Solving the Fruit Cost Problem

This problem involves finding the original costs of oranges and bananas given information about their combined prices under different conditions. We can solve this using algebra by setting up and solving a system of linear equations.

Defining Variables

Let's define the variables:

• Let \(o\) represent the original cost of one orange in Rupees (₹).
• Let \(b\) represent the original cost of one banana in Rupees (₹).

Formulating Equations

We are given two scenarios, which we can translate into two equations:

1. Scenario 1: Arnab buys 9 oranges and 8 bananas for ₹276.

   This gives us the equation:

   \(9o + 8b = 276\) (Equation 1)

2. Scenario 2: The cost of an orange is decreased by 10%, and the cost of a banana remains the same. The cost of 5 oranges and 10 bananas is ₹210.

   If the cost of an orange is decreased by 10%, the new cost is \(o - 0.10o = 0.9o\).

   This gives us the equation:

   \(5(0.9o) + 10b = 210\)

   Simplifying this equation:

   \(4.5o + 10b = 210\) (Equation 2)

Solving the System of Equations

We now have a system of two linear equations:

1. \(9o + 8b = 276\)

2. \(4.5o + 10b = 210\)

To make Equation 2 easier to work with, we can multiply it by 2 to eliminate the decimal:

\(2 \times (4.5o + 10b) = 2 \times 210\)

\(9o + 20b = 420\) (Equation 3)

Now we have a simpler system:

1. \(9o + 8b = 276\)

3. \(9o + 20b = 420\)

We can use the elimination method. Subtract Equation 1 from Equation 3:

\((9o + 20b) - (9o + 8b) = 420 - 276\)

\(9o + 20b - 9o - 8b = 144\)

\(12b = 144\)

Now, solve for \(b\):

\(b = \frac{144}{12}\)

\(b = 12\)

So, the original cost of one banana is ₹12.

Next, substitute the value of \(b\) (12) back into Equation 1 to find \(o\):

\(9o + 8(12) = 276\)

\(9o + 96 = 276\)

\(9o = 276 - 96\)

\(9o = 180\)

Solve for \(o\):

\(o = \frac{180}{9}\)

\(o = 20\)

So, the original cost of one orange is ₹20.

Calculating the Final Cost

The question asks for the original cost of 4 oranges and 3 bananas.

Cost = (Number of oranges \(\times\) Original cost per orange) + (Number of bananas \(\times\) Original cost per banana)

Cost = \(4o + 3b\)

Substitute the values we found for \(o\) and \(b\):

Cost = \(4(20) + 3(12)\)

Cost = \(80 + 36\)

Cost = \(116\)

Conclusion

The original cost of 4 oranges and 3 bananas is ₹116.