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Inequality

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Source: — Data Sufficiency |
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by jeffedwards » Tue May 25, 2010 7:45 pm
The way I solved this problem was by finding numbers that could gives us a true or a false. Since we cannot resolve a definite solution, the answer is E - cannot solve combined. I wrote examples of fractions that would work...just subtract one from the bottom, top, or both depending on the part and you'll see the same.

One - insufficient;
Case one true - ¼
Case two false 2/0.5

Two - insufficient
Case one true - 2/5
Case two false - 1.25/2

Together - insufficient
Case one true - 1/4
Case two false - 1/0.5
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by liferocks » Tue May 25, 2010 9:11 pm
The question is
is 2a<b

From 1
2a<(b-1)..if (b-1)>0

if 2a>=b we will get b<(b-1) which is not possible so 2a<b ,if (b-1)>0

but if (b-1)<0 ,2a>(b-1) and if 2a>=b,b>(b-1) which is absolutely possible.Hence depending on the sign of b-1, a/b<1/2 can be true or false...insufficient

from 2
using the same logic as used for condition 1 it can be shown that (a-1)/b < 1/2 is not sufficient to conclude whether a/b<1/2 is correct.

hence ans option E
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by jeffedwards » Wed May 26, 2010 9:24 am
liferocks wrote:The question is
is 2a<b
Can we really assume the question is asking if 2a<b? We don't know if a or b is negative or not, so when we divide or multiply by one of the variables, we don't know if we would need to flip the sign or not...right

For example

a/b < ½
4/-2 < ½
This statement is true, but if we assume 2a<b we would get
8<-2
Which is obviously not true
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by tpr-becky » Wed May 26, 2010 10:09 am
In an algebraic inequality the main issue is the question of positive negative - if you multiply or divide a negative you must switch the sign. (you can pick numbers but I prefer to look at big concepts that can be brought into the next problem).

Here we have no information as to whether a or b is positive or negative and we have to have that information in order to solve the problem.

statement 1 involves multiplying or dividing by a variable that could be positive or negative so we can't answer the problem. BCE

Statemetn 2 involves the same problem - CE

together, no matter how you do the algebra you still have to multiply or divide by a variable that is undefined in terms of postive negative so the answer must be E.
Becky
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by gmatmillenium » Wed Jun 02, 2010 1:57 am
tpr-becky wrote:In an algebraic inequality the main issue is the question of positive negative - if you multiply or divide a negative you must switch the sign. (you can pick numbers but I prefer to look at big concepts that can be brought into the next problem).

Here we have no information as to whether a or b is positive or negative and we have to have that information in order to solve the problem.

statement 1 involves multiplying or dividing by a variable that could be positive or negative so we can't answer the problem. BCE

Statemetn 2 involves the same problem - CE

together, no matter how you do the algebra you still have to multiply or divide by a variable that is undefined in terms of postive negative so the answer must be E.
Becky,

from 1, we derive that 2a<b-1....whereas for the asked inequality to be true, 2a should be <b.....doesnt 1 hence answer yes in all situations?
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by kvcpk » Wed Jun 02, 2010 3:17 am
gmatmillenium wrote:
tpr-becky wrote:In an algebraic inequality the main issue is the question of positive negative - if you multiply or divide a negative you must switch the sign. (you can pick numbers but I prefer to look at big concepts that can be brought into the next problem).

Here we have no information as to whether a or b is positive or negative and we have to have that information in order to solve the problem.

statement 1 involves multiplying or dividing by a variable that could be positive or negative so we can't answer the problem. BCE

Statemetn 2 involves the same problem - CE

together, no matter how you do the algebra you still have to multiply or divide by a variable that is undefined in terms of postive negative so the answer must be E.
Becky,

from 1, we derive that 2a<b-1....whereas for the asked inequality to be true, 2a should be <b.....doesnt 1 hence answer yes in all situations?
Hi gmatmillenium,

I thought the same way as you are thinking now.. There is a small catch there.
2a/(b-1)<1 doesnt necessarily mean that 2a<b-1

2a/(b-1)<1 means:
2a<(b-1) if (b-1) >0 and
2a>(b-1) if (b-1)<0

So we cant always say 2a<b-1.

I hope this is clear. take a small example and evaluate if you need: 2/3<1 -> 2<3 but 2/-3<1 doesnt mean 2<-3

Regards,
Praveen
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by tpr-becky » Wed Jun 02, 2010 8:05 am
Praveen is correct - when you multiply or divide by a variable you don't know whether that variable is positive or negative which means you don't know whether you have to switch the sign - that is the issue at play in this questions.
Becky
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by boazkhan » Wed Jun 02, 2010 8:47 am
Hi Becky, For such problems where we have to test cases...Many times I am unable to generate cases that give both a Yes and a No...The way you solved the problem is by basically looking for signs for a and b...I could tell that that we are not provided information for whether a and b are positive or negative. Can I take this as a rule of thumb for such inequality questions...If I don't have any information for the sign, I should not waste my time by plugging in numbers?

Thanks!
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by tpr-becky » Wed Jun 02, 2010 9:07 am
While plugging in numbers is a valid technique for some DS I try to avoid it as much as possible becuase it is time consuming and leads to too many incorrect answers. The best techniqe is to define a rule when you have to plug in answers and then use that rule consistently.

Yes, if you are multiplying or dividing by variables in an inequality you need to have information about their signs in order to come up with a single answer.

Best of luck.
Becky
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