Question

Два брата одновременно вышли из дома и направились в школу.Старший брат шел со скоростью 80м/мин Младший брат шел со скоростью на 30м/мин меньшей чем скорость старшего брата поэтому он пришел в школу на 6мин позже За сколько минут дошел до школы старший брат

Answer 1

Пусть s-путь тогда x-скорость,t1-время первого y-скорость,t2-время второго
y=x-30
t1+6=t2
x*t1=y*t2
t2=(x*t1)/y
t1+6=(x*t1)/(x-30)
(t1+6)*(x-30)=x*t1
(t1+6)*50=80t1
50t1+300=80t1
30t1=300
t1=10мин

Answer 2

Calculation of the Time Taken by the Older Brother to Reach School

To calculate the time taken by the older brother to reach school, we need to consider the speed at which he was walking and the time difference between the two brothers.

Given: - Speed of the older brother: 80 m/min - Speed of the younger brother: 30 m/min less than the speed of the older brother - Time difference between the two brothers: 6 minutes

Let's calculate the time taken by the older brother to reach school.

We can set up the following equation:

Distance = Speed × Time

Let's assume the time taken by the older brother to reach school is t minutes.

The distance covered by the older brother is the same as the distance covered by the younger brother plus the time difference:

Distance covered by the older brother = Distance covered by the younger brother + Time difference

Using the equation for distance, we can write:

Speed of the older brother × t = Speed of the younger brother × (t + 6)

Substituting the given values:

80t = (80 - 30) × (t + 6)

Simplifying the equation:

80t = 50t + 300

30t = 300

t = 10

Therefore, the older brother took 10 minutes to reach school.