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saswata4s
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If y<0, which of the following must decrease as y decreases? 14 Mar 2017, 11:44
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

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58% (01:54) correct 42% (01:50) wrong based on 153 sessions

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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
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C
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If y<0, which of the following must decrease as y decreases? 14 Mar 2017, 17:39
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saswata4s
If y<0, which of the following must decrease as y decreases?

I. \(10y -100\)
II. \(\frac{y - 1}{y}\)
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
Show more
Dear saswata4s,

I'm happy to respond. :-) This is a great question! :-)

Statement I is relatively easy. As y decreases, any positive multiple of y will decrease, so 10y decreases. If something decreases, adding or subtracting a constant won't change the pattern, so if 10y decreases, then 10 y - 100 will decrease. Statement 1 is a yes.

Statement III is also relatively straightforward. This is a parabola. Parabolas go up on both sides, so this increase for both positive and negative values of y in the long run. Depending on where the vertex is, it might decrease then increase--we don't need to calculate the vertex. The point is, we KNOW this will not keep decreasing forever and ever. Statement III is a no.

Statement II is the tricky one. \(\frac{y - 1}{y}\) = 1 - \(\frac{1}{y}\). The one doesn't matter: that's another constant. The real question is what happens to \(-\frac{1}{y}\) when y < 0? Well, as y < 0 decreases, we will get negative numbers with larger and large absolute values. The negative sign (technically, an "opposite sign" in this case) would make all those negative positives, so this would have the same result as positive y's increasing, starting from just above zero and going to larger and large positive values. We know, in that case, \(\frac{1}{y}\) would continually decrease. Thus, for decreasing negative numbers, \(-\frac{1}{y}\) continually decreases. Statement II is a yes.

OA = (C)

Great math question!

Does all this make sense?
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If y<0, which of the following must decrease as y decreases? 20 Mar 2017, 06:51
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saswata4s
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
Show more

Let’s test each Roman numeral by using numerical values for y.

I. 10y-100

If y = -1, then 10y - 100 = -10 - 100 = -110, and if y = -2, then 10y - 100 = -20 - 100 = -120.

We see that 10y-100 WILL decrease as y decreases.

II. (y-1)/y

If y = -1, then (y-1)/y = -2/-1 = 2, and if y = -2, then (y-1)/y = -3/-2 = 1.5.

We see that (y-1)/y WILL decrease as y decreases.

III. y^2 - y

If y = -1, then y^2 - y = 1 - (-1) = 2, and if y = -2, then y^2 - y = 4 - (-2) = 6.

We see that y^2 - y WILL NOT decrease as y decreases.

Answer: C
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If y<0, which of the following must decrease as y decreases? 20 Mar 2017, 08:19
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saswata4s
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
Show more

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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If y<0, which of the following must decrease as y decreases? 21 Mar 2017, 04:05
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saswata4s
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
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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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If y<0, which of the following must decrease as y decreases? 20 Oct 2020, 20:31
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