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If x, y, z are chosen from the three numbers –3, 1/2, and 2, what is the largest possible value of the expression x*z^2/y?
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
sanjitscorps18
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17 May 2022, 07:25
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Bunuel
If x, y, z are chosen from the three numbers –3, 1/2, and 2, what is the largest possible value of the expression x*z^2/y?
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
Not particularly given whether we can reuse the numbers or is it a one-to-one mapping. However, the choices are such that only a single combination would provide the largest possible value for the expression hence this is a fair question
x*z^2/y
Maximize x*z^2 and minimize 1/y
x * z^2. If we assign x = 2 and z = -3 we get the value as 18.
We can only assign the negative value to the variable that is squaring else we would end up getting a negative value or a smaller value (if repetition is allowed).
Now y is minimum if we use y = 1/2
Hence to obtain the maximum possible value we get
x*z^2/y = 18 * 2 = 36
Option D
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Schools: Anderson '25 Booth '25 Broad '25 Carlson Carroll CEIBS CBS '25 CUHK Darden '25 Goizueta '25 ESADE Foster '25 Fuqua '25 Pepperdine "25 Haas '25 HBS '25 HKUST '25 IESE '25 IMD Jan'23 intake ISB '24 Cornell '25 Rice Jones '25 Judge '24 Kelley '25 Kellogg '25 Kenan-Flagler '25 LBS '25 Marshall '25 McCombs '25 McDonough '25 Mendoza '25 Merage '25 NTU NUS '25 Moore '25 Owen '25 Said '24 BU Questrom '25 Ross '25 Rotman '25 Erasmus Scheller '25 SDA Bocconi "24 Simon '25 Sloan '25 Smeal Stanford '25 Tepper '25 Tuck '25 WBS Wharton '25 Yale '25
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14 Jun 2022, 09:12
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sanjitscorps18
Bunuel
If x, y, z are chosen from the three numbers –3, 1/2, and 2, what is the largest possible value of the expression x*z^2/y?
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
(A) –3/8
(B) 16
(C) 24
(D) 36
(E) 54
Not particularly given whether we can reuse the numbers or is it a one-to-one mapping. However, the choices are such that only a single combination would provide the largest possible value for the expression hence this is a fair question
x*z^2/y
Maximize x*z^2 and minimize 1/y
x * z^2. If we assign x = 2 and z = -3 we get the value as 18.
We can only assign the negative value to the variable that is squaring else we would end up getting a negative value or a smaller value (if repetition is allowed).
Now y is minimum if we use y = 1/2
Hence to obtain the maximum possible value we get
x*z^2/y = 18 * 2 = 36
Option D
what if we use
x=2
y=1/2
z=-3
we get 169.
