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sgoll
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Can someone please tell me the quick method to solve questions like #1?

For #2, i get the ans A) 2k/15.

Thanks for answering



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PS - airline ticket 13 Feb 2007, 13:36
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if u realize u ll find that u can simplify the given fraction

dividing it by 3twice ie:9

for 2

120...................900

X............................K

x = 120K/900 = 12/90 k = 2/15k............A IS MY ANSWER
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sgoll
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Thanks Yezz. I too get A for #2.
As per the answers I have, the ans is B.

#1, is there a technique to solve such questions or is it just cancelling out common factors?
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Tuneman
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I think trial and error is the best for #1, and yes A for #2
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PS - airline ticket 13 Feb 2007, 22:51
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sgoll
Thanks Yezz. I too get A for #2.
As per the answers I have, the ans is B.

#1, is there a technique to solve such questions or is it just cancelling out common factors?
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That is a good question, i have no idea how to solve these apart from brute force...these kind of questions are really annoying :( esply when you are running out of time, does anyone have any suggestions ??? I just use an LCM kind of method

3|351, 558
-----------
3|117, 186
-----------
| 39,62

no more divisors available so 39/62




and 2k/15 for #2
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if the sum of the numbers is divisible by 3 then 3 is a factor

if the sum of the numbers is divisible by 9 then 9 is a factor

ie

3+5+1=9...divisible by both 3 and 9



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