"Is x<y?"
(1) 2x<3y
(2) xy>0
My answer = C (I tried various combinations of positives and negatives, x < y and x> y, and I was likely too hasty and just miscalculated (I feel more comfortable testing numbers instead of recalling all the rules in these instances). But, the answer is E....why is the answer E?
Thank you! Still have 'em coming...
Factorization
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Target question: Is x<y?ostrowskiamy wrote:"Is x<y?"
(1) 2x<3y
(2) xy>0
Statement 1: 2x<3y
There are several pairs of numbers that meet this condition. Here are two:
Case a: x=5 and y=4, in which case x is greater than y
Case b: x= -2 and y= -1, in which case x is less than y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: xy>0
There are several pairs of numbers that meet this condition. Here are two:
Case a: x=5 and y=4, in which case x is greater than y
Case b: x= -2 and y= -1, in which case x is less than y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined:
There are still several pairs of numbers that meet both conditions. Here are two:
Case a: x=5 and y=4, in which case x is greater than y
Case b: x= -2 and y= -1, in which case x is less than y
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Answer = E
Cheers,
Brent
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- ceilidh.erickson
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When you see inequality problems in DS, you want to think about all of the kinds of numbers that behave differently: positives/negatives, fractions/integers, and 0, 1, and -1.
It's good that you were able to eliminate each statement on its own; this means that you were thinking about positives and negatives. But this statement gives me pause:
- positives and negatives
- fractions and integers
- 0
- 1 and -1
Which each of these, your goal should be to PROVE INSUFFICIENCY. Can we test different cases to get a YES answer and a NO answer? As Brent proved, you can.
When we combine the statements, we know that 2x < 3y, and that xy > 0 (in other words, they have the SAME SIGN). That only rules out one positive, one negative, but there are still plenty of other cases to test. Be systematic in your number testing!
It's good that you were able to eliminate each statement on its own; this means that you were thinking about positives and negatives. But this statement gives me pause:
It's fine to test numbers with inequalities, but you have to be systematic about it! You can't just test numbers at random. You need to ask yourself - have I tested different categories of numbers that behave differently? Your number-testing checklist should have looked like this:I feel more comfortable testing numbers instead of recalling all the rules in these instances
- positives and negatives
- fractions and integers
- 0
- 1 and -1
Which each of these, your goal should be to PROVE INSUFFICIENCY. Can we test different cases to get a YES answer and a NO answer? As Brent proved, you can.
When we combine the statements, we know that 2x < 3y, and that xy > 0 (in other words, they have the SAME SIGN). That only rules out one positive, one negative, but there are still plenty of other cases to test. Be systematic in your number testing!
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
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Statement 1: 2x<3yostrowskiamy wrote:"Is x<y?"
(1) 2x<3y
(2) xy>0
To see the relationship between x and y more clearly, PUT ONE VARIABLE IN TERMS OF THE OTHER:
x < (3/2)y.
If y=2, then x<3.
Case 1: It's possible that y=2 and x=2, in which case x=y.
Case 2: It's possible that y=2 and x=1, in which case x<y.
INSUFFICIENT.
Case 1 and Case 2 satisfy BOTH STATEMENTS.
Thus, even when the two statements are combined, it's possible that x=y or that x<y.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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