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by sachin_yadav » Thu Aug 04, 2011 10:09 am
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

Hi Experts,

please help

I am confused on the square root of 0.345. I have checked all the links and read the O.G explanation. It says if the exponent is lower, the value is greater.

For instance, r = 1/4, s = (1/4)^2, and t = sqrt 1/4

In this case r > s and t > r > s

but if r = .4, s = (.4)^2, and t = sqrt 0.4 then t = 0.2
In this case t < r because 0.2 < 0.4. So, s < t < r (0.16 < 0.2 < 0.4)

I am confused please help. I don't know what i am missing. I even tried one more example

sqrt 0.81 = 0.9. Now 0.9 > 0.81. This is different.

Looking forward to you reply.

Thanks
Sachin
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Source: — Problem Solving |

by Frankenstein » Thu Aug 04, 2011 10:33 am
Hi,
but if r = .4, s = (.4)^2, and t = sqrt 0.4 then t = 0.2
In this case t < r because 0.2 < 0.4. So, s < t < r (0.16 < 0.2 < 0.4)
sqrt(0.04) = 0.2
sqrt(0.4) is around 0.63.
I hope this helps you.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by sachin_yadav » Thu Aug 04, 2011 10:47 am
Frankenstein wrote:Hi,
but if r = .4, s = (.4)^2, and t = sqrt 0.4 then t = 0.2
In this case t < r because 0.2 < 0.4. So, s < t < r (0.16 < 0.2 < 0.4)
sqrt(0.04) = 0.2
sqrt(0.4) is around 0.63.
I hope this helps you.
Oh.......wait how can i miss this.....Oh no.... I am so stupid.:shock:

Thanks Frankenstein
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by GMATGuruNY » Thu Aug 04, 2011 11:33 am
sachin_yadav wrote: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

Hi Experts,

please help

I am confused on the square root of 0.345. I have checked all the links and read the O.G explanation. It says if the exponent is lower, the value is greater.

For instance, r = 1/4, s = (1/4)^2, and t = sqrt 1/4

In this case r > s and t > r > s

but if r = .4, s = (.4)^2, and t = sqrt 0.4 then t = 0.2
In this case t < r because 0.2 < 0.4. So, s < t < r (0.16 < 0.2 < 0.4)

I am confused please help. I don't know what i am missing. I even tried one more example

sqrt 0.81 = 0.9. Now 0.9 > 0.81. This is different.

Looking forward to you reply.

Thanks
Sachin
Ignore the numbers given. We can plug in any fraction between 0 and 1.
Let r = 1/4, s = (1/4)² = 1/16, and t = √(1/4) = 1/2.
The smallest value is s = 1/16, the largest value is t = 1/2.

The correct answer is D.
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by Brian@VeritasPrep » Thu Aug 04, 2011 11:35 am
Nice explanation, Frankenstein!

And, Sachin, another way to look at this one is to just recognize that the GMAT loves to test the properties of numbers. They used 0.345 because it's a ridiculous number to try to calculate, but if you see that the property that that number will hold is one "of a fraction", you can test the relationship much more easily.

If you square a fraction*, it goes down (1/2)^2 = 1/4

So by that same logic, if you take the square root (do the opposite) it should go up: sqrt(1/4) = 1/2.

So the square root should be the highest, the "standard" should be in the middle, and the square should be the lowest. And you can get that without really having to do much math.


One other suggestion - I'd use fractions instead of decimals whenever possible, because the math is generally more intuitive using fractions. When you slipped up on the square root, my guess is that it was because you were using decimals and it's not as intuitive to move a decimal point as it is to say:

sqrt (1/4) = sqrt 1 / sqrt 4 = 1/2

the presence of both numerator and denominator in a fraction really makes you deal with both in a case like this...


*Technically it's not "any fraction", but rather a fraction between 0 and 1. But personally when I'm thinking of properties of numbers, rather than put in a ton of caveats and definitions I just think of "a fraction" as a positive proper fraction for the sake of making the description to myself easier to follow.
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by sachin_yadav » Fri Aug 05, 2011 9:10 am
Thanks Mitch sir and Brian sir for this fantastic explanation.

Regards
Sachin
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