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  1. #1
    Available on PM sumit2goody's Avatar
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    Default Quant Thread for CAT 2009 (Arithmetic)

    Post your queries regarding ratio proportion , averages ,profit/loss, SP CP , time speed distance etc here
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    We have 10 sacks full of balls. All sacks contain balls weighing 2 kgs each, except one of the sacks which contains balls weighing 1 kg each. What would be the minimum no of weighings reqd to identify the sack containing the balls weighing 1 kg ?

    1) 1
    2) 9
    3) 10
    4) 5

  3. #3
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    Another one :

    Gopal borrowed from Varnma Rs 500 on 1st March, on the condition that hw will repay the amount in six equal monthly instalements of Rs 100 each at the beginining of every month, starting from 1st April of that year. Find the effective monthly rate of interest in %.

    1) 6.67
    2) 10
    3) 3.33
    4) 12

    PS : I do not know the answer to this one...sorry

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    Is it option 1? Just guessing.

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    Quote Originally Posted by tusharsem View Post
    Another one :

    Gopal borrowed from Varnma Rs 500 on 1st March, on the condition that hw will repay the amount in six equal monthly instalements of Rs 100 each at the beginining of every month, starting from 1st April of that year. Find the effective monthly rate of interest in %.

    1) 6.67
    2) 10
    3) 3.33
    4) 12

    PS : I do not know the answer to this one...sorry
    i think....
    principle = 500, amount = 600 => interest = 100, time = 6
    using simple interest formula, PRT/100 = 100
    500 * R * 6 = 10000
    R = 20/6 = 3.33%
    so i think it's option 3......

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    Quote Originally Posted by tusharsem View Post
    We have 10 sacks full of balls. All sacks contain balls weighing 2 kgs each, except one of the sacks which contains balls weighing 1 kg each. What would be the minimum no of weighings reqd to identify the sack containing the balls weighing 1 kg ?

    1) 1
    2) 9
    3) 10
    4) 5
    is it 9??
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    Quote Originally Posted by nipunkesari View Post
    i think....
    principle = 500, amount = 600 => interest = 100, time = 6
    using simple interest formula, PRT/100 = 100
    500 * R * 6 = 10000
    R = 20/6 = 3.33%
    so i think it's option 3......
    looks good to me...any other views???
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    Quote Originally Posted by Ruchi.Desai View Post
    Is it option 1? Just guessing.


    Which question ??

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    Quote Originally Posted by sumit2goody View Post
    is it 9??

    No its not. Just read the question again, very carefully.

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    Quote Originally Posted by tusharsem View Post
    We have 10 sacks full of balls. All sacks contain balls weighing 2 kgs each, except one of the sacks which contains balls weighing 1 kg each. What would be the minimum no of weighings reqd to identify the sack containing the balls weighing 1 kg ?

    1) 1
    2) 9
    3) 10
    4) 5
    Going by the options, is it 5?
    Surprisingly, I'm able to weigh out the sack having 1kg balls in 4 weighings also!
    1st-> Divide the sacks into two sets of 5 sacks each and weigh.
    2nd-> Divide the sacks into sets of 2 and 3 (from the 5 lighter ones earlier)
    3rd-> Divide the sacks into sets of 1 and 2 (taking worst case, since minimum is asked)
    4th-> Weigh the remaining 2 sacks separately (again, taking worst case in previous step)

    Am I skipping anything?
    P.S. - I've assumed the sacks have equal number of balls each

  11. #11
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    Quote Originally Posted by apoorv.sharma View Post
    Going by the options, is it 5?
    Surprisingly, I'm able to weigh out the sack having 1kg balls in 4 weighings also!
    1st-> Divide the sacks into two sets of 5 sacks each and weigh.
    2nd-> Divide the sacks into sets of 2 and 3 (from the 5 lighter ones earlier)
    3rd-> Divide the sacks into sets of 1 and 2 (taking worst case, since minimum is asked)
    4th-> Weigh the remaining 2 sacks separately (again, taking worst case in previous step)

    Am I skipping anything?
    P.S. - I've assumed the sacks have equal number of balls each

    Yes I know. This question requires a very different approach.
    And yes you can assume that the sacks have equal no of balls.

    I'll wait for some more tme before posting the solution.

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    Quote Originally Posted by sumit2goody View Post
    looks good to me...any other views???
    Is that it ??
    Was it so simple ??
    Cant believe it !!!

    I guess I got confused because I was doing it by some other method. Will post question as well when I get home...thanks anyways

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    The weighing question.......

    1) Divide those 10 sacks into 2 bundles of 4 each and one bundle of two, ofcourse randomly.

    2) Weigh the two bundles (4 sacks each) individually, and find out if they weigh the same or different. (Two Weighings Done)

    3) If they weigh different, then pick up the lighter bundle. Split the sacks into groups of two each. Weigh them separately. (Another two weighings)

    4) Now you would get the pair of sack, one of which contains the lighter one. Say the pair of sacks were (a,b) and (c,d) where the first pair contains the lighter sack. We know that either a, c, d are of the same weight or b,c,d are of the same weight. Now, weigh (a and c) and compare the same with the weight of (c and d). If both the weights are same, then b is the lighter sack else a is the lighter sack. (Another weighing here)

    5) Hence a total of 5 weighings.

    6) If the two weighings of 4 bundles each were to be same, then we could have weighed the remaining two sacks individually and could have found the lighter sack in less than 5 weighings. But since we need to find the minimum number of weighings, we would go for the worst case scenario.

  14. #14
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    Quote Originally Posted by ameyapatkar View Post
    The weighing question.......

    1) Divide those 10 sacks into 2 bundles of 4 each and one bundle of two, ofcourse randomly.

    2) Weigh the two bundles (4 sacks each) individually, and find out if they weigh the same or different. (Two Weighings Done)

    3) If they weigh different, then pick up the lighter bundle. Split the sacks into groups of two each. Weigh them separately. (Another two weighings)

    4) Now you would get the pair of sack, one of which contains the lighter one. Say the pair of sacks were (a,b) and (c,d) where the first pair contains the lighter sack. We know that either a, c, d are of the same weight or b,c,d are of the same weight. Now, weigh (a and c) and compare the same with the weight of (c and d). If both the weights are same, then b is the lighter sack else a is the lighter sack. (Another weighing here)

    5) Hence a total of 5 weighings.

    6) If the two weighings of 4 bundles each were to be same, then we could have weighed the remaining two sacks individually and could have found the lighter sack in less than 5 weighings. But since we need to find the minimum number of weighings, we would go for the worst case scenario.

    Well the solution looks quite elegant but as I said before this requires a different approach, thats why I thought of posting it.

    ( I guess you are all going to beat me up once I post the solution )

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    Quote Originally Posted by tusharsem View Post
    Well the solution looks quite elegant but as I said before this requires a different approach, thats why I thought of posting it.

    ( I guess you are all going to beat me up once I post the solution )
    so cmon, post it! let's see how badly you get beaten up

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    Quote Originally Posted by apoorv.sharma View Post
    so cmon, post it! let's see how badly you get beaten up
    Actually am in the office..will post it today itself when I reach home

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    Quote Originally Posted by tusharsem View Post
    We have 10 sacks full of balls. All sacks contain balls weighing 2 kgs each, except one of the sacks which contains balls weighing 1 kg each. What would be the minimum no of weighings reqd to identify the sack containing the balls weighing 1 kg ?

    1) 1
    2) 9
    3) 10
    4) 5

    Solution :

    We take one ball from the first sack, 2 balls from the second, 3 from the third and so on until we reach the last sack from which we pick 10 balls.

    1 + 2 + 3 + ... + 10 = 55

    If all balls weighed 2 kgs the weight would be 110 kgs but since 1 sack has balls of 1 kg, the actual weight is 110 - N where N is the no of the sack containing 1 kg balls.

  19. The Following 3 Users Say Thank You to tusharsem For This Useful Post:

    abhi2007b (21-Jul-09), apoorv.sharma (14-Mar-09), tarunbatra84 (15-May-09)

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    Even I think its option 1. Same method as Tushar used. Well think I m getting sumwhr now.

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    Default tusharsem. is 1 answer?

    Another one :

    Gopal borrowed from Varnma Rs 500 on 1st March, on the condition that hw will repay the amount in six equal monthly instalements of Rs 100 each at the beginining of every month, starting from 1st April of that year. Find the effective monthly rate of interest in %.

    1) 6.67
    2) 10
    3) 3.33
    4) 12

    This question...is option 1 the answew?

  22. #20
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    Quote Originally Posted by Ruchi.Desai View Post
    Another one :

    Gopal borrowed from Varnma Rs 500 on 1st March, on the condition that hw will repay the amount in six equal monthly instalements of Rs 100 each at the beginining of every month, starting from 1st April of that year. Find the effective monthly rate of interest in %.

    1) 6.67
    2) 10
    3) 3.33
    4) 12

    This question...is option 1 the answew?
    Well, actually as I had mentioned above I do not know the answer to this question. nipunkeasri and sumit2goody have taken out the answer as option 3.

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