Q.At what rate percent per annum will ₹5,000 yield ₹1,500 as simple interest in 5 years?
Concept understanding — Simple Interest
Simple interest (SI) is interest calculated only on the original principal for the whole period. Unlike compound interest, it never earns interest on interest, so the amount grows by the same fixed sum every year.
Simple interest grows linearly with time — a fixed rupee amount is added each year. Because SI∝N, the interest for any number of years is simply proportional to the interest for one year.
How it works
The interest for one year is a fixed fraction of the principal, and that same amount repeats every year. So over N years the total interest is just the one-year interest multiplied by N. The amount (maturity value) is the principal plus this accumulated interest.
Any problem gives four of the five quantities P,R,N,SI,A and asks for the fifth — substitute and solve.
SI=100P×N×RA=P+SI=P(1+100NR)
where P = principal, R = rate % per annum, N = time in years.
Key quantities
- Principal (P) — the sum originally lent or invested.
- Rate (R) — interest per ₹100 per year.
- Time (N) — in years (convert months: N=months/12).
- Amount (A) — principal plus total interest.
Quick example
Find the simple interest and amount on ₹8,000 at 6% p.a. for 4 years.
- Identify: P=8000, R=6, N=4.
- SI=1008000×6×4=100192000=1920.
- A=P+SI=8000+1920=9920.
Result: interest ₹1,920 and amount ₹9,920.
- Leaving time in months instead of converting to years.
- Using the compound amount formula when the question clearly says simple interest.
- Confusing the rate as a percent (6) with the rate as a decimal — the formula already divides by 100.
If a sum becomes n times itself under simple interest, the interest earned is (n−1)P. Since SI∝N, doubling (interest =P) in t years means quadrupling (interest =3P) takes 3t years.
Here SI, P and T are known and the rate is the unknown, so rearrange SI=100PRT for R.
R=P×T100×SI=5000×5100×1500.
Substituting gives R=25000150000=6, so the rate is 6% per annum.
The rate is 6% per annum.
Given: P=5000, SI=1500, T=5 years; find R.
Method 1 — rearrange the formula.
R=P×T100×SI=5000×5100×1500=25000150000=6.
So the rate is 6% per annum.
Method 2 — independent check. At 6% per annum, one year's interest on ₹5,000 is 6% of ₹5,000 = ₹300. Over 5 years that is 300×5=1500, which is exactly the interest given — confirming R=6%.
The required rate of interest is 6% per annum.
Placing P or T in the numerator instead of the denominator. Always start from the base formula and rearrange it, rather than trying to remember three separate rearranged versions.
- CA Foundation 2026Set jan-20261 markMCQQ.If a sum double itself in 8 years, then in how many years it will becomes four times, assuming that the simple interest is calculated. (A) 16 years (B) 12 years (C) 24 years (D) 20 years
›Reveal solutionSolution
SI to double = 8 years means rate ×8=1; to make it 4 times the interest must be 3P, requiring t=24 years.
Step 1 — condition for doubling
Doubling means the interest earned equals the principal P. With SI=P⋅r⋅t:
P=P⋅r⋅8 ⇒ 8r=1 ⇒ r=81.
Step 2 — condition for becoming four times
Four times the sum means total =4P, so interest =4P−P=3P:
3P=P⋅r⋅t ⇒ 3=rt=8t.
Step 3 — solve for t
t=3×8=24 years.
Under simple interest the multiple grows linearly, so 4× takes 3× the doubling time.
Watch outDo not just double the doubling time to 16 years (option A). Becoming four times means the interest must be 3P, not 2P — so it is three doubling-periods of interest, giving 24 years, not two.
TipUnder simple interest the interest earned is directly proportional to time. Since doubling adds 1P of interest in 8 years, reaching 4P (which needs 3P of interest) simply takes 3×8=24 years — no rate calculation required.
✓Final answer(C) 24 years
- CA Foundation 2026Set jan-20261 markMCQQ.Shiva invested an amount of ₹ 12,000 at the rate of 10% pa simple interest and another amount at the rate of 20% pa simple interest. The total interest earned at the end of the year on total amount invested became 14% pa. Find the total amount invested. (A) ₹ 18,000 (B) ₹ 20,000 (C) ₹ 24,000 (D) ₹ 26,000
›Reveal solutionSolution
1200+0.20y=0.14(12000+y)⇒y=8000; total =12000+8000=₹20,000.
Step 1 — set up
First part ₹12,000 at 10%, second part y at 20%. The combined interest equals 14% of the total invested:
0.10(12000)+0.20y=0.14(12000+y).
Step 2 — expand
1200+0.20y=1680+0.14y.
Step 3 — solve for y
0.06y=480 ⇒ y=8000.
Step 4 — total invested
The question asks for the total amount, so add both parts:
12000+8000=₹20,000.
Watch outy=8000 is only the second amount — option (A) traps you here. The question wants the total, so you must add back the first ₹12,000. Also note 14% is not the plain average of 10% and 20% (that would be 15%); it is a weighted average.
TipThink alligation: 14% lies between 10% and 20%, closer to 10%, so the larger share sits at 10% — consistent with the ₹12,000 being the bigger part.
✓Final answer(B) ₹20,000
- CA Foundation 2026Set may-20261 markMCQQ.The simple interest at the rate of p% per annum for p years will be ₹ p. Then, the principal is ________. (A) ₹ p (B) ₹ 100p (C) ₹ p2100 (D) ₹ p100
›Reveal solutionSolution
Put rate =p%, time =p years and SI=₹p into the simple-interest formula and solve for the principal; the p2 from rate×time cancels one p from the interest, giving P=p100.
Step 1 — Write the simple-interest formula
SI=100P×R×T
Step 2 — Substitute the given data
Rate R=p%, time T=p years, and the interest is SI=₹p:
p=100P×p×p=100Pp2
Step 3 — Solve for the principal P
Multiply both sides by 100 and divide by p2:
P=p2100p=p100
Why the other options are wrong: (A) ₹p ignores the formula entirely; (B) ₹100p forgets to divide by p2; (C) ₹p2100 cancels the p in the interest twice instead of once.
Watch outThe interest itself contains a factor of p (it equals ₹p). Cancel only ONE p against the p2 in the denominator — dropping both gives the wrong answer (C).
TipWhen rate and time are equal (both p), the SI formula carries a p2; keep the algebra symbolic and cancel at the end rather than plugging in a number.
✓Final answer(D) ₹ p100
- CA Foundation 2026Set may-20261 markMCQQ.Ravi deposits some amount in bank for 521 years at the simple interest rate of 7% per annum. Ravi receives ₹ 66,480 at the end of term. Compute the amount of initial deposit by Ravi in the Bank. (A) ₹ 48,000 (B) ₹ 50,000 (C) ₹ 45,000 (D) ₹ 51,000
›Reveal solutionSolution
Maturity value under simple interest =P(1+100RT). Here the multiplier is 1.385, so the deposit is 66480÷1.385=₹48,000.
Step 1 — Amount formula for simple interest
A=P(1+100R×T)
Step 2 — Compute the growth factor
R=7%, T=521=5.5 years:
1+1007×5.5=1+10038.5=1.385
Step 3 — Solve for the principal
P=1.385A=1.38566480=₹48,000
Check: 48000×1.385=66480 ✓.
Why the other options are wrong: (B) 50,000, (C) 45,000 and (D) 51,000 do not reproduce ₹ 66,480 when multiplied by 1.385.
Watch out₹ 66,480 is the MATURITY amount (principal + interest), not the interest. Divide by (1+100RT), not by 100RT.
TipConvert 521 to 5.5 before multiplying — a common slip is to use 5 years and lose the half-year interest.
✓Final answer(A) ₹ 48,000
- CA Foundation 2025Set jan-20251 markMCQQ.A certain amount at a rate of simple interest x, doubles in 5 years. At another rate of simple interest y, it becomes three times in 8 years. Then the difference between these two interest rates is (A) 5% (B) 8% (C) 3% (D) 4%
›Reveal solutionSolution
SI doubling → x=20%; SI tripling in 8 yr → y=25%; difference =5%.
Step 1 — Rate that doubles the money (simple interest)
Doubling means interest earned = principal, so P=P⋅100x⋅5.
x=5100=20%
Step 2 — Rate that triples the money
Tripling means interest earned =2P, so 2P=P⋅100y⋅8.
y=8200=25%
Step 3 — Difference of the rates
y−x=25%−20%=5%
Why the other options are wrong: (D) 4% would arise from wrongly using 100/8 for the tripling case; (B) 8% and (C) 3% do not match either computed rate.
Watch outFor SIMPLE interest, "doubles" means interest =P (multiplier −1=1) and "triples" means interest =2P. Use the multiplier minus one — not the multiplier itself.
TipFor SI, time to become m times at rate r satisfies r⋅t=100(m−1) — plug in m=2 and m=3 directly.
✓Final answer(A) 5%
- CA Foundation 2025Set jan-20251 markMCQQ.Mr. A invested ₹ 20,000 in a bank at the rate of 4.5% p.a. He received ₹ 27,500 after end of term. Find out the period ? (A) 4.50 Yrs (B) 8.34 Yrs (C) 6.50 Yrs (D) 8.10 Yrs
›Reveal solutionSolution
SI=27500−20000=7500; t=P×rSI×100=9007500≈8.34 years.
Step 1 — Find the interest earned
SI=A−P=27500−20000=7500
Step 2 — Apply the simple-interest formula for time
SI=100P⋅r⋅t⇒t=P×rSI×100
Step 3 — Substitute
t=20000×4.57500×100=90000750000=8.33≈8.34 years
Why the other options are wrong: (A) 4.50 Yrs simply echoes the rate; (C) 6.50 Yrs and (D) 8.10 Yrs come from arithmetic slips — only ₹7,500 of interest at ₹900/year gives ≈8.34 years.
Watch outUse the INTEREST (₹7,500), not the maturity amount (₹27,500), in the numerator — a frequent slip that inflates the time.
TipCompute annual interest first (P×r/100=₹900), then divide total interest by it — quicker than the full formula.
✓Final answer(B) 8.34 Yrs
- CA Foundation 2025Set may-20251 markMCQQ.A sum of ₹ 725 is lent in the beginning of a year at a certain rate of simple interest. After 8 months, a sum of ₹ 362.50 more is lent but at the rate twice the former. At the end of the year, ₹ 33.50 is earned as interest from both the loans. What was the original rate of interest ? (A) 3.6% (B) 4.54% (C) 3.46% (D) 4.12%
›Reveal solutionSolution
7.25r+37.25r=33.5⇒r≈3.46%.
Step 1 — Interest on the first loan (full year)
SI=100P⋅R⋅T
SI1=100725⋅r⋅1=7.25r
Step 2 — Interest on the second loan (4 months, double rate)
Lent after 8 months → time = 4 months = 124=31 year, rate = 2r:
SI2=100362.50⋅2r⋅31=37.25r
Step 3 — Add and solve
7.25r+37.25r=7.25r⋅34=329r=33.50
r=2933.50×3=29100.5≈3.46%
Why the other options are wrong: (A) 3.6% and (D) 4.12% come from mistiming the second loan (using 6 months or the full year instead of 4 months).
Watch outThe second loan runs only the remaining 4 months (months 8–12), not a full year — using T=1 there over-counts its interest and gives a wrong rate.
TipNotice 362.50=21×725, so SI2=7.25r×31; factoring 7.25r out keeps the algebra clean.
✓Final answer(C) 3.46%
- CA Foundation 2025Set sep-20251 markMCQQ.If ₹ 2,470 is obtained as an interest in 4 years and 4 months at the rate of 3% per annum simple interest rate in bank deposit, how much amount was deposited in ₹ ? (A) 17,000 (B) 18,000 (C) 19,000 (D) 20,000
›Reveal solutionSolution
SI =10013P=2470⇒P=₹19,000.
Step 1 — Convert the time to years
4 years 4 months =4+124=313 years.
Step 2 — Apply the simple-interest formula
SI=100P⋅r⋅t=100P×3×313=10013P
Step 3 — Solve for the principal
10013P=2,470⇒P=132,470×100=₹19,000.
Watch outConvert months to a fraction of a year (124=31), not a decimal like 4.4 — 4 months is one-third of a year, giving the clean 313.
TipThe 3 in the rate cancels the 3 in 313, leaving the tidy 10013P — easy to invert.
✓Final answer(C) 19,000
- CA Foundation 2024Set sep-20241 markMCQQ.The sum required to earn a monthly interest of ₹ 1,200 at 18% per annum simple interest is : (A) ₹ 50,000 (B) ₹ 60,000 (C) ₹ 80,000 (D) ₹ 66,000
›Reveal solutionSolution
Monthly SI =P×r×121; solve P×0.18×121=1200.
Step 1 — Write the monthly simple interest
Monthly SI=P×100r×121
Here r=18%, so monthly interest =P×10018×121=P×0.015.
Step 2 — Set equal to ₹1,200 and solve
0.015P=1200⇒P=0.0151200=80,000
So the required sum is ₹80,000.
Why the other options are wrong: ₹50,000 gives ₹750/month, ₹60,000 gives ₹900/month, ₹66,000 gives ₹990/month — none reach ₹1,200.
Watch outThe interest is MONTHLY, so divide the annual rate by 12 — using the full 18% annual gives 12× too small a principal.
TipMonthly interest of ₹1,200 = annual interest of ₹14,400; then P=14400/0.18=80,000 — same answer, sometimes faster.
✓Final answer(C) ₹ 80,000
- CA Foundation 2022Set dec-20221 markMCQQ.A farmer borrowed ₹ 3600 at the rate of 15% simple interest per Annum. At the end of 4 years, he cleared this account by paying ₹ 4000 and a cow. The cost of the cow is: (A) ₹ 1000 (B) ₹ 1200 (C) ₹ 1550 (D) ₹ 1760
›Reveal solutionSolution
Amount due = principal + SI = ₹5760; cow = 5760 − 4000 = ₹1760.
Step 1 — Compute the simple interest
SI=100P×R×T=1003600×15×4=₹2160
Step 2 — Total amount to be cleared
A=P+SI=3600+2160=₹5760
Step 3 — Value of the cow
Cow=A−cash paid=5760−4000=₹1760
Watch outDo not equate the cow only to the interest (₹2160 minus something) — the settlement covers the full amount owed (principal + interest), not just interest.
TipFor SI settlement problems, always build A = P + SI first, then subtract whatever cash was paid.
✓Final answer(D) ₹ 1760
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
- CA Foundation 2021Set dec-20211 markMCQQ.An amount is lent at R% simple interest for R years and the simple interest amount was one-fourth of the principal amount. Then R = ________ (A) 5 (B) 6 (C) 5½ (D) 6½
›Reveal solutionSolution
R²/100 = 1/4 ⇒ R = 5 — (A).
Step 1 — Write the simple interest
With rate R% for R years:
SI=100P⋅R⋅R=100PR2
Step 2 — Apply the one-fourth condition and solve
100PR2=4P⇒100R2=41⇒R2=25⇒R=5
Watch outBoth the rate and the time equal R here, so the R appears squared — treating time as a fixed number gives the wrong equation.
TipWhen rate and time are the same unknown, the SI formula produces R², turning the condition into a simple square-root step.
✓Final answer(A) 5
NoteThis answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
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