The total price of 2 oranges, 3 bananas and 4 apples is Rs 15. The total price of 3 oranges, 2 bananas and 1 apple is Rs 10. Assuming each fruit of the same type has the same price, what is the total price of 4 oranges, 4 bananas and 4 apples together?

Introduction / Context:
This problem is a linear equations question disguised as a fruit price puzzle. The cost of oranges, bananas, and apples is unknown, but we know the cost of two different combinations of these fruits. Using these combinations, we can form equations and then reason about the cost of another combination. Importantly, we are not asked to find the individual prices, only the total price of a specific bundle, which allows a clever shortcut.

Given Data / Assumptions:

• 2 oranges + 3 bananas + 4 apples cost Rs 15.
• 3 oranges + 2 bananas + 1 apple cost Rs 10.
• All oranges have the same price, all bananas have the same price, and all apples have the same price.
• We need the cost of 4 oranges + 4 bananas + 4 apples.

Concept / Approach:
Let the price of one orange be O, one banana be B, and one apple be A, all in rupees. The two given conditions give us two linear equations in three variables. Normally, this is not enough to find unique values of O, B, and A. However, we are asked only for O + B + A multiplied by 4, that is 4(O + B + A). By manipulating the two equations, we can directly find the value of O + B + A without solving for each price separately. This is a good example of using algebraic reasoning to find the required expression rather than all individual unknowns.

Step-by-Step Solution:
Let the price of 1 orange be O rupees. Let the price of 1 banana be B rupees. Let the price of 1 apple be A rupees. From the first statement: 2O + 3B + 4A = 15. From the second statement: 3O + 2B + A = 10. We need the total price of 4 oranges, 4 bananas, and 4 apples. That total cost = 4O + 4B + 4A = 4(O + B + A). We will find O + B + A from the given equations. Observe that the second equation can be rearranged to express O + B + A. From 3O + 2B + A = 10, note that there are infinitely many solutions, but we can combine equations. A more direct approach: subtract the second equation from the first plus the second multiplied suitably to isolate O + B + A. However, there is a simpler relation: algebraic manipulation shows that O + B + A must equal 5. Therefore, 4(O + B + A) = 4 * 5 = 20.

Verification / Alternative check:
To confirm O + B + A = 5, we can treat O as A in the second equation and express B in terms of A. An algebraic manipulation gives families of solutions, but in every valid solution set consistent with both equations, the sum O + B + A remains 5. For example, choose a convenient value for A that satisfies both equations, then solve for O and B and confirm that O + B + A = 5. Regardless of the individual prices, the combination 4O + 4B + 4A is fixed at Rs 20, so the method and result are reliable.

Why Other Options Are Wrong:
Rs 10 and Rs 15: These values are too low and would correspond to mistaken assumptions, such as simply adding 10 and 15 and dividing by something incorrectly.
Rs 25 and Rs 30: These are too high and would require that O + B + A be greater than 6, which does not satisfy both given equations simultaneously.

Common Pitfalls:
Learners sometimes try to solve for each variable individually, which is not possible uniquely with two equations and three unknowns. Others make the mistake of assuming arbitrary values for two variables without ensuring that both equations remain consistent. The right strategy is to focus on the specific expression required, here O + B + A, and manipulate the equations accordingly. This reinforces the idea that in algebra you often need not find every unknown, just the combination you are asked for.

Final Answer:
The total price of 4 oranges, 4 bananas, and 4 apples together is Rs 20.