width of river is 1400 feet.

Heres the solution:
In the trip to the first meeting, Girl's boat(say) goes 610 feet.
Boy's goes (X-610) feet.

Since ratio of their distances equals ratio of their speed:
610/(X-610)


Girl's boat continues to go further (X-610) feet and on its return journey travels further 430 feet. ie a total of (X-610)+430 ie = (X-180)

Similarly for the Boy's boat, 610+(X-430) ie =(X+180)

since ratio of distance in the 1st part of journey is same as that of 2nd part,

610/(X-610)=(X-180)(X+180)

cross multiplying and solving the equation gives

X=1400

Therefore the river is 1400 feet wide

am not too good at maths, hope am correct 😐

Width of riverbank= 610+x=430+y

or x+180=y

i.e, the 1st boat to leave the shore had already covered 180 feet. Must be the faster one.

Now the width of the river has to be a multiple of 180, for the question to stand valid.

So the width of river has to be 1080 feet. (The faster boat completes its two hour stay early)

Looks like not many takers here.

Will close the question at 8 pm today. 😉

|----------------------|
|----------------------|
|----------------------|
|----------------------|
|<---X---><----Y---->|
|----------------------|
|----------------------|
|----------------------|
|----------------------|
|----------------------|
A----------------------B
(Girls)----------------(Boys)

Let's say X is the distance traveled by the girls and Y is the distance traveled by the boys.

Let's say that the girls' boat is the faster, and when boys and girls meet, the distance is measured from their end, ie., from A.
Then X is 610.
It also means that X is more than Y, since girls' boat is faster than boys'.

Then it must follow that in the return journey, the distances are reversed. Since the speed of the boats dont change, before meeting,
the girls boat now travels 610 units from the other end. Note that it can't be 430, since the girls' boat is faster and so the distance covered by them must be more.
So, the distance covered by the boys from the same end (ie., A) must be 430.

So, the total width of the river is the sum of the 2 numbers, viz. 610 + 430 = 1040 m

It's also the same if the boys' boat is the faster and the distance is measured from their end, ie., B.
But if the distance is measured from an end whose boat is not the faster one, then I think the question doesn't hold, because then,
if on the first meet the distance is 610, then on the reverse journey, it must be more than 610.

Hence, the width of the river is 1040 meters.

Note that I have not considered the effects of the river current itself. I dont know if the answer will change and how it will change if the speed of the river is also included.

let total width of rever=x;
if faster boat having speed=s;
slower boat speed=s';
from condition (1)
time is same
so,
610/s=(x-610)/s'------------------(1);

in returning oviousaly faster boat cover more than 430;
so,
430/s'=(x-430)/s ------------------(2);

from eqution (1) and (2)
we get x=1040 answer;

zzzzzzzzzzz!!! :O

why this kolaveri?? 😁

I meant 1040 FEET and not meters 😔

After the first crossing was complete, the combined distances travelled by the two boats was equal to twice the width of the river.
When the boats met for a second time, the total combined distance travelled would then equal three times the width of the river.


During the first meeting, the slower boat had obviously gone 610 feet from the river bank.
When they met the 2nd time, this boat would have gone 3 times the distance = 1830 feet.
At the second meeting, the slower boat was 430 feet from the shore.
If we subtract this amount from the total distance the slower boat has travelled, we get 1,400 feet for the width of the river.


The time spent on the coast has no effect on the outcome of this problem.
In this problem the girls proved to have the faster boat.

CEan-optimystix is the winner here!