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: Set up the variables.
Let the man buy \(x\) apples, so the number of oranges is \(40-x\), since he bought 40 fruits in total.
Let the price of one apple be \(a\) rupees and the price of one orange be \(b\) rupees.

Step 2: Write the two payment equations.
For the original purchase, the total bill is \(xa+(40-x)b=17\).
When the counts are interchanged, he buys \(40-x\) apples and \(x\) oranges, so the new bill is \((40-x)a+xb=15\).

Step 3: Add the two equations.
Adding the left sides and the right sides removes \(x\) completely:
\[ xa+(40-x)b+(40-x)a+xb = 17+15 \]
Group the \(a\) terms and the \(b\) terms on the left side:
\[ a(x+40-x)+b(40-x+x) = 32 \]
\[ 40a+40b=32 \]

Step 4: Solve for the pair price and check the options.
Divide both sides by 40:
\[ a+b = \frac{32}{40} = 0.8 \text{ rupees} \]
0.8 rupees is 80 paise, so one apple plus one orange costs 80 paise. Options (A) 70 paise, (B) 60 paise and (D) 1 rupee do not satisfy \(40a+40b=32\), so they are wrong.

Final Answer:
The cost of one apple and one orange together is 80 paise. \[ \boxed{80 \text{ paise}} \]