Understanding the Problem: Cost of Fruits

The question asks us to find the total cost of one apple and one kiwi fruit based on two given scenarios involving the purchase of multiple apples and kiwi fruits. This type of problem can be solved using a system of linear equations.

Setting Up the Equations

Let's define variables for the cost of each fruit:

• Let \(a\) be the cost of one apple (in Rs.).
• Let \(k\) be the cost of one kiwi fruit (in Rs.).

From the problem statement, we can form two equations based on the purchase information:

1. A man buys 2 apples and 3 kiwi fruits for Rs. 37.
   This translates to the equation: \(2a + 3k = 37\) (Equation 1)
2. He buys 4 apples and 5 kiwi fruits for Rs. 67.
   This translates to the equation: \(4a + 5k = 67\) (Equation 2)

Solving the System of Linear Equations

We have a system of two linear equations with two variables. We can solve this using methods like substitution or elimination. Let's use the elimination method to find the values of \(a\) and \(k\).

Step-by-Step Solution using Elimination

Our goal is to eliminate one variable to solve for the other. We can eliminate \(a\) by making its coefficient the same in both equations. The coefficient of \(a\) in Equation 2 is 4. The coefficient of \(a\) in Equation 1 is 2. We can multiply Equation 1 by 2 to make the coefficient of \(a\) in Equation 1 equal to 4.

Multiply Equation 1 by 2:

\(2 \times (2a + 3k) = 2 \times 37\)

\(4a + 6k = 74\) (Equation 3)

Now we have two equations (Equation 2 and Equation 3) where the coefficient of \(a\) is the same:

• Equation 2: \(4a + 5k = 67\)
• Equation 3: \(4a + 6k = 74\)

Subtract Equation 2 from Equation 3 to eliminate \(a\):

\((4a + 6k) - (4a + 5k) = 74 - 67\)

\(4a + 6k - 4a - 5k = 7\)

\((4a - 4a) + (6k - 5k) = 7\)

\(0 + k = 7\)

\(k = 7\)

So, the cost of one kiwi fruit is Rs. 7.

Now that we have the value of \(k\), substitute \(k = 7\) into either Equation 1 or Equation 2 to find the value of \(a\). Let's use Equation 1:

\(2a + 3k = 37\)

\(2a + 3(7) = 37\)

\(2a + 21 = 37\)

Subtract 21 from both sides:

\(2a = 37 - 21\)

\(2a = 16\)

Divide by 2:

\(a = \frac{16}{2}\)

\(a = 8\)

So, the cost of one apple is Rs. 8.

Calculating the Total Cost

The question asks for the total cost of 1 apple and 1 kiwi fruit. This is given by \(a + k\).

Total cost = \(a + k = 8 + 7 = 15\)

The total cost of 1 apple and 1 kiwi fruit is Rs. 15.

Conclusion on Fruit Costs

Based on the given information and solving the system of equations, we found the cost of one apple to be Rs. 8 and the cost of one kiwi fruit to be Rs. 7. The combined cost of one of each fruit is Rs. 15.

Revision Table: Solving Linear Equations

Additional Information: System of Equations Applications

Systems of linear equations are widely used to model and solve problems in various real-world scenarios. They are particularly useful when dealing with situations involving multiple unknown quantities that are related through several conditions.

• Economics: Determining supply and demand equilibrium points.
• Physics: Analyzing circuits, forces, or motion.
• Chemistry: Balancing chemical equations.
• Finance: Calculating investments, loans, and budgets.
• Daily Life: Calculating costs, mixing solutions, or managing resources, similar to this fruit cost problem.

Understanding how to set up and solve systems of equations is a fundamental skill in algebra with broad applicability.