Solving a Fruit Cost Puzzle: The Cost of 3 Oranges, 3 Bananas, and 3 Mangoes
Finding the cost of a combination of fruits using algebraic equations is a common mathematical challenge. Here, we solve the problem of determining the cost of 3 oranges, 3 bananas, and 3 mangoes based on the given cost of individual combinations.
The Problem
We are given the following information:
The cost of 2 oranges, 3 bananas, and 4 mangoes is Rs. 15. The cost of 3 oranges, 2 bananas, and 1 mango is Rs. 10.Step-by-Step Solution
Let's define the cost of an orange as (o), a banana as (b), and a mango as (m).
Step 1: Define the Equations
To solve this, we first establish the equations based on the given information:
2 oranges, 3 bananas, and 4 mangoes cost Rs. 15:
2o 3b 4m 15
3 oranges, 2 bananas, and 1 mango cost Rs. 10:
3o 2b m 10
Step 2: Solve the System of Equations
First, multiply the first equation by 3 and the second equation by 2 to simplify the process:
Multiplying the first equation by 3:
6o 9b 12m 45
Multiplying the second equation by 2:
6o 4b 2m 20
Next, subtract the second equation from the first to eliminate (o):
(6o 9b 12m) - (6o 4b 2m) 45 - 20
This simplifies to:
5b 10m 25
Divide by 5:
b 2m 5
Now, multiply this equation by 3 to get:
3b 6m 15
From the second equation, we can solve for (3o 2b m 10), and multiply by 3:
9o 6b 3m 30
Subtract the simplified first equation from this equation:
(9o 6b 3m) - (6o 9b 12m) 30 - 45
This simplifies to:
3o - 3b - 9m -15
We already know (b 2m 5), so we can substitute:
3o - 3(5 - 2m) - 9m -15
Expanding and simplifying:
3o - 15 6m - 9m -15
3o - 3m - 15 -15
3o - 3m 0
o - m 0
o m
Substitute (o m) into (b 2m 5):
b 2m 5
b 2o 5
b 5 - 2o
Now, substituting (b 5 - 2o), (o m), and (b 5 - 2m) into the equation (2o 3b 4m 15):
2o 3(5 - 2o) 4o 15
2o 15 - 6o 4o 15
15 - o 15
o 0
m 0
b 5
Conclusion
The cost of 3 oranges, 3 bananas, and 3 mangoes is Rs. 15, as derived from the given constraints:
3o 3b 3m 3(0) 3(5) 3(0) 15