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21 May 2019, 02:19
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A
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Difficulty:
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Question Stats:
65% (02:41) correct
35%
(02:39)
wrong
based on 231
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FRESH GMAT CLUB TESTS QUESTION
The length of rectangle B is x percent less than the length of rectangle A, and the width of rectangle B is y percent, greater than the width of rectangle A (where both x and y are greater than 0 and less than 100). Is the area of rectangle A greater than the area of rectangle B?
(1) x = y
(2) x + y = 20
M36-49
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08 Nov 2021, 04:09
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Bunuel
Official Solution:
The length of rectangle B is \(x\) percent less than the length of rectangle A, and the width of rectangle B is \(y\) percent greater than the width of rectangle A (where both \(x\) and \(y\) are greater than 0 and less than 100). Is the area of rectangle A greater than the area of rectangle B?
The length of rectangle B is \(x\) percent less than the length of rectangle A (say \(m\)): \(length_B=m(1-\frac{x}{100})\)
The width of rectangle B is \(y\) percent greater than the width of rectangle A (say \(n\)): \(width_B=n(1+\frac{y}{100})\)
The question: is \(mn > m(1-\frac{x}{100})*n(1+\frac{y}{100})\)?
Reduce by \(mn\): is \(1 > (1-\frac{x}{100})(1+\frac{y}{100})\)?
Simplify: is \(100^2 > (100-x)(100+y)\)?
(1) \(x = y\)
The question becomes: is \(100^2 > (100-x)(100+x)\)?
Is \(100^2 > (100-x)(100+x)\)?
Is \(100^2 > 100^2-x^2\)?
Is \(x^2>0\)?
Since given that \(x > 0\), then the answer to this question is YES. Sufficient.
(2) \(x + y = 20\)
Test extreme cases:
If \(x\) very close to 0 and \(y\) is very close to 20, then \((100-x)(100+y) \approx 100*120 > 100^2\). So, in this case we'd have an NO answer to the question.
If \(x\) very close to 20 and \(y\) is very close to 0, then \((100-x)(100+y) \approx 80*100 < 100^2\). So, in this case we'd have an YES answer to the question.
Not sufficient.
Answer: A
General Discussion
21 May 2019, 03:32
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Bunuel
Let the length of rectangle A = a
So, length of rectangle B = a(1-x/100)
Let the breadth of rectangle A = b
So, breadth of rectangle B = b(1+y/100)
Area of A = a*b
Area of B = a(1-x/100)*b(1+y/100)
= ab(1 + (y-x)/100 - xy/10000)
(1) x = y
Area of B = ab(1 + (x-x)/100 - x^2/10000)
= ab(1 - x^2/10000)
Since, 0<x<100. 0<x^2<10000
So Area of B is less than A
Sufficient
(2) x+y = 20.
Nothing can be said.
Eg: If x=19,y=1
Area of B = ab(0.81)(1.01) < ab
If x=1,y=19
Area of B = ab(0.99)(1.19) > ab
Insufficient
So, option A
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