Simple interest questions, with answers

The questions, with answers

1. 1.What is the simple interest formula, and what is the difference between interest and amount?

   Simple interest = P × R × T / 100, where P is the principal, R the rate per cent per year and T the time in years. The interest is the same every year because it is always computed on the original principal. The amount is what you have at the end: A = P + SI = P × (100 + RT) / 100. The reading error is here: a question that says "amounts to Rs 5,600" is giving A, not the interest, and the interest is A minus P. Rs 6,000 at 7% for 2 years earns 6000 × 7 × 2 / 100 = Rs 840 and amounts to Rs 6,840. If the time is in months, divide by 12 first: 9 months is 0.75 years.

2. 2.How do you find the rate or the time when the interest is known?

   Rearrange: R = 100 × SI / (P × T) and T = 100 × SI / (P × R). If Rs 4,000 earns Rs 720 in 3 years, R = 100 × 720 / (4000 × 3) = 6%. If Rs 2,500 at 4% earns Rs 400, T = 100 × 400 / (2500 × 4) = 4 years. Because the formula is a product, a pair of problems with the rate and time swapped give the same interest — 5% for 4 years and 4% for 5 years both give 20% of the principal — and questions sometimes lean on that. When the rate and time are numerically equal ("the rate equals the number of years"), SI = P × R² / 100, so R² = 100 × SI / P and R is a square root.

3. 3.How do you find the principal from the amount?

   Divide, don't subtract. A = P × (100 + RT) / 100, so P = 100 × A / (100 + RT). A sum that amounts to Rs 5,600 in 4 years at 10% has RT = 40, so P = 100 × 5600 / 140 = Rs 4,000, and the interest was Rs 1,600. The mistake is to compute 10% of 5,600 for four years and subtract it, treating the amount as the principal. The same rearrangement answers "what sum yields Rs 900 interest at 6% in 5 years": SI = P × 30 / 100, so P = 900 × 100 / 30 = Rs 3,000.

4. 4.How long does a sum take to double or triple at simple interest?

   A sum doubles when the interest equals the principal, so RT = 100 and T = 100 / R years — at 5% that is 20 years, at 12.5% it is 8. It triples when the interest is twice the principal, RT = 200, so tripling takes exactly twice as long as doubling, and quadrupling three times as long: the extra multiples arrive at a constant pace, because simple interest is linear. So "a sum doubles in 6 years at simple interest; when does it become four times" is 18 years, not 12 — the sum needs three principals' worth of interest, at one principal per 6 years. Under compound interest the answer would be 12, which is the contrast the question is usually testing.

5. 5.How do you handle a sum lent in two parts at different rates?

   Let one part be x and the other the remainder, write each part's interest, and set the sum equal to the total interest. Rs 10,000 is split between 6% and 9% for one year, earning Rs 720 in total. If x is at 9%: 0.09x + 0.06 × (10000 - x) = 720, so 0.03x = 120 and x = Rs 4,000 at 9%, Rs 6,000 at 6%. Sanity-check with the average rate: 720 on 10,000 is 7.2%, which is closer to 6 than to 9, so more money must be at 6% — and it is. The alligation shortcut gives the same split: distances from 7.2 to 6 and to 9 are 1.2 and 1.8, so the 6% part to the 9% part is 1.8 : 1.2 = 3 : 2, again Rs 6,000 and Rs 4,000.

6. 6.How do you solve a problem where the rate changes partway through?

   Treat each period separately and add the interest, since simple interest on the same principal is additive. Rs 8,000 at 5% for the first 2 years and 8% for the next 3 years earns 8000 × 5 × 2 / 100 + 8000 × 8 × 3 / 100 = 800 + 1,920 = Rs 2,720. The same idea handles a rate that changes every year — sum the yearly rates, then multiply once: rates of 4%, 5% and 6% over three years are 15% in total, so on Rs 8,000 the interest is Rs 1,200. Never average the rates unless the periods are equal; weight each rate by its own duration.

7. 7.What is the difference between simple and compound interest in the first and second years?

   In the first year there is no difference: both give P × R / 100, because compound interest has nothing to compound yet. From the second year on, compound interest is computed on the principal plus the interest already earned, so it pulls ahead — and the difference for two years is exactly the interest on the first year's interest: P × (R/100)². On Rs 10,000 at 10%, both give Rs 1,000 in year one; in year two simple interest adds another 1,000 while compound adds 1,100, so the two-year difference is Rs 100 = 10000 × 0.01. That formula is a standard shortcut: "the difference between CI and SI for 2 years at 5% is Rs 25 — find the sum" is P × 0.0025 = 25, so P = Rs 10,000.

8. 8.How do you find the annual instalment that clears a debt at simple interest?

   Each instalment paid early earns simple interest until the debt's due date, and the sum of the instalments plus that interest must equal the debt. A debt of Rs 6,200 is to be cleared in 2 equal annual instalments at 10% simple interest, the first paid one year from now and the last at the due date two years from now. The first instalment x earns interest for the remaining year, so it is worth x + 0.1x = 1.1x at the due date; the second is paid on the due date and is worth x. So 1.1x + x = 6200, 2.1x = 6200 and x = Rs 2,952.38. In the general form, an instalment of x paid k years early is worth x × (100 + R × k) / 100 at maturity; add them up and equate to the amount due.

How the diagnostic asks it

One question from the Aptitude bank, exactly as a sitting would show it. The bank has 4 on simple interest and 60 across Aptitude.