Algebraic rate and ratio brainteasers.

Question

"A boat is traveling upstream against a current. It takes 45 minutes to travel 30 miles. On the return trip (with current) it takes 30 minutes. What is the speed displayed on the boats speedometer?"

I've backed out a base rate of 60 mph from (30 = 45(r-c) and 30 = 30 = 30(r+c)) where r is the base rate and c is the current. From there, c = 10 mph and r = 50. Just not sure how to find the average speed, my guess is 50 from (60)+(40)/2.

As an aside feel free to include links to places where I can practice this type of problem. Many thanks!

Answer 1

Alternative approach.

Let $s_1, s_2$ = the speed of the boat, and current, respectively.

Then $~ \displaystyle (s_1 + s_2) \times \frac{1}{2} = 30 = (s_1 - s_2) \times \frac{3}{4}.$

Therefore, $~2(s_1 + s_2) = 3(s_1 - s_2) \implies 5s_2 = s_1.$

Therefore, $~\displaystyle 6s_2 \times \frac{1}{2} = 30 \implies s_2 = \frac{30 \times 2}{6} = 10 \implies s_1 = 50.$

Answer 2

Let us work in a "time-space' $t,y$ coordinate plane.

Fig. : The horizontal axis is the time axis ; the vertical axis is the (1D = river) space axis.

The first part, resp. second part, of the travel can be expressed in this way:

$$y=-s_1t \ \text{with} \ s_1=\frac{30}{45} \ \ \ \ \ \text{and resp.} \ \ \ \ \ y=s_2t+k \ \text{with} \ s_2=\frac{30}{30}$$

Now; we could consider that the boat, when arrived at $B$ continues in the same upstream direction with the same speed that when is going downstream, materialized by the dotted black line $BD$ (therefore symmetric of $BC$ with respect to $t$ axis), we can consider that the mean speed is like the one obtained by considering the red line joining $A$ to $D$, i.e.,

$$v_{mean}=\dfrac{distance}{time}=\dfrac{30+30}{45+30}=\dfrac{60 \ miles }{75 \ minutes}= 48 miles/hr$$