Math Problem Statement

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution

This is a classic problem involving relative speed and time. Let us solve it step-by-step.


Problem Breakdown:

  • Distance traveled = 480km480 \, \text{km}
  • Let the train's speed be xkm/hx \, \text{km/h}.
  • If the speed is reduced by 8km/h8 \, \text{km/h}, the time taken increases by 3 hours.

From the formula: Time=DistanceSpeed,\text{Time} = \frac{\text{Distance}}{\text{Speed}}, we can write two equations for the time taken:

  1. At normal speed: 480x\frac{480}{x}
  2. At reduced speed: 480x8\frac{480}{x - 8}

The difference in time is given as 3 hours: 480x8480x=3\frac{480}{x - 8} - \frac{480}{x} = 3


Step 1: Solve the equation

Rearranging and simplifying: 480x8480x=3\frac{480}{x - 8} - \frac{480}{x} = 3 Take the LCM of xx and x8x - 8: 480x(x8)x(x8)=3480 \cdot \frac{x - (x - 8)}{x(x - 8)} = 3 4808x(x8)=3480 \cdot \frac{8}{x(x - 8)} = 3 3840x(x8)=3\frac{3840}{x(x - 8)} = 3 Multiply through by x(x8)x(x - 8) to eliminate the fraction: 3840=3x(x8)3840 = 3x(x - 8) Expand and simplify: 3840=3x224x3840 = 3x^2 - 24x 3x224x3840=03x^2 - 24x - 3840 = 0 Divide through by 3: x28x1280=0x^2 - 8x - 1280 = 0


Step 2: Solve the quadratic equation

Use the quadratic formula: x=b±b24ac2a,x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=8b = -8, c=1280c = -1280.

Substitute: x=(8)±(8)24(1)(1280)2(1)x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-1280)}}{2(1)} x=8±64+51202x = \frac{8 \pm \sqrt{64 + 5120}}{2} x=8±51842x = \frac{8 \pm \sqrt{5184}}{2} x=8±722x = \frac{8 \pm 72}{2}


Step 3: Calculate the two solutions

x=8+722=40(valid speed),x = \frac{8 + 72}{2} = 40 \quad \text{(valid speed)}, x=8722=32(not valid, speed cannot be negative).x = \frac{8 - 72}{2} = -32 \quad \text{(not valid, speed cannot be negative)}.


Final Answer:

The speed of the train is: 40km/h\boxed{40 \, \text{km/h}}


Would you like a more detailed explanation or have any questions? Here are some related questions to consider:

  1. How can we verify the solution to this problem?
  2. What happens to the equation if the distance is changed?
  3. How would the solution change if the time difference was not 3 hours but another value?
  4. How does the quadratic formula help in solving real-world problems like this?
  5. What is the impact of reducing speed further on the time taken?

Tip: When solving such problems, always check units (e.g., km, hours) and ensure they are consistent throughout!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Distance-Speed-Time Relationships

Formulas

Time = Distance / Speed
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

Theorems

Quadratic formula theorem

Suitable Grade Level

Grades 9-10