official quant thread for cat08 - Page 256
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Re: official quant thread for cat08
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implex
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Re: official quant thread for cat08 - 13-05-2008, 08:52 AM

Quote:
Originally Posted by Hi2All View Post
Hi friends...
There are 40 seats in a bus. People agree to share the money for the number of seats. The total money comes to 70.37. How many seats were free?
The trick here is to find a prime number less than 40 which divides 70.37!!
that will suffice
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 08:53 AM

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Originally Posted by linksuresh View Post
also, could someone also explain how to calculate the minimum value for the question 87
minimum can be zero!!
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 09:05 AM

official solution 90)

It is easy to figure out speed of ali is 60 kmph and that of Bobby is 40 ( given) and also the distnace between them can be found out to be 50 kmph
they meet for the first time in 50/100=30 mins
thereafter they need to travel 100 km ( both together) relative speed is 100 so time is 1 hour

the trick here is to see that even if they travel in the same direction for a part of the journey that gets made up by the part they travel in opposite direction!

There is a goof up in the options. I am sorry for that X(4) should be 3:30

This can be seen by this . at 2:30 both ali and bobby will be at town B.
in 50 minutes ali will reach town A and bobby will be close to it
in the next ten minutes ali will come back 10 km and bobby will have reached 40 kms from B now 40+10=50
so they meet at 3:30
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 09:28 AM

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Originally Posted by implex View Post
minimum can be zero!!
hey implex...1 doubt....min value of xy+yz+zx can be zero when x,y,z <= 0
but in d given sum...x,y,z <0 ...what is the min value then??
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 09:30 AM

Quote:
Originally Posted by implex View Post
official solution 90)

It is easy to figure out speed of ali is 60 kmph and that of Bobby is 40 ( given) and also the distnace between them can be found out to be 50 kmph
they meet for the first time in 50/100=30 mins
thereafter they need to travel 100 km ( both together) relative speed is 100 so time is 1 hour

the trick here is to see that even if they travel in the same direction for a part of the journey that gets made up by the part they travel in opposite direction!

There is a goof up in the options. I am sorry for that X(4) should be 3:30

This can be seen by this . at 2:30 both ali and bobby will be at town B.
in 50 minutes ali will reach town A and bobby will be close to it
in the next ten minutes ali will come back 10 km and bobby will have reached 40 kms from B now 40+10=50
so they meet at 3:30
ok..thought so...that x(4) should be 3:30...so all of these is the answer
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 09:31 AM

Quote:
Originally Posted by bhavin422 View Post
hey implex...1 doubt....min value of xy+yz+zx can be zero when x,y,z <= 0
but in d given sum...x,y,z <0 ...what is the min value then??
the thing is if we assume that two of (x,y,z) are very small say h, where h tends to zero
then also xy+yz+zx will be just greater than zero and we will not be wrong if we assume that value to be zero in the limiting case.
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 09:44 AM

Quote:
official solution 90)

It is easy to figure out speed of ali is 60 kmph and that of Bobby is 40 ( given) and also the distnace between them can be found out to be 50 kmph
they meet for the first time in 50/100=30 mins
thereafter they need to travel 100 km ( both together) relative speed is 100 so time is 1 hour

the trick here is to see that even if they travel in the same direction for a part of the journey that gets made up by the part they travel in opposite direction!

There is a goof up in the options. I am sorry for that X(4) should be 3:30

This can be seen by this . at 2:30 both ali and bobby will be at town B.
in 50 minutes ali will reach town A and bobby will be close to it
in the next ten minutes ali will come back 10 km and bobby will have reached 40 kms from B now 40+10=50
so they meet at 3:30
Yeah that will make all of them as true. So, my method was correct. I thought a lot after u said that answer is not correct but couldn't come up with better approach. Thanks by the way for nicce questions. . Keep it up
   
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Re: official quant thread for cat08
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linksuresh
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Re: official quant thread for cat08 - 13-05-2008, 09:55 AM

Quote:
Originally Posted by implex View Post
The trick here is to find a prime number less than 40 which divides 70.37!!
that will suffice
as per ur logic, that wud be 31 commuters.. givin 2.27 per head...

could u elaborate more on why it shd be prime.. as in wats the logic ?
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 10:13 AM

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Originally Posted by linksuresh View Post
as per ur logic, that wud be 31 commuters.. givin 2.27 per head...

could u elaborate more on why it shd be prime.. as in wats the logic ?
as there is equal amount contributed by all, so 70.37 should be divisible by a prime. if it is not prime then we will get many values and then we won't be able to decide how many people are seated and how many are vacant.
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 13-05-2008, 11:19 AM

Quote:
Originally Posted by implex View Post
as there is equal amount contributed by all, so 70.37 should be divisible by a prime. if it is not prime then we will get many values and then we won't be able to decide how many people are seated and how many are vacant.

sorry friend.. i am unable to understand..
u told it should be a prime...why can't it be anyother prime number(i.e 37,29..) other than 31..
please giv me some more explanation..so that i can understand..

Thanxx..


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