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The distance from X to Y is 20 miles, and the distance from

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The distance from X to Y is 20 miles, and the distance from [#permalink] New post 03 May 2003, 16:49
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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
[Reveal] Spoiler: OA
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 [#permalink] New post 03 May 2003, 18:16
B,

Min is 8 Max is 12+20=32
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 [#permalink] New post 03 May 2003, 22:11
Correct.

tzolkin, could you explain why min is 8, e.g. 20-12, and max is 32?

Thanks.
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 [#permalink] New post 14 May 2003, 08:55
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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] New post 09 Oct 2008, 05:08
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] New post 09 Oct 2008, 05:12
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IMO B:

we have two poss. here
1: z-x-y (12-0-20) = 32

2: x-z-y (0-12-8) = 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] New post 09 Oct 2008, 15:15
the sum of any two sides of a triangle is greater that the third side.

side X-Y = 20
side X-Z = 12

Y-Z , 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] New post 10 Oct 2008, 04:00
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] New post 10 Oct 2008, 06:13
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] New post 31 Oct 2008, 16:55
B

Draw a cirlce with center X and radius Z
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Re: PS: Distance Trouble [#permalink] New post 01 Nov 2008, 12:53
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, 20-8 <= d <= 20+12.
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Re: PS: Distance Trouble [#permalink] New post 02 Nov 2008, 22:35
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, 20-8 <= 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] New post 02 Nov 2008, 22:43
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, 20-8 <= 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   [#permalink] 02 Nov 2008, 22:43
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