As we are unaware of relation between efficiencies of men and women, the given data is inadequate to answer the question.

•   Initial units of work = 8 units.

•   Out of which, 7 units are completed in 21 days.

•   So, the number of days that man takes in completing one unit of work = 21/7 = 3 days.

•   Now, amount of new work = 8 + 50% of 8 = 8 + 4 = 12 units.

•   Work remaining = 12 – 7 = 5 units.

•   So, the time taken by the man to complete the rest of the work = 5 × 3 = 15 days.

X can complete 20% of work in 8 days.

So, he will complete 100% work in 40 days.

Similarly, Y can complete 25% of work in 6 days.

So, he will complete 100% work in 24 days.

In 1 day the amount of work they will complete while working together = 1/40 + 1/24 = (3 + 5)/120 = 8/120 = 1/15

So, they need 15 days to complete the whole work.

Hence,  the time needed to complete 40% of the work = 40% of 15 = 6 days

•   Given, 18 adults = 30 children

•   1 adult = 30/18 children

•   So, with 12 adults, 6 more adults can board the lift.

•   Thus, 12 + 6 (30/18) = 22 total people.

•   So, out of 22 people, 10 are children.

•   Ram and Shyam can complete the 60% of the work n 4 days.

•   Therefore, their one day work will be:

•   1/R + 1/S = 60/100 x 1/4

•   Or, 1/R + 1/S = 3/20

•   Remaining 40% of work is completed by Shyam alone in 8 days.

•   So, 100% work can be completed by Shyam in 8/40 x 100= 20 days.

•   Or, 1/R + 1/20 = 3/20

•   Or, 1/R = 3/20 - 1/20 = 2/20

•   Or, 1/R = 1/10

•   Ram alone can complete the job in 10 days.

•   W can do 25% of a work in 30 days. W can do ¼ of a work in 30 days.

•   W can do the complete work in 120 days.

•   X can do1/4 of the work in 10 days.

•   X can do the complete work in 40 days.

•   Y can do 40% of the work in 40 days. 

•   Y can do 100% of the work in 100 days.

•   Y can do the complete work in 100 days.

•   Z can do1/3 of the work in 13 days.

•   Z can do the complete work in 39 days.

•   Hence, it is clear from above results that Z will complete the work first.

1. Correct Answer: Option B: 12 minutes
 2. Explanation:
     To solve this problem, we need to determine the combined rate at which both pipes A and B fill the tank and then calculate the time it takes for them to fill the tank together.
     Step 1: Determine the rate of each pipe.
         ○ Pipe A can fill the tank in 20 minutes. Therefore, the rate of Pipe A is:
       \[
       \text{Rate of Pipe A} = \frac{1}{20} \text{ of the tank per minute}
       \]
         ○ Pipe B can fill the tank in 30 minutes. Therefore, the rate of Pipe B is:
       \[
       \text{Rate of Pipe B} = \frac{1}{30} \text{ of the tank per minute}
       \]
     Step 2: Determine the combined rate of both pipes.
     When both pipes are opened simultaneously, their rates add up. Therefore, the combined rate is:
     \[
     \text{Combined Rate} = \frac{1}{20} + \frac{1}{30}
     \]
     To add these fractions, find a common denominator. The least common multiple of 20 and 30 is 60.
     \[
     \frac{1}{20} = \frac{3}{60} \quad \text{and} \quad \frac{1}{30} = \frac{2}{60}
     \]
     Adding these gives:
     \[
     \text{Combined Rate} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}
     \]
     Step 3: Calculate the time taken to fill the tank.
     The combined rate of \(\frac{1}{12}\) means that both pipes together fill \(\frac{1}{12}\) of the tank in one minute. Therefore, to fill the entire tank, it will take:
     \[
     \text{Time} = 12 \text{ minutes}
     \]
     Thus, the correct answer is Option B: 12 minutes.

Option D: 10 days
Let's break down the problem step by step:
     1. Initial Food Supply: The garrison initially has food for 1000 soldiers for one month (30 days). This means the total food supply is equivalent to 1000 soldiers × 30 days = 30,000 soldier-days of food.
     2. Food Consumption in the First 10 Days: For the first 10 days, there are 1000 soldiers consuming the food. Therefore, the food consumed in these 10 days is 1000 soldiers × 10 days = 10,000 soldier-days of food.
     3. Remaining Food: After 10 days, the remaining food is 30,000 soldier-days - 10,000 soldier-days = 20,000 soldier-days of food.
     4. Increased Number of Soldiers: After 10 days, 1000 more soldiers join, making a total of 2000 soldiers in the garrison.
     5. Calculating Remaining Days: With 2000 soldiers now, we need to determine how long the remaining 20,000 soldier-days of food will last.
            ○ The number of days the remaining food will last is calculated by dividing the remaining soldier-days of food by the number of soldiers:
        \[
        \text{Remaining Days} = \frac{\text{Remaining Soldier-Days of Food}}{\text{Number of Soldiers}} = \frac{20,000}{2000} = 10 \text{ days}
        \]
     Therefore, the soldiers would be able to carry on with the remaining food for 10 days.

Option D: 8
     1. Let's assume the total amount of petrol in the tank is P liters.
     2. Arun's current petrol consumption rate allows the petrol to last for 10 days. Therefore, his daily petrol consumption is:
        $$\text{Daily Consumption} = \frac{P}{10}$$
     3. If Arun starts using 25% more petrol every day, his new daily consumption will be:
        $$\text{New Daily Consumption} = \frac{P}{10} + 0.25 \times \frac{P}{10} = \frac{P}{10} \times 1.25 = \frac{1.25P}{10} = \frac{P}{8}$$
     4. To find out how many days the petrol will last with the new consumption rate, we divide the total petrol by the new daily consumption:
        $$\text{Number of Days} = \frac{P}{\frac{P}{8}} = 8$$
     Therefore, the tank-full petrol will last for 8 days when Arun uses 25% more petrol every day.

The penalty for delay on first day is 200, for second day is 250, for third day is 300 and so on. The penalties form an arithmetic progression (AP) with a common difference of 50 and first element (a) as 200. The sum of penalties for 10 days is thus same as the sum of numbers in the A.P. with n as 10. Sum of AP (formula) = n/2 [2a + (n-1) d] = 10/2 [ 2 X 200 + (10-1) X 50 ] = 4250

Total requirement = 4000 X 150 = 6,00,000 litres Volume (capacity) of tank = L x B x H (for cuboids) = 20 X 15 X 6 = 1800 m3 1m = 1000 liters So, capacity = 18,00, 000 So water will last for 18,00,000/6,00,000 = 3 days.