Question

a bus runs at 40 km per hour to reach its destination but arrives 11 minute late if it runs at 50 km per hour it will be 5 minute late find the correct time in which the bus will not be late in reaching its destination​

Answer 1

Answer:

In 19 minutes, the bus will not be late in reaching its destination.​

Step-by-step explanation:

This question can be solve by 2 methods.

Given -

A bus runs at 40 km per hour to reach its destination.

The bus arrives 11 mins late.

It runs at 50 km per hour it will be 5 minute late.

To Find

The time in which the bus will not be late in reaching its destination​.

• Solution 1 -

Consider the -

Time as - t.

It implies,

Increase in speed by 25% will decrease of 1/5th in time if distance is constant.

So,

$\sf\\ \\\implies \dfrac{t}{5} = (11- 5) = 6\\ \\\implies t = 30m.$

We get the normal time as 30 minutes. But it is given that the bus arrives 11 mins late.

So,

$\sf \implies 30-11 = 19$

∴ The answer is 19 Minutes.

$\rule{300}{1.5}$

• Solution 2 -

Find the Ratio of the speeds & Time.

So,

Ratio of speeds = 40 : 50 = 4 : 5

Now,

If distance is constant then, Time is inversely proportional to Speed.

So,

Time Ratio = 5 : 4

Time saved = $\sf \dfrac{(5 - 4)}{5}=\dfrac{1}{5}$

⇒  30 minutes.

⇒ 30 - 11 = 19 minutes.

∴ The answer is 19 Minutes.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Correct time in which the bus will not be late in reaching its destination = 19 mins

$\bold{\underline{\underline{Step\:by\:step\:explanation:}}}$

Given (part 1) :

• A bus runs at 40 km per hour
• The bus arrives 11 mins late.

Given (part 2) :

• The speed of the bus is 50 km per hour
• The bus arrives 5 mins late.

To find :

• The correct time in which the bus will not be late in reaching its destination.

Solution :

Let x be the distance traveled by bus.

Let y be the correct time in which the bus will not be late in reaching its destination.

Let's first solve the first part.

Speed = 40 km/hr.

Time = y - 11 min.

Formula :

$\bold{\boxed{\large{\sf{\blue{Speed\:=\:{\dfrac{Distance}{Time}}}}}}}$

Block in the values,

$\implies$$\bold{40\:=\:{\dfrac{x}{y-11}}}$

Cross multiplying,

$\implies$$\bold{40(y-11) = x}$

$\implies$$\bold{40y-440=x}$

$\implies$$\bold{x=40y-440}$

$\bold{x-40y\:=\:-440}$ -->(1)

Let's solve for second part :-

Speed = 50 km/hr

Time = y - 5 mins

Block in the values in the formula,

$\implies$$\bold{50\:=\:{\dfrac{x}{y-5}}}$

Cross multiplying,

$\implies$$\bold{50(y-5) = x}$

$\implies$$\bold{50y-250 = x}$

$\implies$$\bold{x= 50y-250}$

$\bold{x-50y=-250}$ --> (2)

Solve equation 1 and equation 2 simultaneously by elimination method.

Subtract equation 2 from equation 1,

x - 40y = - 440 ---> (1)

x - 50y = 250 --> (2)

------------------------

10y = - 190

$\implies$$\bold{y={\dfrac{-190}{10}}}$

$\implies$$\bold{y=-19}$

Since time cannot be negative.

•°• y = 19 mins

•°• the correct time in which the bus will not be late in reaching its destinationis 19 mins.