Can someone help me solve this?

Question

A boat travels 88 km upstream and 88 km back. The time for the round trip is 44 hrs. The speed of the stream is 44 km/hr. What is the speed of the boat in still water?

Answer 1

This is a creative twist on a classic problem, but the set of equations is the same.  Let's call B the speed of the boat in still water (in hm/hr).  Then we just need to define the equations for how we would normally calculate the time up and back.  First, the trip up river:

 

T_up = 88 / (B - 44)          (distance over velocity)

 

T_down = 88 / (B + 44)

 

Now we know that the sum of the times is 44 hours, so:

 

88 / (B - 44)  +  88 / (B + 44)  =  44

 

From here, we just need to work the algebra to solve for B.  Since we know that (B - 44) can't equal zero (or else the boat never would have made it up), we can multiply both sides by (B-44)(B+44) = B² - 44².

 

88 (B+44) + 88(B-44) = 44(B² - 44²)

 

Dividing by 44 will simplify:

 

2(B+44) + 2(B-44) = (B² - 44²)

 

Then we can consolidate to the left side:

 

-B² + 4B+44² = 0

 

Applying the quadratic formula, we get:

 

B = [-4 ± √(16 - 4*(-1)*(44²))]/(2*(-1))

 

= 2 ± 44.05 = 46.05 or -42.05

 

Now, the negative answer doesn't make sense, so the speed of the boat must be 46.05 km/hr.

 

If you have any questions on the explanation, let me know.

Answer 2

d = r · t

d_up = (r - 44)t_up ⇒ 88/(r - 44) = t_up

d_down = (r + 44)t_down ⇒ 88/(r + 44) = t_down

t_up + t_down = 44

88/(r - 44) + 88/(r + 44) = 44

 

Can you solve for r?