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Sorry for my bad English. I was preparing for entry into physical and mathematical School high school in my city. But I am stuck while trying to solve this task:

On a river, at the same time from locations A and B, with the same speed, two boats sail towards each other. The boat coming from location A reached location B two hours after meeting the boat coming from location B, and the boat coming from location B reached location A 4.5 hours after meeting the boat that came from location A. How much more is the speed of the boats as compared to the flow rate?

I was trying to make an equality, but it did not help me to solve the task.

S = 2(x+y) + 4,5(x-y)

x is the speed of the boat, y is flow rate

I do not have an idea how to solve this, thank you for helping.

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1 Answers1

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Assuming units of distance to be $km$ and that of time to be $hrs$.

Let the velocity of the river be $v$, in the direction from $A$ to $B$, and let the speed of the boats be $u$. Then, the velocity of the boat leaving from location $A$ (or boat $A$) will be $u+v$ from $A$ to $B$, and that of the boat leaving from location $B$ (or boat $B$) will be $u-v$ from $B$ to $A$. Let the distance between $A$ and $B$ be $d$. Let $t$ be time at which the two boats meet. Then, as $$\text{Distance}=\text{Speed}\times{Time}$$ For boat $A$, $$\tag1d=(u+v)\times(t+2)$$ For boat $B$, $$\tag2d=(u-v)\times(t+4.5)$$ Thus, $$ut+vt+2u+2v=ut-vt+4.5u-4.5v$$ $$\tag32vt=2.5u-6.5v$$ Moreover, when the boats meet, the sum of distances covered by boat $A$ and boat $B$ is $d$. The time taken to meet is $t$. So, $$\tag4d=(u+v)\times t+(u-v)\times t$$ Using $(1)$ and $(4)$, $$ut+vt+2u+2v=2ut$$ $$ut-vt=2u+2v$$ $$t=\frac{2(u+v)}{u-v}$$ Using in $(3)$, $$\frac{4v(u+v)}{u-v}=2.5u-6.5v$$ $$4uv+4v^2=2.5u^2-6.5uv-2.5uv+6.5v^2$$ $$2.5u^2-13uv+2.5v^2=0$$ $$5u^2-26uv+5v^2=0$$ $$5u^2-25uv-uv+5v^2=0$$ $$5u(u-5v)-v(u-5v)=0$$ $$(5u-v)(u-5v)=0$$ The speed of the boat must be more than that of the river. Thus, $$u=5v$$ Thus, the speed of the boat is $5$ times the flow rate.

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