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If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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12 Sep 2012, 19:09
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If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y? (1) n=m+1 (2) n=2m
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Last edited by Bunuel on 13 Sep 2012, 00:36, edited 1 time in total.
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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13 Sep 2012, 01:32
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saurabhsingh24 wrote: If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?
(1) n=m+1 (2) n=2m The question is in fact "is \(3^n>2\cdot{3^m}\)?" (1) The given inequality becomes \(3^{m+1}>2\cdot{3^m}\). Dividing through by \(3^m\), which is for sure a positive number, we obtain \(3 > 2,\) obviously true. Sufficient. (2) Now the given inequality becomes \(3^{2m}>2\cdot{3^m}\). Dividing through again by \(3^m\), we obtain \(3^m>2\). This inequality holds only for \(m>0\). Not sufficient. Answer A.
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Re: Data Sufficiency [#permalink]
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12 Sep 2012, 20:58
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saurabhsingh24 wrote: If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y? (1) n=m+1 (2) n=2m There will be three cases: when M is positive: option 1 is sufficient to answer. and option 2 also give answer When M is negative: Option 1 is sufficient to answer. But option 2 is not sufficient to answer. When M is Zero: option Option 1 is sufficient to answer But Option 2 is not sufficient to answer. Try with some nos. for these three cases. If you like the post, kindly give me 1 kudos.



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Re: Data Sufficiency [#permalink]
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13 Sep 2012, 00:10
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If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y? (1) n=m+1 (2) n=2m we have to find if x > 2y i.e. 3^n > 2*3^m STAT1: n = m+1 so we have to prove that 3^(m+1) > 2* 3^m => 3 * 3^m > 2*3^m => 3 * 3^m  2*3^m > 0 => 3^m > 0 So ALL integer value of m 3^m will be greater that 0 So, SUFFICIENT STAT2: n=2m so we have to prove that 3^(2m) > 2* 3^m 3^(2m)  2* 3^m > 0 3^m* (3^m  2) > 0 for ALL values of m 3^m will be greater than 0 So, 3^m* (3^m  2) > 0 if (3^m  2) is > 0 (3^m  2) > 0 for all positive values of m (3^m  2) < 0 for 0 and all negative values of m So, Answer will be A Hope it helps!
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?
(1) n=m+1 (2) n=2m
3^n > 2*3^m ? 3^nm >2
s0 nm >=1 A sufficient
2. n =2m
if n=0 m =0 no. if n=2 m =4 yes



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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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05 Sep 2013, 08:42
saurabhsingh24 wrote: If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?
(1) n=m+1 (2) n=2m Basically the Q asks whether 3^n> 2*3^m St 1 substituting for n we get 3^(m+1)  2*3^m > 0 3* 3^m 2 *3^m>0 3^m>0 > m can be 0 or 1 or 2 and so one so n= m+1 (1 or 2 or 3 for corresponding values of m) We see that 3^n > 2*3^m So A is sufficient so ruling out B,C and E From st 2 we have n= 2m We get 3^2m 2*3^m>0 9*3^m  2*3^m> 0 or 7*3^m>0 So m =0, n=0 >putting in the above equation we get (3^2m 2*3^m>0) > 12 =1 not greater than zero But if m=1,n=2 then we get 3^2m 2*3^m>0 >96>0 3>0 so 2 ans choices possible so D ruled out Ans is A
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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18 Sep 2013, 18:35
Here is how I solved it, x = 3^n and y =3^m 1. n = m +1; therefore, x = 3^m+1 => 3 * 3^m, now 3^m = y, therefore x = 3y and hence x > 2y Sufficient 2. n =2m; x = 3^2m; x = y^2; now nothing is given about x and y; therefore for y = 0; x =0 and hence the x is not greater than 2y whereas if y =1, x = 1 again not satisfied, if y = 2, x = 4 ; x = 2y; if y = 3, x = 9; hence NS
Therefore answer is .A.
Please let me know in case my line of reasoning is not correct, especially for the second statement.



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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]
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