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If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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19 Jun 2017, 04:44
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If q, s, and t are all different numbers, is q < s < t ? (1) t  q = t  s + s  q (2) t > q
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Last edited by Bunuel on 19 Jun 2017, 04:59, edited 1 time in total.
Renamed the topic and edited the question.



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If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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If q, s, and t are all different numbers, is q < s < t ?(1) t  q = t  s + s  q Notice that the right hand side is positive (it's the sum of two absolute values, so two nonnegative values, in fact, in our case two positive values, since we know that the variables are distinct). Thus the left hand side must also be positive, which means that t > q. So, we can have 3 cases for s: a. sqtIn this case \(s < q < t\): \(t  s > 0\) and \(s  q < 0\), which would mean that \(t  s = t s\) and \(s  q = (s  q)\) (recall that x = x when x > 0 and x = x when x <= 0). So, \(t  s + s  q = (t s)  (s  q) = t  2s + q\). So, in his case we'd have \(t  q = t  2s + q\) or \(q=s\). But we are told that q, s, and t are all different numbers, so this case is out. b. qstIn this case \(q < s < t\): \(t  s > 0\) and \(s  q > 0\), which would mean that \(t  s = t s\) and \(s  q = s  q\). So, \(t  s + s  q = (t s) + (s  q) = t  q\). This matches the info given in the statement. c. qtsIn this case \(q < t < s\): \(t  s < 0\) and \(s  q > 0\), which would mean that \(t  s = (t s)\) and \(s  q = s  q\). So, \(t  s + s  q = (t s) + (s  q) = t + 2s  q\). So, in his case we'd have \(t  q = t + 2s  q\) or \(t=s\). But we are told that q, s, and t are all different numbers, so this case is out. Only q < s < t case is possible. Sufficient. (2) t > q. Not sufficient. Answer: A.
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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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19 Jun 2017, 13:27
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AbdurRakib wrote: If q, s, and t are all different numbers, is q < s < t ?
(1) t  q = t  s + s  q
(2) t > q With less math: Start with statement 2. This is insufficient, since s could be greater than t, or s could be between q and t. Eliminate answers B and D.Statement 1: When you see x  y, think 'distance between x and y on the number line'. That's all that means. So, this statement says that t  q is equal to the distance between t and s, plus the distance between s and q. In other words, s has to be between t and q. Jot down some diagrams on your paper to convince yourself of that: in order for the distances to make sense, s has to be in the middle. Also, tq has to be positive, since it's the sum of two absolute values. So, t is greater than q. If t is greater than q and s is in the middle, you know that q < s < t. Sufficient.
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If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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22 Aug 2017, 06:38
Could someone please explain the little struggle I'm having with Option B)?
If we want to prove that q < s < t, that means that also q < t (regardless of the placement of s).
Now B tells us, that q > t, hence q < t can't be true at the same time? Can't I conclude that q < s < t doesn't hold when q > t and clearly answer the question with a no?
What mistake am I making here?



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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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22 Aug 2017, 07:10
Bunuel wrote: Vezax27 wrote: Could someone please explain the little struggle I'm having with Option B)?
If we want to prove that q < s < t, that means that also q < t (regardless of the placement of s).
Now B tells us, that q > t, hence q < t can't be true at the same time? Can't I conclude that q < s < t doesn't hold when q > t and clearly answer the question with a no?
What mistake am I making here? We want to answer whether q < s < t and (2) says that t > q, NOT q > t. Oh gosh, how stupid of me... Guess I need a break for the day



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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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17 Nov 2017, 12:48
AbdurRakib wrote: If q, s, and t are all different numbers, is q < s < t ?
(1) t  q = t  s + s  q
(2) t > q We need to determine whether q < s < t, given that q, s, and t are all different numbers. Statement One Alone: t  q = t  s + s  q Since q, s, and t are different numbers, both t  s and s  q are positive quantities, and their sum t  s + s  q will also be positive. This also makes the lefthand side t  q positive. Since t  q > 0, we have t > q. We know t > q, but we still have to determine whether s is between them. That is, is q < s < t? We have three scenarios to consider. (1) If q < s < t, then t > s and s > q, and then: t  q = t  s + s  q t  q = t  q We see that this equation holds true: t  q = t  s + s  q, and furthermore q < s < t. (2) If s < q < t, then t > s and q > s, and thus t s is positive while s  q is negative, and we have: t  s + s  q t  s + [(s  q)] t  s  s + q t  2s + q ≠ t  q Since t  2s + q ≠ t  q, the equation does not hold and we can’t have s < q < t. (3) If q < t < s, then s > t and s > q, and thus t  s is negative while s  q is positive, and we have: t  s + s  q (t  s) + s  q t + s + s  q t + 2s  q Since t + 2s  q ≠ t  q, we see that the equation does not hold, so we can’t have q < t < s. We see that only scenario 1 is true if t  q = t  s + s  q, and we do have q < s < t. Statement one alone is sufficient. Statement Two Alone: t > q We know t > q, but we still have to determine whether s is between them. It’s possible that q < s < t, but it is also possible that s < q < t or q < t < s. Since we don’t know anything about s, we can’t determine which case is valid. Statement two alone is not sufficient. Answer: A
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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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15 Dec 2017, 10:11
the fastest way to solve this is to draw the number line qts look at 1 absolute value is the length of the line section if s is in between, the answer is yes s can not be outside qt because if s is so, we can not have condition 1. remember /sq/= the line of sq. length sq can not be length of sq+length of ts.
that is all.



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Re: If q, s, and t are all different numbers, is q < s < t ? [#permalink]
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23 Dec 2017, 14:26
Bunuel wrote: If q, s, and t are all different numbers, is q < s < t ?
(1) t  q = t  s + s  q
Notice that the right hand side is positive (it's the sum of two absolute values, so two nonnegative values, in fact, in our case two positive values, since we know that the variables are distinct). Thus the left hand side must also be positive, which means that t > q. So, we can have 3 cases for s:
a. sqt In this case \(s < q < t\): \(t  s > 0\) and \(s  q < 0\), which would mean that \(t  s = t s\) and \(s  q = (s  q)\) (recall that x = x when x > 0 and x = x when x <= 0). So, \(t  s + s  q = (t s)  (s  q) = t  2s + q\).
So, in his case we'd have \(t  q = t  2s + q\) or \(q=s\). But we are told that q, s, and t are all different numbers, so this case is out.
b. qst In this case \(q < s < t\): \(t  s > 0\) and \(s  q > 0\), which would mean that \(t  s = t s\) and \(s  q = s  q\). So, \(t  s + s  q = (t s) + (s  q) = t  q\).
This matches the info given in the statement.
c. qts In this case \(q < t < s\): \(t  s < 0\) and \(s  q > 0\), which would mean that \(t  s = (t s)\) and \(s  q = s  q\). So, \(t  s + s  q = (t s) + (s  q) = t + 2s  q\).
So, in his case we'd have \(t  q = t + 2s  q\) or \(t=s\). But we are told that q, s, and t are all different numbers, so this case is out.
Only q < s < t case is possible. Sufficient.
(2) t > q. Not sufficient.
Answer: A. Bunnuel I tried it in this way . Is it right or wrong given If q, s, and t are all different numbers, is q < s < t ? or is 0<sq<tq? Since i am doing the same operation on all the parts on inequality it doesn't violates any rule i suppose ST 1 t  q = t  s + s  q Here tq is positive ( sum of two mod's) and also greater than sq as something is been added to s  q Now this equation can be solved in two ways only either same sign or opposite sign of mods same sign leads tq= ts +sq lhs=rhs therefore 0<sq<tq different signs lead to tq= ts s+q ,,,,,, not possible or tq = t+s+sq ,,,,,, again lhs is not equal to rhs . this means that the case sq<0<tq doesnt holds Hence st 1 is sufficient st 2 clearly not suficient hence a




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