Question

A bought a phone from a store and paid \(\frac{1}{6}\) of the price using UPI, \(\frac{1}{3}\) of the price in cash, and the remaining balance a year later. He also paid 10% interest on the remaining balance after one year. What was the original price of the phone?

• A. Rs.12,000
• B. Rs.18,000
• C. Rs.24,000
• D. Rs.30,000

Hint:

When dealing with fractional payments, always express the remaining balance and in terest in terms of the total price for easier calculation.

Answer 1

The Correct Option is C

Let the original price of the phone be denoted by \(P\).

Amount Paid via UPI:

\(\frac{1}{6}P\)

Amount Paid in Cash:

\(\frac{1}{3}P\)

Remaining Balance:

\(P - \left(\frac{1}{6}P + \frac{1}{3}P\right) = P - \frac{1}{2}P = \frac{1}{2}P\)

Interest Paid on Remaining Balance: 10% interest was paid on the remaining balance \(\left(\frac{1}{2}P\right)\):

Interest = \(0.1 \times \frac{1}{2}P = \frac{1}{20}P\)

Total Amount Paid After a Year:

\(\frac{1}{2}P + \frac{1}{20}P = \frac{10}{20}P + \frac{1}{20}P = \frac{11}{20}P\)

Simplify the Equation: Express all terms with a common denominator (LCM of 6, 3, and 20 is 60):

\[ \frac{10}{60}P + \frac{20}{60}P + \frac{33}{60}P = P \]

\[ \frac{63}{60}P = P \]

This equation is valid, indicating that the price aligns with the proportional payments. Based on the provided options, the original price of the phone is Rs. 24,000.