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Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction.He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction.If Moody decides to stand still on the escalator,then the time,in seconds,needed to finish riding the escalator is

Updated On: Jan 15, 2026
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Solution and Explanation

Moody ascends an escalator. At his regular pace, the escalator ride lasts 30 seconds. When he doubles his walking pace, the ride takes 20 seconds. How long would the ride take if he remained stationary?

Step 1: Define Variables

  • Moody’s normal walking speed = \( W \) units/sec
  • Escalator’s speed = \( E \) units/sec

Combined speed when walking normally: \( W + E \)

Combined speed when walking at double speed: \( 2W + E \)

Step 2: Set Up Equations

Assume the total length of the escalator is 1 unit. Using the formula time = distance / speed, we form two equations:

  • Equation 1: \( (W + E) \times 30 = 1 \Rightarrow 30W + 30E = 1 \)
  • Equation 2: \( (2W + E) \times 20 = 1 \Rightarrow 40W + 20E = 1 \)

Step 3: Solve for Speeds

From Equation 1, isolate \( E \):

\[ E = \frac{1 - 30W}{30} \]

Substitute this expression for \( E \) into Equation 2:

\[ (2W + \frac{1 - 30W}{30}) \times 20 = 1 \]

Simplify the equation:

\[ 2W + \frac{1 - 30W}{30} = \frac{1}{20} \]

Multiply all terms by 30 to eliminate fractions:

\[ 60W + 1 - 30W = 1.5 \]

Combine like terms and solve for \( W \):

\[ 30W + 1 = 1.5 \Rightarrow 30W = 0.5 \Rightarrow W = \frac{1}{60} \]

Step 4: Calculate Escalator Speed

Substitute the value of \( W \) back into the equation for \( E \):

\[ E = \frac{1 - 30 \cdot \frac{1}{60}}{30} = \frac{1 - 0.5}{30} = \frac{0.5}{30} = \frac{1}{60} \]

Step 5: Determine Time if Stationary

If Moody stands still, his speed is solely the escalator's speed, \( E = \frac{1}{60} \) units/sec.

The time taken is calculated as distance / speed:

Time = \( \frac{1}{\frac{1}{60}} = 60 \) seconds

Final Answer:

\[ \boxed{60 \text{ seconds}} \]

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