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bonanova v Phase v BMAD

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BMAD, bonanova, and Phase race each other in a 100 meters race. All of them run at a constant speed throughout the race.

Bmad beats bonanova by 30 meters.
bonanova beats Phase by 20 meters.

How many meters does BMAD beat Phase by?

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5 answers to this question

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Posted · Report post

44 meters

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Posted (edited) · Report post

44 metres

Edited by hhh3
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Posted · Report post

50 m

when bmad at 100 m (the end), bononova is at 70 m because he has been beat by 30m. which at that time puts phase at 50m because he is 20m behind bononova.

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Posted · Report post

Bmad runs 100 m in the time Bonanova runs 70.


BMAD's Speed / Bonanova Speed = 100 / 70
Similarly , Bonanova's Speed / Phase's Speed = 100 /80

So BMAD's speed / Phase's speed = 10000/5600 = 100/56
So when BMAD reached finishing line at 100m, phase was at 56 m and hence 44 m behind

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Posted (edited) · Report post

Got the same answer as the others,

44 meters

Except I used slightly different math:

bonanova ran at 70% of Bmad's speed.
Phase ran at 80% of bonanova's speed.

.70 x .80 = .56

100 meters x .56 =

56 meters, which is then subtracted from the 100 meter race distance to yield the 44 meter answer.

Edited by bgm1961
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