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The figure above shows a rectangular play area in which one child sta
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28 May 2021, 04:52
28 May 2021, 04:52
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83% (01:35) correct
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The figure above shows a rectangular play area in which one child stands at B while another child runs back and forth along the entire side AD. If the running child is in a position randomly located along side AD at a given time, what is the probability that the two children are at most 50 feet apart at that time?
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 1
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Re: The figure above shows a rectangular play area in which one child sta
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29 May 2021, 01:42
29 May 2021, 01:42
Carcass wrote:
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The attachment shot71.jpg is no longer available
The figure above shows a rectangular play area in which one child stands at B while another child runs back and forth along the entire side AD. If the running child is in a position randomly located along side AD at a given time, what is the probability that the two children are at most 50 feet apart at that time?
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 1
Since, both of them have to be 50 ft apart, let us join BD' such that BD' = 50
Apply Pythagoras theorem in triangle ABD':
\(50^2 = 30^2 + (AD')^2\)
\(AD' = 40\)
Now, probability that the two children are at most 50 feet apart = \(\frac{Area (△ABD')}{ Area (△ABD)} = \frac{(30)(40)}{(30)(50)} = \frac{40}{50} = \frac{4}{5}\)
Hence, option D
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gmatclubot
Re: The figure above shows a rectangular play area in which one child sta [#permalink]
29 May 2021, 01:42
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