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59. Is a^2 + b^2 > c^2 ? (1) a^3 + b^3 > c^3 (2) a + b > c

The best way to deal with this problem is to plug numbers. Remember on DS questions when plugging numbers, your goal is to prove that the statement is not sufficient. So you should try to get a YES answer with one chosen number(s) and a NO with another.

Is a^2 + b^2 > c^2 ?

(1) a^3 + b^3 > c^3, both YES and NO answers are easy to get:

If a and b are any positive numbers (for example a=b=1) and c is zero then the answer will be YES: 1^3+1^3>0^3 and 1^2+1^2>0^2; If a and b are some positive numbers (for example a=b=1) and c is big enough negative numbers (for example c=-10) then the answer will be NO: 1^3+1^3>(-10)^3 and 1^2+1^2<(-10)^2; Not sufficient.

(2) a + b > c. The same set of numbers will work for this statement as well:

If a and b are any positive numbers (for example a=b=1) and c is zero then the answer will be YES: 1+1>0^3 and 1^2+1^2>0^2; If a and b are some positive numbers (for example a=b=1) and c is big enough negative numbers (for example c=-10) then the answer will be NO: 1+1>-10 and 1^2+1^2<(-10)^2; Not sufficient.

(1)+(2) We have one set of numbers which gives the answer YES for both statements and another set of numbers which gives the answer NO for both statements. Not sufficient.

Is \(a^2 + b^2 > c^2\)? (1) \(a^3 + b^3 > c^3\) (2) \(a + b > c\)

Even using both statements, it's possible that a and b are both equal to 0, say, and that c is equal to -1,000,000 or any other very negative number. Then each statement will be true, but the answer to the question will be 'no'. Of course it's also easily possible to get a 'yes' answer, so E.
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The only way to solve it is picking numbers. However, is there a method to know which numbers should I pick?

This question can be easily done using numbers. When ever you see such powers, always start testing -ve and +ve numbers.

if a =1 b =5 and c =2... the answer to statement 1 and 2 is yes. if a =-1 b =-2 and c =-5 the answer to statement 1 and 2 is no.

Even if you combine the result is same because in both the cases the above scenario can be used.

Picking number is a habbit of practice.

Thank you, kudos for you! However, how did you know that you had to choose those kind of numbers. Why not a fraction (i.e. 0.5), based on the fact that a fraction has a special behavior when it is squared? Also, there are infinite combinations of integer numbers (positive and negative)? how did you know that you had to choose a combination like that?

Thank you.
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You have to use your gutt feeling when you have multiple choices.

Since the equations had even and odd powers, I decided to go first using -ve and +ve numbers. My second try would have been fractions.
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Took 3 minutes, but got it. Number plugging is the only way here, three numbers are to be tried - all positive integer, all fraction, all negative.

Using AD/BCE strategy, starting with 2 since it looks easier.

2. Equating works for positive int, frac. But breaks for negative integers. insuff. 1. Same as 2. insuff.

Nothing in common, both same scenarios. E.
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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

a) a= +ve; b= +ve; c=+ve integers satisfies a and answer is yes a= +ve; b= +ve; c=-ve integers satisfies a and answer could be no Insufficient b)a= +ve; b= +ve; c=+ve integers satisfies b and answer is yes a= +ve; b= +ve; c=-ve integers satisfies b and answer could be no Insufficient

If the two conditions are insufficient for the exact same reason; together they are also not sufficient! So E..

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Re: Is a^2 + b^2 > c^2 ? (1) a^3 + b^3 > c^3 (2) a + b [#permalink]

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27 Jan 2012, 08:35

+1 E

Using a table to organize the data is ideal in this type of questions!

Bunuel, do you have a method to pick the numbers? For example, I have an order to eliminate possibilities: - First, I analyze positives and negatives scenarios - Then, numbers between 0 and 1, and between -1 and 0. - Finally, fractions (more than 1)

What do you think?
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"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

Using a table to organize the data is ideal in this type of questions!

Bunuel, do you have a method to pick the numbers? For example, I have an order to eliminate possibilities: - First, I analyze positives and negatives scenarios - Then, numbers between 0 and 1, and between -1 and 0. - Finally, fractions (more than 1)

What do you think?

Well in many cases trying positive/negative/proper fraction/zero can give you the result, though it really depends on the question to choose which numbers to use for plug-in method.
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The best way to deal with this problem is to plug numbers. Remember on DS questions when plugging numbers, your goal is to prove that the statement is not sufficient. So you should try to get a YES answer with one chosen number(s) and a NO with another.

Is a^2 + b^2 > c^2 ?

(1) \(a^3 + b^3 > c^3\), both YES and NO answers are easy to get:

If \(a\) and \(b\) are any positive numbers (for example \(a=b=1\)) and \(c\) is zero then the answer will be YES: \(1^3+1^3>0^3\) and \(1^2+1^2>0^2\);

If \(a\) and \(b\) are some positive numbers (for example \(a=b=1\)) and \(c\) is large enough negative number (for example \(c=-10\)) then the answer will be NO: \(1^3+1^3>(-10)^3\) and \(1^2+1^2<(-10)^2\); Not sufficient.

(2) \(a + b > c\). The same set of numbers will work for this statement as well:

If \(a\) and \(b\) are any positive numbers (for example \(a=b=1\)) and \(c\) is zero then the answer will be YES: \(1^3+1^3>0^3\) and \(1^2+1^2>0^2\);

If \(a\) and \(b\) are some positive numbers (for example \(a=b=1\)) and \(c\) is large enough negative number (for example \(c=-10\)) then the answer will be NO: \(1^3+1^3>(-10)^3\) and \(1^2+1^2<(-10)^2\); Not sufficient.

(1)+(2) We have one set of numbers which gives the answer YES for both statements and another set of numbers which gives the answer NO for both statements. Not sufficient.

It is easier if you plug in values. Remember you plug in values to disprove what is given in the option. However, you can sort of get to eliminating option 1 & 2 just by looking at them individually. Option 1: CUBES...So negative values would still be negative although bigger in value. Option 2: If i square on both sides there is chance the sign ma change (if the nos. are negative) therefore it individually wouldn't suffice as well.

Number plugging though simplest of all approaches to me is the scariest. I totally echo metallicafan's apprehensions. How can on the basis of putting in say 2/3 sets of number can be say with certainty that the equation would be true for all the possible number sets.

the power to the variables should tell us everything... We are asked about Power to EVEN number and GIVEN in the choices is ODD power.... so clearly it should be insuff...

Is \(a^2 + b^2 > c^2\).....

lets see the statements- (1) \(a^3 + b^3 > c^3\).. As mentioned above, we are told about ODD powers here and asked about EVEN powers.. It will be insuff...

lets see with an example..

a = 1, b = 2 and c = -10... \(1^3 + 2^3 > -10^3\).. But it will not be TRUE for \(a^2 + b^2 > c^2\)........NO a= 1, b=2 and c = 1.. It will be TRUE for both .............YES

Different answers Insuff

(2) \(a + b > c\) SAME as statement (1) above Insuff