Question

-The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream. Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be​

Answer 1

Answer:

• Speed of the motorboat in upstream is 10 km/hr.

Step-by-step explanation:

Given that:

• The speed of a motor boat is 20 km/hr.
• Let speed of the stream be x km/hr.

• Speed in downstream = (20 + x) km/hr
• Speed in upstream = (20 - x) km/hr

To Find:

• Speed of the motorboat in upstream.

Formula used:

• Time = Distance/Speed

In downstream:

• Speed = (20 + x) km/hr
• Distance = 15 km

• Time = 15/(20 + x) hr

In upstream:

• Speed = (20 - x) km/hr
• Distance = 15 km

• Time = 15/(20 - x) hr

Given:

• For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

According to the question.

⇢ 15/(20 - x) = 15/(20 + x) + 1

⇢ 15/(20 - x) - 15/(20 + x) = 1

Taking 15 common.

⇢ 15{1/(20 - x) - 1/(20 + x)} = 15(1/15)

Cancelling 15 both sides.

⇢ 1/(20 - x) - 1/(20 + x) = 1/15

Taking (20 - x) (20 + x) LCM in LHS and common in RHS.

⇢ (20 + x - 20 + x)/{(20 - x) (20 + x)} = {(20 - x) (20 + x)}/15{(20 - x) (20 + x)}

Cancelling {(20 - x) (20 + x)} both sides.

⇢ 2x = {(20 - x) (20 + x)}/15

Cross multiplication.

⇢ 2x × 15 = 400 - x²

⇢ 30x - 400 + x² = 0

Solving quadratic equation.

⇢ x² + 30x - 400

Splitting 30x.

⇢ x² + 40x - 10x - 400 = 0

Taking common.

⇢ x(x + 40) - 10(x + 40) = 0

⇢ (x - 10) (x + 40) = 0

⇢ x = 10 or x = - 40

We know that:

• The speed of an object is always positive.

• Speed of the stream = 10 km/hr.

• Speed in upstream = (20 - x) = (20 - 10) = 10 km/hr

Hence,

• Speed of the motorboat in upstream is 10 km/hr.

Answer 2

Given :-

The  speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream. Let speed of the stream be x km/hr

To Find :-

Speed of motorboat

Solution :-

Speed of boat when its upstream = 20 - x

Speed of boat when its downstream = 20 + x

We know that

Time = Distance/Speed

Time = 15/(20 - x)[When up stream]

Time = 15/20 + x[When down stream]

$\sf \dfrac{15}{20-x}-\dfrac{15}{20+x} = 1$

$\sf 15 \times\bigg\lgroup\dfrac{1}{20-x} - \dfrac{1}{20+x}\bigg\rgroup = 15\bigg\lgroup\dfrac{1}{15}\bigg\rgroup$

$\sf \dfrac{20+x-20+x}{20-x\times20+x} = \dfrac{20-x\times20+x}{15 (20-x\times20+x)}$

$\sf \dfrac{x+x}{20 - x\times20+x} = \dfrac{20-x\times20+x}{15 (20-x\times20+x)}$

$\sf \dfrac{x+x}{1} = \dfrac{20-x\times 20+x}{15}$

$\sf \dfrac{2x}{1} = \dfrac{20 - x \times 20 + x}{15}$

$\sf \dfrac{2x}{1} = \dfrac{400 - x^2}{15}$

$\sf 15(2x) = 400-x^2$

$\sf 30x=400-x^2$

$\sf 30x-400+x^2=0$

$\sf x^2+30x-400=0$

$\sf x^2+(40x-10x)-400=0$

$\sf x^{2} +40x-10x-400=0$

$\sf x(x+40)-10(x-40)=0$

$\sf (x-10)(x+40)=0$

Either

x - 10 = 0

x = 10 + 0

x = 10 km/h

or

x + 40 = 0

x = 0 - 40

x = -40 km/h

So, speed can't be negative.

Speed = 10 km/h

Speed of boat in upstream = 20 - 10 = 10 km/h