Episode 8 — Aptitude and Reasoning / 8.3 — Simple Interest
8.3.c Solved Examples -- Simple Interest
Problems are organized into three tiers: Basic, Medium, and Advanced. Attempt each problem on your own before reading the solution.
BASIC LEVEL (Problems 1-8)
Problem 1: Finding SI (Direct Application)
Q: Find the Simple Interest on Rs. 6500 at 8% per annum for 5 years.
Solution:
Given: P = 6500, R = 8, T = 5
SI = (P x R x T) / 100
SI = (6500 x 8 x 5) / 100
SI = 260000 / 100
SI = Rs. 2600
Answer: Rs. 2600
Problem 2: Finding Amount
Q: Find the amount when Rs. 12,000 is invested at 7.5% per annum simple interest for 4 years.
Solution:
Given: P = 12000, R = 7.5, T = 4
SI = (12000 x 7.5 x 4) / 100
Quick calculation:
7.5% of 12000 = 900 (per year)
For 4 years: 900 x 4 = 3600
Amount = P + SI = 12000 + 3600 = Rs. 15,600
Answer: Rs. 15,600
Problem 3: Finding Principal
Q: A sum of money earns Rs. 4050 as simple interest in 3 years at 9% per annum. Find the sum.
Solution:
Given: SI = 4050, R = 9, T = 3
P = (SI x 100) / (R x T)
P = (4050 x 100) / (9 x 3)
P = 405000 / 27
P = Rs. 15,000
Answer: Rs. 15,000
Problem 4: Finding Rate
Q: At what rate percent per annum will Rs. 3000 earn Rs. 720 as simple interest in 4 years?
Solution:
Given: P = 3000, SI = 720, T = 4
R = (SI x 100) / (P x T)
R = (720 x 100) / (3000 x 4)
R = 72000 / 12000
R = 6%
Answer: 6% per annum
Problem 5: Finding Time
Q: In how many years will Rs. 8000 amount to Rs. 11,200 at 10% per annum simple interest?
Solution:
Given: P = 8000, A = 11200, R = 10
SI = A - P = 11200 - 8000 = 3200
T = (SI x 100) / (P x R)
T = (3200 x 100) / (8000 x 10)
T = 320000 / 80000
T = 4 years
Answer: 4 years
Problem 6: Time in Months
Q: Find the simple interest on Rs. 4800 at 6% per annum for 8 months.
Solution:
Given: P = 4800, R = 6, T = 8 months = 8/12 = 2/3 years
SI = (4800 x 6 x 2/3) / 100
SI = (4800 x 4) / 100
SI = 19200 / 100
SI = Rs. 192
Answer: Rs. 192
Problem 7: Doubling of Principal
Q: At what rate of simple interest will a sum of money double itself in 12.5 years?
Solution:
When amount doubles: A = 2P, so SI = P
R x T = 100
R x 12.5 = 100
R = 100 / 12.5
R = 8%
Answer: 8% per annum
Problem 8: Finding Principal from Amount
Q: What sum will amount to Rs. 16,800 at 12% per annum simple interest in 5 years?
Solution:
Given: A = 16800, R = 12, T = 5
A = P(100 + RT) / 100
16800 = P(100 + 60) / 100
16800 = P x 160 / 100
16800 = 1.6P
P = 16800 / 1.6
P = Rs. 10,500
Verification:
SI = (10500 x 12 x 5) / 100 = 6300
A = 10500 + 6300 = 16800 [Correct]
Answer: Rs. 10,500
MEDIUM LEVEL (Problems 9-16)
Problem 9: Different Rates for Different Periods
Q: Find the SI on Rs. 10,000 at 6% for the first 2 years, 8% for the next 3 years, and 10% for the remaining 1 year.
Solution:
P = 10000
SI = P x (R1.T1 + R2.T2 + R3.T3) / 100
SI = 10000 x (6x2 + 8x3 + 10x1) / 100
SI = 10000 x (12 + 24 + 10) / 100
SI = 10000 x 46 / 100
SI = Rs. 4600
Answer: Rs. 4600
Problem 10: Finding Two Unknowns (Rate and Time Linked)
Q: A sum of Rs. 5000 amounts to Rs. 7000 at simple interest in a certain time. If the rate of interest is doubled, it would amount to Rs. 9000 in the same time. Find the rate and time.
Solution:
Case 1: P = 5000, A = 7000, Rate = R, Time = T
SI = 7000 - 5000 = 2000
(5000 x R x T) / 100 = 2000
R x T = 40 ... (i)
Case 2: P = 5000, A = 9000, Rate = 2R, Time = T
SI = 9000 - 5000 = 4000
(5000 x 2R x T) / 100 = 4000
2R x T = 80
R x T = 40 ... (ii)
Wait -- both give R x T = 40. Let's re-examine.
From (i): 5000.R.T / 100 = 2000 --> R.T = 40
From Case 2: 5000.(2R).T / 100 = 4000 --> 2RT = 80 --> RT = 40
This is consistent but doesn't uniquely determine R and T.
The problem needs one more condition. Let's re-read:
"amounts to Rs. 9000 in the same time" -- this gives SI = 4000.
SI doubled when R doubled -- which is expected (SI is proportional to R).
Since RT = 40, there are multiple solutions:
R = 8, T = 5 OR R = 10, T = 4 OR R = 20, T = 2, etc.
If the question intends for us to find R specifically from the
doubling clue:
When R doubles, SI becomes 2 x 2000 = 4000 (consistent).
We need additional info. Most likely the intended reading is:
R = 8%, T = 5 years (a common exam answer pair for RT = 40).
Note: In many exam versions, a specific T or R value is given.
With only this information, R x T = 40. If forced to pick:
Answer: R = 8% per annum, T = 5 years (most likely intended answer)
Problem 11: Borrowing and Lending Profit
Q: Suresh borrows Rs. 15,000 at 10% per annum simple interest and lends it at 14% per annum simple interest. Find his gain after 3 years.
Solution:
Interest paid on borrowing:
SI_borrow = (15000 x 10 x 3) / 100 = Rs. 4500
Interest earned on lending:
SI_lend = (15000 x 14 x 3) / 100 = Rs. 6300
Gain = SI_lend - SI_borrow = 6300 - 4500 = Rs. 1800
Shortcut:
Gain = P x (R_lend - R_borrow) x T / 100
Gain = 15000 x (14 - 10) x 3 / 100
Gain = 15000 x 4 x 3 / 100
Gain = Rs. 1800
Answer: Rs. 1800
Problem 12: Split Investment (Two Parts)
Q: Rs. 20,000 is divided into two parts such that the SI on the first part at 6% for 3 years is equal to the SI on the second part at 8% for 4 years. Find the two parts.
Solution:
Let first part = x, second part = (20000 - x)
SI on first part = SI on second part
(x x 6 x 3) / 100 = ((20000 - x) x 8 x 4) / 100
Cancel 100 from both sides:
18x = (20000 - x) x 32
18x = 640000 - 32x
18x + 32x = 640000
50x = 640000
x = 12800
First part = Rs. 12,800
Second part = Rs. 20,000 - 12,800 = Rs. 7,200
Verification:
SI1 = (12800 x 6 x 3) / 100 = 230400 / 100 = 2304
SI2 = (7200 x 8 x 4) / 100 = 230400 / 100 = 2304 [Equal. Correct.]
Answer: Rs. 12,800 and Rs. 7,200
Problem 13: Finding Principal When Amounts at Two Different Times Are Given
Q: A sum at simple interest amounts to Rs. 4480 in 3 years and Rs. 5120 in 5 years. Find the principal and rate.
Solution:
A1 = 4480 at T1 = 3 years
A2 = 5120 at T2 = 5 years
SI per year = (A2 - A1) / (T2 - T1)
SI per year = (5120 - 4480) / (5 - 3)
SI per year = 640 / 2 = 320
SI for 3 years = 320 x 3 = 960
P = A1 - SI for 3 years = 4480 - 960 = 3520
Rate:
R = (SI per year x 100) / P
R = (320 x 100) / 3520
R = 32000 / 3520
R = 9.09% (approximately)
Let's check with exact fraction:
R = 32000/3520 = 2000/220 = 100/11 = 9 (1/11)%
Verification:
SI for 3 yrs = (3520 x 100/11 x 3) / 100 = (3520 x 300) / 1100 = 960
A = 3520 + 960 = 4480 [Correct]
Answer: P = Rs. 3520, R = 100/11 % (approximately 9.09%)
Problem 14: Time for Amount to Become n Times
Q: A sum of money becomes 5/3 of itself in 5 years at simple interest. Find the rate of interest.
Solution:
A = (5/3) x P
SI = A - P = (5/3)P - P = (2/3)P
SI = (P x R x T) / 100
(2/3)P = (P x R x 5) / 100
(2/3) = (R x 5) / 100
(2/3) = 5R / 100
R = (2/3) x (100/5)
R = 200/15
R = 40/3
R = 13.33% or 13 (1/3) %
Answer: 13 1/3 % per annum
Problem 15: Total Interest from Multiple Investments
Q: Ravi invests Rs. 5000 at 8%, Rs. 7000 at 10%, and Rs. 3000 at 12% per annum simple interest. Find his total interest after 2 years.
Solution:
SI1 = (5000 x 8 x 2) / 100 = 800
SI2 = (7000 x 10 x 2) / 100 = 1400
SI3 = (3000 x 12 x 2) / 100 = 720
Total SI = 800 + 1400 + 720 = Rs. 2920
Answer: Rs. 2920
Problem 16: Effective Rate on Combined Investment
Q: Rs. 18,000 is split into two parts. One part is lent at 10% and the rest at 15% per annum. If the total annual interest is Rs. 2100, find each part.
Solution:
Let part at 10% = x
Then part at 15% = 18000 - x
Annual SI:
(x x 10 x 1)/100 + ((18000 - x) x 15 x 1)/100 = 2100
10x/100 + (270000 - 15x)/100 = 2100
(10x + 270000 - 15x) / 100 = 2100
270000 - 5x = 210000
5x = 60000
x = 12000
Part at 10% = Rs. 12,000
Part at 15% = Rs. 6,000
Verification:
SI1 = (12000 x 10) / 100 = 1200
SI2 = (6000 x 15) / 100 = 900
Total = 1200 + 900 = 2100 [Correct]
Answer: Rs. 12,000 at 10% and Rs. 6,000 at 15%
ADVANCED LEVEL (Problems 17-24)
Problem 17: Three-Way Split Investment
Q: A man invests a total of Rs. 30,000 in three different schemes at 6%, 8%, and 10% per annum SI. After one year, he earns Rs. 2400 in total interest. If the amounts invested at 6% and 8% are in the ratio 2:3, find each investment.
Solution:
Let amount at 6% = 2k
Let amount at 8% = 3k
Then amount at 10% = 30000 - 5k
Total annual SI = 2400:
(2k x 6)/100 + (3k x 8)/100 + ((30000-5k) x 10)/100 = 2400
12k/100 + 24k/100 + (300000 - 50k)/100 = 2400
(12k + 24k + 300000 - 50k) / 100 = 2400
(300000 - 14k) / 100 = 2400
300000 - 14k = 240000
14k = 60000
k = 60000/14 = 30000/7
Amount at 6% = 2k = 60000/7 = Rs. 8,571.43
Amount at 8% = 3k = 90000/7 = Rs. 12,857.14
Amount at 10% = 30000 - 5(30000/7) = 30000 - 150000/7 = 60000/7 = Rs. 8,571.43
Verification:
SI1 = (60000/7 x 6) / 100 = 360000/700 = 514.29
SI2 = (90000/7 x 8) / 100 = 720000/700 = 1028.57
SI3 = (60000/7 x 10) / 100 = 600000/700 = 857.14
Total = 514.29 + 1028.57 + 857.14 = 2400 [Correct]
Answer: Rs. 60000/7 each at 6% and 10%, Rs. 90000/7 at 8%
(approximately Rs. 8571, Rs. 12857, Rs. 8571)
Problem 18: Finding the Sum When Difference of SI at Two Rates is Given
Q: The difference between the SI on a certain sum at 8% per annum for 4 years and at 6% per annum for 5 years is Rs. 60. Find the sum.
Solution:
SI at 8% for 4 years = (P x 8 x 4) / 100 = 32P / 100
SI at 6% for 5 years = (P x 6 x 5) / 100 = 30P / 100
Difference:
32P/100 - 30P/100 = 60
2P/100 = 60
P = 60 x 100 / 2
P = Rs. 3000
Verification:
SI1 = (3000 x 8 x 4)/100 = 960
SI2 = (3000 x 6 x 5)/100 = 900
Difference = 960 - 900 = 60 [Correct]
Answer: Rs. 3000
Problem 19: Mixed Borrowing and Lending with Different Times
Q: A man borrows Rs. 8000 at 8% SI for 3 years and Rs. 6000 at 10% SI for 4 years. He lends the entire amount at R% SI for 5 years and gains Rs. 1300 overall. Find R.
Solution:
Total interest paid on borrowing:
SI1 = (8000 x 8 x 3) / 100 = Rs. 1920
SI2 = (6000 x 10 x 4) / 100 = Rs. 2400
Total paid = 1920 + 2400 = Rs. 4320
Total amount lent = 8000 + 6000 = Rs. 14000
Interest earned on lending:
SI_earned = (14000 x R x 5) / 100 = 700R
Gain = SI_earned - Total paid = 1300
700R - 4320 = 1300
700R = 5620
R = 5620 / 700
R = 8.03% (approximately)
Let's express exactly: R = 5620/700 = 562/70 = 281/35 = 8.03%
Hmm, this doesn't give a clean answer. Let me re-read...
Actually, the gain of Rs. 1300 is the net gain. So:
700R = 5620
R = 281/35 ≈ 8.03%
In many exam settings this would be set up to yield a clean rate.
If gain were Rs. 1680: R = 6000/700 ≈ 8.57%
If the total lent for 5 years and gain = 1300:
R = 5620/700 = 8.03% (approx)
Answer: R = 281/35 % per annum (approximately 8.03%)
[Note: This problem illustrates that not all problems yield neat
answers. In exams, the numbers are usually chosen to give clean
values. The method, however, is what matters.]
Problem 20: SI with Annual Deposits (Recurring Deposit Style)
Q: A man deposits Rs. 2000 at the beginning of each year in a bank that pays 10% SI. How much total interest will he earn at the end of 4 years?
Solution:
Each deposit earns SI for a different number of years:
Deposit 1 (Year 1): Rs. 2000 for 4 years
SI1 = (2000 x 10 x 4) / 100 = Rs. 800
Deposit 2 (Year 2): Rs. 2000 for 3 years
SI2 = (2000 x 10 x 3) / 100 = Rs. 600
Deposit 3 (Year 3): Rs. 2000 for 2 years
SI3 = (2000 x 10 x 2) / 100 = Rs. 400
Deposit 4 (Year 4): Rs. 2000 for 1 year
SI4 = (2000 x 10 x 1) / 100 = Rs. 200
Total SI = 800 + 600 + 400 + 200 = Rs. 2000
Shortcut:
Total SI = P x R/100 x (T + (T-1) + ... + 1)
Total SI = 2000 x 10/100 x (4 + 3 + 2 + 1)
Total SI = 200 x 10
Total SI = Rs. 2000
General formula for n annual deposits:
Total SI = P x R/100 x n(n+1)/2
Answer: Rs. 2000
Problem 21: SI and CI Difference for 2 Years
Q: The simple interest on a certain sum for 2 years at 10% per annum is Rs. 2000. Find the difference between the compound interest and simple interest for the same period.
Solution:
SI = 2000, R = 10, T = 2
First find P:
P = (SI x 100) / (R x T) = (2000 x 100) / (10 x 2) = 10000
SI for 2 years = Rs. 2000 (given)
CI for 2 years:
CI = P[(1 + R/100)^2 - 1]
CI = 10000[(1.1)^2 - 1]
CI = 10000[1.21 - 1]
CI = 10000 x 0.21
CI = Rs. 2100
Difference = CI - SI = 2100 - 2000 = Rs. 100
Shortcut formula:
CI - SI (for 2 years) = P x (R/100)^2
= 10000 x (10/100)^2
= 10000 x 0.01
= Rs. 100
Answer: Rs. 100
Problem 22: Finding Time When a Fraction of Principal is Earned as Interest
Q: In what time will the simple interest on Rs. 7500 at 6% per annum be 2/5 of the principal?
Solution:
SI = (2/5) x P = (2/5) x 7500 = Rs. 3000
T = (SI x 100) / (P x R)
T = (3000 x 100) / (7500 x 6)
T = 300000 / 45000
T = 6.67 years = 6 years and 8 months
Alternatively:
T = 20/3 years = 6 (2/3) years = 6 years 8 months
Answer: 6 years 8 months (or 20/3 years)
Problem 23: Two Sums with Related Principals and Different Rates
Q: Two sums of money are in the ratio 3:4. The SI on the first at 7% per annum for 5 years equals the SI on the second at a certain rate for 3 years. Find the rate on the second sum.
Solution:
Let the sums be 3x and 4x.
SI on first = (3x x 7 x 5) / 100 = 105x / 100
SI on second = (4x x R x 3) / 100 = 12Rx / 100
They are equal:
105x / 100 = 12Rx / 100
105 = 12R
R = 105 / 12
R = 8.75%
Answer: 8.75% per annum
Problem 24: Complex Multi-Step Problem
Q: A man has Rs. 40,000. He lends Rs. 15,000 at 10% per annum SI, Rs. 12,000 at 8% per annum SI, and the remaining at R% per annum SI. After 2 years, his total interest earned is Rs. 7600. He then invests the entire Rs. 40,000 plus the earned interest at 9% SI. Find the total amount he has at the end of 5 more years (i.e., 7 years from the start).
Solution:
Step 1: Find R for the third investment.
Remaining amount = 40000 - 15000 - 12000 = Rs. 13000
SI from first: (15000 x 10 x 2) / 100 = Rs. 3000
SI from second: (12000 x 8 x 2) / 100 = Rs. 1920
SI from third: (13000 x R x 2) / 100 = 260R
Total SI = 3000 + 1920 + 260R = 7600
4920 + 260R = 7600
260R = 2680
R = 2680 / 260
R = 10.31% (approximately)
Exact: R = 2680/260 = 268/26 = 134/13 ≈ 10.31%
Step 2: After 2 years, total money = 40000 + 7600 = Rs. 47,600
Step 3: This is invested at 9% SI for 5 years.
SI = (47600 x 9 x 5) / 100
SI = 2142000 / 100
SI = Rs. 21,420
Total amount at end of 7 years = 47600 + 21420 = Rs. 69,020
Answer: Rs. 69,020
Problem 25: Installment Problem
Q: A borrows Rs. 5000 from B at 8% SI per annum. He repays Rs. 2000 at the end of the first year and Rs. 2000 at the end of the second year. How much should he pay at the end of the third year to settle the debt?
Solution:
Year 1:
Amount owed at start = 5000
Interest for year 1 = (5000 x 8 x 1)/100 = 400
Total owed at end of year 1 = 5000 + 400 = 5400
Payment made = 2000
Remaining = 5400 - 2000 = 3400
Year 2:
Amount owed at start of year 2 = 3400
Interest for year 2 = (3400 x 8 x 1)/100 = 272
Total owed at end of year 2 = 3400 + 272 = 3672
Payment made = 2000
Remaining = 3672 - 2000 = 1672
Year 3:
Amount owed at start of year 3 = 1672
Interest for year 3 = (1672 x 8 x 1)/100 = 133.76
Total owed at end of year 3 = 1672 + 133.76 = 1805.76
Final payment = Rs. 1805.76
Answer: Rs. 1805.76 (approximately Rs. 1806)
Note on SI Installment Problems: In some exam conventions, SI is calculated on the original principal for the entire period, and then payments are simply subtracted from the total amount due. The interpretation above (where the remaining balance earns interest) is the more practical and commonly tested approach.
Problem 26: Ratio of SI
Q: The ratio of the SI on a certain sum for 3 years and 5 years at the same rate is what?
Solution:
SI for 3 years = (P x R x 3) / 100
SI for 5 years = (P x R x 5) / 100
Ratio = [(P.R.3)/100] / [(P.R.5)/100]
= 3 / 5
Answer: 3 : 5
Key insight: When P and R are constant, SI is directly
proportional to T, so the ratio of SI = ratio of time periods.
Problem 27: Annual Income from Two Investments
Q: A man invests Rs. 22,000 in two different schemes. Scheme A offers 8% per annum and Scheme B offers 11% per annum simple interest. If the annual income from both investments is Rs. 2090, find how much he invested in each scheme.
Solution:
Let investment in Scheme A = x
Then investment in Scheme B = 22000 - x
Annual income:
(x x 8) / 100 + ((22000 - x) x 11) / 100 = 2090
8x + 242000 - 11x = 209000
-3x = 209000 - 242000
-3x = -33000
x = 11000
Scheme A: Rs. 11,000
Scheme B: Rs. 11,000
Verification:
SI_A = (11000 x 8) / 100 = 880
SI_B = (11000 x 11) / 100 = 1210
Total = 880 + 1210 = 2090 [Correct]
Answer: Rs. 11,000 in each scheme
Summary of Problem Types Covered
+────────────────────────────────────────────────────────────+
| # | Type | Level |
+─────+─────────────────────────────────────────+───────────+
| 1 | Finding SI | Basic |
| 2 | Finding Amount | Basic |
| 3 | Finding Principal from SI | Basic |
| 4 | Finding Rate | Basic |
| 5 | Finding Time from Amount | Basic |
| 6 | Time in months | Basic |
| 7 | Doubling of principal | Basic |
| 8 | Principal from Amount | Basic |
| 9 | Different rates for different periods | Medium |
| 10 | Two unknowns linked | Medium |
| 11 | Borrowing and lending profit | Medium |
| 12 | Split investment (equal SI) | Medium |
| 13 | P and R from amounts at two times | Medium |
| 14 | Amount becomes fraction of itself | Medium |
| 15 | Multiple investments total interest | Medium |
| 16 | Split investment (total SI given) | Medium |
| 17 | Three-way split investment | Advanced |
| 18 | Difference of SI at two rates/times | Advanced |
| 19 | Mixed borrowing/lending different times | Advanced |
| 20 | Recurring deposits with SI | Advanced |
| 21 | SI and CI difference | Advanced |
| 22 | Fraction of principal as interest | Advanced |
| 23 | Ratio-based principal comparison | Advanced |
| 24 | Multi-step compound scenario | Advanced |
| 25 | Installment/debt repayment | Advanced |
| 26 | Ratio of SI at same rate | Medium |
| 27 | Annual income from two schemes | Medium |
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