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If in set S above, x is an integer x is an integer greater than one

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If in set S above, x is an integer x is an integer greater than one  [#permalink]

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New post 15 Oct 2018, 09:51
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Set S = {\(x\), \(\frac{x}{4}\) , \(y^2\), \(5y\), \(x^2\), \(y\)}

If in set S above, \(x\) is an integer greater than one and \(0 < y ≤ 0.7\), which of the following represents the range of the set S?


A) \((x + y)(x − y)\)

B) \(x^2 ​– 3y\)

C) \(y^2\)

D) \(x – y\)

E) \(x^2 − y\)

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Re: If in set S above, x is an integer x is an integer greater than one  [#permalink]

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New post 15 Oct 2018, 10:43
According to the constraints of the variables, \(x^2\) will be the greatest value and \(y^2\) - the lowest
So the range we have got \(x^2-y^2\)

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Ans: A
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Re: If in set S above, x is an integer x is an integer greater than one  [#permalink]

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New post 15 Oct 2018, 11:01
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CAMANISHPARMAR wrote:
Set S = {\(x\), \(\frac{x}{4}\) , \(y^2\), \(5y\), \(x^2\), \(y\)}

If in set S above, \(x\) is an integer greater than one and \(0 < y ≤ 0.7\), which of the following represents the range of the set S?


A) \((x + y)(x − y)\)

B) \(x^2 ​– 3y\)

C) \(y^2\)

D) \(x – y\)

E) \(x^2 − y\)


Nice question!

In order to determine the range, we must determine the greatest and smallest values in the set.

Let's start with the expressions with x in them: x, x/4 and x²
Since x > 1, we know that x/4 < x < x²

Now the expressions with y in them: y, 5y and y²
Since 0 < y ≤ 0.7 y² < y < 5y

From these two inequalities, we must determine the greatest and smallest values
We'll do so by finding some EXTREME values.

First, since x is an INTEGER greater than 1, the smallest possible value of x is 2
So, the smallest possible value of x/4 = 2/4 = 1/2

Next, since y ≤ 0.7, the greatest possible value of y is 0.7
So, the greatest possible value of y² = (0.7)² = 0.49

Notice that the greatest possible value of y² is still less than the smallest possible value of x/4
Since y² < y < 5y, we can conclude that y² is the SMALLEST value in the set.
----------------------------------------

Since x is an INTEGER greater than 1, the smallest possible value of x is 2
So, the smallest possible value of x² = 2²= 4
Next, since y ≤ 0.7, the greatest possible value of 5y is 3.5

Notice that the smallest possible value of x² is still greater than the greatest possible value of 5y
Since x/4 < x < x², we can conclude that x² is the GREATEST value in the set.
----------------------------------------

Range = GREATEST value - SMALLEST value
= x² - y²
= (x + y)(x - y)

Answer: A

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Re: If in set S above, x is an integer x is an integer greater than one   [#permalink] 15 Oct 2018, 11:01
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