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If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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07 Nov 2006, 10:50
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If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of the following is the correct ordering of r, s, and t ? A. r < s < t B. r < t < s C. s < t < r D. s < r < t E. t < r < s OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifr0345s ... 32067.html
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Last edited by Bunuel on 03 Oct 2017, 01:09, edited 3 times in total.
Edited the question and added the OA



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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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07 Nov 2006, 11:26
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agree D, for simplicity u could just replace the decimal with 1/4
r=1/4
s=r^2=1/8
t=sqrt s=1/2
s<r<t



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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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28 Aug 2011, 06:29
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petrifiedbutstanding wrote: Can someone provide a good solution to this? This problem is pretty straightforward, but I'd definitely be confused if it got a lil more complex on the real GMAT... Any help will be greatly appreciated. For, 0<x<1; the higher power of x always be smaller than lower power of x. 0<x<1: \(x^4<x^3<x^2<x^1<x^{\frac{1}{2}}<x^{\frac{1}{4}}\) x=1; \(x^1=x^2=x^3=x^{real numbers}=1\) x>1; \(x^{\frac{1}{4}}<x^{\frac{1}{2}}<x^1<x^2<x^3<x^4\) For x<0; it gets subtle because all even powers become +ve. Try some examples and see what you get. But, there too: consider 0>x>1; x=1; 1>x as three separate case.
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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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11 Apr 2012, 00:47
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dzodzo85 wrote: I am also in the same boat on this question, can someone give a much better detailed solution to this problem? If \(r=0.345\), \(s=0.345^2\), and \(t=\sqrt{0.345}\), which of the following is the correct ordering of \(r\), \(s\), and \(t\) ? A. r < s < t B. r < t < s C. s < t < r D. s < r < t E. t < r < s If \(x\) is in the range \(0<x<1\) then: \(0<x^2<x<\sqrt{x}<1\). Consider the simpler example: if \(x=\frac{1}{4}\) then \(0<\frac{1}{16}<\frac{1}{4}<\frac{1}{2}<1\). So, \(s<r<t\). Answer: D.
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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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26 Jul 2012, 02:58
Hi bunuel, thanks for that example. I had solved it by keeping the fraction itself. Thanks
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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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26 Apr 2013, 05:40
Hi Mr Bunuel.... I did solve this question by replacing .04 to .345...........
Can it be wise to tackle it by replacement theory????



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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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26 Apr 2013, 07:16
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Simply follow this rule, If x < 1, it reduces in value with each power>1. So for \(x < 1, x>x^2>x^3>x^4..\) and so on In the current example, \(t=sqrt(0.345)\) (a value less than 1) \(r = t^2\) \(s = r^2 = t^4\) So, as per our above rule, \(t^4<t^2<t\) implying \(s<r<t\)
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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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22 May 2014, 03:20
Another similar question for practice: \(a = 0.283\) \(b = a^2\) \(c = a^3\) \(d = a^4\) Which of the following options correctly describes the relationship between the 4 numbers? (A) a<b<c<d (B) a<c< d<b (C) c<a<d<b (D) a<c<b<d (E) d<c<b<a SOLUTION The simplest way to solve the question is this: Simplify the value of a to 0.3. So, b= 0.09 c = 0.027 d = 0.0081 So, the correct order will be Option C: c<a<d<b A Tip!If a had been a difficult number like, say a= 0.784, then instead of simplifying it by rounding it off to 0.8, I suggest you simply take an easier negative decimal, say 0.2, and calculate the relationship between a,b,c,d for it. That relationship would hold true for any negative decimal between 0 and 1. This tip will save you time because calculating the cube and fourth power of 0.8 will be timeconsuming.
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Last edited by JapinderKaur on 22 May 2014, 03:25, edited 1 time in total.



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Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of [#permalink]
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03 Oct 2017, 01:09




Re: If r = 0.345, s = (0.345)^2, and t = sqrt(0.345) , which of
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