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Solve for x: 0<x4x<5 = ? [#permalink]
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16 Aug 2009, 00:44
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Solve for x: 0<x4x<5 = ? A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0
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Last edited by Bunuel on 04 Dec 2012, 03:22, edited 2 times in total.
Renamed the topic and edited the question.



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Re: solve for x? [#permalink]
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16 Aug 2009, 02:08
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I am getting E as an answer if x is ve. If x is ve, we get 0 < 5x < 5 Dividing both sides by 5, we flip both the sides and we land up on 0 > x > 1 ie 1 < x < 0. Is that the correct method to solve this problem? Please explain.
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Re: solve for x? [#permalink]
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16 Aug 2009, 13:54
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IMO E:
5 + 4x > x 5 + 4x > x first cndtn 5 >  3x x>5/3
or
5 + 4x > x x > 1
using
4x < x or 4x < x x<0
or 4x < x x<0
thus range lies between 1 to 0
correct me if i m wrong!!



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Re: solve for x? [#permalink]
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16 Aug 2009, 14:10
0<x4x<5 = ?
A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0
i did it this way
split the inquality into 2
a) /x/4x<5 ie: 5<x4x<5 ie: 5<3x<5 ie: 5/3>x>5/3....1
b) /x/4x>0 thus /x/>4x thus eithe x>4x ie: 3x>0 ie: x <0 or x<4x ie: x>4x ie: x>0
and i get lost here??



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Re: solve for x? [#permalink]
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03 Apr 2011, 13:40
0<5x<5
=> 1<x<0
Answer E.



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Re: solve for x? [#permalink]
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03 Apr 2011, 18:15
I'm also getting E. 2 cases x > 0 or x < 0 0< x  4x < 5 => 0< 5x < 5 => x > 1 and x < 0 x  4x < 5 => 3x < 5 => x > 5/3 so x > 0 as x is +ve So 1 < x < 0 Answer E
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Re: solve for x? [#permalink]
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14 Jun 2011, 15:22
hi, apply definition absolute value obtain two possibilites: A. if x≥0 then 0<x4x<5 go 0<x4x<5 go 0<3x<5, go 0>x>5/3. The intersection is empty B. if x<0 then 0<x4x<5 go 0<x4x<5 go 0<5x<5 go 0>x>1. the intersection is 1<x<0
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Re: solve for x? [#permalink]
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14 Jun 2011, 20:24
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puneetj wrote: Got to the correct answer but took too much time...E If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer. 0<x4x<5 = ? A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0 Say, consider option A. If x = 1, x4x = 5 which doesn't satisfy the inequality so A is out. If x = 1/2, x4x is negative so B and C are out. If x = 4/5, x4x = 4 so D is out and E is the answer.
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Re: solve for x? [#permalink]
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03 Dec 2012, 21:52
0<x4x<5 A. x<0 Test: x=5 \(54(5) = 5 + 20\) is not less than 5 FALSE! B. 0<x<1 Test: x=\(\frac{1}{4}\) \(\frac{1}{4}4(\frac{1}{4})=\frac{3}{4} is not greater than 0\) FALSE! C. 3/5<x<1 x<1 as tested with B FALSE! D. 3/5<x<0 E. 1<x<0 We see that D and E are almost the same except for E. covers 3/5 unlike D. Let x=3/5 \(\frac{3}{5}4(\frac{3}{5})=\frac{15}{5}=3\) Answer: E
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Re: solve for x? [#permalink]
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10 May 2013, 08:39
VeritasPrepKarishma wrote: puneetj wrote: Got to the correct answer but took too much time...E If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer. 0<x4x<5 = ? A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0 Say, consider option A. If x = 1, x4x = 5 which doesn't satisfy the inequality so A is out. If x = 1/2, x4x is negative so B and C are out. If x = 4/5, x4x = 4 so D is out and E is the answer. I believe the choice of 4/5 to execlude D is wrong 4/5 is not in the range of 3/5<x<0 ????? accordingly i think both D and E could solve as the right range in my opinion is 5/3 < x < 0??? am i right or wrong plz advise!



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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10 May 2013, 23:46
yezz wrote: Solve for x: 0<x4x<5 = ?
A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0 x4x>0 = x>4x =\(\frac{x}{x}\)<1 > x<0. Again, x4x<5 = x4x<5 = 5x<5>x>1. E.
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Re: solve for x? [#permalink]
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11 May 2013, 04:33
yezz wrote: VeritasPrepKarishma wrote: puneetj wrote: Got to the correct answer but took too much time...E If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer. 0<x4x<5 = ? A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0 Say, consider option A. If x = 1, x4x = 5 which doesn't satisfy the inequality so A is out. If x = 1/2, x4x is negative so B and C are out. If x = 4/5, x4x = 4 so D is out and E is the answer. I believe the choice of 4/5 to execlude D is wrong 4/5 is not in the range of 3/5<x<0 ????? accordingly i think both D and E could solve as the right range in my opinion is 5/3 < x < 0??? am i right or wrong plz advise!
Focus on "and if there is even one value outside the given range that does satisfy the inequality, it is not the answer." given above. 4/5 is a value which satisfies 0 < x4x < 5 since 4/54(4/5) = 4. Since 4/5 does not lie in the range 3/5<x<0 so (D) cannot be the answer. The correct range needs to cover all possible values of x.
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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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11 May 2013, 06:31
@Karishma ... U r absolutely right , been away for years from GMAT ... E is the perfect answer



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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12 May 2013, 07:44
we ve got 3 critical values that are ( 0 from modulus , 5/3& 1 from /x/ = 5+4x) , draw on the number line and test values
.....5/3........1...........0....................
only 1<x<0 is the range where all values of x satisfy the compound inequality 0</x/4x<5



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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Solve for x: 0<x4x<5 = ?
A. x<0 B. 0<x<1 C. 3/5<x<1 D. 3/5<x<0 E. 1<x<0
There are two options here  plugging in values given to us in the answer choices or simplifying the inequality.
0<x4x<5
x>0: 0<x4x<5 0<3x<5 0<x<5/3 5/3<x<0 INVALID as x does not fall within the range of x>0 OR x<0: 0<(x)4x<5 0<5x<5 0<x<1 1<x<0 VALID as x falls within the range of x<0
There is only one valid solution: 1<x<0. (E)



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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29 Jan 2014, 08:05
Would be happy to hear some comments on whether this approach is correct
x4x>0
So we have two cases
If x>0 then x4x>0 3x>0 x<0, this contradicts and hence is not a valid solution
If x<0 then 5x>0 x<0, this solution is valid
So we get that 1<x<0 replacing in the original inequality
E



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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jlgdr wrote: Would be happy to hear some comments on whether this approach is correct
x4x>0
So we have two cases
If x>0 then x4x>0 3x>0 x<0, this contradicts and hence is not a valid solution
If x<0 then 5x>0 x<0, this solution is valid
So we get that 1<x<0 replacing in the original inequality
E Knowing only x < 0, how do you choose between (A), (D) and (E)? You need to consider x  4x < 5 too When x < 0, x 4x < 5 5x < 5 x > 1 That's how you get 1 < x < 0 Or work on the whole inequality in one go 0 < x  4x < 5 When x < 0, 0< x  4x < 5 0 < 5x < 5 0 < x < 1 0 > x > 1 which is the same as 1 < x < 0
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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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25 Apr 2015, 21:50
Hey, So I had a doubt. For the equaltiy: 0<x4x<5 if I try and solve it algebraically, i first take x<0 In that case won't this equality be: 0<x4(x)<5. I can't just substitute mod x with x and leave the other x be can I? Please help! Thanks!



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Re: Solve for x: 0<x4x<5 = ? [#permalink]
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natashakumar91 wrote: Hey, So I had a doubt. For the equaltiy: 0<x4x<5 if I try and solve it algebraically, i first take x<0 In that case won't this equality be: 0<x4(x)<5. I can't just substitute mod x with x and leave the other x be can I? Please help! Thanks! Say you have an inequality: 4x < 5 and you know that x must be negative. How will you solve the inequality? Will you say that the inequality becomes 4x < 5? No. You are given that 4x < 5. Without changing the inequality, you can write this as x < 5/4. x needs to be negative. All negative values will be less than 5/4. Why do you substitute x in place of x? You cannot solve an equation/inequality with x in it. You need to remove the absolute value sign. You know that x = x if x is positive and x = x if x is negative. Since you know that x is negative, you can write x in place of x without changing the inequality. If you change the simple x to x in the inequality, the inequality changes. Check out this post for a more detailed explanation: http://www.veritasprep.com/blog/2014/06 ... thegmat/
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