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Re: As j increases from 135 to 136 [#permalink]
08 Aug 2012, 10:23

Since as per Qs J is a positive no. We can try a pretty small no and check all options. Let say j=2 and it increase to 3 i) decrease from -3 to -8 ii) decrease from -2 to -4 iii) decrease from 1/4 to 1/9 Answer: E _________________

I y = 1 - x^2 is the equation of a downward parabola, intercepting the x axis at -1 and 1. So, far away from the roots, the values of the function are decreasing as x increases. YES

II y=x-x^2 is also the equation of a downward parabola, roots 0 and 1. Again, far away from the roots, the values of the function are decreasing as x increases. YES

III Definitely decreases, the larger the denominator, the smaller the value of the fraction. YES

Answer E _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: As j increases from 135 to 136 [#permalink]
08 Aug 2012, 10:53

SOURH7WK wrote:

Since as per Qs J is a positive no. We can try a pretty small no and check all options. Let say j=2 and it increase to 3 i) decrease from -3 to -8 ii) decrease from -2 to -4 iii) decrease from 1/4 to 1/9 Answer: C

The values of j are defined: 135 and 136! _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: As j increases from 135 to 136 [#permalink]
27 Aug 2012, 08:13

1

This post received KUDOS

take a smaller number like 2 and 3 and solve. We get the ans to be E _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: As j increases from 135 to 136 [#permalink]
10 Dec 2012, 00:31

I. 1-j^2 Of course, a larger portion is taken from 1. Definitely the value will decrease. II. j - j^2 I thought of smaller positive consecutive numbers such as 2 and 3. -2 vs. -6 shows a decrease. \frac{1}{j^2} The larger the denominater, the smaller the fraction.

Re: As j increases from 135 to 136, which of following must [#permalink]
02 Mar 2013, 23:04

I. As j increases, j^2 will increase too. Therefore 1-j^2 will decrease. II. j-j^2 = j(1-j). As j increases, 1-j will decrease and be negative. Multiplied by the increased j, this gives us a more negative figure. Therefore j-j^2 will decrease too. III. Simple reciprocal. As j increases from 135 to 136, 1/j^2 is bound to decrease

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