A person carries Rs. 500 and wants to buy apples and oranges out of it. If the cost of one apple is Rs. 5 and the cost of one orange is Rs. 7 then what is the number of ways in which a person can buy both apples and oranges using total amount?
14
Understanding the Problem Details
The question asks us to find the number of different combinations of apples and oranges a person can buy with exactly Rs. 500. We are given the cost of one apple and the cost of one orange, and the condition that the person must buy at least one of each fruit.
- Total money the person has: Rs. 500
- Cost of one apple: Rs. 5
- Cost of one orange: Rs. 7
- Constraint: The person must buy both apples and oranges (meaning the number of apples purchased must be at least 1, and the number of oranges purchased must be at least 1).
- Interpretation: Based on the options provided, the phrasing "using total amount" is interpreted as spending exactly Rs. 500.
Setting up the Mathematical Equation and Constraints
Let 'a' be the number of apples purchased and 'o' be the number of oranges purchased.
The total cost of buying 'a' apples and 'o' oranges is \(5a + 7o\).
Since the person spends exactly Rs. 500, the equation is:
\(5a + 7o = 500\)
The constraints given are that the person must buy both apples and oranges. In terms of 'a' and 'o', this means:
- \(a \ge 1\)
- \(o \ge 1\)
Also, the number of fruits must be non-negative integers. Since we already have \(a \ge 1\) and \(o \ge 1\), this implies that 'a' and 'o' must be positive integers.
Solving the Linear Diophantine Equation
We need to find the number of positive integer solutions (a, o) to the equation \(5a + 7o = 500\). This is a linear Diophantine equation.
Rearrange the equation to isolate one variable:
\(5a = 500 - 7o\)
For 'a' to be an integer, \(500 - 7o\) must be divisible by 5.
Since 500 is divisible by 5, \(7o\) must also be divisible by 5. Because 7 and 5 are coprime (their greatest common divisor is 1), 'o' must be divisible by 5.
Let \(o = 5k\) for some positive integer 'k' (since \(o \ge 1\)).
Substitute \(o = 5k\) into the equation \(5a + 7o = 500\):
\(5a + 7(5k) = 500\)
\(5a + 35k = 500\)
Divide the entire equation by 5:
\(a + 7k = 100\)
Now, we can express 'a' in terms of 'k':
\(a = 100 - 7k\)
So, the general integer solution for the equation \(5a + 7o = 500\) is \(a = 100 - 7k\) and \(o = 5k\), where 'k' is any integer.
Applying the Minimum Purchase Constraints (\(a \ge 1, o \ge 1\))
We know that the person must buy at least one apple and at least one orange. We use the expressions for 'a' and 'o' in terms of 'k' and apply these constraints:
Constraint 1: \(a \ge 1\)
\(100 - 7k \ge 1\)
\(100 - 1 \ge 7k\)
\(99 \ge 7k\)
\(k \le \frac{99}{7}\)
\(k \le 14.14...\)
Since 'k' must be an integer, this implies \(k \le 14\).
Constraint 2: \(o \ge 1\)
\(5k \ge 1\)
\(k \ge \frac{1}{5}\)
\(k \ge 0.2\)
Since 'k' must be an integer, this implies \(k \ge 1\).
Combining both constraints, the possible integer values for 'k' are \(1 \le k \le 14\).
Calculating the Number of Ways
The integer values that 'k' can take are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14.
Each valid integer value of 'k' corresponds to a unique pair of (a, o) that satisfies the equation \(5a + 7o = 500\) and the constraints \(a \ge 1\) and \(o \ge 1\).
The number of such integer values of 'k' is the number of ways the person can buy both apples and oranges for exactly Rs. 500.
Number of values of k = (Last value) - (First value) + 1
Number of values of k = \(14 - 1 + 1 = 14\).
Therefore, there are 14 different ways the person can buy apples and oranges spending exactly Rs. 500 while buying at least one of each.
Let's check a couple of values for k:
- If \(k=1\): \(a = 100 - 7(1) = 93\), \(o = 5(1) = 5\). Cost = \(5(93) + 7(5) = 465 + 35 = 500\). Valid.
- If \(k=14\): \(a = 100 - 7(14) = 100 - 98 = 2\), \(o = 5(14) = 70\). Cost = \(5(2) + 7(70) = 10 + 490 = 500\). Valid.
All integer values of k from 1 to 14 yield valid combinations.
Conclusion
The number of ways a person can buy both apples and oranges using a total amount of Rs. 500 is 14, assuming "using total amount" means spending exactly Rs. 500.
Revision Table: Key Concepts Used
| Concept | Description | Application in Problem |
|---|---|---|
| Linear Equation | An equation involving variables raised to the power of 1. | Representing the total cost: \(5a + 7o = 500\). |
| Diophantine Equation | An equation where only integer solutions are sought. | We need integer values for 'a' and 'o'. |
| Divisibility Rules | Rules to determine if one number is divisible by another. | Used to find the general form of 'o' (\(o=5k\)). |
| Integer Constraints | Conditions that variables must be integers within a certain range. | \(a \ge 1\), \(o \ge 1\) are used to find the valid range for 'k'. |
Additional Information: Linear Diophantine Equations
A linear Diophantine equation is an equation of the form \(Ax + By = C\), where A, B, and C are integers, and we seek integer solutions for x and y. A solution exists if and only if the greatest common divisor of A and B (gcd(A, B)) divides C.
In our problem, the equation is \(5a + 7o = 500\). Here, A=5, B=7, and C=500. The gcd(5, 7) is 1. Since 1 divides 500, integer solutions exist. Our method of finding the general solution \(a = 100 - 7k, o = 5k\) is a standard way to solve such equations.
If the question had asked for the number of ways to spend at most Rs. 500 while buying both (\(5a + 7o \le 500\), \(a \ge 1, o \ge 1\)), the approach would involve summing up the possible values for one variable while iterating through the other, using floor functions as shown in the scratchpad. The interpretation of the question phrasing is crucial in problems like this.
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