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rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator
then p/q < (p+a)/(q+a)
similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always
Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3
You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values. for x,y = 6,1;...... x/y = 6; for x,y = 10, 5 ......x/y = 2
stmt 2: sufficient. -5x < -14 y or x/y > 14/5
Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3
Last edited by rashminet84 on 12 Jul 2009, 20:18, edited 1 time in total.
rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator
then p/q < (p+a)/(q+a)
similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always
Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3
You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values. for x,y = 6,1;...... x/y = 6; for x,y = 5,10 ......x/y = 2
stmt 2: sufficient. -5x < -14 y or x/y > 14/5
Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3
No inconvenience, at all. We are all helping each other!
Oh yes, ofcourse
but sometimes people try a lot to make sense of a wrong post, thereby wasting time and effort. It might lead to frustration also when one comes to know that it was the mistake of the one who posted.
I think ans is D.What is the OA? Both stmnts are sufficient.
Ankur,for proving the sufficiency of stmnt1 you can plug in the below values y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true
You can also plug in fractions to double check the ans. Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6 This time also 3x>7y
I think ans is D.What is the OA? Both stmnts are sufficient.
Ankur,for proving the sufficiency of stmnt1 you can plug in the below values y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true
You can also plug in fractions to double check the ans. Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6 This time also 3x>7y
Yes Shruti, OA is D. Somehow plugging doesnt work for me sometimes in DS, hence I was trying to solve stmt 1 to check whether it is sufficient. Anyway thanks a lot for explanation. Plugging will work for this problem. Thanks again.
given x >y + 4 . Can any one verify if we can substitute x > y+4 in the to be proved inequation ??
can we say this ? 3(y+4) > 7(y) --> stmt 1
if yes then A makes sense..because stmt 1 will end up with y < 3 and for all values of 0 < y < 3 , 3x > 7y.
or is it wrong to approach this problem that way ? Please help !
The problem with your approach is that it is already assuming both conditions to be true ie, 3x>7y and that x > y+4
and after doing this, you get that these two conditions will be satisfied only when y<3, which is true. As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y. But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.
The problem with your approach is that it is already assuming both conditions to be true ie, 3x>7y and that x > y+4
and after doing this, you get that these two conditions will be satisfied only when y<3, which is true. As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y. But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.
OA is D but I'm getting B after solving both the statements Thats why I logged this question in the forum So far I havnt got the conclusive solution why statement 1 is sufficient. Or is it the OA is wrong?
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