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# The distance from planet Olcott to planet Zabar is 70 light years. The

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Re: The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
bhoot80 wrote:
sanahazari wrote:
I feel the answer should be (E) 140.

As the length of the third side of a triangle can't be more than the sum of the other two sides and can't be less than the difference of the other two sides, the other options are acceptable.

Hence the distance would be $${20} \leq distance \leq {120}$$

It's should be d. 120..law of triangles leg

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hi.

the minimum distance will be WHEN D is in between O and Z then OZ is 70 and DZ is 50, so OD will be 70-20=50.. O-20-D-50-Z
MAX when again all three are in a straight line and D is on other side... O-70-Z-50-D OD is 50+70=120

140>120 not possible

E
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The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
chetan2u wrote:
bhoot80 wrote:
sanahazari wrote:
I feel the answer should be (E) 140.

As the length of the third side of a triangle can't be more than the sum of the other two sides and can't be less than the difference of the other two sides, the other options are acceptable.

Hence the distance would be $${20} \leq distance \leq {120}$$

It's should be d. 120..law of triangles leg

Sent from my SM-J700F using GMAT Club Forum mobile app

hi.

the minimum distance will be WHEN D is in between O and Z then OZ is 70 and DZ is 50, so OD will be 70-20=50.. O-20-D-50-Z
MAX when again all three are in a straight line and D is on other side... O-70-Z-50-D OD is 50+70=120

140>120 not possible

E

chetan2u , I used your approach, because as far as I understand it, the triangle inequality theorem, and what is sometimes called its converse, are literally inequalities. One side cannot equal the sum of the lengths of two other sides, and one side cannot equal the difference between the two other sides.

You didn't mention that the $$\leq$$ sign was impermissible, i.e., you did NOT say that it should be just a < sign.

I understand that there can be degenerate triangles with area zero. I also understand that the GMAT does not test them.

I used the straight-line method for values 20 and 120 (to allow them) because I thought that those two values would be impermissible with the triangle inequality theorem.

Is it true that the third side could be equal to the sum of the other two? Or equal to the difference between the other two? On the GMAT?
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Re: The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
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genxer123 -- i think u misinterpreted the question .No where in the question it is mention that 3 point must form a triangle,so it can be in a straight line also..
.But 140 is not possible in either case ...
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The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
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sobby wrote:
genxer123 -- i think u misinterpreted the question .No where in the question it is mention that 3 point must form a triangle,so it can be in a straight line also..
.But 140 is not possible in either case ...

sobby , thanks for responding.

I initially drew two sides of a triangle because I thought it might be a hypotenuse problem. As soon as I realized there was nothing about right angles in the prompt, (though a triangle DOES allow for 50 and 70), I drew a straight line.

Because I saw no explicit rejection of the equation with "$$\leq$$" to which he attached his response, I could not understand whether or not chetan2u was endorsing the idea that the triangle INequality theorem allowed for -- well, equality, I guess, for the numbers 20 (= 70 - 50) and 120 (50 + 70). I think his "use a straight line" reply means "no endorsement."
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Re: The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
Consider three planets are points of triangle or points in straight line
Then all point except E is possible
Case 1. Considering planets are in Triangle points
70-50<Distance<70+50
Case 2: Considering Straight line
20, 120

So 140 cannot possible hence answer is E
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Re: The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
I draw a circle point Z 5 units radius. Any point on this circle is a D possibility.
Max: O -> D Furthest Distance = Diameter: 70+50=120
Min: O->D Shortest Distance 70-50=20

=> Can't be 140.

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The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
My thought was that because the question implies the distance is not a straight line, in which case the distance between A and D would simply be 120, it would have to form a triangle. Because the sum of each possible pair of lengths must be greater than the length of the third side, A would be the odd man out.

Ex:
70+50=120 which is greater than 70 and 50.
50+20=70 which is not greater than 70 and therefore would not form a triangle.

Going with A and hoping this rule does not only apply to right triangles
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Re: The distance from planet Olcott to planet Zabar is 70 light years. The [#permalink]
Bunuel wrote:
The distance from planet Olcott to planet Zabar is 70 light years. The distance from planet Zabar to planet Devra is 50 light years. Which of the following could NOT be the distance in light years between planet Olcott and planet Devra?

(A) 20
(B) 50
(C) 70
(D) 120
(E) 140

The distance, d, between Olcott and Devara must be no more than the sum of 70 and 50 and no less than the difference between 70 and 50. That is,

70 - 50 ≤ d ≤ 70 + 50

20 ≤ d ≤ 120

Therefore, we see that all the numbers in the answer choices can be the distance between the two planets except 140.

Alternate Solution:

If the planets are placed on a line, we have: O…...…Z…...D, we see that the maximum distance that they can span is 70 + 50 = 120 light years. Answer choice E is, therefore, impossible.