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# If q, s, and t are all different numbers, is q < s < t ?

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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, 03:44
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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
[Reveal] Spoiler: OA

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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

Last edited by Bunuel on 19 Jun 2017, 03:59, 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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19 Jun 2017, 04:28
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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 non-negative 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. ---s---q-------t-------
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. -------q---s---t-------
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. -------q-------t---s---
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.

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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, 12:27
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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, t-q 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, 05:38
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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22 Aug 2017, 06:08
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.
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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, 06:10
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, 11:48
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 left-hand 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.

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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, 09:11
the fastest way to solve this is to draw the number line
q------------t-----s
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 /s-q/= 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, 13:26
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 non-negative 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. ---s---q-------t-------
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. -------q---s---t-------
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. -------q-------t---s---
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.

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<s-q<t-q?
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 t-q is positive ( sum of two mod's) and also greater than s-q 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 t-q= t-s +s-q lhs=rhs therefore 0<s-q<t-q

different signs lead to t-q= t-s -s+q ,,,,,, not possible

or t-q = -t+s+s-q ,,,,,, again lhs is not equal to rhs . this means that the case s-q<0<t-q doesnt holds

Hence st 1 is sufficient
st 2 clearly not suficient hence a

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Re: If q, s, and t are all different numbers, is q < s < t ?   [#permalink] 23 Dec 2017, 13:26
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