Question

Ласточка пролетела 420 км а синица 300км .Ласточка была в пути на 2 часа больше чем синица. Сколько времени была в пути каждая птица если они летели с одинаковой скоростью?

Answer 1

420-300=120 разница в км-х
120:2=60 скорость км\час
420:60=7 ласточка
300:60=5 синица

Answer 2

1)420-300=120 разница полёта
в км
2) 120:2=60км общая скорость
3)420:60=7 ласточка
4) 300:60=5 синица

Answer 3

Problem Analysis

We are given that a swallow flew 420 km and a tit flew 300 km. The swallow was in the air for 2 hours longer than the tit. We need to determine the time each bird was in the air if they were flying at the same speed.

Solution

Let's assume that the speed of both birds is x km/h.

We know that the distance traveled by the swallow is 420 km and the distance traveled by the tit is 300 km. We can use the formula distance = speed × time to find the time traveled by each bird.

For the swallow: 420 km = x km/h × (t + 2) hours (where t is the time traveled by the tit)

For the tit: 300 km = x km/h × t hours

We can solve this system of equations to find the values of x and t.

Calculation

Let's solve the system of equations:

From the equation for the swallow: 420 = x(t + 2)

From the equation for the tit: 300 = xt

We can rewrite the first equation as: 420 = xt + 2x

Now we can solve for t: 420 - 2x = xt t = (420 - 2x) / x

Substituting this value of t into the equation for the tit: 300 = x((420 - 2x) / x) 300 = 420 - 2x 2x = 420 - 300 2x = 120 x = 60

Now we can substitute the value of x back into the equation for t: t = (420 - 2(60)) / 60 t = (420 - 120) / 60 t = 300 / 60 t = 5

So, the swallow was in the air for 5 hours and the tit was in the air for 5 - 2 = 3 hours.

Answer

If the swallow and the tit were flying at the same speed, the swallow was in the air for 5 hours and the tit was in the air for 3 hours.