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A boy can row a boat at a constant rate of 5 miles per hour in still
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01 Aug 2024, 01:30
01 Aug 2024, 01:30
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Question Stats:
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27%
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A boy can row a boat at a constant rate of 5 miles per hour in still water. In a river with the constant current rate, he rows upstream for 15 minutes and then rows downstream, returning to his starting point in another 12 minutes. What is the rate of the current, in miles per hour?
(A) 5/9
(B) 5
(C) 9
(D) 25
(E) 45
(A) 5/9
(B) 5
(C) 9
(D) 25
(E) 45
A boy can row a boat at a constant rate of 5 miles per hour in still
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01 Aug 2024, 04:03
01 Aug 2024, 04:03
Method 1:
Lets the speed of the current be \(x\).
\(T_{DS} = 12 min = \dfrac{12}{60} hr = \dfrac{1}{5} hr\) , whereas DS = downstream
\(T_{US} = 15 min = \dfrac{15}{60} hr = \dfrac{1}{4} hr\), whereas US = upstream
\(RS_{DS} = 5 + x\), where as RS is relative speed, and Speed in still watrer = 5 m/hr.
\(RS_{US} = 5 - x\)
\(\dfrac{RS_{us}}{RS_{ds}} = \dfrac{T_{ds}}{T_{us}}\)
\(\dfrac{5 - x}{5+x} = \dfrac{\frac{1}{5}}{\frac{1}{4}} = \dfrac{4}{5}\)
\(25 - 5\cdot x = 20 + 4\cdot x\)
\(x = \dfrac{5}{9}\)
A
Lets the speed of the current be \(x\).
\(T_{DS} = 12 min = \dfrac{12}{60} hr = \dfrac{1}{5} hr\) , whereas DS = downstream
\(T_{US} = 15 min = \dfrac{15}{60} hr = \dfrac{1}{4} hr\), whereas US = upstream
\(RS_{DS} = 5 + x\), where as RS is relative speed, and Speed in still watrer = 5 m/hr.
\(RS_{US} = 5 - x\)
Quote:
NOTE: \(D = {S_{US}}\times {T_{US}} = {S_{DS}}\times {T_{DS}}\)
Rewrite the above equation:\(\dfrac{RS_{us}}{RS_{ds}} = \dfrac{T_{ds}}{T_{us}}\)
\(\dfrac{5 - x}{5+x} = \dfrac{\frac{1}{5}}{\frac{1}{4}} = \dfrac{4}{5}\)
\(25 - 5\cdot x = 20 + 4\cdot x\)
\(x = \dfrac{5}{9}\)
A
A boy can row a boat at a constant rate of 5 miles per hour in still
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01 Aug 2024, 07:54
01 Aug 2024, 07:54
lets assume a speed of flow of water and b speed of boy.
Given - b = 5mph and he rows upstream for 15 minutes and then rows downstream, returning to his starting point in another 12 minutes.
no need to convert in hrs as the distance is same it will cancel out from both sides.
\(15(5-a) = 12(5+a)\)
\(25-5a = 20 + 4a\)
\(a = \frac{5}{9}\)
To find - What is the rate of the current, in miles per hour \(= a = 5/9\)
Answer is Option A.
Given - b = 5mph and he rows upstream for 15 minutes and then rows downstream, returning to his starting point in another 12 minutes.
no need to convert in hrs as the distance is same it will cancel out from both sides.
\(15(5-a) = 12(5+a)\)
\(25-5a = 20 + 4a\)
\(a = \frac{5}{9}\)
To find - What is the rate of the current, in miles per hour \(= a = 5/9\)
Answer is Option A.
