Question

A boat goes 30 km upstream and 44 downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. On this information some student guessed the speed of the boat in still water as 8.5 km/h and speed of the boat in still water as 8.5 km/h and speed of the stream as 3.8 km/h. Do you agree with their guess?
Explain what do we learn from the incident?

Answer 1

Let the speed of boat = x km/hr.
Let the speed of stream = y km/hr.
Net Speed of boat in upstream = (x - y)km/hr.
Net Speed of boat in downstream = (x + y)km/hr.
Time taken to cover 30 km upstream $=\frac{30}{x-y}$ hrs.
Time taken to cover 40 km downstream $=\frac{44}{x+y}$ hrs.
According to question,
Total time taken = 10 hrs.
$\frac{30}{x-y}+\frac{44}{x+y}=10$ ...(i)
Now, Time taken to cover 55 km downstream
$=\frac{55}{x+y}hrs.$
Time taken to cover 40 km upstream $=\frac{40}{x-y}$ hrs.
Total time taken = 13 hrs.
$\frac{40}{x-y}+\frac{55}{x+y}=13$ ...(ii)
Solving eq. (i) and eq. (ii).
Let $\frac{1}{x-y}=u,\frac{1}{x+y}=v$
$30u+44v=10$ ...(iii)
$40u+55v=13$ ...(iv)
Multiplying eq. (iii) by 4 and eq. (iv) by 3, and subtracting we get
$120u+176v=40120u+165v=39---11v=1v=\frac{1}{11}$
Putting the value of v in eq. (iii)
$30u+44v=10\Rightarrow 30u+44\times \frac{1}{11}=10\Rightarrow 30u+4=10\Rightarrow 30u=6\Rightarrow u=\frac{6}{30}$
or $u=\frac{1}{5}$
Now,
$v=\frac{1}{11}$
$\Rightarrow \frac{1}{x+y}=\frac{1}{11}$
$\Rightarrow x+y=11$ ...(v)
And $u=\frac{1}{5}$
$\Rightarrow \frac{1}{x-y}=\frac{1}{5}$
$\Rightarrow x-y=5$ ...(vi)
On solving eq. (v) and (vi)
$x+y=11x-y=5\underset{–}{}2x=16orx=8$
Put the value of x in eq. (v)
8 + y = 11
y = 11 - 8
y = 3
The speed of boat in still water = 8 km/hr.
The speed of stream = 3 km/hr.
We learn that the speed of boat is slow in upstream and fast in downstream