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20 Feb 2015, 01:35
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (01:58) correct
39%
(02:01)
wrong
based on 122
sessions
History
Date
Time
Result
Not Attempted Yet
20 Feb 2015, 13:13
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Hi rajthakkar,
We can solve this DS question by TESTing VALUES.
We're told that (P+Q) > 3. We're asked if Q > 1. This is a YES/NO question.
Fact 1: P > 0
IF...
P = 1
Q > 2
The answer to the question is YES.
IF...
P = 4
Q > -1
IF....
Q = 0
The answer to the question is NO.
Fact 1 is INSUFFICIENT.
Fact 2: P - 1 < Q
You might find it easier to rewrite this inequality:
P < Q + 1
Looking at this, and considering that (P+Q) > 3, we know that there is a minimum value for Q.
For example, using the information in Fact 2....
IF....
Q = 1
P < 2
P+Q < 3
But this does NOT fit with the information given at the beginning (so Q CANNOT be 1). By extension, if we make Q smaller, then P gets smaller (and this also does not fit the prompt). So Q MUST be > 1 and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
We can solve this DS question by TESTing VALUES.
We're told that (P+Q) > 3. We're asked if Q > 1. This is a YES/NO question.
Fact 1: P > 0
IF...
P = 1
Q > 2
The answer to the question is YES.
IF...
P = 4
Q > -1
IF....
Q = 0
The answer to the question is NO.
Fact 1 is INSUFFICIENT.
Fact 2: P - 1 < Q
You might find it easier to rewrite this inequality:
P < Q + 1
Looking at this, and considering that (P+Q) > 3, we know that there is a minimum value for Q.
For example, using the information in Fact 2....
IF....
Q = 1
P < 2
P+Q < 3
But this does NOT fit with the information given at the beginning (so Q CANNOT be 1). By extension, if we make Q smaller, then P gets smaller (and this also does not fit the prompt). So Q MUST be > 1 and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
07 Aug 2015, 07:59
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Bookmarks
St1: P > 0;
Also, p+q>3 which is q>3-p. If P=1 then q>1, if P=2 - No. Not sufficient;
ST2: p-1<q;
Rearrange: p-1<q to p-q<1 and to -p+q>-1. Given signs of both inequalities are the same we can sum -p+q>-1 with p+q>3. It gives us 2q>2 and q>1. Sufficient.
Also, p+q>3 which is q>3-p. If P=1 then q>1, if P=2 - No. Not sufficient;
ST2: p-1<q;
Rearrange: p-1<q to p-q<1 and to -p+q>-1. Given signs of both inequalities are the same we can sum -p+q>-1 with p+q>3. It gives us 2q>2 and q>1. Sufficient.
