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Math Revolution GMAT Instructor V
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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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[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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1
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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=>

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.

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GMAT 1: 700 Q48 V38 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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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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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.

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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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.

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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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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Posts: 7958
If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t  [#permalink]

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1
1
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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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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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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Joined: 02 Aug 2009
Posts: 7958
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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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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Thanks chetan2u, well understood.

Posted from my mobile device 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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