A boat takes 2 hours to travel downstream a river from port A to port B , and 3 hours to return to port A . Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is
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Official Correct Answer: D. Okay, so I'm trying to solve this CAT question about boats and rivers. It's a time, speed, and distance problem. Let me read the question again carefully.
A boat takes 2 hours to travel downstream from port A to port B, and 3 hours to return upstream to port A. Another boat takes a total of 6 hours to travel from B to A and back to B. We need to find the time taken by the slower boat to go from A to B.
Hmm, okay. So, there are two boats here. Let's denote the first boat as Boat 1 and the second as Boat 2. The river has a constant current, so the speed of the river (let's call it 'r') is the same for both boats. The speeds of the boats themselves are also constant, so let's denote the speed of Boat 1 as 'v1' and Boat 2 as 'v2'.
When a boat is going downstream, its effective speed is (v + r), and when it's going upstream, it's (v - r).
For Boat 1:
- Downstream time: 2 hours. So, distance from A to B is (v1 + r) * 2.
- Upstream time: 3 hours. So, distance from B to A is (v1 - r) * 3.
Since the distance between A and B is the same, we can set these equal:
(v1 + r) * 2 = (v1 - r) * 3.
Let me write that equation out:
2(v1 + r) = 3(v1 - r)
Expanding both sides:
2v1 + 2r = 3v1 - 3r
Let me bring like terms together:
2v1 - 3v1 = -3r - 2r
- v1 = -5r
So, v1 = 5r.
Okay, so the speed of Boat 1 is 5 times the river's speed. That makes sense because it's going downstream faster and upstream slower.
Now, let's think about Boat 2. It takes a total of 6 hours to go from B to A and back to B. Let's denote the time taken by Boat 2 to go downstream as t2 and upstream as (6 - t2).
Wait, actually, no. Wait, the total time is 6 hours for the round trip. So, the time taken to go from B to A (upstream) is t2, and the time to return downstream is (6 - t2).
But wait, actually, the direction might be different. Let me clarify: When Boat 2 starts from B, it's going upstream to A, and then downstream back to B. So, the time upstream is t2, and downstream is (6 - t2).
So, the distance from B to A is (v2 - r) * t2, and the distance from A to B is (v2 + r) * (6 - t2). Again, since the distance is the same, we can set these equal:
(v2 - r) * t2 = (v2 + r) * (6 - t2)
But wait, we don't know t2 yet. Hmm, maybe there's another way. Alternatively, since the distance is the same as for Boat 1, which we found earlier as (v1 + r)*2, which is equal to (v1 - r)*3.
We know v1 = 5r, so Computing the distance between A and B.
Distance = (v1 + r)*2 = (5r + r)*2 = 6r*2 = 12r.
Alternatively, Distance = (v1 - r)*3 = (5r - r)*3 = 4r*3 = 12r. Yep, that matches.
So, the distance between A and B is 12r.
Now, for Boat 2, the total time is 6 hours for the round trip. Let's denote the time taken by Boat 2 to go upstream (from B to A) as t2, and downstream (from A to B) as (6 - t2).
So, the distance upstream is (v2 - r)*t2 = 12r.
The distance downstream is (v2 + r)*(6 - t2) = 12r.
So, we have two equations:
1. (v2 - r)*t2 = 12r
2. (v2 + r)*(6 - t2) = 12r
Let me write these out:
(v2 - r)t2 = 12r ...(1)
(v2 + r)(6 - t2) = 12r ...(2)
Now, we have two equations with two variables: v2 and t2. Let's try to solve for v2 and t2.
From equation (1):
(v2 - r)t2 = 12r
From equation (2):
(v2 + r)(6 - t2) = 12r
Expanding equation (2):
6(v2 + r) - t2(v2 + r) = 12r
Which is:
6v2 + 6r - t2v2 - t2r = 12r
Now, let's rearrange terms:
6v2 - t2v2 + 6r - t2r = 12r
Factor terms:
v2(6 - t2) + r(6 - t2) = 12r
Notice that (6 - t2) is a common factor:
(6 - t2)(v2 + r) = 12r
Wait, but from equation (2), we know that (v2 + r)(6 - t2) = 12r, which is the same as equation (2). So, that doesn't help much.
Maybe we can express t2 from equation (1) and substitute into equation (2).
From equation (1):
(v2 - r)t2 = 12r
So, t2 = 12r / (v2 - r)
Now, substitute this into equation (2):
(v2 + r)(6 - (12r / (v2 - r))) = 12r
Compute the term inside the parentheses:
6 - (12r / (v2 - r)) = [6(v2 - r) - 12r] / (v2 - r)
Compute numerator:
6v2 - 6r - 12r = 6v2 - 18r
So, the term becomes (6v2 - 18r)/(v2 - r)
So, equation (2) becomes:
(v2 + r) * (6v2 - 18r)/(v2 - r) = 12r
Multiply both sides by (v2 - r):
(v2 + r)(6v2 - 18r) = 12r(v2 - r)
Let