Question

The speed of the current is 5 km / hour. A motorboat goes 10 km upstream and back again to the starting point in 50 minutes. The speed (in kmf hour) of the motorboat in still water is
A). 20
B). 26
C). 25
D). 28​

Answer 1

QuEsTiOn :-

The speed of the current is 5 km / hour. A motorboat goes 10 km upstream and back again to the starting point in 50 minutes. The speed (in km/h) of the motorboat in still water is

A). 20

B). 26

C). 25

D). 28

Given :-

Speed of the current = 5 km/h

Time taken = 50 min

Distance = 10km

$\huge{ \underline{ \underline{ \green{ \sf{ SoLuTiOn :- }}}}}$

Let the speed of the motorboat = xkm/h

Speed of motorboat upstream = x-5

Speed of motorboat when it back again= x+5

We know that

${\purple{\boxed{\large{\bold{</u><u>T</u><u>ime = \frac{</u><u>D</u><u>istance \: }{</u><u>S</u><u>peed} }}}}}$

According to question

$\implies \frac{10}{x - 5} + \frac{10}{x + 5} = \frac{50}{60} \\ \\ \implies \frac{10(x + 5) + 10(x - 5)}{(x - 5)(x + 5)} = \frac{5}{6} \\ \\ \implies 10x + 50 + 10x - 50 = \frac{5}{6} (x - 5)(x + 5) \\ \\ \implies20x = \frac{5}{6} ( {x}^{2} - 25) \\ \\ \implies20x \times 6 = 5( {x}^{2} - 25) \\ \\ \implies120x = 5 {x}^{2} - 125 \\ \\ \implies5 {x}^{2} - 120x - 125 = 0 \\ \\ \implies {x}^{2} - 24x - 25 = 0 \\ \\ \implies {x}^{2} - 25x + x - 25 = 0 \\ \\ \implies \: x(x - 25) + 1(x - 25) = 0 \\ \\ \implies(x - 25)(x + 1) = 0 \\ \\ \implies \: x = 25 \: and \: x = - 1$

Speed never be negative.

x= 25

Hence ,

${\purple{\boxed{\large{\bold{Speed \: of \:motorboat \: in \: still water = 25km/h}}}}}$

Option C is correct.

Answer 2

Let speed of boat be x km/h.

Given speed of current = 5 km/h

Therefore, Upstream speed of boat = (x-5) km/h

Downstream speed of boat = (x + 5) km/h

According to the question,

10x−5+10x+5=5060

10(x+5+x−5x2−25)=56

12×2x=x2−25

x2−24x−25=0

x2−25x+x−25=0

(x−25)(x+1)=0

Therefore, x = 25 [because,x≠−1]

option c is a right answer