Math Problem Statement

Mike and Bryan walk up a moving escalator. The escalator moves at a constant speed. Mike takes five steps for every four of Bryan’s. Mike gets to the top of the escalator after having taken 30 steps; while Bryan takes only 27 steps to reach the top. If the escalator was turned off, how many steps would they each have to take to reach the top?

Solution

To solve this problem, we need to determine the total number of steps in the escalator when it is turned off.

Step 1: Variables and relationships

Let:

  • NN be the total number of steps on the escalator when it is turned off.
  • vv be the speed of the escalator in steps per second.
  • sMs_M and sBs_B be the speeds of Mike and Bryan in steps per second, respectively.

Given:

  1. Mike takes 30 steps while the escalator moves 30+N30sM30 + N - 30s_M.
  2. Bryan takes 27 steps while the equatorm will applyTo solve the problem systematically, we establish the relationships between the variables:

Definitions and Setup

  • Let NN be the total number of steps on the escalator when it is stationary.
  • The escalator moves at a constant rate of vv steps per second.
  • Mike's step rate is 5x5x steps per second.
  • Bryan's step rate is 4x4x steps per second.

When the escalator is moving:

  • Mike takes 30 steps to reach the top.
  • Bryan takes 27 steps to reach the top.

The total number of steps covered by Mike or Bryan is the sum of:

  1. Steps they take themselves.
  2. Steps the escalator moves during their travel.

Step 1: Express the total distance

The time Mike and Bryan take to reach the top can be calculated from their respective step counts and rates:

  • Mike's travel time: tM=305x=6xt_M = \frac{30}{5x} = \frac{6}{x}.
  • Bryan's travel time: tB=274xt_B = \frac{27}{4x}.

The escalator moves at vv steps per second, so the distance moved by the escalator during this time is:

  • For Mike: vtM=v6xv \cdot t_M = v \cdot \frac{6}{x}.
  • For Bryan: vtB=v274xv \cdot t_B = v \cdot \frac{27}{4x}.

The total distance covered is equal to the stationary escalator length NN:

  • For Mike: 30+v6x=N30 + v \cdot \frac{6}{x} = N.
  • For Bryan: 27+v274x=N27 + v \cdot \frac{27}{4x} = N.

Step 2: Solve for vv and NN

From the two equations:

  1. N=30+v6xN = 30 + v \cdot \frac{6}{x},
  2. N=27+v274xN = 27 + v \cdot \frac{27}{4x}.

Equating the two expressions for NN: 30+v6x=27+v274x.30 + v \cdot \frac{6}{x} = 27 + v \cdot \frac{27}{4x}.

Simplify: 3=v(274x6x).3 = v \left( \frac{27}{4x} - \frac{6}{x} \right).

Combine terms inside the parentheses: 274x6x=27244x=34x.\frac{27}{4x} - \frac{6}{x} = \frac{27 - 24}{4x} = \frac{3}{4x}.

Substitute back: 3=v34x.3 = v \cdot \frac{3}{4x}.

Solve for vv: v=4x.v = 4x.

Step 3: Substitute vv back to find NN

Using v=4xv = 4x in Mike’s equation for NN: N=30+6x4x.N = 30 + \frac{6}{x} \cdot 4x.

Simplify: N=30+24=54.N = 30 + 24 = 54.

Step 4: Verify for Bryan

For Bryan: N=27+274x4x.N = 27 + \frac{27}{4x} \cdot 4x.

Simplify: N=27+27=54.N = 27 + 27 = 54.

Final Answer

When the escalator is stationary, the total number of steps is: 54 steps.\boxed{54 \text{ steps.}}

Additional Notes

  • Mike and Bryan would both need to climb 54 steps to reach the top without the escalator moving.

Let me know if you want a detailed breakdown or further explanation!


Related Questions:

  1. How would the solution change if Mike and Bryan walked at the same speed?
  2. What is the relationship between the step counts and escalator speed in general?
  3. How can we use relative rates to solve similar problems?
  4. If the escalator moved faster, how would their step counts change?
  5. How could this problem be modeled graphically?

Tip: Breaking a problem into small steps and validating results helps avoid calculation errors!

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Math Problem Analysis

Mathematical Concepts

Algebra
Proportions
Relative Motion

Formulas

Distance = Speed × Time
Steps taken = Total steps - Escalator contribution

Theorems

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Suitable Grade Level

Grades 9-12