A train travels 150 km at a certain speed. If the speed were increased by 5 km/hr, the journey would have taken 1 hour less. Find the original speed.

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Let's solve this step by step. We have a train traveling 150 kilometers at some unknown speed. The problem tells us that if we increase the speed by 5 kilometers per hour, the journey would take 1 hour less time. We need to find the original speed. First, let's define our variables clearly. Let S be the original speed of the train in kilometers per hour. We know the distance is 150 kilometers. Using the formula that time equals distance divided by speed, the original time taken is 150 divided by S hours. Now let's set up our equation. If the speed increases by 5 kilometers per hour, the new speed becomes S plus 5. Since the journey takes 1 hour less, the new time is T minus 1. Using the distance formula, 150 equals the new speed times the new time. Substituting our expression for T, we get our main equation. Now let's solve the equation step by step. First, we multiply both sides by S to eliminate the fraction. Then we expand the right side and rearrange to get a quadratic equation: S squared plus 5S minus 750 equals zero. We can factor this as S plus 30 times S minus 25 equals zero. This gives us S equals negative 30 or S equals 25. Since speed cannot be negative, the original speed is 25 kilometers per hour. Let's solve this classic speed and distance problem. A train travels 150 kilometers at a certain speed. If the speed were increased by 5 kilometers per hour, the journey would have taken 1 hour less. We need to find the original speed. First, let's define our variables. Let v be the original speed in kilometers per hour. The increased speed would be v plus 5 kilometers per hour. The distance is 150 kilometers. We know that time equals distance divided by speed. Now let's write the time equations. The time taken at the original speed is 150 divided by v. The time taken at the increased speed is 150 divided by v plus 5. According to the given condition, the difference between these times is 1 hour. Therefore, we have the equation: 150 over v minus 150 over v plus 5 equals 1. Let's solve this equation step by step. First, we find a common denominator for the fractions. This gives us 150 times v plus 5 minus 150v, all over v times v plus 5, equals 1. Simplifying the numerator, we get 750 over v times v plus 5 equals 1. Cross-multiplying gives us 750 equals v squared plus 5v. Rearranging, we get the quadratic equation: v squared plus 5v minus 750 equals zero. Now we solve the quadratic equation v squared plus 5v minus 750 equals zero using the quadratic formula. We get v equals negative 5 plus or minus the square root of 3025, all divided by 2. This simplifies to v equals negative 5 plus or minus 55, divided by 2. This gives us v equals 25 or v equals negative 30. Since speed cannot be negative, the original speed is 25 kilometers per hour.