Question

two cars leave Delhi at 8am travelling in opposite directions. the car travelling north runs 5km per hour faster than the car travelling south. after four hour they are 420 km apart how fast is each car travelling.
guys please tell me correct answer. with explanation.​

Answer 1

Answer:-

• The speed of car 1 is 40 km/hr.
• The speed of car 2 is 45 km/hr.

Explaination:-

★ Given:-

Two cars leave Delhi from a point at the same time, travelling in the opposite direction. If the average speed of one car is 5 km/hr more than that of the other and they are 425 km apart at the end of 5 hours.

★ To find:-

The average speeds of the two car

Let the speed of car 1 = x km/h

The speed of car 2 = x + 5 km/h

They are moving away from each other

Combined speed = x + x + 5 = 2x + 5 km/h

Time = 5 hours

Distance = 425 km

Distance = Speed × Time

425 = 2x + 5 × 5

425 = 10x + 25

10x = 425 - 25

10x = 400

x = 400 / 10

x = 40

The speed of car 1 is 40 km/hr.

The speed of car 2 is 40 + 5 = 45 km/hr.

Answer 2

$\huge \mathfrak \red{answer}$

45 km/hr

________________

Question:

》Two cars leave Delhi at the same time travelling in oppoisite directions. If the average speed off the car is 5km/hr more than that of the other and they are 425km apart at the end of 5hrs which is the average speed of each?

_____________________

》$\sf{car \: 1 \times x kmph}$

》$\sf{car 2 x + 5kmph}$

》so they r moving eachother speed

》$\sf \red{x + x + 5}$

》$\sf \red{2x + 5}$

__________________________

》$\sf{5hours \: is \: time}$

》$\sf{425 \: km \: is \: distance}$

》$\tt \red{distance = speed \times time}$

》$\tt{(2x + 5) \times 5 = 425}$

》$\tt{10x + 25 = 425}$

》$\tt{10x = 425 - 25}$

》$\tt{ = 400}$

》$\tt{x = \frac{400}{10}}$

》$\tt{x = 40 \: km/ph}$

》$\tt{car2\: 40 + 5 = 45km/h}$

i hope its help uh♡

45 km/hr