To maintain 8 cows for 60 days, a milkman spends Rs. 6400. To maintain 5 cows for n days, he spends Rs. 4800. What is the value of n?
72 days
Calculating Days for Cow Maintenance Cost
This problem involves understanding the relationship between the number of cows, the number of days they are maintained, and the total cost. We can assume that the cost is directly proportional to both the number of cows and the number of days. This means if you double the cows or double the days, the cost will double.
Let the cost per cow per day be a constant value, say \(k\). Then, the total cost can be expressed as:
Total Cost = \(k \times\) Number of Cows \(\times\) Number of Days
Analyzing the First Scenario
In the first scenario, we are given:
- Number of Cows = 8
- Number of Days = 60
- Total Cost = Rs. 6400
Using the formula, we can write:
\(6400 = k \times 8 \times 60\)
Now, let's solve for \(k\), the cost per cow per day:
\(6400 = k \times 480\)
\(k = \frac{6400}{480}\)
\(k = \frac{640}{48}\)
We can simplify this fraction:
\(k = \frac{640 \div 16}{48 \div 16} = \frac{40}{3}\)
So, the cost per cow per day is Rs. \(\frac{40}{3}\).
Analyzing the Second Scenario
In the second scenario, we are given:
- Number of Cows = 5
- Number of Days = \(n\) (which we need to find)
- Total Cost = Rs. 4800
We use the same formula with the value of \(k\) we just found:
\(4800 = k \times 5 \times n\)
Substitute \(k = \frac{40}{3}\):
\(4800 = \frac{40}{3} \times 5 \times n\)
\(4800 = \frac{40 \times 5}{3} \times n\)
\(4800 = \frac{200}{3} \times n\)
Now, we need to solve for \(n\). To do this, multiply both sides by \(\frac{3}{200}\):
\(n = 4800 \times \frac{3}{200}\)
\(n = \frac{4800 \times 3}{200}\)
We can simplify this calculation. Divide 4800 by 200:
\(\frac{4800}{200} = \frac{48}{2} = 24\)
So, the equation becomes:
\(n = 24 \times 3\)
\(n = 72\)
Thus, the value of \(n\) is 72 days.
Summary of Calculation
| Scenario | Cows | Days | Cost (Rs.) | Formula: Cost = \(k\) \(\times\) Cows \(\times\) Days |
|---|---|---|---|---|
| 1 | 8 | 60 | 6400 | \(6400 = k \times 8 \times 60\) \(\implies k = \frac{6400}{480} = \frac{40}{3}\) |
| 2 | 5 | \(n\) | 4800 | \(4800 = \frac{40}{3} \times 5 \times n\) \(\implies n = \frac{4800 \times 3}{200} = 72\) |
The value of \(n\) is 72 days.
Revision Table: Cow Maintenance Problem
| Key Concept | Description | Application in Problem |
|---|---|---|
| Direct Proportionality | When two quantities increase or decrease together at a constant rate. Here, cost is directly proportional to cows and days. | Cost \(\propto\) Cows \(\times\) Days, leading to Cost = \(k\) \(\times\) Cows \(\times\) Days. |
| Finding Constant (k) | Using known values from one situation to find the constant of proportionality. | Using 8 cows, 60 days, Rs. 6400 to find \(k = \frac{40}{3}\). |
| Using Constant to Find Unknown | Using the calculated constant and new values to find the missing quantity. | Using \(k = \frac{40}{3}\), 5 cows, Rs. 4800 to find the unknown number of days, \(n\). |
Additional Information: Direct vs. Inverse Proportionality
Understanding proportionality is key to solving many word problems. There are two main types:
- Direct Proportionality: As one quantity increases, the other quantity increases proportionally. As one quantity decreases, the other decreases proportionally. Example: The cost of apples is directly proportional to the number of apples bought. \( \text{Cost} = k \times \text{Number of Apples} \).
- Inverse Proportionality: As one quantity increases, the other quantity decreases proportionally. As one quantity decreases, the other increases proportionally. Example: The time taken to complete a job is inversely proportional to the number of workers (assuming they work at the same rate). \( \text{Time} = \frac{k}{\text{Number of Workers}} \).
In this problem, the total cost of maintaining cows increases if you have more cows or if you maintain them for more days, demonstrating a direct proportionality relationship with the product of cows and days.
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