Kim’s wristwatch always shows the correct time, including ‘am’ and ‘pm’. Jim’s watch is identical to Kim’s watch in all aspects except its pace, which is slower than the pace of Kim’s watch. At 12 noon on January 1st, Jim sets his watch to the correct time, but an hour later, it shows 12:57 pm. At 12 noon on the next June 1st, Jim resets his watch to the correct time. On how many instances between, and including 12 noon on the two dates mentioned, do Jim’s and Kim’s watches show the exact same time, including the ‘am’ and the ‘pm’?
- 7
- 17
- 10
- 9
- 15
The Correct Option is D
Approach Solution - 1
Jim’s watch is slower than Kim’s.
After 1 real hour, Jim’s watch shows 12:57 pm instead of 1:00 pm, so it loses 3 minutes per hour.
Rate of loss
Loss per hour = 3 minutes
Loss per day = 24 × 3 = 72 minutes = 1.2 hours
Jim’s watch will show the exact same time (including am/pm) as Kim’s watch whenever it is behind by a whole number of 24 hours, since the time display repeats every 24 hours.
Time to lose 24 hours
24 ÷ 1.2 = 20 days
So, Jim’s and Kim’s watches coincide every 20 days.
Counting the instances
From 12 noon on January 1st to 12 noon on June 1st, the total duration is:
150 days
The coincidences occur on days:
0, 20, 40, 60, 80, 100, 120, 140
This gives 8 instances, including 12 noon on January 1st.
On 12 noon on June 1st, Jim resets his watch to the correct time, giving one additional instance.
Final Answer
9
Hence, the given answer 9 is correct.
Approach Solution -2
To solve the problem, we must first determine the rate at which Jim's watch loses time compared to Kim's. According to the problem, in one real hour, Jim's watch shows 57 minutes instead of 60 minutes. Let's calculate how much time Jim's watch loses per hour:
1 real hour = 60 minutes;
Jim's watch shows = 57 minutes;
Time lost = 60 - 57 = 3 minutes/hour.
Now, we need to calculate the total time lost by Jim's watch from January 1st to June 1st:
Number of days from January 1st to June 1st:
- January: 31 days
- February: 28 days (non-leap year)
- March: 31 days
- April: 30 days
- May: 31 days
Total days = 31 + 28 + 31 + 30 + 31 = 151 days.
Total hours = 151 days × 24 hours/day = 3624 hours.
Total time lost = 3 minutes/hour × 3624 hours.
Convert minutes to hours:
Time lost = (3 × 3624) / 60 = 181.2 hours.
This means that by June 1st at 12 noon, Jim’s watch lags behind Kim’s by 181.2 hours, which is equivalent to 181 hours and 12 minutes. Since both watches show the same time at 12 noon on January 1st and June 1st, we need to find additional instances when their times align between these two dates.
An alignment occurs every time Jim's watch reaches a full hour that Kim's already passed. Specifically, we consider every full 12-hour cycle, because both watches would match at this point.
181 hours and 12 minutes translates into 15 full 12-hour cycles. However, we should consider both the starting and ending points of this period:
Instances of same time (besides starting and ending): (181 hours / 12 hours) = 15.083 cycles approximately.
Since we include the start and end time alignments including 12 noon on both days, subtract the partial cycle and consider only completed cycles between cycles:
Cycles completed every 12 hours from noon on January 1 to noon on June 1.= 9 full cycles.
Therefore, the number of instances when Jim's and Kim's watches show the exact same time is 9.
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