Question: A man can row 50 km upstream and 72 km downstream in 9 hours. He can also row 70 km upstream and 90 km downstream in 12 hours. Find the rate of current.
- 2 kmph
- 3 kmph
- 4 kmph
- 6 kmph
- 8 kmph
“A man can row 50 km upstream and 72 km downstream in 9 hours”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”.
To decipher GMAT Problem Solving questions, the candidate must possess analytical knowledge of basic math concepts. GMAT quant is designed to examine the candidate’s qualitative skills. This GMAT quantitive section includes arithmetic, geometry and algebra. This section of the GMAT includes a question with a series of options. The candidate must select the correct option by solving it with mathematical calculations. The GMAT Quant topic includes mathematical problems in the problem-solving part that are required to be cracked with proper mathematical facts.
Solution and Explanation:
Approach Solution 1:
The problem statement informs that
Given:
- A man can row 50 km upstream and 72 km downstream in 9 hours.
- He can also row 70 km upstream and 90 km downstream in 12 hours.
Find Out:
- The rate of current
Let’s assume the speed of the current and man be r and m km/hr respectively.
We can infer two equations from the given two scenarios:
50/(m-r) + 72/(m+r) = 9 ... (i) -That is the sum of the time taken by the man to move upstream and downstream is 9 hours.
70/(m-r) + 90/(m+r) = 12 ... (ii) - That is the sum of the time taken by the man to move upstream and downstream is 12 hours.
Here, assuming (m+r) = v and (m-r) = u, where ‘v’ is the downstream speed and ‘u’ is the upstream speed.
Therefore, by solving v and u we get,
50/u + 72/v = 9
70/u + 90/v = 12
=> u = 10 and v = 18
=> m - r = 10 and m + r = 18
Therefore, by simplifying, we can get the value of ‘r’ that is equal to 4.
Hence, option C is the correct answer.
Correct Answer: C
Approach Solution 2:
The problem statement discloses that
Given:
- A man goes 50 km upstream and 72 km downstream in 9 hours.
- He further goes 70 km upstream and 90 km downstream in 12 hours.
Find Out:
- The speed of the current
We can say,
The man takes 12 hours to go 70 km upstream (U) and 90 km downstream (D)....(i)
Also, the man takes 9 hours to go 50 km upstream and 72 km downstream
If the time is increased by one-third of its value, then it would become 12 hours (in the first scenario); in that case, the distance also needs to be increased by ⅓ rd of the respective values.
=> 200/3 km upstream (U) and 96 km downstream (D) takes 12 hours ... (ii)
Thus equating the times from equations (i) and (ii):
200/3 (U) + 96 (D) = 70 (U) + 90 (D)
=> 10/3 (U) = 6 (D)
=> 10 (U) = 18 (D)
=> Time required to cover 10 km upstream is equal to the time taken to go 18km downstream.
Let the speed of the current and man be r and m km/hr respectively.
=> 10/(m-r) = 18/(m+r)
=> m : r = 7 : 2
=> m = 7k and r = 2k
Thus we can derive from the first scenario:
50/5k + 72/9k = 9
=> k = 2
=> Therefore, rate of current = 2k = 4 km/hr
Hence, option C is the correct answer.
Correct Answer: C
Approach Solution 3:
The problem statement suggests that
Given:
- A man can row 50 km upstream and 72 km downstream in 9 hours.
- He can also row 70 km upstream and 90 km downstream in 12 hours.
Asked: Find the speed of the current.
Let the speed of rowing be r and the speed of current be c.
A man can row 50 km upstream and 72 km downstream in 9 hours
Therefore, we get, 50/(r-c) + 72/(r+c) = 9 …(i)
He can also row 70 km upstream and 90 km downstream in 12 hours,
Therefore, we can say, 70/(r-c) + 90/(r+c) = 12
7/(r-c) + 9/(r+c) = 1.2 ….(ii)
By multiplying equation no. (ii) with 8, we get,
56/(r-c) + 72/(r+c) = 9.6 …(iii)
By subtracting equation no. (iii) to (i), we can infer,
6/(r-c) = .6
r-c = 10
r = c + 10 …. (iv)
Putting the value of r in equation (ii)
.7 + 9/(2c+10) = 1.2
9/(2c+10) = .5
18 = 2c + 10
c = 4 km/hr,
Therefore, the speed of the current is equal to 4km/hr
Hence option C is the correct answer.
Correct Answer: C
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