Question

a motor boat whose speed is 20 km/hr in still water takes 1 hr more to go 21 km upstream than to return downstream to the same spot. find the speed of the stream

Answer 1

Given :-

• Speed of motor Boat in still water = 20km/h.
• Distance cover = 21km/h.
• Time in upstream = 1 Hour + Time in Downstream.

To find :-

• Speed of stream ?

Concept used :-

• Downstream Speed = (Speed of Boat in Still water + Speed of Water in The river).
• Upstream Speed = (Speed of Boat in Still water - Speed of Water in The river).

Solution :-

Let us Assume That, Speed of Stream is x km/h.

Than,

→ Downstream Speed = (20 + x) km/h.

→ Distance covered = 21km

→ Time Taken = D/S = 21/(20+x) Hours.

And,

→ Upstream Speed = (20 - x) km/h.

→ Distance covered = 21km

→ Time Taken = D/S = 21/(20 - x) Hours.

A/q,

→ Upstream Time - Downstream = 1

→ 21/(20 - x) - 21/(20 + x) = 1

→ { 21(20+x) - 21(20 - x) } / (20 - x)(20+x) = 1

→ 21*20 - 21*20 + 21x + 21x = (20² - x²)

→ 42x = 400 - x²

→ x² + 42x - 400 = 0

→ x² + 50x - 8x - 400 = 0

→ x(x + 50) - 8(x + 50) = 0

→ (x +50)( x - 8) = 0

→ x = (-50) or 8. [ Negative value ≠ ]

Hence, Speed of Stream is 8km/h..

Answer 2

Answer:

• Speed of Motor Boat = 20 km/hr
• Let Speed of Stream = y km/hr
• Fixed Distance = 21 km
• Time Taken more = 1 km

$\rule{120}{1}$

$\boxed{\bf{\mid{\overline{\underline{\bigstar\:According\:to\:the\:Question :}}}}\mid}$

$:\implies\sf Time_{(upstream)}-Time_{(downstream)}=1\:hr\\\\\\:\implies\sf \dfrac{Distance}{Upstream}-\dfrac{Distance}{Downstream}=1\:hr\\\\\\:\implies\sf Distance\Bigg\lgroup\dfrac{1}{Upstream}-\dfrac{1}{Downstream}\Bigg\rgroup=1\:hr\\\\\\:\implies\sf 21\Bigg\lgroup\dfrac{1}{(20-y)}-\dfrac{1}{(20+y)}\Bigg\rgroup=1\:hr\\\\\\:\implies\sf \Bigg\lgroup \dfrac{20 + y - 20 + y}{(20)^2 - (y)^2} \Bigg\rgroup = \dfrac{1}{21}\\\\\\:\implies\sf \Bigg\lgroup \dfrac{2y}{400 - y^2}\Bigg\rgroup = \dfrac{1}{21} \\\\\\:\implies\sf 42y = 400 - y^2\\\\\\:\implies\sf y^2 + 42y - 400 = 0\\\\\\:\implies\sf y^2 + 50y - 8y - 400 = 0\\\\\\:\implies\sf y(y + 50) - 8(y + 50) = 0\\\\\\:\implies\sf (y - 8)(y + 50) = 0\\\\\\:\implies\sf \green{y = 8\:km/hr} \quad or \quad \red{y = - \:50 \:km/hr}$

⠀

$\therefore\:\underline{\textsf{Speed of the Stream will be \textbf{8 km/hr}}}.$