Question

Two ghats are located on a riverbank and are 21
km apart. Leaving one of the ghats for the other, a
motorboat returns to the first ghat in 270 minutes,
spending 40 min of that time in taking the passengers
at the second ghat. Find the speed of the boat in still
water if the speed of the river flow is 2.5 km/h?
(a) 10.4 km/h (b) 12.5 km/h
(c) 22.5 km/h Ad) 11.5 km/h​

Answer 1

Given :

The distance between ghat = D = 21 km

Total time in travelling upward and downward = t = 270 min = 4.5 hours

The speed of river flow = y = 2.5 km/h

To Find :

The speed of the boat in still  water

Solution :

Let the the speed of the boat in still  water = x km/h

∵  Time = $\dfrac{Distance}{Speed}$

For upward motion

 Speed = ( x - y ) km/h

For downward motion

 Speed = ( x + y ) km/h

So, Total time taken = Time taken in upward + Time taken in downward

Or,  4.5 = $\dfrac{21}{x-y}$ + $\dfrac{21}{x+y}$

Or,  $\dfrac{4.5}{21}$  = $\dfrac{1}{x-2.5}$ + $\dfrac{1}{x+2.5}$  

Or, 0.21 =  $\dfrac{x+2.5+x-2.5}{(x-2.5)(x+2.5)}$

Or,  0.21 ( x - 2.5 ) ( x + 2.5 ) = 2 x

Or,  0.21 × ( x² - 6.25 ) = 2 x

Or, 0.21 x² - 2 x - 1.31 = 0

Solving this quadratic eq , we get

 x = $\dfrac{2\pm \sqrt{(-2)^{2}-4\times (0.21)\times (-1.31))}}{2 \times 0.21}$

∴  x = 10.13  ,  - 0.615

So, Speed =  x = 10.13 km/h

Hence, The speed of boast in still water is 10.13 km/h   Answer

Answer 2

Answer: The answer is option d) i.e 11.5 hours

Step-by-step explanation:

Since the motorboat halts for 40mins at the other end so the total time of travel will be 270-40=230mins

Convert into hours

230/60=3.83hours

Form an equation

Let the speed of boat be v

For upward motion

 Speed = ( v - 2.5 ) km/h  

For downward motion

Speed = ( v + 2.5) km/h

Now ATQ

$21/(v+2.5) + 21/(v-2.5) =3.83$

Solving using the options

when v=11.5 then

$21/14 + 21/9 =3.83$