Question

A motor boat can travel 30 km upstream and 28 km downstream in 7 h. It can travel 21 km upstream and return in 5 h. find the speed of the boat in still water and the speed of the stream.

Answer 1

Let the speed of the motorboat in still water and the speed of the stream are u km/h and v km/h respectively
Then, a motor boat speed in downstream = (u – v) km/h
Motorboat has taken time to travel 30 km upstream
$t_1=\frac{30}{u-v}h$
and motor boat has taken time to travel 28 km downstream
$t_2=\frac{28}{u+v}h$
by first condition, a motor boat can travel 30 km upstream and 28 km down stream in 7 h

i.e., $t_1+t_2$ = 7h
$\Rightarrow \frac{30}{u-v}+\frac{28}{u+v}=7$ ...eq(i)
Now, motor boat has taken time to travel 21 km upstream and return i.e., $t3=\frac{21}{u-v}$ (upstream)

(for downstream)
And $t4=\frac{21}{u+v}$

By second condition, $t_4+t_3=5h$
$\Rightarrow \frac{21}{u-v}+\frac{21}{u+v}=5$ ...eq(ii)
$Letx=\frac{1}{u-v}andy=\frac{1}{u+v}$
Eqs. (i) and (ii) becomes 30x + 28y = 7 ...eq(iii)
and 21x + 21y = 5
$\Rightarrow x+y=\frac{5}{21}$ ...eq(iv)
Now, multiplying in Eq. (iv) by 28 and then subtracting from Eq. (iii), we get

30x + 28y = 7

28x + 28y = $\frac{140}{21}$
$2x=7-\frac{20}{3}=\frac{21-20}{3}$
$2x=\frac{1}{3}\Rightarrow x=\frac{1}{6}$
On putting the value of x in Eq, (iv), we get
$\frac{1}{6}+y=\frac{5}{21}$
$\Rightarrow y=\frac{5}{21}-\frac{1}{6}=\frac{10-7}{42}=\frac{3}{42}\Rightarrow \frac{1}{14}$
$\therefore x=\frac{1}{u-v}=\frac{1}{6}\Rightarrow u-v=6$ ...eq(v)
$Andy=\frac{1}{u+v}=\frac{1}{14}$
$\Rightarrow u+v=14$ ...eq(vi)
Now, adding Eqs. (v) and (vi), we get

2u = 20 $\Rightarrow$ u = 10

On putting the value of u in eq (v), we get

10 - v = 6

$\Rightarrow$ v = 4

Hence, the speed of the motorboat in still water is 10 km/h and the speed of the stream 4 km/h