Consider the following statements:
1. Between 3:16 pm. and 3:17 pm, both hour hand and minute hand 'coincide.
2. Between 4:58 pm. and 4:59 p.m.. both minute hand and second hand coincide
Which of the above statements is/are correct?
UPSC - 2022
Between 6 PM and 7 PM the minute hand of clock will be ahead of the hour hand by 3 minutes at
UPSC - 2015
To solve this clock problem, we need to understand the relative speeds of the hour and minute hands. We'll calculate the angular distance covered by each hand and use their relative speed to find the time when the minute hand is 3 minutes ahead of the hour hand between 6 PM and 7 PM.
Relative Speeds of Hands:
Initial Position at 6 PM:
Target Position:
Calculating the required relative distance to cover:
Time Calculation:
C) 6:36 PM is the correct answer. At this time, the minute hand will be 3 minutes ahead of the hour hand.
If the 3rd day of a month is Monday, which one of the following will be the fifth day from 21st of this month?
UPSC - 2014
This problem involves determining a specific day of the week based on a given reference day. We will use the concept of odd days (remainder when the number of days is divided by 7) to find the day of the week.
The number of days between the 3rd and the 21st is \(21 - 3 = 18\) days.
To find the day of the week for the 21st, we need to find the odd days in 18. Divide 18 by 7:
\[ 18 \div 7 = 2 \text{ with a remainder of } 4 \]So, there are 4 odd days. Starting from Monday (the 3rd), we count 4 days forward:
Therefore, the 21st day of the month is Friday.
The question asks for the fifth day from the 21st. This means we need to find the day of the week for \(21 + 5 = 26\)th of the month.
Since the 21st is Friday, we count 5 days forward from Friday:
Alternatively, we can find the odd days from the 21st to the 26th. The number of days is 5. So, we add 5 odd days to Friday.
Friday + 5 days = Wednesday.
Correct Option: C) Wednesday
Assume that
1. the hour and minute hands of a clock move without jerking.
2. the clock shows a time between 8 o'clock and 9 o'clock.
3. the two hands of the clock are one above the other.
After how many minutes ( nearest integer) will the two hands be again lying one above the other?
UPSC - 2014
This problem involves understanding the relative speeds of the hour and minute hands of a clock. When the two hands are one above the other, they are at 0 degrees relative to each other. We need to find the time interval after which this alignment occurs again.
C) 65
The hands align approximately every \(65 \frac{5}{11}\) minutes. The question asks for the nearest integer, which is 65 minutes.
A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 8 o'clock in the morning, at what other time will they all ring together?
UPSC - 2014
To find out when the three bells will ring together again, we need to find the least common multiple (LCM) of the time intervals at which each bell rings. The LCM will give us the minimum time after which all three events (bells ringing) will coincide again.
First bell rings every 18 minutes.
Second bell rings every 24 minutes.
Third bell rings every 32 minutes.
First, find the prime factorization of each number:
\[ 18 = 2 \times 3^2 \]\[ 24 = 2^3 \times 3 \]\[ 32 = 2^5 \]To find the LCM, take the highest power of each prime factor present in the factorizations:
\[ \text{LCM}(18, 24, 32) = 2^5 \times 3^2 \]\[ \text{LCM}(18, 24, 32) = 32 \times 9 \]\[ \text{LCM}(18, 24, 32) = 288 \]The LCM is 288 minutes. This means the bells will ring together again after 288 minutes.
There are 60 minutes in an hour. Divide 288 by 60:
\[ 288 \text{ minutes} = \frac{288}{60} \text{ hours} \]\[ 288 \text{ minutes} = 4 \text{ hours and } 48 \text{ minutes} \]The bells last rang together at 8:00 AM.
Add 4 hours and 48 minutes to 8:00 AM:
\(8:00 \text{ AM} + 4 \text{ hours } 48 \text{ minutes} = 12:48 \text{ PM}\)
Correct Option: B) 12:48 hrs
A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is
UPSC - 2022
10 minutes slow
5 minutes slow
10 minutes fast
5 minutes fast
This problem involves calculating travel times and comparing clock readings. We need to determine the actual time spent on the journey to and from the village and then use this information to find the discrepancy in the village clock's time.
The man started from home at 14:30 hours and returned home at 16:00 hours.
\[ \text{Total actual time elapsed} = 16:00 - 14:30 = 1 \text{ hour } 30 \text{ minutes} = 90 \text{ minutes} \]The total time consists of three parts:
So, \(T_1 + 25 \text{ minutes} + T_2 = 90 \text{ minutes}\).
\[ T_1 + T_2 = 90 - 25 = 65 \text{ minutes} \]Let the distance to the village be \(D\).
Let the speed to the village be \(S\).
Then, the time taken to the village is \(T_1 = \frac{D}{S}\).
For the return journey:
The time taken to return home is \(T_2 = \frac{1.25D}{2S}\).
We can express \(T_2\) in terms of \(T_1\):
\[ T_2 = \frac{1.25}{2} \times \frac{D}{S} = \frac{1.25}{2} T_1 = 0.625 T_1 \]We have the equations:
\[ T_1 + T_2 = 65 \]\[ T_2 = 0.625 T_1 \]Substituting the second equation into the first:\[ T_1 + 0.625 T_1 = 65 \]\[ 1.625 T_1 = 65 \]\[ T_1 = \frac{65}{1.625} = 40 \text{ minutes} \]Now, find \(T_2\):\[ T_2 = 0.625 \times 40 = 25 \text{ minutes} \]The man started at 14:30 and took \(T_1 = 40\) minutes to reach the village.
\[ \text{Actual arrival time at village} = 14:30 + 40 \text{ minutes} = 15:10 \text{ hours} \]The village clock indicated 15:15 hours when he arrived.
Actual time: 15:10 hours
Village clock time: 15:15 hours
The village clock is \(15:15 - 15:10 = 5\) minutes fast.
Correct Option: D) 5 minutes fast
