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How to assess combine statements in Yes/No DS

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How to assess combine statements in Yes/No DS

by vanquish » Sat May 22, 2010 12:22 am
Hello all
I am struggling when i have to assesse statement 1 & 2 combine. This is where i fall down everytime; i am so frastrated having narrowed the answer down to either C or E yet can't game it at this point - all the time!

More specifically, say if i have 2 statments that says:
S1: (X^3) - (X^5) < 0
S2: (X^2) -1 < 0
Since S1 & 2 are individually insffuicient to determine if x is negative, in assessing the statements combine, can i add the two equations together to get (X^3)-(X^5)+(X^2) < 1? I.e use this to determine whether ans is choice C or E.
I thought bcos we can add 2 inequal eqn if the inequal sign is pointing same direction, i did just this. However it produce ans E (bcos if x is either 1/2 or -1/2, my combined eqn would still be satisfied); x can either be negative or positive. Unfortunately, the correct answer was C (i.e. in combining S1 & S2, x can only be negative).

I used picking values -2, -1/2, 0, 1/2, 2 to solve this question. So i tested S1, S2 with these values and found each statement to be insufficient. Then i test (X^3)-(X^5)+(X^2) < 1 to determine whether C or E is the answer. It gave me ans E but this is wrong.

I don't know where i have gone wrong - can anyone kindly help me?
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by sumanr84 » Sat May 22, 2010 11:26 pm
I donot know why no one has shown interest in this Q till now, anyway lets solve this one.

Q. Is X < 0 ?
S1: (X^3) - (X^5) < 0
S2: (X^2) -1 < 0

You are correct in pointing out that each statement alone is not sufficient in getting the solution. I will use number line to arrive at the solution.

S1. X^3 ( 1 - X^2) < 0
This gives us 2 eqns. X^3 < 0 OR (1 - X^2) < 0
Alternately , X < 0 OR X < -1 and X > 1
Let us plug some values to check this inequality,
We know that (X^3) - (X^5) > 0
how about for X < -1 e.g X = -2
(-2^3) - (-2^5) = -8 - ( - 32) = 24 > 0
So, clearly X < -1 is not correct as it does not satisfy S1
Representing the above on number line,

-------------------------(-1)xxxxxxxx(0)-----------(1)xxxxxxxxxxxxxxxxxxxx [ x represents valid values satisfying eq in S1 )

S2: (X^2) -1 < 0
Simplifies to , -1 < X < 1

-------------------------(-1)xxxxxxxx(0)xxxxxxxx(1)----------------------------- [ x represents valid values satisfying eq in S2 )

Combining both the statements S1 and S2, we see that only values b/w -1 and 0 satisfy both S1 and S2.
Therefore, X must be < 0, to be precise -1 < X < 0

So, Correct Answer is C
I am on a break !!
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by thephoenix » Sun May 23, 2010 4:06 am
I don't know where i have gone wrong - can anyone kindly help me?
the error u must have done is when u selected x=1/2 or 2 u must have not have examined that these value does not satisfies either of the two given statements
i.e for x=1/2 the expression x^3-x^5+x^2<1 is true however x^3-x^5<0 is not true
and for x=2 the expression x^3-x^5+x^2<1 is true however x^2-1<0 is not true
hence combining we get range for x as -1<x<0--->x<0
hth
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by vanquish » Mon May 24, 2010 2:20 am
sumanr84 wrote:I donot know why no one has shown interest in this Q till now, anyway lets solve this one.

Q. Is X < 0 ?
S1: (X^3) - (X^5) < 0
S2: (X^2) -1 < 0

You are correct in pointing out that each statement alone is not sufficient in getting the solution. I will use number line to arrive at the solution.

S1. X^3 ( 1 - X^2) < 0
This gives us 2 eqns. X^3 < 0 OR (1 - X^2) < 0
Alternately , X < 0 OR X < -1 and X > 1
Let us plug some values to check this inequality,
We know that (X^3) - (X^5) > 0
how about for X < -1 e.g X = -2
(-2^3) - (-2^5) = -8 - ( - 32) = 24 > 0
So, clearly X < -1 is not correct as it does not satisfy S1
Representing the above on number line,

-------------------------(-1)xxxxxxxx(0)-----------(1)xxxxxxxxxxxxxxxxxxxx [ x represents valid values satisfying eq in S1 )

S2: (X^2) -1 < 0
Simplifies to , -1 < X < 1

-------------------------(-1)xxxxxxxx(0)xxxxxxxx(1)----------------------------- [ x represents valid values satisfying eq in S2 )

Combining both the statements S1 and S2, we see that only values b/w -1 and 0 satisfy both S1 and S2.
Therefore, X must be < 0, to be precise -1 < X < 0

So, Correct Answer is C
Thanks for the reply Sumanr84, it certainly looks a lot faster to solve the question the way you do by manipulating the eqns. It took me a long time to use "picking numbers" method to test each statement, and it is error prone.

Thephoenix has very accurately spotted that I had made a mistake when it tried x=1/2 for S1: x^3-x^5<0.

However, could you (or anyone indeed) kindly help me with the following?

How do you get from X^3 < 0 to X < 0
Did you cube root both side of the equation?
Is so, can you actually use root operation or exponent operation on variables with unknown sign (i.e we dun kno x is positive or negative) inequality?

Can anyone explain do we need to "flip" inequal sign if we apply root or exponent operations to both sides? I know if we multiple or divide by a negative value in equality, we "flip" the inequal sign. Is this even a relevant question??

Again, how do you get from X^2 > 1 to X < -1 and X > +1

Also, do mean (1 - X^2) < 0 to X < -1 AND X > 1, or do you
mean (1 - X^2) < 0 to X^2 > 1 to X < -1 OR X > +1
Do we use "and" when we refer to a continuous range, and use "or" when the range is not continuous such as -1 > X > 1

Finally, how do you get from S1: X^3 ( 1 - X^2) < 0 to (X^3) - (X^5) > 0

Now for the combined statement analysis:
How do you read a combination of Statement 1: X < 0 OR X < -1 and X > 1, with Statement 2: -1 < X < 1
Is it best to draw nr line as you have done or is there a more effective way in manipulating an equation, perhaps one formed by combining S1 & S2?
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by vanquish » Mon May 24, 2010 2:25 am
Hey thephoneix

Well spotted!

my error indeed is in evaluating the "picked" no. 1/2 on statement 1: x^3-x^5 < 0.

That's very sharp of u

ta
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by thephoenix » Mon May 24, 2010 2:25 am
hey vanquish just remember this for a even power of a variable the sign may change........i.e for x^2 ; x can be +ve or -ve
but for odd powers the sign never changes

for x^2<1---->+x<1 or -x<1
or x>-1 so ineq is -1<x<1
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by vanquish » Mon May 24, 2010 2:30 am
Errr... i dun get the last sentence. I can see x > -1 ----> -x < 1 (since i multiple both sides by -1) but how do you get to your senteence below?

thephoenix wrote: or x>-1 so ineq is -1<x<1
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by vanquish » Mon May 24, 2010 2:35 am
oh i think i got it now....

x^2<1 ----> +x<1 or -x<1 ----> so ineq is -1<x<1

Is tis right? ihope

sorry, i m a bit slow.... dah
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by thephoenix » Mon May 24, 2010 3:34 am
yes its correct
for -x<1 when u multiply bth side by -1 the sign changes so it becomes x>-1
[rule multiplying an ineq with a negative no. changes the sign]
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by vanquish » Mon May 24, 2010 4:27 am
Hey Phoenix, really appreicate your reply.

Are you able to help with a really basic question - and explain why.
I am embarrass to say i've spent hours trying to work out what's correct.

Q: Is a > c?
S1: 5a > 6c

My thos were:
S1 sufficient bcos a > 1.2c. So a will always be bigger than c for the statement to hold true.
However someone sugguested that S1 is NOT sufficient bcos there are 2 variables "a" and "c" and i dun have any info about either of them. Is this right?

What about
Q: Is a > c?
S1: 5a < 6c

I really struggled with this one!
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by Stacey Koprince » Tue May 25, 2010 8:47 am
Received a PM asking me to reply. I can't discuss the above question specifically because no source has been cited, but I can address the question I was asked in the PM:
i would be grateful if you could provide some advice on how to correctly and efficiently approach a yes/no DS question. Is systematically picking values e.g. -2, -1/2, 0, 2, 1/2 to test a good way or not?
Yes, these can be tricky, and yes, there is a pretty systematic approach you can use.

For yes/no questions, the general rules are:
- an "always yes" answer IS sufficient
- an "always no" answer IS sufficient
- a "sometimes yes, sometimes no" answer is NOT sufficient

When you have a "theory" type question - one in which you could solve using theory or you could solve by testing the statements with real numbers - it is better to use theory when you know exactly how to think through the theory. But if you don't know the theory well enough to do that, then you are going to have to test the theory, and the easiest way to do that is to start testing numbers.

The big drawback with testing numbers is that you don't know whether you're choosing the right numbers. You might try 10 different numbers, and get "yes" answers for all of them, but you happened not to try the number, or kind of number, that would give you a "no" and so you get the problem wrong. That drawback will always exist when testing numbers on yes/no DS questions, but there are things that you can do to minimize the risk.

First, you generally want to test numbers that fall into different "number property" categories. For example: negative integers, negative fractions between -1 and 0, zero, fractions between 0 and 1, one, positive integers. Each of those categories has certain characteristics that are different from the other categories.

Second, as you test numbers, try to understand what's happening and why it's happening. When you tested zero in the problem, why did you get the response that you got? How do you think choosing something smaller or larger might change the outcome? At the least, this will help you to figure out what numbers you should try next. In some cases, this may actually allow you to figure out the theory behind what the question is testing, and then you can use theory to answer the question more confidently.

Third, have a "yes/no" mentality as you test numbers. If you keep getting the same response (for example, I've tried 5 numbers and gotten 5 yes answers), then you never know when you're done testing, right? (Because maybe I just haven't found the number that will give me a no.) So, after I try the first number, I see what answer I got (let's say I got "yes"). Then, I actively try to figure out what I could try that would give me the opposite answer: no. Because as soon as I have at least one yes and at least one no, I'm done - I know that this info is insufficient. So start thinking about how the calculation worked when you tried your first number and actively try to find a number that you think will give you the opposite answer.

If you are actively TRYING to get the opposite answer, and you do that three or four times, and you STILL keep getting the same answer, then you can feel more confident in saying the info is sufficient (though you still don't really know for sure, unless you figure out the theory).
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by vanquish » Tue May 25, 2010 8:01 pm
Hey Stacey,

Thank you for your kind reply, really appreciate it.

The above questions that i have are not from anywhere, they are just 2 questions i made up to ask myself testing my own understanding. I am honest; as you can see the two statements cannot co-exist together in a question as they contradict each other. e.g. You can't have a question which says: Is a > c? S1: 5a > 6c ; S2: 5a < 6c

What promted me to ask this question was the 2nd example given in:
https://www.beatthegmat.com/mba/gmat-data-sufficiency.
There the main lesson was you cannot multiple or divide variables with unknown neg or pos sign. Now that's fine, that i have understood.
But having asked myself "my own questions" i found that i dun really understand "something" since i can't ans my own question. However, i don't know what was it that i am missing and have been scratching my head for hours!?!?!?

Would you kindly unravel my own questions for me?
vanquish wrote:Hey Phoenix, really appreicate your reply.

Are you able to help with a really basic question - and explain why.
I am embarrass to say i've spent hours trying to work out what's correct.

Q: Is a > c?
S1: 5a > 6c
My thos were:
S1 sufficient bcos a > 1.2c. So a will always be bigger than c for the statement to hold true.
However someone sugguested that S1 is NOT sufficient bcos there are 2 variables "a" and "c" and i dun have any info about either of them. Is this right?

What about
Q: Is a > c?
S1: 5a < 6c
I really struggled with this one!
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by thephoenix » Tue May 25, 2010 10:19 pm
vanquish wrote:Hey Phoenix, really appreicate your reply.

Are you able to help with a really basic question - and explain why.
I am embarrass to say i've spent hours trying to work out what's correct.

Q: Is a > c?
S1: 5a > 6c

My thos were:
S1 sufficient bcos a > 1.2c. So a will always be bigger than c for the statement to hold true.
However someone sugguested that S1 is NOT sufficient bcos there are 2 variables "a" and "c" and i dun have any info about either of them. Is this right?

What about
Q: Is a > c?
S1: 5a < 6c

I really struggled with this one!
well i will tell u my approach for such problems , which involve variables and ineq
#1check info available in Q stem(whether variables are integers or any real no.), if integers then it will stated in the stem. If its not then the variable can be any thing a +ve int, a -ve int, a +ve fraction, a -ve fraction
#2 if its integer then we need to check with a +ve value or -ve value and in some cases a even no. or odd no.
#3 if its not then we have to try with a +ve int, a -ve int, a +ve fraction, a -ve fraction,even odd depending upon the case
#4 when nothing is stated about the variables then many a times we can decide on the basis of a +ve or -ve value,
but in case of exponents one should also consider of fractional values
#zero is important number , in many cases the ineq fails for zero, while picking numbers be aware of zero
# all the above are important for yes no type DS
# for Ds where a value is asked always check the ineq by putting the values back into it
# sometimes if asked whether a>b then instead of proving the ineq try disproving it . remember disproving an ineq such as a>b does no always means that one has to show a<b , proving a=b will also work
if i am incorrect then pls correctme

the above problem Is a>c? since it is not stated that whether a and c are int or not we have to careful for all condition before deciding
always start with easy numbers , chose numbers which hold the relationship true in the given statement
s1)5a>6c
for a=2 and c=1 5a>6c and a>c...............YES
and for a=-2 and c=-3 5a>6c and a>c..............YES
but for a=c=-1 5a>6c ,but a is not greater than c , infact its equal..........NO
hence two answers and therefore insufficient
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by Stacey Koprince » Thu May 27, 2010 11:40 am
But having asked myself "my own questions" i found that i dun really understand "something" since i can't ans my own question. However, i don't know what was it that i am missing and have been scratching my head for hours!?!?!?

Q: Is a > c?
S1: 5a > 6c
My thos were:
S1 sufficient bcos a > 1.2c. So a will always be bigger than c for the statement to hold true.
However someone sugguested that S1 is NOT sufficient bcos there are 2 variables "a" and "c" and i dun have any info about either of them. Is this right?
As you've noted, we can't do this question as a full, regular DS problem because the two statements contradict each other. But let's just take a look at the question and St1 only.

This is a yes/no question. If a statement can answer the question "always yes" OR "always no," then that statement is sufficient. If the statement only tells us that the answer is sometimes yes and sometimes no, then that statement is not sufficient.

We can divide the statement by 5, because we know the sign of that number. (Positive, so I don't have to switch the inequality sign.) So, yes, the simplified inequality is a > (6/5)c or a > 1.2c.

This is an inequality, not an equals sign, so it makes it a little bit trickier to interpret. Whatever c is, it is multiplied by positive 1.2. Various things happen, depending upon what c is in the first place.

If c is positive, then multiplying it by positive 1.2 makes the resulting number BIGGER than it was. For example, if c=2, then (1.2)c = 2.4. If c is 0.5, then (1.2)c = 0.6.

If c is zero, then multiplying it by positive 1.2 doesn't change c; the resulting number is still zero.

If c is negative, then multiplying it by positive 1.2 makes the resulting number SMALLER than it was. For example, if c= -2, then (1.2)c = -2.4. If c is -0.5, then (1.2)c = -0.6.

Statement 1 tells us that a is greater than 1.2c. Let's pick random numbers from our different categories to test here. Let's say c=2. If c=2, then 1.2c = 2.4. a would then have to be something greater than 2.4, which would also be greater than c=2. So in this case, yes, a is always greater than c.

But what about when c is zero or negative? Let's try when c is negative, because that had the opposite effect on the multiplication: it made the resulting number smaller. If c= -2, then 1.2c = -2.4. a has to be something greater than -2.4. a could be -2.3; that's greater than -2.4. But -2.3 is NOT greater than -2. In this case, a is smaller than c. (Note: a could also be 3; that's greater than -2.4. In this case a is greater than c=-2.)

We have a "sometimes yes / sometimes no" scenario now, so we know the statement is NOT sufficient.
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