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City B is 5 miles east of City A. City C is 10 miles south

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City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A. 11
B. 12
C. 13
D. 14
E. 15
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Feb 2014, 10:40, edited 1 time in total.
Edited the OA.

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City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 27 Feb 2011, 03:37
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Madelaine88 wrote:
City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A/ 11
B/ 12
C/ 13
D/ 14
E/ 15


Refer to the diagram below:
Attachment:
untitled.PNG
untitled.PNG [ 3.76 KiB | Viewed 6581 times ]


From 45-45-90 right triangle BDC: \(BD=DC=5\sqrt{2}\);

\(AC^2=AD^2+DC^2=(AB+BD)^2+DC^2=(5+5\sqrt{2})^2+(5\sqrt{2})^2\approx{(5+7)^2+7^2}=193\);

\(AC=\sqrt{193} \approx{14}\).

Answer: D.
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Re: Distance [#permalink]

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New post 27 Feb 2011, 03:43
This is easy to do using coordinate geometry. Let B be at origin, then A would lie on X axis with coordinates (-5,0). Since C is south east of B, it will be along the line y=-x. and the coordinate of a point 10 units from origin on this line would be (5*2^0.5,-5*2^0.5) (calculated as follows - if the coordinates are k,-k then distance is given by 10 = (k^2+k^2)^0.5)

Now distance between A and C would be given by d = ((5*2^0.5)^2+(5*(2^0.5+1))^2))^0.5 which would be ~(195.7)^0.5 which is closest to 14 among the choices. so answer D

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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 24 Feb 2014, 10:23
I am bit confuse with answer. Spoiler OA says answer is C and bunuel says answer is D.

Moreover, Bunuel how can you surely say that the BD=DC or BDC is an isoceles right angle triangle.

My conclusion to answer is C.

As C is in south-east direction with 10. SO BC=10 and AB=5

A. With Pythagoras theorem 10^2 +5^2 =sqrt(125) =11. So A is not an answer.
B. It is closer to 11 and more toward extreme
C. C is correct answer logically.
D. It is closer to 15 and more toward extreme
E. Sum of two side of triangle is always greater than 3rd side so 10+5=15 . So E is not an answer

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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 24 Feb 2014, 10:43
vikrantgulia wrote:
I am bit confuse with answer. Spoiler OA says answer is C and bunuel says answer is D.

Moreover, Bunuel how can you surely say that the BD=DC or BDC is an isoceles right angle triangle.

My conclusion to answer is C.

As C is in south-east direction with 10. SO BC=10 and AB=5

A. With Pythagoras theorem 10^2 +5^2 =sqrt(125) =11. So A is not an answer.
B. It is closer to 11 and more toward extreme
C. C is correct answer logically.
D. It is closer to 15 and more toward extreme
E. Sum of two side of triangle is always greater than 3rd side so 10+5=15 . So E is not an answer


Edited the OA, it's D, not C.

BDC is an isosceles triangle because south-east of city B means 45 degrees angle from AB.
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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 22 Jul 2015, 13:10
Bunuel

Just clarifying this point to avoid making the same mistake again (better risk making a stupid question now than getting it wrong on the GMAT).

When i see directions (North, East, Northeast, Southeast, etc) on GMAT i should consider EXACTLY that direction (Southeast = Exactly between South and East)?

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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 22 Jul 2015, 15:15
Bunuel wrote:
Madelaine88 wrote:
City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A/ 11
B/ 12
C/ 13
D/ 14
E/ 15

Refer to the diagram below:
Attachment:
untitled.PNG

From 45-45-90 right triangle BDC: \(BD=DC=5\sqrt{2}\) --> \(AC^2=AD^2+DC^2=(AB+BD)^2+DC^2=(5+5\sqrt{2})^2+(5\sqrt{2})^2\approx{(5+7)^2+7^2}=193\) --> \(AC\approx{14}\).

Answer: D.


Hello Bunuel,

I tried it this way -

1) If we drop a perpendicular from B to AC (lets call the point X) we get 2 rt triangles - AXB and BXC
2) Triangle AXB has one side and 5 so we look for Pyth. Triplets and fill in 3 and 4 for AX and BX.
3)Triangle BXC - we have 2 sides now - so we again apply Pyth. theorem and find the side XC ( which is \sqrt{10^2-4^2} = ~9)

Side AC = 3+9 = 12

I know I am wrong but can you please point out where.

Thank you in advance.
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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 22 Jul 2015, 19:38
aimtoteach wrote:
Bunuel wrote:
Madelaine88 wrote:
City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A/ 11
B/ 12
C/ 13
D/ 14
E/ 15

Refer to the diagram below:
Attachment:
untitled.PNG

From 45-45-90 right triangle BDC: \(BD=DC=5\sqrt{2}\) --> \(AC^2=AD^2+DC^2=(AB+BD)^2+DC^2=(5+5\sqrt{2})^2+(5\sqrt{2})^2\approx{(5+7)^2+7^2}=193\) --> \(AC\approx{14}\).

Answer: D.


Hello Bunuel,

I tried it this way -

1) If we drop a perpendicular from B to AC (lets call the point X) we get 2 rt triangles - AXB and BXC
2) Triangle AXB has one side and 5 so we look for Pyth. Triplets and fill in 3 and 4 for AX and BX.
3)Triangle BXC - we have 2 sides now - so we again apply Pyth. theorem and find the side XC ( which is \sqrt{10^2-4^2} = ~9)

Side AC = 3+9 = 12

I know I am wrong but can you please point out where.

Thank you in advance.


This is the same way I did it. Can you explain how the triangle made a 45 degree angle?

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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 22 Jul 2015, 20:00
aimtoteach wrote:
Bunuel wrote:
Madelaine88 wrote:
City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A/ 11
B/ 12
C/ 13
D/ 14
E/ 15

Refer to the diagram below:
Attachment:
untitled.PNG

From 45-45-90 right triangle BDC: \(BD=DC=5\sqrt{2}\) --> \(AC^2=AD^2+DC^2=(AB+BD)^2+DC^2=(5+5\sqrt{2})^2+(5\sqrt{2})^2\approx{(5+7)^2+7^2}=193\) --> \(AC\approx{14}\).

Answer: D.


Hello Bunuel,

I tried it this way -

1) If we drop a perpendicular from B to AC (lets call the point X) we get 2 rt triangles - AXB and BXC
2) Triangle AXB has one side and 5 so we look for Pyth. Triplets and fill in 3 and 4 for AX and BX.
3)Triangle BXC - we have 2 sides now - so we again apply Pyth. theorem and find the side XC ( which is \sqrt{10^2-4^2} = ~9)

Side AC = 3+9 = 12

I know I am wrong but can you please point out where.

Thank you in advance.


Hi,
where you are going wrong is by assuming the sides to be an integer..
hyp is 5, the other sides can be 1 and \(\sqrt{99}\)
and many more ways ...
Hope it helped
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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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New post 22 Jul 2015, 20:02
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roopika2990 wrote:
aimtoteach wrote:

I tried it this way -

1) If we drop a perpendicular from B to AC (lets call the point X) we get 2 rt triangles - AXB and BXC
2) Triangle AXB has one side and 5 so we look for Pyth. Triplets and fill in 3 and 4 for AX and BX.
3)Triangle BXC - we have 2 sides now - so we again apply Pyth. theorem and find the side XC ( which is \sqrt{10^2-4^2} = ~9)

Side AC = 3+9 = 12

I know I am wrong but can you please point out where.

Thank you in advance.


This is the same way I did it. Can you explain how the triangle made a 45 degree angle?


Hi roopika2990,
the angle is 45 degree because it is given south east and south east will make an angle of 45 degree
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Re: City B is 5 miles east of City A. City C is 10 miles south [#permalink]

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Re: City B is 5 miles east of City A. City C is 10 miles south   [#permalink] 18 Oct 2017, 17:39
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