500 ps

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500 ps

by dunkin77 » Tue Apr 03, 2007 5:42 pm
Hi,

The answer is A) and hope someone could give me explanation..thax


There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?
(A) 101
(B) 103
(C) 106
(D) 107
(E) 109

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by ajith » Tue Apr 03, 2007 5:51 pm
The numbers following the first conidtion in the given range ie remainder of 2 when divided by 3 are 101, 104 and 107
The numbers following the second conidtion in the given range ie remainder of 1 when divided by 4 are 101, 105,109

when you look for the number which follows both the conditions it should be in both the sets , ie the number is 101

There are more complex methods, but this is the easiest for these kind of problems, I will explain one difficult method in the next post.
Always borrow money from a pessimist, he doesn't expect to be paid back.

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by dunkin77 » Tue Apr 03, 2007 8:40 pm
Thank you!! it's very clear now :D

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by Cybermusings » Wed Apr 04, 2007 3:44 am
If the cards are counted 3 at a time and 2 are left over. This indictaes that when the total number of cards is divided by the numeral 3, the remainder is 2. Factors of 3 between 100 and 110 include 102, 105 and 108. Hence the numbers could be 101 (not possible), 104 and 107.

Fact 2 : If counted 4 at a time there is one left over. That means when the total number of cards are divided by 4, the remainder is 1. This is possible only in the case of 101

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by Neo2000 » Wed Apr 04, 2007 8:08 am
Umm maybe you could have simply divided your answer choices by 3 and then 4 to see if you got remainders 2 and 3 respectively. In this case, 101 straightaway gave you the required remainders :)