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03 Sep 2012, 06:08
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:40) correct
18%
(01:48)
wrong
based on 2063
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The weights of all dishes of type X are exactly the same, and the weights of all dishes of type Y are exactly the same. Is the weight of 1 dish of type X less than the weight of 1 dish of type Y ?
(1) The total weight of 3 dishes of type X and 2 dishes of type Y is less than the total weight of 2 dishes of type X and 4 dishes of type Y.
(2) The total weight of 4 dishes of type X and 3 dishes of type Y is less than the total weight of 3 dishes of type X and 4 dishes of type Y.
(1) The total weight of 3 dishes of type X and 2 dishes of type Y is less than the total weight of 2 dishes of type X and 4 dishes of type Y.
(2) The total weight of 4 dishes of type X and 3 dishes of type Y is less than the total weight of 3 dishes of type X and 4 dishes of type Y.
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03 Sep 2012, 06:08
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SOLUTION
The weights of all dishes of type X are exactly the same, and the weights of all dishes of type Y are exactly the same. Is the weight of 1 dish of type X less than the weight of 1 dish of type Y ?
Say the weight of 1 dish of type X is \(x\) and the weight of 1 dish of type Y is \(y\). The question asks whether \(x<y\).
(1) The total weight of 3 dishes of type X and 2 dishes of type Y is less than the total weight of 2 dishes of type X and 4 dishes of type Y --> \(3x+2y<2x+4y\) --> \(x<2y\). If \(x=1\) and \(y=2\), then the answer is YES but if \(x=3\) and \(y=2\), then the answer is NO. Not sufficient.
(2) The total weight of 4 dishes of type X and 3 dishes of type Y is less than the total weight of 3 dishes of type X and 4 dishes of type Y --> \(4x+3y<3x+4y\) --> \(x<y\). Sufficient.
Answer: B.
The weights of all dishes of type X are exactly the same, and the weights of all dishes of type Y are exactly the same. Is the weight of 1 dish of type X less than the weight of 1 dish of type Y ?
Say the weight of 1 dish of type X is \(x\) and the weight of 1 dish of type Y is \(y\). The question asks whether \(x<y\).
(1) The total weight of 3 dishes of type X and 2 dishes of type Y is less than the total weight of 2 dishes of type X and 4 dishes of type Y --> \(3x+2y<2x+4y\) --> \(x<2y\). If \(x=1\) and \(y=2\), then the answer is YES but if \(x=3\) and \(y=2\), then the answer is NO. Not sufficient.
(2) The total weight of 4 dishes of type X and 3 dishes of type Y is less than the total weight of 3 dishes of type X and 4 dishes of type Y --> \(4x+3y<3x+4y\) --> \(x<y\). Sufficient.
Answer: B.
General Discussion
03 Sep 2012, 06:33
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Bunuel
Let X & Y are respective weights of Dish 1 & Dish 2. So Question reduces to is X<Y?
St 1: Insufficient: 3X + 2Y < 2X + 4Y,
=> X<2Y, Insufficient, X=4 & Y=5 or X=5 & Y=4 both conditions satisfy the equation.
St 2: Sufficient: 4X + 3Y < 3X + 4Y
=> X < Y, clearly sufficient.
Hence Answer is Option B.
