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Bunuel
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If x, y, z are chosen from the three numbers 3, 1/2, and 2, what is 17 May 2022, 05:30
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88% (00:58) correct 12% (01:51) wrong based on 92 sessions

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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
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D
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sanjitscorps18
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If x, y, z are chosen from the three numbers 3, 1/2, and 2, what is 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
Show more


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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If x, y, z are chosen from the three numbers 3, 1/2, and 2, what is 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


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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what if we use
x=2
y=1/2
z=-3

we get 169.
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sanjitscorps18
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If x, y, z are chosen from the three numbers 3, 1/2, and 2, what is 24 Jun 2022, 10:36
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WilliamKane
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


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.
Show more


Apologies for the delayed response. This is the combination that provides the largest number for the expression. Taking these numbers as variable values we get x*z^2/y as 2*(-3)^2 / (1/2)
= 2 * 9 / (1/2)
= 18 / (1/2)
= 18 x 2 = 36

Hope it's clear now!
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