Q1. What is compound interest?

Ans: Interest charged on principal plus accumulated interest (interest on interest).

Q2. Write the compound amount formula for annual compounding.

Ans: \(A = P\left(1+\dfrac{R}{100}\right)^N.\)

Q3. If \(P=1000\), \(R=10\%\), \(N=1\), what is \(A\)?

Ans: \(A=1000\left(1+\dfrac{10}{100}\right)=1000\times1.1=1100.\)

Q4. Define principal.

Ans: The initial sum of money invested or borrowed.

Q5. If \(A=1210\), \(P=1000\), what is CI after 1 year?

Ans: CI = \(A-P = 1210-1000=210.\)

Q6. What is amount of \$1 after one year at \(R\)%?

Ans: \(1+\dfrac{R}{100}.\)

Q7. For R = 12.5%, write the one-year multiplier.

Ans: \(1+\dfrac{12.5}{100}=1.125=\dfrac{9}{8}.\)

Q8. If CI for 1 year on \(P\) at rate \(R\) is \(I\), what is new principal for next year?

Ans: \(P+I = P\left(1+\dfrac{R}{100}\right)\).

Q9. If \(P=4000\), \(R=12.5\%\), \(N=3\), state A formula (no calc).

Ans: \(A=4000\left(\dfrac{9}{8}\right)^3.\)

Q10. What is the difference between simple and compound interest?

Ans: Simple interest computed on original principal only; compound interest is computed on principal + accumulated interest.

Q11. If rate is compounded half-yearly, what rate per period is used?

Ans: \(R/2\) per half-year.

Q12. If annual rate \(R\) is 10% compounded annually, what is effective 2-year multiplier?

Ans: \((1.10)^2=1.21.\)

Q13. If amount doubles in 3 years, what is multiplier per year?

Ans: Annual multiplier = \(2^{1/3}\).

Q14. CI is always _____ than or equal to SI for same P, R, N. Fill.

Ans: Greater than or equal to (equal only when \(N=1\) or \(R=0\)).

Q15. If monthly compounding, how many periods in \(N\) years?

Ans: \(12N\) periods.

Q16. Write CI formula rearranged to find principal \(P\).

Ans: \(P = \dfrac{A}{\left(1+\dfrac{R}{100}\right)^N}.\)

Q17. Principal \(P\) is 5000, rate 10%, N = 2. Find CI (quick).

Ans: \(A=5000(1.1)^2=5000\times1.21=6050\Rightarrow CI=1050.\)

Q18. If depreciation at 20% per year, what's value after 2 years on \(P\)?

Ans: \(A=P(0.8)^2=P\times0.64.\)

Q19. When interest compounded annually, CI for 0 years equals?

Ans: \(0\) (A = P).

Q20. If A = P at rate R for N years, what does that imply?

Ans: Either \(R=0\) or \(N=0\).

Q1. Compute CI on ₹2000 at 5% for 2 years (annual compounding).

Ans: \(A=2000(1.05)^2=2000\times1.1025=2205.\) CI = \(2205-2000=₹205.\)

Q2. Find amount on ₹5000 at 8% p.a. for 3 years.

Ans: \(A=5000(1.08)^3=5000\times1.259712=6298.56.\)

Q3. Principal becomes ₹6655 in 3 years at 10% p.a. Find P.

Ans: \(P=\dfrac{6655}{(1.1)^3}=\dfrac{6655}{1.331}=5000.\)

Q4. If ₹9000 yields CI ₹1890 at 10% per annum, how many years?

Ans: A = 9000+1890=10890. Solve \(10890=9000(1.1)^N\Rightarrow (1.1)^N=1.21\Rightarrow N=2.\)

Q5. Convert 12% annual, compounded half-yearly: rate per half-year?

Ans: \(12\%/2=6\%\) per half-year.

Q6. ₹4000 invested 3 years at \(12.5\%\) p.a. amount = ? (use fraction).

Ans: Rate = \(12.5\%=1/8\) → multiplier \(=9/8\). So \(A=4000\times(9/8)^3=4000\times729/512=4000\times1.423828125=5695.3125\) → ₹5695.31.

Q7. If monthly compound at 12% p.a., multiplier per month?

Ans: Monthly rate = \(12/12=1\%\). Multiplier per month \(=1.01.\)

Q8. A decreases by 20% yearly. Express multiplier.

Ans: Multiplier = \(1-0.20=0.8.\)

Q9. If P grows to \(P\times(1+r)^2\) in 2 years; express annual % if total multiplier = 1.44.

Ans: \((1+r)^2=1.44\Rightarrow 1+r=1.2\Rightarrow r=0.2=20\%.\)

Q10. Compare SI and CI on ₹10000 at 8% for 2 years.

Ans: SI = \(10000\times8\times2/100=1600\). CI: \(A=10000(1.08)^2=11664\Rightarrow CI=1664.\)

Q11. If amount after 2 years is 1.21P, what is R?

Ans: \((1+R/100)^2=1.21\Rightarrow 1+R/100=1.1\Rightarrow R=10\%.\)

Q12. How to compute amount for \(N\) years with half-yearly compounding?

Ans: \(A=P\left(1+\dfrac{R/2}{100}\right)^{2N}.\)

Q13. If ₹12500 at 12% p.a. for 3 years (annual compounding) — amount?

Ans: \(A=12500(1.12)^3=12500\times1.404928=17561.6.\)

Q14. Shalaka takes ₹8000 at 10.5% p.a. for 2 years; CI = ?

Ans: Multiplier \(=1+0.105=1.105\). \(A=8000(1.105)^2=8000\times1.221025=9768.2\). CI = \(9768.2-8000=1768.2\).

Q15. Effective yearly rate if 12% compounded semi-annual?

Ans: Effective = \((1+0.06)^2-1=1.1236-1=12.36\%.\)

Q16. If A = 291600 after 2 years, P=250000, find R per year.

Ans: \((1+R/100)^2 = 291600/250000 =1.1664\Rightarrow 1+R/100=\sqrt{1.1664}=1.08\Rightarrow R=8\%.\)

Q17. Depreciation: machine ₹250000, 10% p.a. depreciation, value after 2 years?

Ans: \(V=250000(0.9)^2=250000\times0.81=202500.\)

Q18. If A after N years = P(1+r)^N, solve for N given A/P and r.

Ans: \(N=\dfrac{\ln(A/P)}{\ln(1+r)}\) (use logs).

Q19. If 320 workers increase by 25% yearly, number after 2 years?

Ans: \(320\times(1.25)^2=320\times1.5625=500\) workers.

Q20. If ₹700 grows to ₹847 at 10% compound yearly, how many years?

Ans: Solve \((1.1)^N=847/700=1.21\Rightarrow N=2.\)

Q1. Find CI on ₹4000 for 3 years at 12½% p.a. (show steps).

Ans: \(R=12.5\%=1/8\) so multiplier \(=1+\dfrac{1}{8}=\dfrac{9}{8}.\) \[ A=4000\left(\dfrac{9}{8}\right)^3=4000\cdot\frac{729}{512}=4000\cdot1.423828125=5695.3125. \] CI = \(5695.3125-4000=1695.3125\Rightarrow ₹1695.31.\)

Q2. Show how to find principal if \(A\), \(R\), \(N\) given (example: A=6655, R=10%, N=3).

Ans: Use \(P=\dfrac{A}{(1+R/100)^N}=\dfrac{6655}{1.1^3}=\dfrac{6655}{1.331}=5000.\)

Q3. Find number of years if CI on ₹9000 is ₹1890 at 10%.

Ans: \(A=9000+1890=10890.\) Solve \((1.1)^N = 10890/9000 = 1.21\Rightarrow N=2.\)

Q4. A machine costs ₹250000; depreciation 10% p.a. Find depreciation after 2 years.

Ans: Value after 2 years \(=250000(0.9)^2=202500\). Depreciation = \(250000-202500=47500.\)

Q5. Monthly compounding: ₹10000 at 12% for 1 year — compute A.

Ans: monthly rate \(=1\%\). \(A=10000(1.01)^{12}=10000\times1.126825=11268.25.\)

Q6. If effective 2-year multiplier is \(1.1025\), find annual rate R (approx).

Ans: \((1+R/100)^2=1.1025\Rightarrow 1+R/100=\sqrt{1.1025}=1.05\Rightarrow R=5\%.\)

Q7. Show that CI > SI for P,R>0 and N>1 by a simple inequality.

Ans: SI = \(PRN/100\). CI = \(P[(1+R/100)^N -1]\). For \(N>1\) and \(R>0\), binomial (or expansion) shows \((1+R/100)^N > 1+NR/100\) so CI > SI.

Q8. Application: population 250000 increases 8% each year. Find population after 2 years.

Ans: \(A=250000(1.08)^2=250000\times1.1664=291600.\)

Q9. Rehana's scooter ₹60000 depreciates 20% p.a. Find price after 2 years.

Ans: \(A=60000(0.8)^2=60000\times0.64=38400.\)

Q10. Loan ₹12500 at 12% compounded annually for 3 years — how much to repay?

Ans: \(A=12500(1.12)^3=12500\times1.404928=17561.6\) → ₹17561.60.

Q11. Derive formula for amount when interest is compounded m times a year.

Ans: For m compounding periods per year, rate per period = \(R/m\), total periods = \(mN\): \[ A=P\left(1+\dfrac{R}{100m}\right)^{mN}. \]

Q12. Find principal if amount after 2 years at 18% p.a. is 13924.

Ans: \(P=\dfrac{13924}{(1.18)^2}=\dfrac{13924}{1.3924}=10000.\)

Q13. A sum amounts to 4036.80 after 2 years at 16% p.a. Find CI.

Ans: First find P: \(P=\dfrac{4036.80}{1.16^2}=\dfrac{4036.80}{1.3456}=3000.\) CI = \(4036.80-3000=1036.80.\)

Q14. A loan ₹15000 at 12% for 3 years — find A & CI.

Ans: \(A=15000(1.12)^3=15000\times1.404928=21073.92.\) CI = \(21073.92-15000=6073.92.\)

Q15. Principal amounts to ₹13924 in 2 years at 18% — compute P (shown earlier).

Ans: \(P=13924/(1.18^2)=10000.\)

Q16. If population 16000 increases to 17640 in 2 years, find annual r.

Ans: \((1+r)^2=17640/16000=1.1025\Rightarrow 1+r=1.05\Rightarrow r=5\%.\)

Q17. How many years for ₹700 → ₹847 at 10%? (shown earlier)

Ans: \((1.1)^N=847/700=1.21 \Rightarrow N=2.\)

Q18. Difference between SI and CI on ₹20000 at 8% for 2 years?

Ans: SI = \(20000\times0.08\times2=3200.\) CI = \(20000[(1.08)^2-1]=20000(0.1664)=3328.\) Difference = \(3328-3200=128.\)

Q19. If an amount doubles in 5 years with annual compounding, what is approximate R?

Ans: \((1+R)^5=2\Rightarrow 1+R=2^{1/5}\approx1.1487\Rightarrow R\approx14.87\%.\)

Q20. Show reduction in value (depreciation) using compound formula (example).

Ans: For depreciation rate \(d\)%, value after N years: \(V=P(1-d/100)^N\). Example: ₹60000 at 20% for 2 years → \(60000(0.8)^2=38400\).

Ex 1 No.1: P=2000, R=5%, N=2. Find A and CI.

Solution: \(A=2000(1.05)^2=2000\times1.1025=2205.\) CI = \(2205-2000=₹205.\)

Ex 1 No.2: P=5000, R=8%, N=3.

Solution: \(A=5000(1.08)^3=5000\times1.259712=6298.56.\) CI = \(6298.56-5000=₹1298.56.\)

Ex 1 No.3: P=4000, R=7.5%, N=2.

Solution: \(A=4000(1+0.075)^2=4000(1.075)^2=4000\times1.155625=4622.5.\) CI = \(4622.5-4000=₹622.50.\)

Ex 2: Sameerrao took loan ₹12500 at 12% for 3 years; amount to pay?

Solution: \(A=12500(1.12)^3=12500\times1.404928=17561.6.\) He should pay ₹17,561.60.

Ex 3: Shalaka borrowed ₹8000 at 10½% (10.5%) for 2 years — CI?

Solution: \(A=8000(1.105)^2=8000\times1.221025=9768.20.\) CI = \(9768.20-8000=₹1768.20.\)

PS 14.2 Q1: Workers 320, increase 25% every year. After 2 years?

Solution: \(N=2\Rightarrow 320(1.25)^2=320\times1.5625=500\) workers.

PS 14.2 Q2: Shepherd 200 sheep, increase 8% yearly for 3 years.

Solution: \(200(1.08)^3=200\times1.259712=251.9424\approx252\) sheep (rounded).

PS 14.2 Q3: 40000 trees, increase 5% for 3 years.

Solution: \(40000(1.05)^3=40000\times1.157625=46305\) (rounded).

PS 14.2 Q4: Machine ₹250000, depreciation 10% for 2 years — depreciation amount?

Solution: Value after 2 years \(=250000(0.9)^2=202500\). Depreciation \(=250000-202500=₹47500.\)

PS 14.2 Q5: Amount after 2 years is ₹4036.80 at 16% p.a. Find CI.

Solution: \(P=\dfrac{4036.80}{1.16^2}=\dfrac{4036.80}{1.3456}=3000.\) CI = \(4036.80-3000=₹1036.80.\)

PS 14.2 Q6: Loan ₹15000 at 12% for 3 years — amount?

Solution: \(A=15000(1.12)^3=15000\times1.404928=21073.92.\)

PS 14.2 Q7: Principal amounts to 13924 in 2 years at 18% — find principal.

Solution: \(P=13924/(1.18)^2=13924/1.3924=10000.\)

PS 14.2 Q8: Population 16000 → 17640 in 2 years. Find rate.

Solution: \((1+r)^2=17640/16000=1.1025\Rightarrow 1+r=1.05\Rightarrow r=5\%.\)

PS 14.2 Q9: How many years for ₹700 → ₹847 at 10%?

Solution: \((1.1)^N=847/700=1.21\Rightarrow N=2\).

PS 14.2 Q10: Difference between SI and CI on ₹20000 at 8% p.a. for 2 years?

Solution: SI = \(20000\times0.08\times2=3200.\) CI = \(20000[(1.08)^2-1]=20000\times0.1664=3328.\) Difference = ₹128.