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Asifpirlo
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In 1937, the ship of the great sailor SINBAD left an Egyptia 15 Aug 2013, 15:03
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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35% (medium)

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77% (02:45) correct 23% (02:50) wrong based on 133 sessions

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In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
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A
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jaituteja
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In 1937, the ship of the great sailor SINBAD left an Egyptia 15 Aug 2013, 20:13
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Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
Show more

A tough one initially.. Took me around 3 minutes to solve..
But it is logic based.. No rocket science quant required...

The answer should be A...

WEST ( Sindbad after 6 months at Xmiles/month) -------------------------------------
Egyptian Port
|
|
|
(5000 miles) |
|
|
|
|
 
CHENG I SAO ship ay Y miles/ month (SOUTH)

Now, since CHENG's ship started one month after that of Sindabad.

So, when Sindabad travels for 6 months,i.e he travels 6X distance, CHENG would travel for 5 months, i.e 5Y distance

Now, when CHENG travels for 5 months, he covers some distance of 5000, so the distance from the orgin/Egyptian Port will be (5000-5Y).

CHENG will be 6X to west and Sindabad will be 5000-5Y to south.

Applying pythagorus theorum,

the distance b/w them will be { (6x)^2 + (5000-5y)^2 }^1/2



I have a doubt, what if 5y exceeds 5000...???????

Thanks,
Jai
 
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In 1937, the ship of the great sailor SINBAD left an Egyptia 16 Aug 2013, 00:47
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Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
Show more

distance travelled by sindabad in 6 months =\(6X\)
as pirates started 1 month late==>distance travelled by pirates in 5 month =\(5Y\)

See diag we need to calculate distance WT
WT =\(\sqrt{(5000-5Y)^2+(6X)^2}\)...(USING PYTHAGORAS THEOREM)

hence A.
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Asifpirlo
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In 1937, the ship of the great sailor SINBAD left an Egyptia 16 Aug 2013, 04:08
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Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
Show more
Nice try friends. Right you are.
here is my solution :
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Asifpirlo
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In 1937, the ship of the great sailor SINBAD left an Egyptia Updated on: 17 Aug 2013, 03:59
Originally posted by Asifpirlo on 16 Aug 2013, 15:01.
Last edited by Asifpirlo on 17 Aug 2013, 03:59, edited 1 time in total.
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jaituteja
Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2

A tough one initially.. Took me around 3 minutes to solve..
But it is logic based.. No rocket science quant required...

The answer should be A...

WEST ( Sindbad after 6 months at Xmiles/month) -------------------------------------
Egyptian Port
|
|
|
(5000 miles) |
|
|
|
|
CHENG I SAO ship ay Y miles/ month (SOUTH)

Now, since CHENG's ship started one month after that of Sindabad.

So, when Sindabad travels for 6 months,i.e he travels 6X distance, CHENG would travel for 5 months, i.e 5Y distance

Now, when CHENG travels for 5 months, he covers some distance of 5000, so the distance from the orgin/Egyptian Port will be (5000-5Y).

CHENG will be 6X to west and Sindabad will be 5000-5Y to south.

Applying pythagorus theorum,

the distance b/w them will be { (6x)^2 + (5000-5y)^2 }^1/2



I have a doubt, what if 5y exceeds 5000...???????

Thanks,
Jai

Please give KUDOS if it helped..!!!! :)
Show more


Special Thanks to you for properly citing about this problem. This may seem impossible but it is just the simple one indeed..... Thanks jai...

solution of the doubt :
the distance b/w them will be { (6x)^2 +(5000-5y)^2 }^1/2
I have a doubt, what if 5y exceeds 5000...???????
yes it could be, but no problem at all. because (a-b)^2 = (b-a)^2

they are inside whole square..... You will have the same value no matter who is greater..............
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In 1937, the ship of the great sailor SINBAD left an Egyptia 04 Jan 2015, 09:28
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@Asifpirlo...Ans is A.....and I am saying so based on the following logic:-
The chinese ship sailed for 5 months i.e. 5Y miles @Y miles/month, therefore distance from Egyptian port is 5000-5Y. Since A is the only option with this value ...therefore A is the answer
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In 1937, the ship of the great sailor SINBAD left an Egyptia 29 Feb 2024, 03:55
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How do we know that the trajectories of the boats are perpendicular?
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In 1937, the ship of the great sailor SINBAD left an Egyptia 29 Feb 2024, 05:31
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jaituteja

Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
A tough one initially.. Took me around 3 minutes to solve..
But it is logic based.. No rocket science quant required...

The answer should be A...

WEST ( Sindbad after 6 months at Xmiles/month) -------------------------------------


Egyptian Port
|
|
|
(5000 miles) |
|
|
|
|
CHENG I SAO ship ay Y miles/ month (SOUTH)

Now, since CHENG's ship started one month after that of Sindabad.

So, when Sindabad travels for 6 months,i.e he travels 6X distance, CHENG would travel for 5 months, i.e 5Y distance

Now, when CHENG travels for 5 months, he covers some distance of 5000, so the distance from the orgin/Egyptian Port will be (5000-5Y).

CHENG will be 6X to west and Sindabad will be 5000-5Y to south.

Applying pythagorus theorum,

the distance b/w them will be { (6x)^2 + (5000-5y)^2 }^1/2



I have a doubt, what if 5y exceeds 5000...???????

Thanks,
Jai

Please give KUDOS if it helped..!!!! :)


 
Show more
­Hey jaituteja !

Your answer is indeed such a helpful method but I'd like to help me understand a specific part. In the highlighted text, why the distance and direction of each boat switched? (CHENG has 5.000 - 5Y distance and is located south and accordingly Sindbad is west with 6X distance).­

Could anyone chip in? JeffTargetTestPrep KarishmaB manasp35
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In 1937, the ship of the great sailor SINBAD left an Egyptia 17 Apr 2024, 22:52
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Gmatguy007

jaituteja

Asifpirlo
In 1937, the ship of the great sailor SINBAD left an Egyptian port heading Ivory coast at X mile/month .
One month later another ship of the greatest pirate ever CHENG I SAO was 5000 miles due south of the same Egyptian port and heading due north at Y miles/month. Six months later after the leaving of Sinbad’s ship from Egyptian port, how far apart were the ships? [Ivory coast is in the west of Egypt]

(A) { (6x)^2 + (5000-5y)^2 }^1/2
(B) { (36x)^2 + (5000-7y)^2 }^1/2
(C) { (16x)^2 + (5000-7y)^2 }^1/2
(D) { (7x)^2 + (5200-7y)^2 }^1/2
(E) { (2x)^2 + (300-7y)^2 }^1/2
A tough one initially.. Took me around 3 minutes to solve..
But it is logic based.. No rocket science quant required...

The answer should be A...

WEST ( Sindbad after 6 months at Xmiles/month) -------------------------------------



Egyptian Port
|
|
|
(5000 miles) |
|
|
|
|
CHENG I SAO ship ay Y miles/ month (SOUTH)

Now, since CHENG's ship started one month after that of Sindabad.

So, when Sindabad travels for 6 months,i.e he travels 6X distance, CHENG would travel for 5 months, i.e 5Y distance

Now, when CHENG travels for 5 months, he covers some distance of 5000, so the distance from the orgin/Egyptian Port will be (5000-5Y).

CHENG will be 6X to west and Sindabad will be 5000-5Y to south.

Applying pythagorus theorum,

the distance b/w them will be { (6x)^2 + (5000-5y)^2 }^1/2



I have a doubt, what if 5y exceeds 5000...???????

Thanks,
Jai

Please give KUDOS if it helped..!!!! :)



 
­Hey jaituteja !

Your answer is indeed such a helpful method but I'd like to help me understand a specific part. In the highlighted text, why the distance and direction of each boat switched? (CHENG has 5.000 - 5Y distance and is located south and accordingly Sindbad is west with 6X distance).­

Could anyone chip in? JeffTargetTestPrep KarishmaB manasp35
Show more
­

Nothing switched. Jaituteja just reversed the names mistakenly while writing that. 
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In 1937, the ship of the great sailor SINBAD left an Egyptia 18 Apr 2024, 08:28
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KarishmaB

Gmatguy007
­Hey jaituteja !

Your answer is indeed such a helpful method but I'd like to help me understand a specific part. In the highlighted text, why the distance and direction of each boat switched? (CHENG has 5.000 - 5Y distance and is located south and accordingly Sindbad is west with 6X distance).­

Could anyone chip in? JeffTargetTestPrep KarishmaB manasp35
Nothing switched. Jaituteja just reversed the names mistakenly while writing that. 
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­That's what I thought at first, but I wanted to be sure that i didn't miss something.
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