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Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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21 Mar 2016, 04:31
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Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1 * A solution will be posted in two days.
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Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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21 Mar 2016, 06:35
MathRevolution wrote: Is 1+x+x^2+x^3+x^4<1/(1x)?
1) x>0 2) x<1
* A solution will be posted in two days. INFO from Q stem \(1+x+x^2+x^3+x^4<\frac{1}{(1x)}\).. \(1+x+x^2+x^3+x^4\frac{1}{(1x)}<0\).. \(\frac{{(1+x+x^2+x^3+x^4)(1x)1}}{(1x)}<0\).. \(\frac{{1+x+x^2+x^3+x^4xx^2x^3x^4x^5)1}}{(1x)}<0\).. \(\frac{x^5}{(1x)}<0\).. \(\frac{x^5}{(x1)}<0\).. so two conditions1) if x>0, x<1 2) if x<0, x>1 ...... Not possible lets check the statements 1) x>0 if x<1.. ans is YES if x>1.. ans is NO Insuff
2) x<1 if 0<x<1.. ans is YES if x<0.. ans is NO Insuff
combined.. Suff as per point 1, ans is always YES C
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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22 Mar 2016, 17:59
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1 The question is (1x^5)/(1x)<1/(1x)? and –x^5/(1x)<0?, x^5/(x1)<0? is derived when delete 1/(1x) from the both equations of 1/(1x)x^5/(1x)<1/(1x)?. When you multiply (x1)^2 to the both equations, x^5(x1)<0? is derived. Divide it with x^4 and it becomes x(x1)<0?. That is, 0<x<1? is derived, which makes C the answer.
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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26 Mar 2016, 21:18
Is 1+x+x^2+x^3+x^4<1/(1x)?
Although the series on L.H.S. is small otherwise we culd hav used G.P. for the sum .. 1(x^51)/(x1) <1/(1x) x^5/(x1)<0
(+ve)0(ve)1(+ve) So if x lies between 0 and 1 then we have YES for the above equation else NO.
1) x>0. Insufficient. 2) x<1. Insufficient.
Combining Sufficient. C is the answer.



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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25 Apr 2016, 01:16
MathRevolution wrote: Is 1+x+x^2+x^3+x^4<1/(1x)?
1) x>0 2) x<1
* A solution will be posted in two days. I took x=1,2 and x=1/2,0 C
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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05 Sep 2017, 06:24
Ans is Cis \(1+x+x^2+x^3+x^4\) < \(\frac{1}{(1x)}\) Multiply and divide by 1x on both numerator and denominator on L.H.S. equation becomes is \(\frac{(1x^5)}{(1x)}\) < \(\frac{1}{(1x)}\) or \(\frac{x^5}{(1x)}\) > 0 Taking 3 points 1, 0 and 1 on number line value above 1 i.e. x>1 will not hold this equation x<1 but >0 will hold this equation x<0 but > 1 will again not hold equation x< 1 will again satisfy equation so soln is x< 1 and x should be from (0,1) open brackets i.e. 0 and 1 not included. 1) x>0 is insufficient as it includes x>1 also which is not part of solution. A,D eliminated 2) x<0 is insufficient as it includes 1<x<0 which is not part of solution B eliminated combine both 0<x<1 is our solution and Ans is C
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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18 Sep 2017, 00:14
I have a query  If I take (C) > it means x = 0.5, 0.3, 0.2 With x = 0.5 1+ x + x sq + x^3 + x^4 < 1/1x 1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/10.5 1.93< 2 SUFF if x= 0.3 1+0.3+0.09 + 0.027 + 0.0081 = < 1/1  0.3 1.42 no less than 0.7 if x = 0.2 1+0.2+0.04 + 0.008 + 0.0016 <1/ 10.2 1.24 not less than 0.8 So I chose E. Can anyone explain the issue in my reasoning here?



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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18 Sep 2017, 00:26
Madhavi1990 wrote: I have a query  If I take (C) > it means x = 0.5, 0.3, 0.2 With x = 0.5 1+ x + x sq + x^3 + x^4 < 1/1x 1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/10.5 1.93< 2 SUFF if x= 0.3 1+0.3+0.09 + 0.027 + 0.0081 = < 1/1  0.3 1.42 no less than 0.7 if x = 0.2 1+0.2+0.04 + 0.008 + 0.0016 <1/ 10.2 1.24 not less than 0.8 So I chose E. Can anyone explain the issue in my reasoning here? It's \(\frac{1}{1  0.3}\) NOT \(\frac{1}{1} 0.3\). \((1 + 0.3 +0.3^2 + 0.3^3 + 0.3^4 = 1.4251)< (\approx 1.428=\frac{1}{1  0.3})\) \((1 + 0.2 +0.2^2 + 0.2^3 + 0.2^4 = 1.2496)< (1.25=\frac{1}{1  0.3})\)
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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18 Sep 2017, 00:41
hi, sorry I wasn not clear on my post earlier. So after x = 0.2 I got 1.42 < 1/0.7 > 1.42 < 10/7 which is approx 1.4 Thus 1.42 < 1.44 (TRUE) 1.24 < 1/0.8 1.24 < 5/4 1.24 < 1.25 (TRUE) if x = 0.6 then 2.3 < 2.5 (TRUE) Thank you, the above cleared it up for me. Ans C



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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03 Dec 2017, 16:05
MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Is 1+x+x^2+x^3+x^4<1/(1x)?
1) x>0 2) x<1
The question is (1x^5)/(1x)<1/(1x)? and –x^5/(1x)<0?, x^5/(x1)<0? is derived when delete 1/(1x) from the both equations of 1/(1x)x^5/(1x)<1/(1x)?. When you multiply (x1)^2 to the both equations, x^5(x1)<0? is derived. Divide it with x^4 and it becomes x(x1)<0?. That is, 0<x<1? is derived, which makes C the answer. Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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04 Dec 2017, 11:22
Buttercup3 wrote: MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Is 1+x+x^2+x^3+x^4<1/(1x)?
1) x>0 2) x<1
The question is (1x^5)/(1x)<1/(1x)? and –x^5/(1x)<0?, x^5/(x1)<0? is derived when delete 1/(1x) from the both equations of 1/(1x)x^5/(1x)<1/(1x)?. When you multiply (x1)^2 to the both equations, x^5(x1)<0? is derived. Divide it with x^4 and it becomes x(x1)<0?. That is, 0<x<1? is derived, which makes C the answer. Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in? Here algebra is slightly better but yes, it can be done by plugging in numbers. Let me try: (1) x > 0 Now RHS of the question stem is 1/(1x). Since its 1x in the denominator, and this statement says x > 0, we should immediately look for two cases: one where 0<x<1 and one where x>1 (x cannot be 1 as for that RHS would be undefined). So lets take a value of x between 0 and 1, say 0.5. RHS becomes 1/(10.5) = 2 LHS becomes 1+0.5+0.25+0.125+0.0625=1.9375. LHS is less than RHS If we try any other value in the same range say 0.2 or 0.9, still RHS would be greater. Now lets take another case where x > 1, in this case clearly we can see that 1x would be negative, so RHS would be negative while LHS positive. So LHS would become greater. So not sufficient. (2) x < 1 We have got to take cases where 0<x<1, where 1<x<0, and where x<1 If 0<x<1, RHS would be greater as discussed in first statement. If 1<x<0, lets say x=0.5. RHS = 1/(1+0.5) = 0.666.. and LHS = 10.5+0.250.125+0.0625 = 0.6875. LHS is greater We dont even need to look at third case because we can see that this statement is not sufficient. Combining the two statements: 0<x<1. In that case RHS would be greater as discussed. Sufficient. Hence C answer



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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29 Jan 2019, 04:20
I would be endlessly grateful, if any expert could help me see a mistake in my method. For me the answer is A. Is 1+x+x^2+x^3+x^4<1/(1x)? 1. x>0 Additionally, x cannot be 0 because in case we plug in 0 we get 1<1; Therefore, we can devide both sides of the inequality by x. 1+x+x^2+x^3+x^4<1/(1x) x+x^2+x^3+x^4<1/(1x)1 x+x^2+x^3+x^4<x/(1x) (devide both sides by x) 1+x+x^2+x^3<1/(1x) x+x^2+x^3<1/(1x)1 x+x^2+x^3<x/(1x) (devide both sides by x) 1+x+x^2<1/(1x) x+x^2<1/(1x)1 x+x^2<x/(1x) (devide both sides by x) 1+x<1/(1x) x<1/(1x)1 x<x/(1x) (devide both sides by x) 1<1/(1x) 0<x/(1x) th values x can accept here 0<x/(1x) are within 0<x<1 A. sufficient
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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05 Sep 2019, 02:13
ShukhratJon wrote: I would be endlessly grateful, if any expert could help me see a mistake in my method. For me the answer is A.
Is 1+x+x^2+x^3+x^4<1/(1x)?
1. x>0
Additionally, x cannot be 0 because in case we plug in 0 we get 1<1; Therefore, we can devide both sides of the inequality by x.
1+x+x^2+x^3+x^4<1/(1x) x+x^2+x^3+x^4<1/(1x)1 x+x^2+x^3+x^4<x/(1x) (devide both sides by x) 1+x+x^2+x^3<1/(1x) x+x^2+x^3<1/(1x)1 x+x^2+x^3<x/(1x) (devide both sides by x) 1+x+x^2<1/(1x) x+x^2<1/(1x)1 x+x^2<x/(1x) (devide both sides by x) 1+x<1/(1x) x<1/(1x)1 x<x/(1x) (devide both sides by x) 1<1/(1x) 0<x/(1x)
th values x can accept here 0<x/(1x) are within 0<x<1 A. sufficient We can only divide on both sides of an equation if we know the sign of the variable. So, we can't just divide by x on both sides. Hope this helps. Cheers



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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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23 Sep 2019, 22:31
anud33p wrote: We can only divide on both sides of an equation if we know the sign of the variable. So, we can't just divide by x on both sides.
Hope this helps. Cheers
Hi anud33p, Actually we can while analyzing St1, as I did in my previous post. St 1 says that x>0, so we are sure that x doesn’t affect the inequality. I’m dividing both sides by x, considering St 1, so that’s fine. I keep dividing both sides until I get much much simple inequality x/(x1)<0. What I failed to do in my rather old post is to infer the correct conclusion. Since x/(x1)<0 and x*(x1)<0 have the same set of solutions, the answer is 0<x<1. We can see that for the original inequality to hold true x needs to be both greater than 0 and less than 1. We get that constraint only when we combine both statements, so the correct answer is C. To my mind that’s a much easier way to come to the correct answer. We just need to keep dividing both sides by x until we get a simple inequality to manipulate.
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Re: Is 1+x+x^2+x^3+x^4<1/(1x)? 1) x>0 2) x<1
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