Question

A man purchased 40 fruits: apples and oranges for ₹17. Had he purchased as many oranges as apples and as many apples as oranges (i.e., swapped the quantities), he would have paid ₹15. Find the cost of one pair (one apple + one orange).

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

Hint:

When two “swapped-quantity” bills are given, add them to get $(a+o)(x+y)$ directly. If $a+o$ is known, it yields the price of one pair $x+y$ in one step.

Answer 1

The Correct Option is C

Step 1: Define variables.
Let $a$ = number of apples,\ $o$ = number of oranges (so $a+o=40$).
Let $x$ = price of one apple (in rupees), $y$ = price of one orange (in rupees).
Step 2: Form two cost equations.
Given purchase: $ax + oy = 17$.
After swapping quantities: $ox + ay = 15$.
Step 3: Add the two equations to isolate $x+y$.
\[ (ax+oy) + (ox+ay) = (a+o)(x+y) = 17+15 = 32. \] Since $a+o=40$, we get \[ 40(x+y)=32 ⇒ x+y=\frac{32}{40}=0.8\ \text{rupees} = 80\ \text{paise}. \] \[ \boxed{80\ \text{paise}} \]