answer：First, I need to determine how many trains arrive at the station in one hour. Since a train arrives every 5 minutes, there are 60 minutes divided by 5 minutes per train, which equals 12 trains per hour.Each train takes on 320 passengers and drops off 200 passengers. To find the total number of passengers boarding the trains in an hour, I multiply the number of trains by the passengers boarding each train: 12 trains × 320 passengers = 3,840 passengers boarding.Similarly, to find the total number of passengers alighting from the trains in an hour, I multiply the number of trains by the passengers alighting each train: 12 trains × 200 passengers = 2,400 passengers alighting.Therefore, within one hour, 3,840 different passengers board the trains and 2,400 different passengers alight at the station.

answer：First, I need to determine the total capacity of water that each truck can carry. Each truck has three tanks, and each tank can hold 150 liters of water.So, for one truck, the total capacity is:3 tanks × 150 liters per tank = 450 litersNext, since the farmer has three trucks, I'll calculate the total capacity for all three trucks:3 trucks × 450 liters per truck = 1350 litersTherefore, the farmer can carry a total of 1350 liters of water in his trucks.

answer：Okay, so I have this problem about Petya and his three classmates running a 100-meter race. Let me try to figure out how to solve it step by step. First, let me restate the problem to make sure I understand it correctly. Petya and three classmates start a 100-meter race together. Petya finishes first. Twelve seconds after the race started, no one had finished yet, and all four of them had collectively run a total of 288 meters. When Petya finished the race, the other three participants had a combined distance of 40 meters left to the finish line. I need to find out how many meters Petya ran in the first 12 seconds.Alright, so each participant is running at a constant speed. That means their speeds don't change throughout the race. Let me denote Petya's speed as ( v_p ) meters per second, and the speeds of his three classmates as ( v_1 ), ( v_2 ), and ( v_3 ) meters per second. The race is 100 meters, so the time it takes Petya to finish the race is ( t_p = frac{100}{v_p} ) seconds. Now, twelve seconds after the race started, no one had finished yet. That means that in 12 seconds, each runner had covered less than 100 meters. The total distance covered by all four runners in those 12 seconds is 288 meters. So, the sum of the distances each runner covered in 12 seconds is 288 meters.Let me write that as an equation:( 12v_p + 12v_1 + 12v_2 + 12v_3 = 288 )I can factor out the 12:( 12(v_p + v_1 + v_2 + v_3) = 288 )Divide both sides by 12:( v_p + v_1 + v_2 + v_3 = 24 )So, the sum of all four runners' speeds is 24 meters per second.Next, when Petya finished the race, the other three participants had a combined distance of 40 meters left. That means, when Petya finished, each of the other three runners had some distance left, and the sum of those distances is 40 meters.Let me denote the time it took Petya to finish the race as ( t_p ). So, ( t_p = frac{100}{v_p} ). In that same time ( t_p ), each of the other runners ran a distance of ( v_1 t_p ), ( v_2 t_p ), and ( v_3 t_p ) respectively. Therefore, the distance each had left when Petya finished is ( 100 - v_1 t_p ), ( 100 - v_2 t_p ), and ( 100 - v_3 t_p ).The sum of these distances is 40 meters:( (100 - v_1 t_p) + (100 - v_2 t_p) + (100 - v_3 t_p) = 40 )Simplify this:( 300 - (v_1 + v_2 + v_3) t_p = 40 )So,( (v_1 + v_2 + v_3) t_p = 300 - 40 = 260 )Therefore,( (v_1 + v_2 + v_3) t_p = 260 )But from earlier, we know that ( v_p + v_1 + v_2 + v_3 = 24 ). So, ( v_1 + v_2 + v_3 = 24 - v_p ).Substituting into the equation above:( (24 - v_p) t_p = 260 )But ( t_p = frac{100}{v_p} ), so substitute that in:( (24 - v_p) times frac{100}{v_p} = 260 )Let me write this equation:( frac{100(24 - v_p)}{v_p} = 260 )Multiply both sides by ( v_p ):( 100(24 - v_p) = 260 v_p )Expand the left side:( 2400 - 100 v_p = 260 v_p )Combine like terms:( 2400 = 260 v_p + 100 v_p )( 2400 = 360 v_p )Divide both sides by 360:( v_p = frac{2400}{360} )Simplify:Divide numerator and denominator by 60:( v_p = frac{40}{6} )Simplify further:( v_p = frac{20}{3} ) meters per second.So, Petya's speed is ( frac{20}{3} ) m/s.Now, the question is asking how many meters Petya ran in the first 12 seconds. Since speed is constant, distance is speed multiplied by time.So, distance ( d = v_p times t )Here, ( t = 12 ) seconds.So,( d = frac{20}{3} times 12 )Calculate this:( frac{20}{3} times 12 = 20 times 4 = 80 ) meters.So, Petya ran 80 meters in the first 12 seconds.Let me just double-check my steps to make sure I didn't make a mistake.1. I set up the total speed equation correctly: ( v_p + v_1 + v_2 + v_3 = 24 ).2. Then, for the combined distance left when Petya finished, I set up the equation correctly: ( (v_1 + v_2 + v_3) t_p = 260 ).3. Substituted ( v_1 + v_2 + v_3 = 24 - v_p ) into the equation, and substituted ( t_p = 100 / v_p ).4. Solved for ( v_p ) and got ( 20/3 ) m/s, which is approximately 6.666... m/s.5. Then, calculated the distance in 12 seconds: ( (20/3) * 12 = 80 ) meters.Seems solid. Let me just verify with another approach.Alternative approach:Let me denote the time Petya took to finish the race as ( t_p = 100 / v_p ).In that time, each of the other runners ran ( v_i t_p ) meters, so the distance left for each is ( 100 - v_i t_p ). The sum is 40 meters:( sum_{i=1}^{3} (100 - v_i t_p) = 40 )Which simplifies to:( 300 - (v_1 + v_2 + v_3) t_p = 40 )So,( (v_1 + v_2 + v_3) t_p = 260 )We know that ( v_p + v_1 + v_2 + v_3 = 24 ), so ( v_1 + v_2 + v_3 = 24 - v_p ). Therefore:( (24 - v_p) t_p = 260 )But ( t_p = 100 / v_p ), so:( (24 - v_p) * (100 / v_p) = 260 )Which is the same equation as before. Solving gives ( v_p = 20/3 ) m/s.Therefore, in 12 seconds, Petya ran 80 meters.Another way to think about it: If Petya's speed is ( 20/3 ) m/s, then in 12 seconds, he runs ( 20/3 * 12 = 80 ) meters. That seems consistent.Let me check if the other runners' speeds make sense.We know that ( v_p = 20/3 approx 6.6667 ) m/s.So, ( v_1 + v_2 + v_3 = 24 - 20/3 = (72/3 - 20/3) = 52/3 approx 17.3333 ) m/s.So, the total speed of the other three is 52/3 m/s.When Petya finishes the race in ( t_p = 100 / (20/3) = 15 ) seconds.So, in 15 seconds, the other three runners have collectively run ( (52/3) * 15 = 52 * 5 = 260 ) meters. So, each of them has run 260 / 3 ≈ 86.6667 meters on average. Therefore, each has 100 - 86.6667 ≈ 13.3333 meters left. But since there are three runners, the total left is 3 * 13.3333 ≈ 40 meters, which matches the problem statement. So, that checks out.Also, in the first 12 seconds, Petya ran 80 meters, and the other three ran a total of ( (52/3) * 12 = 52 * 4 = 208 ) meters. So, total distance covered by all four in 12 seconds is 80 + 208 = 288 meters, which is exactly what the problem stated. So, that also checks out.Therefore, I'm confident that the answer is 80 meters.**Final Answer**Petya ran boxed{80} meters in the first 12 seconds.

answer：To determine the real part of the complex number ( i(1 + i) ), I start by expanding the expression.First, I distribute the ( i ) across the terms inside the parentheses:[i times 1 + i times i = i + i^2]Next, I recognize that ( i^2 ) is equal to (-1). Substituting this value into the expression gives:[i + (-1) = -1 + i]Finally, I identify the real part of the resulting complex number (-1 + i), which is (-1).

answer：First, I need to determine the current age of the younger son. Since the elder son is 40 years old and there is a 10-year difference between the two sons, the younger son must be 40 - 10 = 30 years old now.Next, I calculate the younger son's age 30 years from now. Adding 30 years to his current age of 30 years gives 30 + 30 = 60 years.Therefore, the younger son will be 60 years old in 30 years.

answer：To determine the value of ( a ) such that the lines ( x + 2y - 1 = 0 ) and ( ax - y - 1 = 0 ) are perpendicular, I start by finding the slopes of each line.For the first line, ( x + 2y - 1 = 0 ), I can rewrite it in slope-intercept form ( y = mx + b ). Solving for ( y ), I get:[2y = -x + 1 quad Rightarrow quad y = -frac{1}{2}x + frac{1}{2}]So, the slope ( m_1 ) is ( -frac{1}{2} ).For the second line, ( ax - y - 1 = 0 ), I also rewrite it in slope-intercept form. Solving for ( y ), I get:[y = ax - 1]Thus, the slope ( m_2 ) is ( a ).Since the lines are perpendicular, the product of their slopes must be ( -1 ). Therefore:[m_1 times m_2 = -1 quad Rightarrow quad -frac{1}{2} times a = -1]Solving for ( a ):[-frac{a}{2} = -1 quad Rightarrow quad a = 2]