A TV set is available for ₹19,650 in cash or for ₹3,100 as a cash down payment and three equal annual instalments. If the shopkeeper charges interest at the rate of 10% per annum compounded annually, calculate the amount of each instalment.

Correct — C. The instalments pay off the balance, and each is discounted back at 10% a year.

Balance: 19,650 − 3,100 = ₹16,550
Present value of three instalments of x:
x(10⁄11) + x(100⁄121) + x(1000⁄1331) = 16,550

Over 1331: x(1210 + 1100 + 1000)⁄1331 = 3310x⁄1331
So 3310x⁄1331 = 16,550
x = 16,550 × 1331 ÷ 3310 = 5 × 1331 = ₹6,655 → option (c)

Why the others are wrong

• (a)
  ₹ 9,655 — Three payments of ₹9,655 total ₹28,965, far more than ₹22,028.05, the amount the whole ₹16,550 debt would reach if left unpaid for all three years.
• (b)
  ₹ 7,655 — Three payments of ₹7,655 total ₹22,965, above ₹22,028.05, which is what ₹16,550 grows to at 10% compound over three full years. Paying earlier can only cost less.
• (d)
  ₹ 8,655 — Discounted at 10%, three instalments of ₹8,655 are worth about ₹21,524 today, well above the ₹16,550 still owed after the down payment.

Instalments are valued at the time of purchase. An instalment of x paid after n years is worth x ÷ 1.1ⁿ today at 10% compound interest, and the three present values must add up to the amount financed.

The amount financed is the cash price minus the down payment: ₹16,550, not ₹19,650.

The fractions 10⁄11, 100⁄121 and 1000⁄1331 are 1 ÷ 1.1, 1 ÷ 1.1² and 1 ÷ 1.1³.

Key facts

• Amount financed = cash price − down payment = 19,650 − 3,100 = ₹16,550.
• At 10% compounded annually, ₹1 due in n years is worth 1 ÷ 1.1ⁿ today.
• 10⁄11 + 100⁄121 + 1000⁄1331 = 3310⁄1331.

Common traps

• Using the full cash price ₹19,650 instead of the ₹16,550 left after the down payment.
• Dividing the grown balance ₹22,028.05 by 3. Each instalment earns interest for a different number of years, so the divisor is 3.31, not 3.

No directly related past PYQ was found.