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If a=15+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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02 Apr 2018, 23:22
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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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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02 Apr 2018, 23:27
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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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05 Apr 2018, 03:34
=> 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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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25 Apr 2018, 14:38
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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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30 Jun 2018, 08:17
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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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30 Jun 2018, 08:45
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. Posted from my mobile device



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Re: If a=15+1312 , b=1513+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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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30 Jun 2018, 20:46
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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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12 Jul 2018, 13:29
Forget about the roots; a=1/5+1/3_1/2=1/30 b=1/51/3+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+1312 , b=1513+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/√31/√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. Posted from my mobile device



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Re: If a=15+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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22 Oct 2018, 18:23
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/√31/√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. Posted from my mobile deviceHi 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+1312 , b=1513+12 , c=15+13+12 , which of the following is t
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23 Oct 2018, 17:22
Thanks chetan2u, well understood. Posted from my mobile device




Re: If a=15+1312 , b=1513+12 , c=15+13+12 , which of the following is t &nbs
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