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# m9 q34

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Re: m9 q34 [#permalink]

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17 Sep 2011, 19:15
I got B as answer...
My logic was

AB + AC> 10
hence AB > (10- AC)

Since A, B and C are three different point, AB, BC,and AC are all positive values.
so AC will have a value defines as 0 < AC < 10.
because if AC> 10, AB becomes negative.
SO this inturn gives the value of AB as : 10 > AB > 0.
So AB is less than 10.

Pls tell me where I went wrong!!
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Re: m9 q34 [#permalink]

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11 Sep 2012, 04:10
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sset009 wrote:
$$A$$ , $$B$$ , and $$C$$ are points on the plane. Is $$AB \lt 10$$ ?

1. $$AC + BC = 10$$
2. $$AB + AC \gt 10$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Couldnt understand the official explanation

New edition of this question reads:

If A, B, and C are distinct points on the number line. Is the length of the line segment AB less than 10?

(1) The sum of the lengths of line segments AC and BC is 10
(2) The sum of the lengths of line segments AB and AC is more than 10

Even when we consider both statements together we can not have a definite answer. Consider two examples below:
Attachment:

Number line.png [ 8.54 KiB | Viewed 2380 times ]
Answer: E.
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Re: m9 q34 [#permalink]

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11 Sep 2012, 04:37
OMG

I 'm wondering where you were Bunuel
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Re: m9 q34 [#permalink]

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11 Sep 2012, 04:39
Could not understand the question

None of he options seem interesting

Choose E! - strategic guess
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Re: m9 q34 [#permalink]

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11 Sep 2012, 04:55
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A , B , and C are points on the plane. Is AB < 10 ?

1. AC + BC = 10
2. AB + AC > 10

Since we already know that individual statements are insufficient, let's combine:

From 1, we know that AC= 10 - BC. Let's substitute this into 2.
AB + 10 - BC > 10 =====> AB > BC

From 1, we know that BC < 10, since AC must be some value greater than zero.
Combining AB > BC with BC < 10, we have AB > 0, and this is not enough to tell us whether AB < 10.

Therefore, E.

Cheers,
Der alte Fritz.
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Re: m9 q34 [#permalink]

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11 Sep 2012, 04:59
It's a tough question because I got to the point where it was between C and E. I didn't know what cases to choose to rule out C convincingly.

This is my attempt at a methodical approach or ruling out answers.

Lets take two cases:
1) Where C is in between A and B

--A-----------------C-------------------B-----
----------4-----------------6------------------

AC+BC = 10 YES
AB+AC>10 YES
AB<10?? NO

2) Where C is outside of A and B

--A-----------------B-------------------C-----
----------2-----------------4------------------

AC+BC = 10 YES
AB+AC>10 NO
AB<10?? YES

But this case is invalid because in this case :AB+AC>10 NO

2) Where C is outside of A and B

--A-----------------B-------------------C-----
----------2-----------------8------------------

AC+BC = 10 YES
AB+AC>10 YES
AB<10?? YES

Choose E and move on.
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Re: m9 q34 [#permalink]

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11 Sep 2012, 05:23
E is the answer.

(1) does not give any relation about AB at all
(2) gives vague information about AB as we donot know about AC

so AD and B are not the answers
C .. when combined also we cannot land on a perfect solution, so C is also not the choice.

Hence E.
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Re: m9 q34 [#permalink]

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11 Sep 2012, 08:57
Answer:E

Because the points are on plane so they can be anywhere .
Look at the scenario when the points are in line A____C____B
in that case

If AC + BC =10 then AB =10 which is not less than 10
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Re: m9 q34 [#permalink]

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11 Sep 2012, 09:03
the answer is E, and a quick dirty way to do it is by plugging in numbers. First thing to note is that it the original question does not say A B or C are distinct, implying that A could equal B and so on. Also, the way the equations are set up it is just easier to think about the points as numbers.

so is stmt 1 enough?
AC + BC = 10. plug in A=2, B=0, C=5
in this case AB<10
now plug in A = 5, B =5, C=1
in this case AB>10.
Therefore stmnt 1 is not enough.

is stmt 2 enough?
AB +AC >10
A = 11, B = 1, C = 1 and AB >10
A = 1, B = 1, C =10 and AB<10
therefore stmnt 2 is not enough

Together?
AC + BC = 10 and AB+AC>10
A=3, B = 2, C =2 and AB<10
A=18, B = 2, C =.5 and AB>10

therefore together not enough
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Re: M09 #34 [#permalink]

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11 Sep 2012, 09:22
matt3030 wrote:
Interesting. I assume that if the question prompt was something like "A, B and C are distinct points on a plane" that the answer would be (a)?

I assume, irrespective of points are distinct or not, the ans is E

The ans is A only if it is given that the 3 points don't lie on the same line.
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Re: M09 #34 [#permalink]

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11 Sep 2012, 09:31
giridharkumar wrote:
matt3030 wrote:
Interesting. I assume that if the question prompt was something like "A, B and C are distinct points on a plane" that the answer would be (a)?

I assume, irrespective of points are distinct or not, the ans is E

The ans is A only if it is given that the 3 points don't lie on the same line.

yes, even if they were distinct A is not enough:

AC + BC = 10.
plug in A=2, B=0, C=5
in this case AB<10
now plug in A = 14, B =6, C=.5
in this case AB>10.
Therefore stmnt 1 is not enough.
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Re: m9 q34 [#permalink]

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11 Sep 2012, 09:53
OldFritz wrote:
A , B , and C are points on the plane. Is AB < 10 ?

1. AC + BC = 10
2. AB + AC > 10

Since we already know that individual statements are insufficient, let's combine:

From 1, we know that AC= 10 - BC. Let's substitute this into 2.
AB + 10 - BC > 10 =====> AB > BC

From 1, we know that BC < 10, since AC must be some value greater than zero.
Combining AB > BC with BC < 10, we have AB > 0, and this is not enough to tell us whether AB < 10.

Therefore, E.

Cheers,
Der alte Fritz.

the Bunuel approach of course is the faster........but yours is pretty good
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Re: m9 q34 [#permalink]

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11 Sep 2012, 23:48
My pick is also E.
If we have been provided with some additional information on the location of the points, for example whether it is a triangle, a straight line or something then provided statements could have been sufficient.
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Re: m9 q34 [#permalink]

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12 Sep 2012, 23:50
But if we solve for the inequality in equation 2, we get AB>10-AC. Since AC cannot be negative or zero, AB will be < 10 no matter if the points are in a triangle or collinear. Isn't the second statement enough and answer B?
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Re: m9 q34 [#permalink]

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10 Sep 2013, 09:44
Hi! I've got E too since we do not know in what order the points are in the plan (whether a triangle or another shape)!

==> Do someone know the difficulty of this question plz?

Thank you very much!
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Re: m9 q34 [#permalink]

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11 Sep 2013, 01:56
Paris75 wrote:
Hi! I've got E too since we do not know in what order the points are in the plan (whether a triangle or another shape)!

==> Do someone know the difficulty of this question plz?

Thank you very much!

The difficulty level is ~650.

Revised version (m9-q34-70894-20.html#p1120622) is a bit easier.
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Re: m9 q34 [#permalink]

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27 Jun 2014, 12:50
Bunuel wrote:
sset009 wrote:
$$A$$ , $$B$$ , and $$C$$ are points on the plane. Is $$AB \lt 10$$ ?

1. $$AC + BC = 10$$
2. $$AB + AC \gt 10$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Couldnt understand the official explanation

New edition of this question reads:

If A, B, and C are distinct points on the number line. Is the length of the line segment AB less than 10?

(1) The sum of the lengths of line segments AC and BC is 10
(2) The sum of the lengths of line segments AB and AC is more than 10

Even when we consider both statements together we can not have a definite answer. Consider two examples below:
Attachment:
Number line.png
Answer: E.

This was a tough question for me. Since there are three distinct points, there are 6 ways to arrange those three points. What is an efficient approach to this problem?
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Re: m9 q34 [#permalink]

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27 Aug 2014, 02:11
My answer is E

1. AC+BC = 10

It is written that A,B, C are on a plane. So they can be anywhere on the plane.
Let's assume they are on a straight line and B is the middle of A and C

A--------B---------C

In that case if AC+BC = 10 then AB is definitely less than 10 (evident from diagram above). So it concludes AB<10
Now let's assume in that scenario C is the middle point between A and B

A--------C---------B

From this image if AC+BC =10 then AB =10 i.e. AB is not <10

So more than 1 answer is possible for 1st option. Hence A not sufficient.

2. AB+AC>10
=> AB>10-AC

Let's assume AB = 12 and AC = 2 then upper equation stands true i.e 12>8. In this case AB is not <10
Now assume AB = 3 and AC = 8 then also upper equations stands true i.e. 3>2 But in this case AB<10

So more than 1 answer is possible. Thus not sufficient.

Similarly if we consider both options we will get different results where AB <10 and AB is not less than 10 i.e not sufficient.

Thus answer will be E.
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Re: m9 q34 [#permalink]

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27 Aug 2014, 15:40
Statement I , AC+BC= 10. From this we can know that AC should be between 1 to 9

Substituting this value in Statement II, AC+AB > 10

Case 1 : When AC = 1, we can have AB = 10 ( to satisfy this condition ) to infinity.

Case 2 : When AC = 9, we can have AB = 2 to infinity.

So, we cannot surely say that AB is greater than 10.

Regards
Amir
Re: m9 q34   [#permalink] 27 Aug 2014, 15:40

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