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i picked numbers and i solved it in 1 min and 50 seconds but my answer is C
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Hi,, is it possible explain ths question without no line..

Thanks in advance




VeritasPrepKarishma
jamifahad
If x<p<q<y Is |q-x|<|q-y|?
(1) |p-x| < |p-y|
(2) |q-x| < |p-y|

As always, do this under 2mins!

The question stem makes me use the number line. I place all variables in the desired order at equal intervals to begin with. The red line is the distance between q and x and the black line is the distance between q and y.
Attachment:
Ques2.jpg
Question: Is the red line shorter than black line?

Then I consider the 2 statements.
Statement 1: The distance between p and x should be less than the distance between p and y. Green line should remain shorter than blue line.
In the 2 diagrams below Stmnt 1, green line is shorter than blue line. But in one of these red line is shorter than black and in the other it is longer than black. Not sufficient.
Attachment:
Ques3.jpg
Statement 2: The distance between q and x should be less than the distance between p and y. Green line should remain shorter than blue line.
In the 2 diagrams below Stmnt 2, green line is shorter than blue line. But in one of these red line is shorter than black and in the other it is longer than black. Not sufficient.

We see that both the same cases are used to prove the two statement insufficient. So even when we take both statements together, they will remain insufficient. (You can use the same two figures for both together case too)
Answer E.
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Thanks Karishma. Wonderful explanation. Much appreciated.

soul123
Hi,, is it possible explain ths question without no line..

Here is OE.
Explanation: Since all of the expressions are contained within absolute value signs, we are talking about distances here. It may be easier to think of the four variables as landmarks along a highway, and the question as asking whether x or y is farther from q.
Statement (1) is insufficient. If y is farther from p than x is from p, we know that p is to the left of the midpoint between x and y. q could also be to the left of the midpoint (in which case it is closer to x), but it could just as easily be to the right of the midpoint.
Statement(2)is also insufficient. We know that |p-y| > |q-y|, which tells us that both |q-y| and |q-x| are less than |p-y|, but not which is greater.

Taken together, the statements are still insufficient. We have several pieces of information:
|p-x| < |p-y|
|q-x| < |p-y|
|q-y| < |p-y|
|p-x| < |q-x|
To put what we can in the correct order: |p-x| < |q-x| < |p-y|
and
|q-y| < |p-y|
Still, we have no way to deduce the relationship between |q-x| and |q-y|.
Choice (E) is correct.

Thanks.
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Hi,, is it possible explain ths question without no line..

Thanks in advance


The best way to deal with a mod question is through the number line. It helps you see the relationship. Anyway, jamifahad has given you the OE.
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Quote:

The best way to deal with a mod question is through the number line. It helps you see the relationship. Anyway, jamifahad has given you the OE.

Agree. The best way is through the number line. Great explanation Karishma.
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@subhajeet:
The values I picked were some random numbers. I just wanted x<p<q<y relation to hold.
Then, while considering statement 1, I kept x, p and y constant so that the green line remains less than the blue line and moved the q around. q could move anywhere between p and y making the red line longer/shorter than the black one. I did not focus on the values I took. Just on the range that q could vary in.
I did the same thing for statement 2 as well.
The values I took have no role in the solution. I just put some numbers so that it is easy to visualize. Try and think of the solution in terms of lengths of various lines.
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The number line approach is best when |a-b| inequalities are given.
To make it easier we can put the variable values for the distance between the points.
Attachment:
1.jpg
1.jpg [ 8.26 KiB | Viewed 10903 times ]

Here d = a+b-c

now question : is a<b

1. c<d => c<a+b-c => 2c<a+b

Not sufficient alone.

2. a<d =. a<a+b-c => c<b

Not sufficient alone.

Now take both of them together.

Following cases are possible:
1. a....c.....b => b>a
2. c.....b.....a => a>b


Thus one possibility gives answer No and the other Yes.
Hence E
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I did this in 41 seconds and the answer is E. However after looking at the options, I have a doubt on my approach:

We all agree that it could not be A, B and D.

Combining both statements:
lp-xl < lp-yl ...........(1)
lq-xl < lp - yl...........(2)

The right hand side of both the equations is equal but lp-xl can be equal to lq-xl or less than or greater than the other. We can not establish any concrete relationship. Hence E.

Can someone comment on this approach. If I am going wrong, where am I going wrong?
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jamifahad
If x<p<q<y Is |q-x|<|q-y|?
(1) |p-x| < |p-y|
(2) |q-x| < |p-y|

As always, do this under 2mins!

Given,
x<p<q<y
|q-x|<|q-y|

1) |p-x| < |p-y|
it means x is closer to p than y is.
but, we don't know whether q is closer to x or y

2) This does not help us in any way

E
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1) |p-x| < |p-y|
it means x is closer to p than y is.
but, we don't know whether q is closer to x or y

2) This does not help us in any way

E
tnx lot dear
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This problem is better solve graphing number line and assigning measures to each tranche

For instance,

We would have something like

X----(A)-----P----(B)----Q----(C)----Y-----(D)

Question is asking is c+d > a+b?

Statement 1

We know that b+c+d >a

c+d > a-b?

Is it more than a+b? Not sufficient

Statement 2

We know that b+c+d >a+b
c+d > a

What about 'b'? We don't know

Both together

We have that c+d > a-b
and that c+d > a

Adding 2c + 2d > 2a - b

Impossible to know whether c+d > a+b

Therefore, answer is E

Hope it clarifies
Cheers
J :)
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I believe the answer should be obvious, even if both side is positive, then we can remove the absolute sign change move x and y on one side,
now, we can see even if 2 statements are combined, there is not enough to answer the question
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equation 1: p-x <y-p => 2p <y+x
equation 2: q-x< y-p => p+q<y+x
qn whether q-x<y-q => 2q <y+x ??
Ans: May or may not be true .so E
trial 4+1 <6
2x 1 <6
but 2 x 4 < 6 is false.
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