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Re: If y<0, which of the following must decrease as y decreases? [#permalink]
saswata4s wrote:
If y<0, which of the following must decrease as y decreases?

I. 10y-100
II. (y-1)/y
III. y^2 - y

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) II and III only


Assume that we have \(0>y_1 > y_2\)

(I) \(10y-100\)
We have \((10y_1-100) -(10y_2-100)=10(y_1-y_2) > 0\).
Hence, if \(y\) decreases, then \(10y-100\) decreases. (I) is true.

(II) \(\frac{y-1}{y}=1-\frac{1}{y}\)
We have \(\big ( 1 - \frac{1}{y_1} \big ) - \big ( 1 - \frac{1}{y_2} \big ) = \frac{1}{y_2} - \frac{1}{y_1} = \frac{y_1-y_2}{y_1y_2}\)

Since \(0>y_1>y_2\) we have \(y_1-y_2>0\) and \(y_1y_2>0\), hence \(\frac{y_1-y_2}{y_1y_2}>0\).

Hence, if \(y\) decreases, then \(\frac{y-1}{y}\) decreases. (II) is true.

(III) \(y^2-y\)

We have \((y_1^2-y_1)-(y_2^2-y_2)=(y_1^2-y_2^2)-(y_1-y_2)=(y_1-y_2)(y_1+y_2-1)\)

Note that \(y_1-y_2>0\). Also \(0>y_1>y_2 \implies 0 > y_1 + y_2 \implies y_1+y_2-1<0\). Hence \((y_1-y_2)(y_1+y_2-1) < 0\).

When \(y_1 > y_2\), then \(y_1^2-y_1 < y_2^2-y_2\). Hence if \(y\) decreases then \(y^2-y\) increases. (III) is not true.

The answer is C.
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Re: If y<0, which of the following must decrease as y decreases? [#permalink]
saswata4s wrote:
If y<0, which of the following must decrease as y decreases?

I. 10y-100
II. (y-1)/y
III. y^2 - y

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) II and III only


I. 10(y-10). Will decrease as y decreases
II. (y-1)/y = 1 - 1/y. As y decreases, 1/y will increase and 1- 1/y will decrease
III. y^2 - y = y(1-y). Not necessarily decrease. it will decrease till y = 1/2 and then increase

option C
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Re: If y<0, which of the following must decrease as y decreases? [#permalink]
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Re: If y<0, which of the following must decrease as y decreases? [#permalink]
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