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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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02 Apr 2018, 23:22
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Difficulty:
45% (medium)
Question Stats:
68% (01:14) correct 32% (01:05) wrong based on 254 sessions
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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?
Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t
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02 Apr 2018, 23:27
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, \(\sqrt{1/2}>\sqrt{1/3}>\sqrt{1/5}\)
Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t
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30 Jun 2018, 19:40
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
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30 Jun 2018, 20:46
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
Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t
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22 Oct 2018, 14:33
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.
Re: If a=15+13-12 , b=15-13+12 , c=-15+13+12 , which of the following is t
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22 Oct 2018, 18:23
1
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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68% (01:14) correct 
