Question

Два брата одновременно вышли из дома и направились в школу. Старший брат шёл со скоростью 80 м /мин.Младший брат шёл со на 30 м/мин меншей,чем скоростью брата,по этому он пришёл в школу на 6 мин позже.За сколько минут дошёл до школы старший брат?

Answer 1

Пусть х мин -время в пути старшего брата, тогда х+6 мин время пути младшего брата
1) 80-30=50 м/мин - скорость младшего брата
составим уравнение:
т.к. оба брата прошли одинаковое расстояние, тогда
х*80=(х+6)*50
80х=50х+300
80х-50х=300
30х=300
х=300:30
х=10 мин -время пути старшего брата
Ответ: 10 мин.

Answer 2

Младший шёл со скоростью 80-30 = 50 м в мин.
Старший дощёл за x минут, младший за x+6 минут. Старший прошёл 80x метров, младший (x+6)*50. Оба они прошли одно и то же расстояние, т. е.
80x = (x+6)*50
80x = 50x + 300
30x = 300
х= 10 мин

Answer 3

Problem Analysis

We have two brothers who left home at the same time and headed to school. The older brother walked at a speed of 80 m/min, while the younger brother walked at a speed 30 m/min slower than the older brother. The younger brother arrived at school 6 minutes later than the older brother. We need to determine how many minutes it took for the older brother to reach the school.

Solution

Let's assume that the time taken by the older brother to reach the school is t minutes. Since the younger brother arrived 6 minutes later, his time taken to reach the school would be t + 6 minutes.

We know that the distance traveled by both brothers is the same, as they started from the same point and ended at the same destination (school).

The distance traveled by the older brother can be calculated using the formula: distance = speed × time.

The distance traveled by the older brother is equal to the distance traveled by the younger brother. Therefore, we can set up the following equation:

80t = (80 - 30)(t + 6)

Simplifying the equation:

80t = 50(t + 6)

Expanding and rearranging the equation:

80t = 50t + 300

30t = 300

t = 10

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

Answer

The older brother took 10 minutes to reach the school.

Explanation

The older brother walked at a speed of 80 m/min, while the younger brother walked at a speed 30 m/min slower than the older brother. Since the younger brother arrived at school 6 minutes later than the older brother, we can set up an equation to solve for the time taken by the older brother to reach the school. By equating the distances traveled by both brothers, we find that the older brother took 10 minutes to reach the school