Question

два насоса работая одновременно могут откачать воду из котлована за 3 часа 36 минут. Один первый насос затратит на эту работу на 3 часа больше, чем второй. за какое время может выкачать воду каждый насос.

Answer 1

1)36/3=12Ч(2-ОЙ НАСОС)

12УМНОЖИТЬ НА 3 =36

Answer 2

Calculation of Pumping Time

To determine the time it takes for each pump to pump out the water, we can set up the following equation:

Let x be the time taken by the second pump (in hours). Then, the time taken by the first pump would be x + 3 hours.

According to the given information, when both pumps work together, they can pump out the water from the pit in 3 hours and 36 minutes, which is equivalent to 3.6 hours.

Using the concept of work rates, we can set up the following equation:

1/(x + 3) + 1/x = 1/3.6

To solve this equation, we can multiply through by the least common denominator, which is (x + 3) * x * 3.6:

3.6 * x + 3.6 * (x + 3) = (x + 3) * x

Simplifying the equation:

3.6x + 3.6x + 10.8 = x^2 + 3x

Rearranging the equation:

x^2 - 4.2x - 10.8 = 0

Now, we can solve this quadratic equation to find the value of x.

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = -4.2, and c = -10.8.

Calculating the discriminant:

D = b^2 - 4ac = (-4.2)^2 - 4(1)(-10.8) = 17.64 + 43.2 = 60.84

Since the discriminant is positive, we have two real solutions.

Using the quadratic formula:

x = (-(-4.2) ± √(60.84)) / (2 * 1) x = (4.2 ± √60.84) / 2 x = (4.2 ± 7.8) / 2

The two solutions are:

x1 = (4.2 + 7.8) / 2 = 12 / 2 = 6 x2 = (4.2 - 7.8) / 2 = -3.6 / 2 = -1.8

Since time cannot be negative, we discard the negative solution.

Therefore, the second pump takes 6 hours to pump out the water, and the first pump takes 6 + 3 = 9 hours to pump out the water.

Answer:

- The first pump can pump out the water from the pit in 9 hours. - The second pump can pump out the water from the pit in 6 hours.

Please note that the above calculations are based on the given information and the assumption that the pumps work at a constant rate throughout the pumping process.