Mr. Kumar drives to work at an average speed of 48 km per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. How far is his office?
UPSC - 2012
This problem involves concepts of speed, distance, and time. We are given the average speed and a relationship between the time taken for two parts of the journey. We need to find the total distance to the office.
Let the total distance to the office be \(D\) km.
The average speed of Mr. Kumar is 48 km/h.
The first 60% of the distance is \(0.60D\) km. The remaining distance is \(100\% - 60\% = 40\%\) of the total distance, which is \(0.40D\) km.
Let \(t_1\) be the time taken to cover the first 60% of the distance and \(t_2\) be the time taken to cover the remaining 40% of the distance.
Using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), we can write:
\(t_1 = \frac{0.60D}{48}\) hours
\(t_2 = \frac{0.40D}{48}\) hours
We are given that the time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. First, convert 10 minutes to hours:
\[10 \text{ minutes} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours}\]
Now, set up the equation based on the given condition:
\[t_1 = t_2 + \frac{1}{6}\]
Substitute the expressions for \(t_1\) and \(t_2\):
\[\frac{0.60D}{48} = \frac{0.40D}{48} + \frac{1}{6}\]
To solve for \(D\), first subtract \(\frac{0.40D}{48}\) from both sides:
\[\frac{0.60D}{48} - \frac{0.40D}{48} = \frac{1}{6}\]
\[\frac{(0.60 - 0.40)D}{48} = \frac{1}{6}\]
\[\frac{0.20D}{48} = \frac{1}{6}\]
Now, multiply both sides by 48:
\[0.20D = \frac{48}{6}\]
\[0.20D = 8\]
Finally, divide by 0.20 to find \(D\):
\[D = \frac{8}{0.20}\]
\[D = \frac{8}{\frac{20}{100}}\]
\[D = \frac{8 \times 100}{20}\]
\[D = \frac{800}{20}\]
\[D = 40 \text{ km}\]
B) The total distance to his office is 40 km.
How many seconds in total are there in x weeks, x days, x hours, x minutes and x seconds?
UPSC - 2022
To find the total number of seconds, we need to convert each given unit of time (weeks, days, hours, minutes, and seconds) into seconds and then sum them up. We will express each conversion in terms of 'x' and then combine the terms.
There are 7 days in a week, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute.
\[ \text{x weeks} = \text{x} \times 7 \text{ days/week} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]\[ = \text{x} \times 7 \times 24 \times 60 \times 60 \text{ seconds} \]\[ = \text{x} \times 604800 \text{ seconds} \]This is already in seconds, so it is just \(\text{x}\) seconds.
Total seconds \( = 604800\text{x} + 86400\text{x} + 3600\text{x} + 60\text{x} + \text{x} \)
Total seconds \( = (604800 + 86400 + 3600 + 60 + 1)\text{x} \)
Total seconds \( = 694861\text{x} \)
D) \(694861\text{x}\)
As calculated, the sum of seconds from x weeks, x days, x hours, x minutes, and x seconds is \(694861\text{x}\).
Which date of June 2099 among the following is Sunday?
UPSC - 2022
To find the day of the week for a specific date, we use the concept of "odd days". Odd days are the number of days remaining after dividing the total number of days by 7. We will calculate the total odd days from a known reference date (e.g., January 1, 0001) or use a standard calendar algorithm that accounts for centuries, years, and months.
Reference Point: We need a known reference point. Let's use the standard method of calculating odd days from the year 0. The number of odd days in 100 years is 5, in 200 years is 3, in 300 years is 1, and in 400 years is 0.
Odd days up to 2000 years:
\[ 2000 \text{ years} = (5 \times 400) \text{ years} \]Number of odd days in 400 years is 0. So, for 2000 years, the number of odd days is \(5 \times 0 = 0\).
Odd days in 199 years (from 2001 to 2099 - 1): We need to calculate odd days for the years 2001 to 2098 (i.e., 98 years). In these 98 years, there are leap years and ordinary years.
Odd days in the year 2099 up to June 7:
Total odd days for 2099 up to June 7: \(3 + 0 + 3 + 2 + 3 + 7 = 18\) days.
Convert to odd days: \(18 \pmod 7 = 4\) odd days.
Total Odd Days: Sum of odd days from all parts:
\[ \text{Total odd days} = (\text{odd days up to 2000}) + (\text{odd days in 98 years}) + (\text{odd days in 2099 up to June 7}) \]\[ \text{Total odd days} = 0 + 3 + 4 = 7 \text{ odd days} \]Convert to final odd days: \(7 \pmod 7 = 0\) odd days.
Day of the Week: The number of odd days corresponds to the day of the week:
Since the total number of odd days is 0, June 7, 2099, is a Sunday.
D) 7 — As calculated, June 7, 2099, corresponds to 0 odd days, which means it is a Sunday.
If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year?
UPSC - 2017
To find the minimum number of working days, we need to maximize the number of holidays. The problem states that the second and fourth Saturdays and all Sundays are holidays. We need to consider a month with the fewest total days and arrange the holidays such that their count is maximized.
Identify the shortest month: The shortest month is February, which has 28 days in a common year.
Maximize Sundays: To maximize the number of Sundays, we should assume the month starts on a Sunday. If February 1st is a Sunday, then the Sundays will be on the 1st, 8th, 15th, 22nd. This gives 4 Sundays.
Identify Saturdays: If February 1st is a Sunday, then February 7th will be a Saturday. The Saturdays will be on the 7th, 14th, 21st, 28th. This gives 4 Saturdays.
Count specific Saturday holidays: The problem states that the second and fourth Saturdays are holidays. In this arrangement, the second Saturday is the 14th, and the fourth Saturday is the 28th. So, there are 2 Saturday holidays.
Calculate total holidays:
Calculate minimum working days: \[ \text{Minimum working days} = \text{Total days in month} - \text{Total holidays} \] \[ \text{Minimum working days} = 28 - 6 = 22 \]
Consider other scenarios (optional check):
Correct Option: B) 22
Mr 'X' has three children. The birthday of the 15. first child falls on the 5th Monday of April, that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December?
UPSC - 2019
This problem involves determining the day of the week for a specific date. We are given two reference birthdays with their respective days of the week and dates. We need to use this information to find the day of the week for the third child's birthday on December 20th.
The key is to establish a fixed day of the week for a known date within the year, then count the number of odd days from that reference point to December 20th.
B) Thursday
Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively. If on 1 st January both of them went for a swim together, when will they go together next?
UPSC - 2019
7th January
8th January
12th January
13th January
This problem can be solved by finding the Least Common Multiple (LCM) of the intervals at which Seeta and Geeta go for a swim. The LCM will represent the number of days after which they will both go for a swim together again.
If they swim on January 1st, and Seeta swims after a gap of 2 days, her next swim is on January 4th (1 + 2 days gap + 1 day for the swim). Her subsequent swims are on January 7th, 10th, 13th, etc. Geeta swims after a gap of 3 days, so her next swim after January 1st is on January 5th (1 + 3 days gap + 1 day for the swim). Her subsequent swims are on January 9th, 13th, etc. The next day they both swim together is January 13th.
