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The distance from X to Y is 20 miles, and the distance from [#permalink]
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03 May 2003, 17:49
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This topic is locked. If you want to discuss this question please repost it in the respective forum. The distance from X to Y is 20 miles, and the distance from X to Z is 12 miles. If d is the distance, in miles, between Y and Z, then d is indicated by: A. 8 ≤ d ≤ 20 B. 8 ≤ d ≤ 32 C. 12 ≤ d ≤ 20 D. 12 ≤ d ≤ 32 E. 20 ≤ d ≤ 32
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B,
Min is 8 Max is 12+20=32



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Correct.
tzolkin, could you explain why min is 8, e.g. 2012, and max is 32?
Thanks.



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Side YZ of triangle XYZ is the distance between the towns Y and Z.
YZ is shortest when angle YXZ approaches 0, i.e, XZ almost overlaps XY
In this case, since Z is a point on XY,
YZ = XY  YZ = 20  12 = 8
YZ is longest when angle YXZ approaches 180, i.e, XZ extends outwards in a straight line from XY
In this case,
YZ = XY + YZ = 20 + 12 = 32



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The distance from X to Y is 20 miles, and the distance from [#permalink]
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09 Oct 2008, 06:08
The distance from X to Y is 20 miles, and the distance from X to Z is 12 miles. If d is the distance, in miles, between Y and Z, then d is indicated by: A. 8 ≤ d ≤ 20 B. 8≤ d ≤ 32 C. 12≤d ≤ 20 D. 12≤d ≤ 32 E. 20 ≤d ≤32
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Re: PS: Distance Trouble [#permalink]
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09 Oct 2008, 06:12
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IMO B:
we have two poss. here 1: zxy (12020) = 32
2: xzy (0128) = 20
so distance min. distance will be 8 and max distance will be 32
8<= d <= 32



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Re: PS: Distance Trouble [#permalink]
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09 Oct 2008, 16:15
the sum of any two sides of a triangle is greater that the third side.
side XY = 20 side XZ = 12
YZ , distance d, shouldn't be equal to nether 8 nor 32 becuase: 12+8 is not > 20 20+12 is not > 32
therefore the answer is C



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Re: PS: Distance Trouble [#permalink]
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10 Oct 2008, 05:00
amitdgr wrote: The distance from X to Y is 20 miles, and the distance from X to Z is 12 miles. If d is the distance, in miles, between Y and Z, then d is indicated by:
A. 8 ≤ d ≤ 20 B. 8≤ d ≤ 32 C. 12≤d ≤ 20 D. 12≤d ≤ 32 E. 20 ≤d ≤32 x.........z..............y z........x.................y y.............z.......x b



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Re: PS: Distance Trouble [#permalink]
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10 Oct 2008, 07:13
Draw two perimeters around X with radii 12 (distance to Z) and 20 (distance to Y). Hence min distance is 8 and max distance is 32.



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Re: PS: Distance Trouble [#permalink]
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31 Oct 2008, 17:55
B
Draw a cirlce with center X and radius Z



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Re: PS: Distance Trouble [#permalink]
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01 Nov 2008, 13:53
Another approach: Look at three points as forming a triangle. Third side of triangle will be >= difference between the two sides and <= sum of two sides.
Hence, 208 <= d <= 20+12.



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Re: PS: Distance Trouble [#permalink]
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02 Nov 2008, 23:35
scthakur wrote: Another approach: Look at three points as forming a triangle. Third side of triangle will be >= difference between the two sides and <= sum of two sides.
Hence, 208 <= d <= 20+12. Caution ! Third side of the triangle will be greater than the difference between the two sides and less than the sum of two sides. (Not >= or <=) In this case, it is not mentioned that the 3 points are not on the same line. Hence the '=' scenario comes into play.



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Re: PS: Distance Trouble [#permalink]
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02 Nov 2008, 23:43
kandyhot27 wrote: scthakur wrote: Another approach: Look at three points as forming a triangle. Third side of triangle will be >= difference between the two sides and <= sum of two sides.
Hence, 208 <= d <= 20+12. Caution ! Third side of the triangle will be greater than the difference between the two sides and less than the sum of two sides. (Not >= or <=) In this case, it is not mentioned that the 3 points are not on the same line. Hence the '=' scenario comes into play. Good point. Thanks for pointing out. When the area of triangle becomes zero, all the three vertices lie on a straight line and then only equality sign will hold true.




Re: PS: Distance Trouble
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02 Nov 2008, 23:43






