This question is a classic example of a problem involving inverse proportion. The amount of provisions is fixed, so if the number of people (boys) decreases, the duration the provisions last will increase, and vice versa. We are given the initial number of boys and how long the provisions would last. After a certain period, some boys leave, and we are told how long the remaining provisions last for the remaining boys. We need to find out how many boys left.
Initially, there are 350 boys, and the provisions are sufficient for 25 days. The total amount of provisions can be thought of in terms of "boy-days" – the total number of boys that can be fed for a certain number of days.
Total initial provisions = Number of boys $\times$ Number of days
Total initial provisions = $350 \text{ boys} \times 25 \text{ days} = 8750 \text{ boy-days}$
For the first 10 days, all 350 boys were present and consuming provisions.
Provisions consumed in 10 days = Number of boys $\times$ Number of days
Provisions consumed in 10 days = $350 \text{ boys} \times 10 \text{ days} = 3500 \text{ boy-days}$
After 10 days, the remaining provisions are the total initial provisions minus the provisions consumed.
Remaining provisions = Total initial provisions - Provisions consumed
Remaining provisions = $8750 \text{ boy-days} - 3500 \text{ boy-days} = 5250 \text{ boy-days}$
The problem states that the remaining provisions will last for 21 more days. This means the 5250 boy-days of provisions are sufficient for the remaining number of boys for a period of 21 days.
Let 'R' be the number of boys remaining in the hostel after 10 days.
Remaining provisions = Number of remaining boys $\times$ Duration remaining provisions will last
$5250 \text{ boy-days} = R \text{ boys} \times 21 \text{ days}$
To find R, we rearrange the equation:
$R = \frac{5250 \text{ boy-days}}{21 \text{ days}}$
$R = 250 \text{ boys}$
So, there are 250 boys remaining in the hostel.
The number of boys shifted to another hostel is the initial number of boys minus the number of boys remaining.
Number of boys shifted = Initial number of boys - Number of remaining boys
Number of boys shifted = $350 \text{ boys} - 250 \text{ boys}$
Number of boys shifted = $100 \text{ boys}$
| Description | Calculation | Value |
|---|---|---|
| Initial Boys | 350 | |
| Initial Provision Days | 25 | |
| Total Initial Provisions (boy-days) | $350 \times 25$ | 8750 |
| Days Passed | 10 | |
| Provisions Consumed (boy-days) | $350 \times 10$ | 3500 |
| Remaining Provisions (boy-days) | $8750 - 3500$ | 5250 |
| Duration Remaining Provisions Last (days) | 21 | |
| Number of Remaining Boys | $5250 / 21$ | 250 |
| Number of Boys Shifted | $350 - 250$ | 100 |
The number of boys shifted to another hostel is 100.
| Key Concept | Explanation |
|---|---|
| Inverse Proportion | When the number of people increases, the resources last for a shorter time (and vice versa), assuming the total resource amount is fixed. This problem uses this concept. |
| Boy-Days | A unit representing the total consumption of provisions, calculated by multiplying the number of boys by the number of days they consume provisions. Useful for comparing provision amounts under different scenarios. |
| Remaining Provisions | The amount of provisions left after a certain period, calculated by subtracting the consumed provisions from the initial total provisions. |
When tackling problems involving provisions, food, or work completion by a group of people, the "total work" or "total provisions" can often be calculated as the product of the number of workers (or consumers) and the time taken (or duration the provisions last). This helps in comparing different scenarios. Here are some tips:
A is 120% of B and B is 65% of C. If the sum of A, B and C is 121.5, then the value of C - 2B + A is:
The price of an item is reduced by 20%. As a result, customers can get 2 kg more of it for ₹360. Find the original price (in ₹) per kg of the item.
A tyre has 3 punctures. The first puncture alone would have made the tyre flat in 9 minutes, the second alone would have done it in 18 minutes, the third alone would have done it in 6 minutes. If the air leaks out at a constant rate, then how long (in minutes) does it take for all the punctures together to make it flat?
If a positive number ‘k’ when multiplied by 30% of itself gives a number which is 170% more than the number ‘k’, then the number ‘k’ is equal to :