Question

An ant climbing up a vertical pole ascends 12 meters and slips down 5 meters in every alternate hour. If the pole is 63 meters high how long will it take it to reach the top?

• A. 18 hours
• B. 17 hours
• C. 16 hours 35 minutes
• D. 16 hours 40 minutes

Hint:

For "climbing and slipping" problems, always calculate the time to reach a height just short of the top (Total Height - One Climb). Then, calculate the time for the final climb separately, as there is no slip after reaching the destination.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The ant's movement happens in a 2-hour cycle. In the first hour, it climbs up. In the second hour, it slips down. We need to find the total time to reach the top of the 63-meter pole.
Step 2: Key Formula or Approach:
The key to solving this type of problem is to realize that once the ant reaches the top, it won't slip back down. So, we should calculate the time taken to reach a point just below the top, from where the final ascent will take it to the target height.
Step 3: Detailed Explanation:
Analyze the cycle:

Hour 1: Climbs 12 m.
Hour 2: Slips 5 m.
Net ascent in a 2-hour cycle = 12 m - 5 m = 7 m.
Calculate progress before the final climb: Total height of the pole = 63 m.
Ascent in the climbing hour = 12 m.
We first calculate the time it takes for the ant to climb to a height from which the final 12 m climb will take it to the top. This height is at least \(63 - 12 = 51\) m.
Let's find the time to climb as close as possible to 51 m using the 2-hour cycles. In each 2-hour cycle, the ant climbs 7 m. To find how many cycles are needed to cover a height near 51 m, we can divide 51 by 7. \[ \frac{51}{7} \approx 7.28 \] Let's take 7 full cycles. Height covered in 7 cycles = \(7 \times 7\) m = 49 m.
Time taken for 7 cycles = \(7 \times 2\) hours = 14 hours.
Now let's track the progress hour by hour from this point:

After 14 hours, the ant is at a height of 49 m.
The 15th hour is a climbing hour. Position at the end of the 15th hour = \(49 + 12 = 61\) m.
The 16th hour is a slipping hour. Position at the end of the 16th hour = \(61 - 5 = 56\) m.
The final ascent: At the beginning of the 17th hour (which is a climbing hour), the ant is at a height of 56 m.
The remaining height to reach the top = \(63 - 56 = 7\) m.
The ant climbs at a rate of 12 meters per hour.
Time required to climb the remaining 7 m = \(\frac{\text{Distance}}{\text{Speed}} = \frac{7}{12}\) hours.
To convert this fraction of an hour into minutes: \[ \frac{7}{12} \times 60 \, \text{minutes} = 7 \times 5 \, \text{minutes} = 35 \, \text{minutes}. \] Total time = 16 full hours + 35 minutes.
Step 4: Final Answer:
The total time it will take for the ant to reach the top is 16 hours and 35 minutes.