Question

A man purchased 40 fruits (apples and oranges) for Rs. 17. Had he purchased as many oranges as apples and as many apples as oranges (i.e., interchanged the counts), he would have paid Rs. 15. Find the cost of one pair consisting of one apple and one orange.

• A. 70 paise
• B. 60 paise
• C. 80 paise
• D. 1 rupee

Hint:

Add the two bill equations together, since apples and oranges bought across both trips always total 40 each.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept.
This is a swap-the-quantities problem. Instead of solving two equations formally, look at what happens when the two purchases are added together as one combined purchase.

Step 2: Key idea.
In the first trip he buys $x$ apples and $40-x$ oranges. In the second trip (after the swap) he buys $40-x$ apples and $x$ oranges.
Adding the apples bought across both trips: $x+(40-x)=40$ apples in total.
Adding the oranges bought across both trips: $(40-x)+x=40$ oranges in total.
So across the two trips together, he effectively bought exactly 40 apples and 40 oranges, no matter what $x$ is.

Step 3: Use the total money spent.
Money spent across both trips $= 17+15 = 32$ rupees.
This 32 rupees paid for 40 apples and 40 oranges together, so:
\[ 40a+40b = 32 \]
where $a$ and $b$ are the prices of one apple and one orange.

Step 4: Find the price of one pair.
Divide by 40 on both sides:
\[ a+b = \frac{32}{40} = 0.8 \text{ rupees} = 80 \text{ paise} \]
This is the cost of one apple plus one orange together, matching option (C).
\[ \boxed{80 \text{ paise}} \]