Question

A car cover one third distance with speed 40 km/h one third with speed 20km/h one third with speed 60km/h find average speed

Answer 1

Answer:

32.727272 km/h

Explanation:

Given :

• Car covers 1/3 distance with speed 40 km/hr

• Car covers 1/3 distance with speed 20 km/hr

• Car covers the last 1/3 distance with speed 60 km/hr

To find :

• Average speed of the car

$average \: speed \: = \frac{total \: distance}{total \: time}$

Let the total distance be X

Time = distance / speed

Time in 1st 1/3 distance = x/3/40 = x/120

Time in 2nd 1/3 distance = x/3/20 = x/60

Time in 3rd 1/3 distance = x/3/60 = x/180

$average \: speed = \frac{x}{ \frac{x}{120} + \frac{x}{60} + \frac{x}{180} }$

Taking lcm of 120,60,180 = 360

$average \: speed = \frac{x}{ \frac{3x}{360} + \frac{6x}{360} + \frac{2x}{360} }$

$average \: speed = \frac{x}{ \frac{11x}{360} }$

$\frac{x}{11x} \times 360$

= 360/11

= 32.7272 km/h

The average speed is equal to 32.727272 km/h

Answer 2

AnswEr :

Given that car covers one third distance with speed of 40 km/h, one third with 20 km/h, one third with 60 km/h.

We have to find average speed of car.

Let the ⅓ of whole distance be y km.

We know,

Distance = Speed × Time

Time = Distance/Speed

Therefore,

→ Time in 1st journey = y/40 h

→ Time in 2nd journey = y/20 h

→ Time in 3rd journey = y/60 h

Now we know that,

Average speed = Total distance/Total time

→ Av. speed = (y + y + y)/(y/40 + y/20 + y/60)

→ Av. speed = 3y/[ (3y + 6y + 2y)/120 ]

→ Av. speed = 3y/(11y/120)

→ Av. speed = 3y × (120/11y)

→ Av. speed = 360/11

→ Av. speed = 32.72 km/h

Average speed of car = 32.72 km/h [Ans.]