Simple Interest

This is the starting point. P, R and T are given, and you have to find the interest, the amount, or something built on them.

Use SI = P × R × T100, then A = P + SI. In your head, though, think R × T = total percent. 6% for 3 years is 18% of P. 9% for 3 years is 27%. 5% for 212 years is 12.5%, which is 18 of P.

Easy calculation. Find 1% of P first (move the decimal point two places to the left), then multiply by RT. For ₹1300 at 6% for 4 years: 1% of 1300 = 13, RT = 24, so SI = 13 × 24 = ₹312.

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Amount in one step. A = P × 100 + RT100. For ₹2500 at 8% for 5 years, RT = 40%, so A = 2500 × 1.4 = ₹3500.

Story questions. After the amount, the question often adds one more step: a processing fee is charged, a motorcycle is handed over as part of the payment, or the yearly interest is shared among scholarships. Solve the SI part first and then do the extra step calmly. Read the last line again to see exactly what is asked: SI, amount, or something else.

Say Radhika lends ₹2400 at 712% for 4 years. RT = 30%, so SI = 30% of 2400 = ₹720 and the amount = ₹3120. If her friend pays ₹1000 in cash and gives a gift worth ₹2000, then ₹3000 of the ₹3120 is cleared and ₹120 is still pending.

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The formula does not change here. Only the time is not a whole number of years, so the whole job is this: convert the time into years correctly, then use RT as before.

Same date in two months. 5 February to 5 August is exactly 6 months = 12 year. 1 March to 1 August is 5 months = 512 year. Don't count days in such cases; counting months is cleaner and exact.

Different dates. Now you count days: the remaining days of the first month (leaving out the starting day), all the full months in between, and the days of the last month. 18 April to 30 June gives 12 + 31 + 30 = 73 days = 73365 = 15 year. If the question says both days are included, add 1.

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Mixed time. 2 years 4 months = 213 years, 2 years 8 months = 223 years, 15 months = 114 years.

Buying on credit. Suppose a buyer "pays ₹20,000 at once and the rest after 9 months with interest". Interest is charged only on the unpaid part (the rest), not on the full price. Total payment = down payment + rest + interest on the rest.

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One example to tie it together. Radhika borrows ₹7300 on 10 May and returns it on 22 July at 10% p.a. Days = 21 (May) + 30 (June) + 22 (July) = 73 days = 15 year, so SI = 7300 × 10% × 15 = ₹146.

Now the question gives the interest (or the amount) and hides one of P, R and T. The formula SI = PRT100 has four quantities, so if three are known, the fourth comes out in one line:

P = SI × 100R × T, R = SI × 100P × T, T = SI × 100P × R

When the amount is given, there are two ways:

• If P is known, find SI = A − P first, then use the formula.
• If P itself is not known, use A = P × 100 + RT100, which gives P = A × 100100 + RT.

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The percent picture. This is the fastest way: link RT% with the SI.

• Finding P: 9% for 4 years = 36% of P = ₹720, so 1% = 20 and P = ₹2000.
• Finding R: ₹324 on ₹1800 is 3241800 = 18% in total. Spread over 4 years, that is 4.5% per year.
• Finding T: a rate of 12.5% = 18, so one year's interest on ₹6400 is ₹800. To earn ₹2400 it needs 3 years.

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Decimals and approximations. Answers like 5.21% or 3.33 years are common. Don't be afraid of them. Divide carefully, or multiply back to see which option fits.

Radhika's ₹2500 grows to ₹3100 at 6% p.a. Here SI = ₹600 = 6002500 = 24% of P. At 6% a year, 24% needs 4 years.

Here the question gives no rupees at all, or very few. It says things like "the interest is 45 of the sum", "principal and interest are in the ratio 10 : 7", or "the rate is equal to the number of years". All of these are solved by one line from the foundation: SI = (R × T)% of P.

Interest as a fraction of P. If SI = 35 of P, then RT = 35 × 100 = 60. Over 6 years, R = 10%. If SI = 90% of P in 6 years, R = 15%.

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A ratio is given. P : SI = 5 : 2 means SI = 40% of P. P : A = 4 : 5 means the interest is 1 part out of 4, that is 25% of P.

Interest equal to the principal. SI = P means RT = 100. At 5%, it takes 20 years. If after 5 years SI = 40% of P, then each year gives 8%, and 100% needs 1008 = 12.5 years.

Interest more or less than the sum by some rupees. At 25% for 8 years, SI = 200% of P, so SI is more than P by 100% of P. If that excess is ₹600, then P = ₹600.

Rate and time linked. "The rate is equal to the number of years" means R = T, so RT = R². If SI = 1625 of P, then R² = 64 and R = T = 8. If "the rate is 5 times the number of years", put R = 5T, so RT = 5T².

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Try one with Radhika. Her interest is 916 of her principal, and the rate equals the time. RT = 916 × 100 = 56.25 = R², so R = 7.5% and T = 7.5 years.

"A sum doubles in 8 years." "It becomes 5 times in 20 years." Before any formula, see what is really happening in such questions.

If the amount becomes n times the principal, then the interest = nP − P = (n − 1) times the principal. So:

• Doubles (n = 2) ⇒ SI = P ⇒ RT = 100
• Becomes 3 times ⇒ SI = 2P ⇒ RT = 200
• Becomes 5 times ⇒ SI = 4P ⇒ RT = 400
• Becomes 85 times ⇒ SI = 35P ⇒ RT = 60

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So R × T = (n − 1) × 100. Rate = (n − 1) × 100T and time = (n − 1) × 100R.

From one multiple to another (same rate). Count the interest in units of P. "It doubles in 9 years. When will it become 3 times?" Doubling means one P of interest in 9 years. Three times needs two P of interest, so 18 years. In general:

n₁ − 1T₁ = n₂ − 1T₂

"4 times in 18 years" means 3P of interest in 18 years, so 1P in 6 years. "7 times" needs 6P, which takes 36 years.

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Same time, different rates. If in a fixed time the sum becomes n₁ times at rate R₁ and n₂ times at R₂, then n₁ − 1R₁ = n₂ − 1R₂.

"Becomes 260% of itself." This means n = 2.6, so SI = 160% of P.

Radhika's money becomes 4 times in 15 years. That is 3P of interest in 15 years, or 1P in 5 years, so the rate = 20%. It will become 7 times (6P of interest) in 30 years.

"A sum amounts to ₹1080 in 2 years and to ₹1200 in 5 years." Here the principal and the rate are both hidden. Recall the staircase from the foundation: in SI, every year adds the same interest. So the difference of the two amounts is the interest of the extra years.

First, one year's interest. From year 2 to year 5 is 3 years, and the amount rose by 1200 − 1080 = ₹120. So one year's interest = ₹40.

Then the principal. Go back from the first amount. In 2 years the interest was 2 × 40 = ₹80, so P = 1080 − 80 = ₹1000.

The rate, if asked. R = one year's SIP × 100 = 401000 × 100 = 4%.

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There is also a direct formula. It is good to know, though the steps above are easier to trust:
P = A₁T₂ − A₂T₁T₂ − T₁

Ratio of two amounts. Suppose "the amount after 6 years is 43 times the amount after 2 years". Use A = (100 + RT)% of P: 100 + 6R = 43(100 + 2R), so 300 + 18R = 400 + 8R and R = 10%.

Radhika's deposit becomes ₹5600 in 3 years and ₹6400 in 5 years. 2 years bring ₹800, so 1 year brings ₹400. Then P = 5600 − 1200 = ₹4400, and the rate = 4004400 × 100 = 9111%.

SI = P × R × T100. The 100 never changes, so SI changes in exactly the same proportion as the product P × R × T. From this one fact come three quick rules:

• Same P and R ⇒ SI ∝ T. SI for 4 years : SI for 10 years = 4 : 10 = 2 : 5.
• Same R and T ⇒ SI ∝ P. If ₹8,000 earns ₹960, then ₹10,000 earns 960 × 100008000 = ₹1,200.
• Same P ⇒ SI ∝ R × T. If 5% for 4 years (20%) gives ₹300, then 6% for 5 years (30%) gives 300 × 3020 = ₹450.

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Equal interest from two sums. If two sums give equal interest, then P₁R₁T₁ = P₂R₂T₂. With A at 5% for 3 years and B at 6% for 5 years, 15A = 30B, so A : B = 2 : 1.

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Carrying the rate across. Often a question gives one complete case ("₹50 amounts to ₹62 in 3 years") only to tell you the rate. Get the rate, or just the percent per year, from that case and use it on the new sum.

Radhika's ₹500 becomes ₹600 in 4 years. That is 20% in 4 years, or 5% per year. At the same rate, her brother's ₹1200 for 6 years earns 30% of 1200 = ₹360.

Relation between X, Y and Z. If Y is the interest on X, and Z is the interest on Y for the same rate and time, then Y = kX and Z = kY with the same k. So YX = ZY, that is, Y² = XZ.

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"Had it been put at a 4% higher rate for 5 years, it would have fetched ₹900 more." In such questions only the change matters. The old interest is (R × T)% of P and the new interest is (R' × T')% of P, so the extra interest is the difference of the two total percents, taken on the same P.

Rate changes, time same. Extra % = (change in rate) × time. 4% higher for 5 years gives 20% of P = ₹900, so P = ₹4500. The old rate was never needed.

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Time changes, rate same. Extra % = rate × (change in time). At 12%, the interest for 9 years and for 7 years differs by 24% of P. If that difference is ₹360, then P = ₹1500.

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Both change. Subtract the total percents. 8% for 9 years (72%) and 10% for 6 years (60%) differ by 12% of P.

New amount after a change. Find the old rate from the given data, then the new rate, then the new amount. Or add the extra interest to the old amount.

Time in months. 15% for 20 months = 15 × 2012 = 25%. Convert each case to a total percent first, then subtract.

Radhika lends some money for 4 years. Had the rate been 3% more, she would have got ₹360 more. Extra = 3 × 4 = 12% of P = 360, so P = ₹3000.

"4% for the first 2 years, 6% for the next 3 years and 8% after that." The principal stays the same throughout and only the rate changes with time. Simple interest never joins the principal, so each period adds its own percent.

Total percent = R₁T₁ + R₂T₂ + R₃T₃ + …

For 6 years in the example above: 4% × 2 + 6% × 3 + 8% × 1 = 8% + 18% + 8% = 34% of P.

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Rates in successive years. "7% in the first year and 8% in the second year" makes 15% in total, so A = 1.15 × P.

A rate rising every year. "It starts at 6% and increases by 2% every year, for 4 years" gives 6 + 8 + 10 + 12 = 36%. Add the yearly rates one by one.

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Careful with beyond. "6% for the first 2 years and 9% beyond 2 years, total 5 years" means the 9% runs for 5 − 2 = 3 years only.

When the principal also changes in between. Take "₹10,000 at r%; after 3 months ₹5,000 more is borrowed and the rate on the whole becomes 1.5r". Break the year into pieces: ₹10,000 at r for 3 months, then ₹15,000 at 1.5r for 9 months. Each piece is a separate SI, and you add them.

Radhika deposits ₹5000 at 6% for 2 years and then at 9% for 3 more years. Total = 12% + 27% = 39% of 5000 = ₹1950.

A person lends to two or three people, or puts money into two or three schemes. The total interest is the sum of the separate interests, but the question hides one thing: the common rate, the common time or one of the sums. Three ideas handle all of them.

1. Same rate, different sums and times: think in rupee-years. ₹6,000 for 3 years and ₹4,000 for 2 years at the same rate behave like ₹(6,000 × 3 + 4,000 × 2) = ₹26,000 for one year. If the total interest is ₹2,080, the rate is 208026000 × 100 = 8%.

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2. Same time, different sums and rates: find one year's total interest. ₹4,000 at 5% and ₹6,000 at 7% give 200 + 420 = ₹620 in a year. If the total interest is ₹1,860, the time is 3 years.

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3. One part unknown. Find the interest from the known parts and subtract it from the total. What is left belongs to the unknown part.

Loans growing to equal amounts. Two debts at different rates become equal after t years when P₁(1 + R₁t100) = P₂(1 + R₂t100). The smaller debt can catch up only if it grows faster, that is, at the higher rate.

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Radhika lends ₹4,000 to one friend for 3 years and ₹2,000 to another for 2 years at the same rate, and gets ₹1,280 in all. Rupee-years = 12,000 + 4,000 = 16,000, so the rate = 128016000 × 100 = 8%.

Here the sum is split into parts, but the parts are not given in rupees. They come as a ratio ("2 : 3 : 5"), as a relation ("one-fourth of the first equals one-fifth of the second", "the second is ₹2000 more than the first"), or as a percentage of each other. Once you write every part in terms of one unit k, the question becomes a Type 10 question.

First, write the parts.

• Ratio 2 : 3 : 5 ⇒ parts 2k, 3k, 5k.
• "13 of A = 14 of B" ⇒ A : B = 3 : 4 (the part with the smaller fraction is the bigger part).
• "C is 150% of A and 200% of B" ⇒ choose C so that both divisions are easy: C = 6k, so A = 4k and B = 3k.

Next, the interest of each part. Each part × its own (R × T)%.

Last, match with the given total (or difference) and find k.

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Weighted rate, when the time is common. For parts 3 : 2 : 5 at 4%, 6% and 10%, the average yearly rate on the whole = 3 × 4 + 2 × 6 + 5 × 1010 = 7.4%. For 2 years, the whole sum earns 14.8%.

Radhika splits her savings in the ratio 1 : 3 and invests them at 8% and 4% for 5 years, earning ₹1,000. With parts k and 3k, interest = 40% of k + 20% of 3k = k, so k = 1000 and her savings = 4k = ₹4,000.

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"₹12,000 is lent in two parts, one at 5% and the other at 10%. The yearly interest is ₹840. Find the part at 10%." This is the standard parts question, and it has a neat short method.

Basic method. Let the 10% part = x, so the 5% part = 12,000 − x.
10x100 + 5(12000 − x)100 = 840 ⇒ 5x = 24,000 ⇒ x = ₹4,800.

Average rate method. The whole ₹12,000 earned ₹840 in a year, so the whole sum behaved as if it were at 84012000 × 100 = 7%. This 7% must lie between 5% and 10%, and it lies closer to the rate on which more money is put. Now use alligation. You will study it fully in the Mixture chapter; this much is enough here:

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Part at lower rate : Part at higher rate = (higher − average) : (average − lower)

(10 − 7) : (7 − 5) = 3 : 2, so the 10% part = 25 × 12,000 = ₹4,800.

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When the times are different. Use the total percent of each part. 5% for 2 years = 10% and 8% for 5 years = 40%. If the whole sum earns 22% in all, then first part (10%) : second part (40%) = (40 − 22) : (22 − 10) = 18 : 12 = 3 : 2.

Difference of interests given. "The second part earned ₹1,500 more than the first." Write one equation in x and solve it.

Interchanged amounts. "If the amounts were interchanged, he would get ₹50 more." Write both equations. Adding them gives the total of the parts; subtracting gives their difference.

Radhika splits ₹6000 between 5% and 8% and gets ₹420 a year. The average is 7%, so the ratio = (8 − 7) : (7 − 5) = 1 : 2. That puts ₹2000 at 5% and ₹4000 at 8% (check: 100 + 320 = 420).

"₹4200 is invested in three schemes at 5%, 6% and 10%, and at the end of a year all three give the same interest." Same interest means P × R × T is the same for every part. So a part at a higher rate needs less money.

Equal interest ⇒ P ∝ 1R × T

For 5%, 6% and 10% (same time): P₁ : P₂ : P₃ = 15 : 16 : 110. Multiply by the LCM 30 to get 6 : 5 : 3. Then 14 parts = ₹4200, 1 part = ₹300, and the 6% part = 5 × 300 = ₹1500.

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Same rate, different times. Three children get equal interest after 2, 3 and 4 years. Here P ∝ 1T, so the parts are 12 : 13 : 14 = 6 : 4 : 3.

Equal amounts, not equal interest. Sometimes a father divides money so that each child gets the same amount (principal + interest) when they grow up. Then P × (100 + RT) is the same for every part:

Equal amounts ⇒ P ∝ 1100 + RT

Radhika divides some money between her two nephews at 10% p.a. so that the elder gets his amount after 2 years and the younger after 5 years, and the two amounts are equal. P₁ : P₂ = 1100 + 20 : 1100 + 50 = 1120 : 1150 = 5 : 4. If she divides ₹9,000, the elder gets ₹5,000 and the younger ₹4,000, and both receive ₹6,000 at the end.

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This last type brings together the everyday uses of SI. Each one is the same formula seen from a new angle.

1. Present worth. ₹1320 is due after 4 years at 8%. What is it worth today? Today's money P grows to 1320: P × 132100 = 1320, so P = ₹1000. In general, Present worth = Amount due × 100100 + RT.

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2. Cash price vs credit price. "12 copies of a book can be bought in cash for a sum, but only 10 copies if the same sum is paid after a year." The sum paid after a year is the amount on the cash price of 10 copies. So the cash price of 10 copies grows to the cash price of 12 copies in one year. Interest = 2 copies on 10 copies, so R = 210 × 100 = 20%.

3. Equal annual instalments. A debt is cleared by equal yearly instalments x. Each instalment paid before the end earns interest for the years that remain. For a debt whose value at the end of n years is A:

A = x + x(1 + R100) + x(1 + 2R100) + … = x[n + n(n − 1)R200]

If the question gives the money borrowed today (P), first find A = P × 100 + nR100. If it gives the sum to be paid in instalments, that sum is A itself.

Take Radhika: she borrows ₹3300 at 10% for 3 years and repays it in 3 equal yearly instalments. A = 3300 × 1.3 = ₹4290. The first instalment earns 2 years' interest (1.2x), the second 1 year's (1.1x) and the last none (x). So 3.3x = 4290 and x = ₹1300.

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4. Part-payment. If some principal is returned in between, interest from that point runs only on the principal that remains. The interest already earned is still owed.

5. Interest added to the principal (reinvestment). "After 2 years the interest is added to the principal." From then on, the new principal is P + SI, and each block of years is a fresh SI question. This is the first step towards Compound Interest.

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