Simple Interest • Topic 3 of 3

SI Applications

This is where simple interest earns CAT marks: instalments, mixed-rate splits and n-fold growth. For equal annual instalments repaying a loan under SI, each instalment’s repayment carries its own interest for the years it stays unpaid, so a loan P repaid in n equal annual instalments x at rate R% satisfies P(1 + nR/100) = x·n + x·R/100·[(n−1) + (n−2) + … + 0]. For two-part splits — a sum divided so different rates give equal or related interest — set the interests equal and solve the resulting ratio; assigning the whole sum a convenient value such as the LCM keeps the arithmetic clean. For growth problems, the master relation is R×T = (n−1)×100: a sum becomes n times its value when the total interest equals (n−1) principals. This single line answers most "doubles / trebles / becomes 5 times" questions in seconds.

✅ Solved examples

1. A sum of money becomes 4 times itself in 15 years at simple interest. Find the rate.
Becomes n times ⇒ R×T = (n−1)×100. Here n = 4, T = 15: R × 15 = 300 ⇒ R = 20% per annum.
2. A sum doubles itself in 10 years at simple interest. In how many years will it become 5 times itself?
Doubles ⇒ R×10 = 100 ⇒ R = 10%. For 5 times: R×T = 400 ⇒ 10×T = 400 ⇒ T = 40 years.
3. ₹15,000 is split into two parts; one is lent at 8% and the other at 10% per annum simple interest. After 2 years the total interest is ₹2,640. Find the part lent at 8%.
Let the 8% part be x, so the 10% part is 15000 − x. Interest = x×8×2/100 + (15000 − x)×10×2/100 = 2640 ⇒ 0.16x + 0.20(15000 − x) = 2640 ⇒ 3000 − 0.04x = 2640 ⇒ 0.04x = 360 ⇒ x = ₹9,000.
4. A man borrows ₹2,000 and agrees to repay it in 2 equal annual instalments at 10% per annum simple interest. Find each instalment.
Owed after 2 years = 2000(1 + 2×10/100) = 2000 × 1.2 = ₹2,400. The first instalment x (paid at end of year 1) earns interest for 1 more year, adding x×10/100. So x + x + x×10/100 = 2400 ⇒ 2.1x = 2400 ⇒ x = ₹1,142.86 (≈ ₹1,142.86).

✏️ Practice — try these, take hints as needed

1. A sum trebles itself in 20 years at SI. Find the rate.
n = 3 ⇒ R×T = (n−1)×100 = 200.
R × 20 = 200.
Solve for R.
10% per annum
2. A sum doubles in 8 years at SI. In how many years will it become 4 times?
Doubles ⇒ R×8 = 100 ⇒ R = 12.5%.
4 times ⇒ R×T = 300.
12.5 × T = 300.
24 years
3. At what rate does a sum become 5 times itself in 16 years at SI?
n = 5 ⇒ R×T = 400.
R × 16 = 400.
Divide.
25% per annum
4. ₹12,000 is split into two parts lent at 5% and 8% per annum SI. After 1 year total interest is ₹810. Find the part at 8%.
Let 8% part = x.
0.05(12000 − x) + 0.08x = 810.
600 + 0.03x = 810.
₹7,000
5. A man repays a loan of ₹1,100 in 2 equal annual instalments at 10% per annum SI. Find each instalment.
Owed after 2 yrs = 1100 × 1.2 = 1320.
First instalment earns 1 extra year: x + x + 0.1x = 1320.
2.1x = 1320.
₹628.57 (≈ ₹628.57)

📝 Topic test — 8 questions

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