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Simple & Compound Interest

Quantitative Aptitude / DI · Arithmetic Applications

Useful for: SSC CGL, Railway RRB, IBPS Banking, OSSC RHT — Regular High School Teacher, UPSC NDA/NA, OSEPA Primary Teacher / JT Category-I, OSEPA Upper Primary — Science, OSSC Mathematics

and 6 more exams

OSSSC Govt Physical Education Teacher (PET), OSSSC Govt TGT Arts, OSSSC Govt TGT Science — CBZ, OSSSC Govt TGT Science — PCM, SSB Lecturer (Degree Colleges), SSB TGT PCM

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Overview

Simple interest grows linearly on the principal; compound interest grows on principal plus earlier interest — the difference grows with time.

Key points

SI = P × R × T / 100; amount = P + SI.
CI amount = P(1 + R/100)^T (annual compounding).
Half-yearly compounding: rate R/2 and time 2T; quarterly: R/4 and 4T.
Difference between CI and SI for 2 years = P(R/100)².
For 3 years: CI − SI = P(R/100)²(3 + R/100).

Explanation

A sum doubling in T years at SI: R = 100/T.
A sum becoming n times in T years at CI becomes n² times in 2T years.
Successive-% shortcut: 10% CI for 2 years = 10 + 10 + 1 = 21%.
Instalment questions: equate present values of instalments to the principal.

Step by step

Identify SI or CI and the compounding period.
Apply the formula.
Subtract the principal when interest (not amount) is asked.

Examples

P = ₹5,000, R = 10%, T = 2 years: SI = ₹1,000; CI = 5,000 × 1.21 − 5,000 = ₹1,050; difference = ₹50 = 5,000 × (0.1)².
A sum doubles in 8 years at SI → R = 100/8 = 12.5%.
At CI a sum becomes 2 times in 5 years → 4 times in 10 years, 8 times in 15 years.

Traps and advanced points

Trap: Using half-yearly time without halving the rate.
Trap: Applying the 2-year CI−SI formula to 3 years.
Trap: Adding interest to the principal in a simple-interest question.

Quick revision

CI − SI (2 yrs) = P(R/100)².
Double in T yrs at SI → R = 100/T.
10% CI, 2 yrs = 21%.

Practice MCQs

Find the simple interest on ₹7400 at 12% per annum for 3 years.
1. 2397.6
2. 3196.8
3. 2930.4
4. 2664
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Answer: (d) 2664. SI = P×R×T/100.
Find the simple interest on ₹8500 at 5% per annum for 2 years.
1. 935
2. 765
3. 1020
4. 850
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Answer: (d) 850. SI = P×R×T/100.
Find the simple interest on ₹9600 at 6% per annum for 4 years.
1. 2764.8
2. 2304
3. 2073.6
4. 2534.4
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Answer: (b) 2304. SI = P×R×T/100.
Find the simple interest on ₹10200 at 8% per annum for 5 years.
1. 4488
2. 4896
3. 3672
4. 4080
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Answer: (d) 4080. SI = P×R×T/100.
Find the simple interest on ₹11300 at 10% per annum for 3 years.
1. 3729
2. 3390
3. 3051
4. 4068
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Answer: (b) 3390. SI = P×R×T/100.
Find simple interest on ₹8000 at 5% per annum for 2 years.
1. 800
2. 640
3. 8800
4. 960
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Answer: (a) 800. SI = PRT/100 = 8000×5×2/100 = ₹800.
Find the simple interest on ₹11400 at 12% per annum for 3 years.
1. 4104
2. 4514.4
3. 3693.6
4. 4924.8
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Answer: (a) 4104. SI = P×R×T/100.
Find the simple interest on ₹12500 at 5% per annum for 2 years.
1. 1250
2. 1375
3. 1500
4. 1125
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Answer: (a) 1250. SI = P×R×T/100.
Find the simple interest on ₹3600 at 6% per annum for 4 years.
1. 1036.8
2. 950.4
3. 777.6
4. 864
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Answer: (d) 864. SI = P×R×T/100.
Find the simple interest on ₹4700 at 8% per annum for 2 years.
1. 902.4
2. 676.8
3. 827.2
4. 752
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Answer: (d) 752. SI = P×R×T/100.
Find the simple interest on ₹5800 at 10% per annum for 4 years.
1. 2552
2. 2088
3. 2784
4. 2320
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Answer: (d) 2320. SI = P×R×T/100.
Find the simple interest on ₹4500 at 5% per annum for 2 years.
1. 400
2. 550
3. 450
4. 500
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Answer: (c) 450. SI = P×R×T/100.
Find the simple interest on ₹4200 at 8% per annum for 5 years.
1. 1512
2. 1680
3. 1848
4. 2016
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Answer: (b) 1680. SI = P×R×T/100.
Find the simple interest on ₹11700 at 8% per annum for 4 years.
1. 4118.4
2. 3744
3. 3369.6
4. 4492.8
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Answer: (b) 3744. SI = P×R×T/100.
Find the simple interest on ₹5900 at 12% per annum for 4 years.
1. 3115.2
2. 3398.4
3. 2832
4. 2548.8
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Answer: (c) 2832. SI = P×R×T/100.