The quickest and most accurate way to solve these problems is through the method of tabular representation.
Note for table : The values in
black are those that have been given and the values in
blue are those that have been calculated.
Attachment:
P1.png [ 21.2 KiB | Viewed 6792 times ]
Since we want to find out the
time taken by the raft to go from Town A to Town B, let us
assign it as variable 'x'.Now let us
assume speed of boat in calm water to be 'b' and
speed of current to be 's'. This gives us the expressions we require for speed in all three cases.
Note : The first thing that should strike us in this problem is that the distances are all the same. This implies that the solution will in all probability lie in equating the expression for the distances.
In
row 1, we are given the time and we know that since the boat is traveling upstream, it's speed must be 'b - s'. Thus we can form an expression for the distance.
Distance(1) = Speed(1) * Time(1) = (b - s)*4.5
In
row 2, we are given the time and we know that since the boat is traveling downstream, it's speed must be 'b + s'. Thus we can form an expression for the distance.
Distance(2) = Speed(2) * Time(2) = (b + s)*3
Now let us use our knowledge of the distances being equal in order to get an expression for 'b' in terms of 's'. This can be done by equating Distance(1) and Distance(2) since they are the same.
(b - s)*4.5 = (b + s)*3 ---> b = 5s
Now lets move on to
row 3. We have assumed the time taken to be 'x' and we know the speed is that of the current 's'. Thus we can obtain an expression in for the distance in terms of 's' and 'x'.
Distance(3) = Speed(3) * Time(3) = s*x
Again, we know that this must be equal to both Distance(1) and Distance(2). So let us equate it with any one of them to obtain an expression for 'x' in terms of 'b' and 's'.
Equating it to Distance(2) we get :
(b + s)*3 = s*x ---> 3b + 3s = s*x ---> Substituting b = 5s ---> 18s = s*x ---> x = 18 hours.OR
In case we would have equated it to Distance(1) we would still have got the same result :
(b - s)*4.5 = s*x ---> 4.5b - 4.5s = s*x ---> Substituting b = 5s ---> 22.5s - 4.5s = s*x ---> x = 18 hours.
Answer : 18 hours.
junker wrote:
Say the distance between A and B is 4.5 miles.
Moving from A to B and 1 mile/hr (4.5 hours expended).
Move from B to A at 1.5 mile/hr (3 hours expended).
So from B to A, the downstream acceleration provided by water is 0.5 miles/hour. So if the engines were shut-off the boat would float down 4.5 miles at 0.5 m/h - so it would take 9 hours.
9 is not a choice in the answers provided, why is my reasoning above wrong?
Moving from A to B and 1 mile/hr (4.5 hours expended)
Move from B to A at 1.5 mile/hr (3 hours expended)
Your reasoning is correct till this point. However,
you cannot subtract one speed from the other to get the speed of the current. In one case it is the speed of the boat - speed of current while in the other case it is the speed of boat + speed of current.
Thus you will have the following two expressions :
b + s = 1.5
b - s = 1Solving them you will get
b = 1.25 mph and
s = 0.25 mphThen you can calculate the time taken by raft to be = 4.5/0.25 = 18 hours.
Although this method is not wrong, there is a lot of scope for silly mistakes (as you would have realized) if your concepts are not one hundred percent clear.
Personally, I believe that the most fool proof way to solve these problems is through tabular representation. (At least until your concepts become strong and maybe even then).
You can check out my post on these types of word problems (you'll find the link below). You might find it helpful.
Cheers.
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