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If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t

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If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 03 Apr 2018, 00:22
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[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c

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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 03 Apr 2018, 00:27
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MathRevolution wrote:
[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c


1/2>1/3>1/5
So,
\(\sqrt{1/2}>\sqrt{1/3}>\sqrt{1/5}\)

option B
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 05 Apr 2018, 04:34
=>

Since \(√\frac{1}{5} < √\frac{1}{3} < √\frac{1}{2}\), we have \(√\frac{1}{5} + √\frac{1}{3} - √\frac{1}{2} < √\frac{1}{5} - √\frac{1}{3} + √\frac{1}{2} < -√\frac{1}{5} + √\frac{1}{3} + √\frac{1}{2}.\)
So \(\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}} < \sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}} < \sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\).

Thus \(c < b < a.\)

Therefore, C is the answer.

Answer: C
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 25 Apr 2018, 15:38
MathRevolution what is the answer to this?? you put OA as B and then an instructor said C
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 30 Jun 2018, 09:17
MathRevolution wrote:
=>

Since \(√\frac{1}{5} < √\frac{1}{3} < √\frac{1}{2}\), we have \(√\frac{1}{5} + √\frac{1}{3} - √\frac{1}{2} < √\frac{1}{5} - √\frac{1}{3} + √\frac{1}{2} < -√\frac{1}{5} + √\frac{1}{3} + √\frac{1}{2}.\)
So \(\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}} < \sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}} < \sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\).

Thus \(c < b < a.\)

Therefore, C is the answer.

Answer: C


MathRevolution What is the answer? OA says B but in your solution you said C.

Thanks.
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 30 Jun 2018, 09:45
msurls wrote:
MathRevolution wrote:
=>

Since \(√\frac{1}{5} < √\frac{1}{3} < √\frac{1}{2}\), we have \(√\frac{1}{5} + √\frac{1}{3} - √\frac{1}{2} < √\frac{1}{5} - √\frac{1}{3} + √\frac{1}{2} < -√\frac{1}{5} + √\frac{1}{3} + √\frac{1}{2}.\)
So \(\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}} < \sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}} < \sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\).

Thus \(c < b < a.\)

Therefore, C is the answer.

Answer: C


MathRevolution What is the answer? OA says B but in your solution you said C.

Thanks.


The answer is B. Look carefully. He derives answer B but in the last step designate the equation with wrong variables.

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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 30 Jun 2018, 20:40
MathRevolution wrote:
[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c


a = √6+√10-√15 , b = √6-√10+√15 , c = -√6+√10+√15

Hence, \(c>b>a\)
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If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 30 Jun 2018, 21:46
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MathRevolution wrote:
[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c



In very simple term..
The three variables are playing around with three square roots ---\(\sqrt{2},\sqrt{3},\sqrt{5}\)
NOW \(\sqrt{2}<\sqrt{3}<\sqrt{5}\)..
so \(√\frac{1}{5}<√\frac{1}{3}<√\frac{1}{2}\) ...
so
1) if you subtract the largest number ( \(√\frac{1}{2}\) ) from sum of two smaller numbers ( \(√\frac{1}{5}+√\frac{1}{3}\) ), it will turn out to be the smallest, hence a is smallest
2) if you subtract middle number ( \(√\frac{1}{3}\) ) from sum of largest and smallest numbers ( \(√\frac{1}{5}+√\frac{1}{2}\) ) , it will turn out to be the next smallest, hence b is next
3) if you subtract smallest number ( \(√\frac{1}{5}\) ) from sum of two larger number ( \(√\frac{1}{3}+√\frac{1}{2}\) ), it will turn out to be the largest, hence c is largest
a<b<c

B
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 12 Jul 2018, 14:29
Forget about the roots;
a=1/5+1/3_1/2=1/30
b=1/5-1/3+1/2=11/30
c:-1/5+1/3+1/2=19/30
so a<b<c
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 22 Oct 2018, 15:33
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c



In very simple term..
The three variables are playing around with three square roots ---\(\sqrt{2},\sqrt{3},\sqrt{5}\)
NOW \(\sqrt{2}<\sqrt{3}<\sqrt{5}\)..
so \(√\frac{1}{5}<√\frac{1}{3}<√\frac{1}{2}\) ...
so
1) if you subtract the largest number ( \(√\frac{1}{2}\) ) from sum of two smaller numbers ( \(√\frac{1}{5}+√\frac{1}{3}\) ), it will turn out to be the smallest, hence a is smallest
2) if you subtract middle number ( \(√\frac{1}{3}\) ) from sum of largest and smallest numbers ( \(√\frac{1}{5}+√\frac{1}{2}\) ) , it will turn out to be the next smallest, hence b is next
3) if you subtract smallest number ( \(√\frac{1}{5}\) ) from sum of two larger number ( \(√\frac{1}{3}+√\frac{1}{2}\) ), it will turn out to be the largest, hence c is largest
a<b<c

B

Hi chetan2u, my reasoning is quite similar with what you have here however, I reasoned that if a=√(1/√5+1/√3-1/√2), the value of the operation within the parenthesis will be smallest but the root of the value given by the operations within the parenthesis will be largest, hence I got a>b>c. Please chetan2u where am I getting it wrong? Thanks.

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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 22 Oct 2018, 19:23
Kem12 wrote:
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(a=\sqrt{√\frac{1}{5}+√\frac{1}{3}-√\frac{1}{2}}\) , \(b=\sqrt{√\frac{1}{5}-√\frac{1}{3}+√\frac{1}{2}}\) , \(c=\sqrt{-√\frac{1}{5}+√\frac{1}{3}+√\frac{1}{2}}\) , which of the following is true?

A. a<c<b
B. a<b<c
C. c<b<a
D. c<a<b
E. b<a<c



In very simple term..
The three variables are playing around with three square roots ---\(\sqrt{2},\sqrt{3},\sqrt{5}\)
NOW \(\sqrt{2}<\sqrt{3}<\sqrt{5}\)..
so \(√\frac{1}{5}<√\frac{1}{3}<√\frac{1}{2}\) ...
so
1) if you subtract the largest number ( \(√\frac{1}{2}\) ) from sum of two smaller numbers ( \(√\frac{1}{5}+√\frac{1}{3}\) ), it will turn out to be the smallest, hence a is smallest
2) if you subtract middle number ( \(√\frac{1}{3}\) ) from sum of largest and smallest numbers ( \(√\frac{1}{5}+√\frac{1}{2}\) ) , it will turn out to be the next smallest, hence b is next
3) if you subtract smallest number ( \(√\frac{1}{5}\) ) from sum of two larger number ( \(√\frac{1}{3}+√\frac{1}{2}\) ), it will turn out to be the largest, hence c is largest
a<b<c

B

Hi chetan2u, my reasoning is quite similar with what you have here however, I reasoned that if a=√(1/√5+1/√3-1/√2), the value of the operation within the parenthesis will be smallest but the root of the value given by the operations within the parenthesis will be largest, hence I got a>b>c. Please chetan2u where am I getting it wrong? Thanks.

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Hi
If x is positive and x<1 .....√x>x
But if x and y are positive and both <1 and x>y....√x>√y...

So Squareroots greater are only when compared to the same variable. The same dies nit hold when the values under the square root are different..
For example...0.09>0.04.....√0.09=0.3 and √0.04=0.2 so √0.09>√0.04 too
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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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New post 23 Oct 2018, 18:22
Thanks chetan2u, well understood.

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Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t   [#permalink] 23 Oct 2018, 18:22
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