do not lie along the same straight line?
_ _ _ _
l_l_l_l_l
l_l_l_l_l
l_l_l_l_l
1) 144
2) 200
3) 204
4) 208
5) 216
edited by aaamresh on 9/15/2008
edited by aaamresh on 9/15/2008 edited by aaamresh on 9/16/2008 Solid sometimes Liquid but never Gaseous. 

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the question for the DS question discussed above will be 5 only.......
as far as no. of votes question go ...i cant seem to leave 53....
can you post the original solution of the problem ...probaly that will clear things up


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i am assuming that with straight lines you are just considering horizontal and vertical and not diagonal
answer is option (1) 144
first ball can be placed in 12 ways in any of the squares (refer to the image that i have attached alongside)
now second ball cant be placed in the same row vertically or horizontally as the first ball...so there are 6 ways to place it..
and lastly the third ball can be placed in 2 ways
so total no. of ways = 12*6*2 = 144


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A new question
there are 4 letters and 4 envelopes. The no. of ways in which all the letters can be put in the wrong envelope is
a) 8
b) 9
c) 16
d) none of these edited by sumit2goody on 9/16/2008 

Invisible
Let''s Bell The CAT Together
yes sumit you have to consider the diagonal arrangement also. that is the main point of the question. if u gonna miss it, your answer will be wrong.
there are 4 letters and 4 envelops. first letter can be placed in 4 ways, second can be placed in 3 ways since one envelope is occupied now, third can be placed in 2 ways and last one can be placed in 1 way.
therefore total no. of placing the letters are = 4*3*2*1= 24
no. of ways of placing the letters in their respective envelopes is 1.
hence required answer is = 24  1 = 23.
hence ans is option 4. edited by aaamresh on 9/16/2008 Solid sometimes Liquid but never Gaseous. 

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ok i will try the diagonal condition also and will try it out
the answer you have posted for letters problem is wrong because you are only subtracting the case in which all the letters are in their correct positions , the question statement says that all letters are at wrong places...so consider that also and try again


Invisible
Let''s Bell The CAT Together
sorry i got it now
total no.of ways= 24
all 4 can be placed correctly in 1 way.
3 of them can be placed correctly in 4 ways.
2 of them can be placed correctly in 6 ways.
1 of them can be placed correctly in 4 ways.
this make 15 ways.
hence required answer is= 24  15 = 9
hence option (b).
is it correct????????
Solid sometimes Liquid but never Gaseous. 

Invisible
Let''s Bell The CAT Together
one more.........
There are 6 different letters and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes?
1) Zero
2) 1/6
3) 1/2
4) 5/6
5) none of these
Solid sometimes Liquid but never Gaseous. 

Let''s Bell The CAT Together
Aamresh wrote:
sorry i got it now
total no.of ways= 24
all 4 can be placed correctly in 1 way.
3 of them can be placed correctly in 4 ways.
2 of them can be placed correctly in 6 ways.
1 of them can be placed correctly in 4 ways.
this make 15 ways.
hence required answer is= 24  15 = 9
hence option (b).
is it correct????????
I have a doubt here  how can we consider the case that when 3 of them is placed correctly because if 3 are placed correctly then the 4th one will automatically be correct...

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