Interest (SI & CI)

Reading rates out of amounts

When a sum amounts to given figures after whole years, the growth between consecutive years tells the story.

At compound interest — the ratio trick

Consecutive yearly amounts differ by the year's interest, and $\frac{A_{T+1}}{A_T} = 1 + \frac{R}{100}$ Amounts ₹4,840 (after 2 years) and ₹5,324 (after 3 years): $\frac{5324}{4840} = 1.1$ → 10%. The principal itself is not needed. From the rate, walk back to P: $\frac{4840}{1.21} = ₹4{,}000$.

At simple interest — the constant difference

SI adds the same rupees every year, so consecutive amounts differ by exactly one year's interest: $A_{T+1} - A_T = \frac{PR}{100}$ ₹4,500 and ₹5,000: yearly interest ₹500; the sum at the start $= 4500 - 500 = ₹4{,}000$; rate $= \frac{500 \times 100}{4000} = 12.5\%$. Non-consecutive years: divide the amount gap by the year gap first. ₹6,800 in 3 years and ₹8,000 in 5 years → $\frac{8000 - 6800}{2} = ₹600$ per year.

Two SIs on one sum

If the SI for 2 years is 1,200 and for 3 years 1,800, the difference (₹600) is one year's interest — and $\frac{PR}{100} = 600$. That is one equation in two unknowns, so the question must supply P or R as well: P ₹10,000 → rate 6%; rate 6% → P ₹10,000. The two SIs together never determine both by themselves.

Amounts spanning 2y and 3y at SI — the classic pair

'A sum amounts to A in 2 years and B in 3 years at SI': yearly interest $= B - A$; P $= A - 2(B - A) = 3A - 2B$; rate $= \frac{(B-A) \times 100}{P}$. ₹5,200 (3 years) and ₹5,600 (4 years): yearly ₹400, P $= 5200 - 3 \times 400 = ₹4{,}000$, rate 10%.

Finding P from an amount at CI

Divide by the chips: amount ₹6,655 after 3 years at 10% → $\frac{6655}{1.331} = ₹5{,}000$. Recognise $1.1^3 = 1.331$, $1.1^2 = 1.21$, $1.2^2 = 1.44$, $1.25^2 = 1.5625$.

A quick worked case

CI, amounts 4,840 → 5,324: ratio 1.1, so 10% — and P $= \frac{4840}{1.21} = ₹4{,}000$ if asked. The same pair at SI would mean yearly interest ₹484, P $= 4356$, rate $= \frac{48400}{4356} = 11\frac{1}{9}\%$ — different answers from the same numbers, because the growth laws differ. Decide CI or SI first; the reading of the gap depends on it.

Where the marks go

• Reading a CI gap as a rupee amount (it grows every year) or an SI gap as a ratio (it is constant) — the two standard swaps.
• Forgetting to divide non-consecutive gaps by the year count: 3 → 5 years is TWO years of interest.
• Discounting only once from a 3-year amount, or compounding when the question says simple.

Quick revision

• CI: $\frac{A_{T+1}}{A_T} = 1 + \frac{R}{100}$ — rate needs no P.
• SI: $A_{T+1} - A_T =$ one year's interest, constant every year.
• Non-consecutive: divide the gap by the year gap.
• P from SI amounts: subtract the accrued interest years × yearly interest.
• P from CI amounts: divide by the chip power.