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# Michael drives x miles due north at arrives at Point A. He then heads

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Manager
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Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Jul 2017, 23:04
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Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

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Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Jul 2017, 23:39
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jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

So total distance travelled by him must be 5+12+13 = 30 miles
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Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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26 Aug 2017, 22:01
VeritasPrepKarishma wrote:
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

Hi is there any alternative way to solve this question ?
If No, is it advisable to learn the pythagorean triplets ? But there are so many triplets possible.
How is one supposed to solve it under 2 min ?

Regards
Sandy DA Silva
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Posts: 187
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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26 Aug 2017, 23:44
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

How do we know that triplet will be like that?
Intern
Joined: 14 Jul 2017
Posts: 21
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 02:31
can you please explain how you got 12 and 13 for other 2 sides?
Manager
Joined: 12 Mar 2015
Posts: 82
GPA: 3.9
WE: Information Technology (Computer Software)
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 08:02
RashmiT wrote:
can you please explain how you got 12 and 13 for other 2 sides?

Hi RashmiiT,

Pythogoras triplet - 3:4:5. So we know shortest cannot be hypotenuse.

We have x^2 + 5^2 = z^2.

All 3 are integers. By testing various values for x we see that (12)^2 + (5)^2 = (13)^2.

So total distance is 12+ 5+ 13= 30
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Joined: 27 Dec 2016
Posts: 232
Concentration: Marketing, Social Entrepreneurship
GPA: 3.65
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Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 08:39
VeritasPrepKarishma wrote:
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

So total distance travelled by him must be 5+12+13 = 30 miles

Hi VeritasPrepKarishma , I answered it 12 miles because I think "THE SHORTEST LEG" is the hypotenuse from the starting point to the last destination.

How must we translate this kind of word question?
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Posts: 187
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 09:27
septwibowo wrote:
VeritasPrepKarishma wrote:
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

So total distance travelled by him must be 5+12+13 = 30 miles

Hi VeritasPrepKarishma , I answered it 12 miles because I think "THE SHORTEST LEG" is the hypotenuse from the starting point to the last destination.

How must we translate this kind of word question?

Hypotenuse is always the longest side in a RIGHT TRIANGLE
Manager
Joined: 27 Dec 2016
Posts: 232
Concentration: Marketing, Social Entrepreneurship
GPA: 3.65
WE: Marketing (Education)
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 10:00
rocko911 wrote:
septwibowo wrote:
VeritasPrepKarishma wrote:
[quote="jedit"]Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

So total distance travelled by him must be 5+12+13 = 30 miles

Hi VeritasPrepKarishma , I answered it 12 miles because I think "THE SHORTEST LEG" is the hypotenuse from the starting point to the last destination.

How must we translate this kind of word question?

Hypotenuse is always the longest side in a RIGHT TRIANGLE[/quote]

Sure, I understand that hypotenuse is the longest side of the triangle. But, we often face similar problem who assume the hypotenuse as a shortcut - so although that is the longest side, we can go from start to finish via hypotenuse rather than via its legs.

Sent from my iPhone using GMAT Club Forum mobile app
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Posts: 187
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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27 Aug 2017, 10:10
rocko911 wrote:
septwibowo wrote:
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

Due north, due east and then back makes a right triangle. All sides of the right triangle need to be integers which means we are looking for pythagorean triplets. The triplet with the shortest leg as 5 is 5, 12, 13.

So total distance travelled by him must be 5+12+13 = 30 miles

Hi VeritasPrepKarishma , I answered it 12 miles because I think "THE SHORTEST LEG" is the hypotenuse from the starting point to the last destination.

How must we translate this kind of word question?

Hypotenuse is always the longest side in a RIGHT TRIANGLE

Sure, I understand that hypotenuse is the longest side of the triangle. But, we often face similar problem who assume the hypotenuse as a shortcut - so although that is the longest side, we can go from start to finish via hypotenuse rather than via its legs.

Hypotenuse is definitely a shortcut but when ITS GIVEN THAT THE PATH MAKES A RIGHT TRIANGLE then it means the hypotenuse WILL ALWAYS BE THE LONGEST SIDE

If it would not be a right triangle then surely it was hard to tell if Hypotenuse is the longest side or not

Thanks
CEO
Joined: 12 Sep 2015
Posts: 3848
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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24 Aug 2018, 17:06
1
Top Contributor
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

At the risk of being that guy, I believe the answer is E. Here's why:

We already know that, if we start at a place on the equator and walk 40,000 km (the approximate circumference of Earth) due east, we will end up at the same place we started.
If we start at a place further north (say Los Angeles) and walk due east, we will return to our starting place in less than 40,000
In fact, the further north we move our starting point, the less the distance one must walk due east to return to the starting point.

So, there must exist a point (very close to the North Pole) where, if we walk due east, we will return to our starting place in 5 miles.
Let's call this point Point Q.
To reiterate, if we start at Point Q and walk due east for 5 miles, we end up at the exact point we started (Point Q).

So, if we start at a point that is 6 miles due south of Point Q, then Michael's journey goes like this:
Michael drives 6 miles due north at arrives at Point A (aka Point Q). He then heads due east for 5 miles (at which point, he arrives back at Point Q) . Finally, he drives 6 miles in a straight line (due south) until he reaches his starting point.

So, the length of the 3 legs of his journey are: 5 miles, 6 miles and 6 miles (the shortest leg being 5 miles)
So, the total trip was 17 miles.

Of course, there's also the option where the total trip is 30 miles.

Since we cannot definitively answer the question, the correct answer must be E

Cheers,
Brent
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Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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23 Nov 2018, 17:26
the shortest leg = 5
z=hypotenuse
x=3rd side

5^2+x^2=z^2
25=z^2-x^2
25=(z-x)(z+x)

z-x=25 or z+x=25
z=25-x or z=25+x

30 is ruled out and cannot be 5 as doesn't satisfy Pythagoras theorem. The only option left is 12.

Hence b=12
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Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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23 Nov 2018, 17:30
GMATPrepNow wrote:
jedit wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A - 5 miles

B - 12 miles

C - 25 miles

D - 30 miles

E - Cannot be determined by the information given.

At the risk of being that guy, I believe the answer is E. Here's why:

We already know that, if we start at a place on the equator and walk 40,000 km (the approximate circumference of Earth) due east, we will end up at the same place we started.
If we start at a place further north (say Los Angeles) and walk due east, we will return to our starting place in less than 40,000
In fact, the further north we move our starting point, the less the distance one must walk due east to return to the starting point.

So, there must exist a point (very close to the North Pole) where, if we walk due east, we will return to our starting place in 5 miles.
Let's call this point Point Q.
To reiterate, if we start at Point Q and walk due east for 5 miles, we end up at the exact point we started (Point Q).

So, if we start at a point that is 6 miles due south of Point Q, then Michael's journey goes like this:
Michael drives 6 miles due north at arrives at Point A (aka Point Q). He then heads due east for 5 miles (at which point, he arrives back at Point Q) . Finally, he drives 6 miles in a straight line (due south) until he reaches his starting point.

So, the length of the 3 legs of his journey are: 5 miles, 6 miles and 6 miles (the shortest leg being 5 miles)
So, the total trip was 17 miles.

Of course, there's also the option where the total trip is 30 miles.

Since we cannot definitively answer the question, the correct answer must be E

Cheers,
Brent

Is this within the scope of GMAT?
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Joined: 12 Sep 2015
Posts: 3848
Re: Michael drives x miles due north at arrives at Point A. He then heads  [#permalink]

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23 Nov 2018, 19:13
Top Contributor
1
dehumaniser wrote:
Is this within the scope of GMAT?

Ha!
Most definitely not!
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Re: Michael drives x miles due north at arrives at Point A. He then heads   [#permalink] 23 Nov 2018, 19:13
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