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A planed named "ship" is flying in a clock wise circular path above

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A planed named "ship" is flying in a clock wise circular path above  [#permalink]

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New post 03 Oct 2018, 07:31
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  55% (hard)

Question Stats:

59% (01:17) correct 41% (01:51) wrong based on 32 sessions

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A planed named "ship" is flying in a clock wise circular path above Town X. A plane named "car" is also flying in a clockwise circular path above town X at a different altitude and speed from Plane named ship. if the circumference of the path of plane Ship is equal to the circumference of the path of plane Car, and at 5pm are directly above the same location over town X, how many circular paths must the planes make so that both planes are directly above the same location at the same time above Town X.

1.The circumference of the circular path traveled by both planes is 4 pi miles ( pi = 3.14)
2. The plane ship travels at a constant rate of \(\frac{1}{8}\) of the length of circular path per hour, and plane car travels at a constant rate of \(\frac{1}{6}\) of the length of circular path per hour.

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A planed named "ship" is flying in a clock wise circular path above  [#permalink]

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New post 03 Oct 2018, 07:59
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A planed named "ship" is flying in a clock wise circular path above Town X. A plane named "car" is also flying in a clockwise circular path above town X at a different altitude and speed from Plane named ship. if the circumference of the path of plane Ship is equal to the circumference of the path of plane Car, and at 5pm are directly above the same location over town X, how many circular paths must the planes make so that both planes are directly above the same location at the same time above Town X.

We just require the speed of two to answer or the ratio of two is also sufficient

1.The circumference of the circular path traveled by both planes is 4 pi miles ( pi = 3.14)
Circumference does not effect as both travel same distance
Insufficient

2. The plane ship travels at a constant rate of \(\frac{1}{8}\) of the length of circular path per hour, and plane car travels at a constant rate of \(\frac{1}{6}\) of the length of circular path per hour.
SHIP travels 1/8 th of distance in one hour so would complete a circle in 8 hours..
CAR travels 1/6 th of distance in one hour so would complete a circle in 6 hours..
Therefore they will meet at same place after LCM of 6 and 8 so 24hours
Thus both planes are directly above the same location at the same time above Town X next day at 5pm and by then SHIP would have done 24/8 or 3 circles and CAR 24/6=4 circles..
Sufficient

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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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