Asifpirlo wrote:
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
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(5000 miles) |
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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..!!!! Special Thanks to you for properly citing about this problem. This may seem impossible but it is just the simple one indeed..... Thanks jai...
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..............