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x is an integer and x raised to any odd integer is greater [#permalink]
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08 Jun 2010, 01:14
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x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity 7^(x1)5^x? (1) z < 25 and w=7^x (2) x = 4
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Last edited by Bunuel on 20 May 2012, 11:57, edited 3 times in total.
Edited the question and added the OA



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Re: DS problem [#permalink]
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ykaiim wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity \(7^x^^1  5^x\)?
1) z < 25 and w = 7x 2) x = 4
Note: The above expression is 7^(x1) 5^x. I guess statement (1) should be: \(z<25\) and \(w=7^x\). "x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\). Q: is \(wz>5(7^{x1}5^x)\)? (1) \(z<25\) and \(w=7^x\) > is \(7^xz>5(7^{x1}5^x)\)? > is \(7^xz>5*7^{x1}5^{x+1}\)? > is \(7^x5*7^{x1}+5^{x+1}>z\)? > as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x5*7^{x1}+5^{x+1}=75+25=27\) > so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) > hence \(7^x5*7^{x1}+5^{x+1}>z\) is true. Sufficient. (2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient. Answer: A.
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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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20 May 2012, 09:33
Could you please reply with the detailed explanation for the above question. i am confused as to how just Statement I is sufficient to answer the question(A), without knowing the value of X?



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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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20 May 2012, 09:52
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Hello, let me try and explain.
paraphrasing the questions gives us, is wz > 5.7^(x1)  5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.
taking statement 1, you have the question as 7^xz > 5.7^(x1)  5^(x+1)
Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.
So, 7^x > 5.7^x1 and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:
7^x  z > 5.7^x1  5^(x+1),
Hence answer A.



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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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20 May 2012, 10:21
the question is not simple but if you think a bit abstract and test number you can arrive to the solution. The FIRST important thing, breaking the problem, is to understand tha our X is positive. Infact, if a number raised to power of 3 is >0 that means X itself is positive because the odd powr maintain the original sign of the number, so X must be positive. At this point using the exponent rules we have w  z > 5*7^x1  5^x+1 > 7^x  24 (from stem x<25) > 5*7^x1  5^x+1 testing number 1 > 7 24 > 1  25 >  17 > 24 this is TRUE. also if you test some value < 25 positive but also negative the result is the same. 1) sufficient x = 4 > w  z > 5*7^3  5^5..........at this point you know nothing about w and z. 2) insufficient A is the answer.
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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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03 Jun 2012, 01:27
Bunuel wrote: carcass wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity 7^x1  5^x?
1 z < 25 and w = 7^x 2 x = 4 Merging similar topics. Please ask if anything remains unclear. why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ?



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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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03 Jun 2012, 02:38
Kaps07 wrote: Bunuel wrote: carcass wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity 7^x1  5^x?
1 z < 25 and w = 7^x 2 x = 4 Merging similar topics. Please ask if anything remains unclear. why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ? We are told that integer x raised to to any odd integer is greater than zero: \(x^{odd}>0\). Now, if \(x=0\) then \(x^{odd}=0^{odd}=0\) (for odd>0), which violates given condition, so \(x\) cannot be zero. Hope it's clear.
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Re: Is w  z greater than 5 times the quantity 7^x1  5^x [#permalink]
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03 Jun 2012, 03:02
Thanks, I misunderstood the question. I thought it was odd int raised to the power x. It is clear now.



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Re: DS problem [#permalink]
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06 Sep 2012, 04:59
Bunuel wrote: ykaiim wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity \(7^x^^1  5^x\)?
1) z < 25 and w = 7x 2) x = 4
Note: The above expression is 7^(x1) 5^x. I guess statement (1) should be: \(z<25\) and \(w=7^x\). "x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\). Q: is \(wz>5(7^{x1}5^x)\)? (1) \(z<25\) and \(w=7^x\) > is \(7^xz>5(7^{x1}5^x)\)? > is \(7^xz>5*7^{x1}5^{x+1}\)? > is \(7^x5*7^{x1}+5^{x+1}>z\)? > as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x5*7^{x1}+5^{x+1}=75+25=27\) > so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) > hence \(7^x5*7^{x1}+5^{x+1}>z\) is true. Sufficient. (2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient. Answer: A. I don't understand the function of statement z<25. If lowest value of LHS is 27 which is more than z, then it doesn't fulfill the condition given?



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Re: DS problem [#permalink]
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02 May 2013, 12:45
Bunuel wrote: ykaiim wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity \(7^x^^1  5^x\)?
1) z < 25 and w = 7x 2) x = 4
Note: The above expression is 7^(x1) 5^x. I guess statement (1) should be: \(z<25\) and \(w=7^x\). "x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\). Q: is \(wz>5(7^{x1}5^x)\)? (1) \(z<25\) and \(w=7^x\) > is \(7^xz>5(7^{x1}5^x)\)? > is \(7^xz>5*7^{x1}5^{x+1}\)? > is \(7^x5*7^{x1}+5^{x+1}>z\)? > as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x5*7^{x1}+5^{x+1}=75+25=27\) > so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) > hence \(7^x5*7^{x1}+5^{x+1}>z\) is true. Sufficient. (2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient. Answer: A. Please explain why the lowest value of X is 1.



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Re: DS problem [#permalink]
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03 May 2013, 04:03
Rajkiranmareedu wrote: Bunuel wrote: ykaiim wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity \(7^x^^1  5^x\)?
1) z < 25 and w = 7x 2) x = 4
Note: The above expression is 7^(x1) 5^x. I guess statement (1) should be: \(z<25\) and \(w=7^x\). "x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).Q: is \(wz>5(7^{x1}5^x)\)? (1) \(z<25\) and \(w=7^x\) > is \(7^xz>5(7^{x1}5^x)\)? > is \(7^xz>5*7^{x1}5^{x+1}\)? > is \(7^x5*7^{x1}+5^{x+1}>z\)? > as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x5*7^{x1}+5^{x+1}=75+25=27\) > so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) > hence \(7^x5*7^{x1}+5^{x+1}>z\) is true. Sufficient. (2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient. Answer: A. Please explain why the lowest value of X is 1. See the highlighted part: x is an integer and x raised to any odd integer is greater than zero means that x is a positive integer, thus its lowest possible value is 1. Hope it's clear.
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Re: x is an integer and x raised to any odd integer is greater [#permalink]
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16 Jun 2013, 21:48
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ykaiim wrote: x is an integer and x raised to any odd integer is greater than zero; is w  z greater than 5 times the quantity 7^(x1)5^x?
(1) z < 25 and w=7^x (2) x = 4 Responding to a pm: What is the meaning of: x is an integer and x raised to any odd integer is greater than zero? It is just a convoluted way of saying 'x is a positive integer'. Since x raised to any odd power is positive, x must be positive. x can take values 1, 2, 3, 4 and so on. Question: Is \(w  z > 5*7^{x1}5^{x+1}\) Using statement 1: Is \(7^x  25 > 5*7^{x1}  5^{x+1}\) Even though we are given that z is less than 25, let's assume it to be 25. In case we can prove that left hand side is greater than right hand side for z = 25, we can prove that left hand side will remain greater than right hand side for any other value of z. If z has a smaller value, say 9, left had side will become even larger (since a smaller number will be subtracted) than the right hand side. Now note that \(7^x\) will be greater than \(5*7^{x  1}\) since \(7^x\) can be written as \(7*7^{x  1}\) (x is a positive integer). Also note that minimum value of x is 1 so \(5^{x + 1}\) will always be equal to or larger than 25 (x is a positive integer). So comparing the left and right hand side expressions, \(7^x\) is greater than \(5*7^{x  1}\) and 25 is less than or equal to \(5^{x+1}\). So left hand side must be greater than the right hand side in all cases when x is a positive integer. Answer (A)
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Re: x is an integer and x raised to any odd integer is greater [#permalink]
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05 May 2014, 04:24
paraphrasing the questions gives us, is wz > 5.7^(x1)  5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0. taking statement 1, you have the question as 7^xz > 5.7^(x1)  5^(x+1) Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1. So, 7^x > 5.7^x1 and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get: 7^x  z > 5.7^x1  5^(x+1), Hence answer A. tnx for this
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Re: x is an integer and x raised to any odd integer is greater [#permalink]
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18 Jul 2015, 03:24
x raised to any odd integer is greater than 0 > x = integer >0 > x>=1
(1) > wz > 7^x  5^2 = 7.7^(x1)  5^2 > 5.7^(x1)  5^2
in order to wz> 5.7^(x1)  5^(x+1), x+1 >=2 > x >=1 > sufficient.
(2) insufficient
=> A.



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