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gmapt prep - square root

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by bradley281 » Thu Aug 21, 2008 11:51 am
substituting numbers for x and y helps me.
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by pepeprepa » Thu Aug 21, 2008 12:03 pm
You should consider it is true and start the different inequalities: Z > W
and then try to work on it to see whether it is true or not. Use squares, cross-multiplying...
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by bha » Thu Aug 21, 2008 12:45 pm
pepeprepa wrote:You should consider it is true and start the different inequalities: Z > W
and then try to work on it to see whether it is true or not. Use squares, cross-multiplying...
..thanks..that looks like a good idea though it will be time consuming as we have to evaluate 3 eqns..thats like solving 3 problems in 1 problem time...anyway something is better than nothing...thanks!!
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by vishubn » Thu Aug 21, 2008 1:59 pm
Hi pepeprepa
Intersting approach ! i would be glad to see if u could solve the very problem mentioned !!

I would take it as an example

Vishu
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by pepeprepa » Thu Aug 21, 2008 2:40 pm
bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.

I.
1/sqrt(x+y) < sqrt(x+y)/2x
2x<x+y
x<y
It can be wrong, don't keep it.

II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y
That must be true

III.
1/sqrt(x+y) < (sqrt(x) - sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) - sqrt(y)
x+y < x - 2*sqrt(y)*sqrt(x) + y
That's wrong, don't keep it
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by uncbeers » Thu Aug 21, 2008 9:26 pm
pepeprepa wrote:bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.

I.
1/sqrt(x+y) < sqrt(x+y)/2x
2x<x+y
x<y
It can be wrong, don't keep it.

II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y
That must be true

III.
1/sqrt(x+y) < (sqrt(x) - sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) - sqrt(y)
x+y < x - 2*sqrt(y)*sqrt(x) + y
That's wrong, don't keep it
For number II and III can you say how you went from step 1 to step 2? For soe reason I am not seeing that but definitely see step 2 to step 3.

Thanks!
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by 4meonly » Fri Aug 22, 2008 7:21 am
pepeprepa wrote: II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
x+y < x + 2*sqrt(y)*sqrt(x) + y
Could you please dicribe this reasoning step-by-step

Thanx!

I solve this question by using 16 for x and 9 for y
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by pepeprepa » Fri Aug 22, 2008 7:40 am
II.
1/sqrt(x+y) < (sqrt(x) + sqrt(y))/(x+y)
I multiply both sides by something positive: (x+y)
sqrt(x+y) < sqrt(x) + sqrt(y)
I put on/at ?? square both sides (they are positive)
x+y < x + 2*sqrt(y)*sqrt(x) + y
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by bha » Fri Aug 22, 2008 10:07 am
pepeprepa wrote:bha, I tam sorry if my way does not satisfy you! If I look at the problem some seconds and I have a method which can enable me to solve it in less than 2 minutes I do it. Unfortunately, we cannot always have a better choice than what we think during the first seconds. But if you can have the right answer and not lose time that's enough.
No complaints mate... :D
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by LSB » Sat Aug 23, 2008 6:03 pm
Wow tough to see that one in less than 2 mins. BHA - out of curiosity ... how were you doing prior to this question. Trying to gauge the difficulty.
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by andes1 » Sun Aug 24, 2008 9:47 am
in less than a minute. you should evaluate all 4 ecuations.

pretend that x=1 and y=1.

the result of the question is 1/2

now the issue is to find something bigger than 1/2

the first one gives you 1/2 so is no bigger.
the second one gives you 2,8/4 so is bigger.
the trird one gives to you o/something, so is not bigger.

the logical answer is the second one.
LEARNING ENGLIS H
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