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12 Aug 2020, 16:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:23) correct
39%
(01:35)
wrong
based on 141
sessions
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Time
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Not Attempted Yet
Hi all,
As we are looking for the minimum of x*y^2, let's first think about how we can get this for y and x (and the overall). We know that all numbers are negative and the multiplication will be negative because y will always be positive (then we need to look for the greatest value in absolute terms, as it is negative, the greatest one in absolute terms will be the one more "on the left" of the lines of numbers
x: we need the greatest number -> this is x=-1/2
y: we need the greatest number -> this is y=-1/4
Then -1/2*(-1/4)^2 = -1/(2^5)= -1/32
Answer D
As we are looking for the minimum of x*y^2, let's first think about how we can get this for y and x (and the overall). We know that all numbers are negative and the multiplication will be negative because y will always be positive (then we need to look for the greatest value in absolute terms, as it is negative, the greatest one in absolute terms will be the one more "on the left" of the lines of numbers
x: we need the greatest number -> this is x=-1/2
y: we need the greatest number -> this is y=-1/4
Then -1/2*(-1/4)^2 = -1/(2^5)= -1/32
Answer D
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12 Aug 2020, 20:19
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We have -\(\frac{1}{2}\) ≤ x ≤ -\(\frac{1}{9}\) and -\(\frac{1}{4}\) ≤ y ≤ -\(\frac{1}{25}\)
Squaring for y, we get \(\frac{1}{16}\) ≤ \(y^2\) ≤ \(\frac{1}{625}\)
To get \(x * y^2\), we multiply and get all possible values with the endpoints of x and the endpoints of \(y^2\)
The possible values are:
(i) \(x * y^2\) = -\(\frac{1}{2}\) * \(\frac{1}{16}\) = -\(\frac{1}{32}\)
(ii) \(x * y^2\) = -\(\frac{1}{2}\) * \(\frac{1}{625}\) = -\(\frac{1}{1350}\)
(iii) \(x * y^2\) = -\(\frac{1}{9}\) * \(\frac{1}{16}\) = -\(\frac{1}{144}\)
(iv) \(x * y^2\) = -\(\frac{1}{9}\) * \(\frac{1}{625}\) = -\(\frac{1}{5625}\)
Since we want the minimum value and all the values are negative, we for denominator with the least value (if all were positive, we would look for the denominator with the maximum value).
The denominator with the least value is -\(\frac{1}{32}\)
Option D
Arun Kumar
Squaring for y, we get \(\frac{1}{16}\) ≤ \(y^2\) ≤ \(\frac{1}{625}\)
To get \(x * y^2\), we multiply and get all possible values with the endpoints of x and the endpoints of \(y^2\)
The possible values are:
(i) \(x * y^2\) = -\(\frac{1}{2}\) * \(\frac{1}{16}\) = -\(\frac{1}{32}\)
(ii) \(x * y^2\) = -\(\frac{1}{2}\) * \(\frac{1}{625}\) = -\(\frac{1}{1350}\)
(iii) \(x * y^2\) = -\(\frac{1}{9}\) * \(\frac{1}{16}\) = -\(\frac{1}{144}\)
(iv) \(x * y^2\) = -\(\frac{1}{9}\) * \(\frac{1}{625}\) = -\(\frac{1}{5625}\)
Since we want the minimum value and all the values are negative, we for denominator with the least value (if all were positive, we would look for the denominator with the maximum value).
The denominator with the least value is -\(\frac{1}{32}\)
Option D
Arun Kumar
20 May 2022, 23:50
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CrackverbalGMAT chetan2u Bunuel
I solved this question correctly when I multiplied x by y^2, but not when I multiplied xy by y.
Can you help me understand what's wrong with the second approach?
I solved this question correctly when I multiplied x by y^2, but not when I multiplied xy by y.
Can you help me understand what's wrong with the second approach?
