Maharashtra Board Class 12 Maths Part 2 Chapter 2 Insurance and Annuity 2.2 Solutions

Question 1. Find the accumulated (future) value of annuity of Rs. 800 for 3 year at interest rate 8% compounded annually.

Answer: Solution:
\( \therefore \) C = Rs. 800
\( \therefore \) n = 3 years
\( \therefore \) r = 8% p.a.
\( \therefore i = \frac{r}{100} = \frac{8}{100} = 0.08 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore A = \frac{800}{0.08} [(1+0.08)^3 -1] \)
\( \therefore A = 10,000[(1.08)^3 - 1] \)
\( \therefore A = 10,000[1.2597 - 1] \)
\( \therefore A = 10,000 \times 0.2597 \)
\( \therefore A = \text{Rs. } 2,597 \) In simple words: To find the future value of this annuity, we calculate the interest rate per period (i), then use the annuity accumulation formula with the given payment (C), number of periods (n), and interest rate. This gives us the total amount accumulated over 3 years.

🎯 Exam Tip: Always correctly identify C, n, r, and i from the problem statement. Pay attention to whether the interest is compounded annually, semi-annually, or quarterly, as this affects 'i' and 'n'.

Question 2. A person invested Rs. 5,000 every year in finance company that offered him interest compounded at 10% p.a., what is the amount accumulated after 4 years?

Answer: Solution:
\( \therefore \) C = Rs. 5,000
\( \therefore \) r = 10% p.a.
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore n = 4 \text{ years} \)
\( \therefore A = \frac{C}{i} [(1+i)^n - 1)] \)
\( = \frac{5,000}{0.1} [(1+0.1)^4 -1] \)
\( = 50,000[(1.1)^4 - 1] \)
\( = 50,000[1.4641 - 1] \)
\( = 50,000 \times 0.4641 \)
\( = \text{Rs. } 23,205 \) In simple words: We calculate the future value of an ordinary annuity by first determining the effective interest rate per period and the total number of periods. Then, we apply the annuity accumulation formula to find the total amount saved after 4 years.

🎯 Exam Tip: Ensure precise calculation of the factor `(1+i)^n - 1`. Errors in this part often lead to incorrect final answers. Double-check given values for `(1+i)^n`.

Question 3. Find the amount accumulated after 2 years if a sum of Rs. 24,000 is invested every six months at 12% p.a. compounded half yearly.

Answer: Solution:
\( \therefore \) C = Rs. 24,000
\( \therefore \) n = 2 years
But invested half yearly
\( \therefore \) n = 2 \( \times \) 2 = 4
\( \therefore \) r = 12% p.a. compounded half yearly
\( \therefore r = \frac{12}{2} = 6\% \)
\( \therefore i = \frac{r}{100} = \frac{6}{100} = 0.06 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( = \frac{24,000}{0.06} [(1+0.06)^4 -1] \)
\( = 4,00,000[(1.06)^4 - 1] \)
\( = 4,00,000[1.2625 - 1] \)
\( = 4,00,000 \times 0.2625 \)
\( = \text{Rs. } 1,05,000 \) In simple words: For half-yearly compounding, the number of periods (n) is doubled, and the annual interest rate (r) is halved to get the periodic interest rate (i). Using these adjusted values, we calculate the accumulated future value of the annuity.

🎯 Exam Tip: When compounding periods are not annual (e.g., half-yearly, quarterly), remember to adjust both 'n' (number of periods) and 'r' (rate per period) accordingly before calculating 'i'.

Question 4. Find the accumulated value after 1 year of an annuity immediate in which Rs. 10,000 are invested every quarter at 16% p.a. compounded quarterly.

Answer: Solution:
\( \therefore \) C = Rs. 10,000
\( \therefore \) n = 1 year
But invested every quarterly
\( \therefore \) n = 1 \( \times \) 4 = 4
\( \therefore \) r = 16% p.a. compounded quarterly
\( \therefore r = \frac{16}{4} = 4\% \)
\( \therefore i = \frac{r}{100} = \frac{4}{100} = 0.04 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( = \frac{10,000}{0.04} [(1+0.04)^4 -1] \)
\( = \frac{10,00,000}{4} [(1.04)^4 -1] \)
\( = 2,50,000 [1.1699 - 1] \)
\( = 2,50,000 \times 0.1699 \)
\( = \text{Rs. } 42,475 \) In simple words: For quarterly compounding, the number of periods (n) is multiplied by 4, and the annual interest rate (r) is divided by 4 to find the periodic interest rate (i). These adjusted values are then used in the annuity accumulation formula to determine the final amount.

🎯 Exam Tip: Accurately adapting 'n' and 'r' for different compounding frequencies (e.g., quarterly) is vital. A common mistake is using the annual rate directly without division.

Question 5. Find the present value of an annuity immediate of Rs. 36,000 p.a. for 3 years at 9% p.a. compounded annually.

Answer: Solution:
\( \therefore \) C = Rs. 36,000
\( \therefore \) n = 3 years
\( \therefore \) r = 9% p.a.
\( \therefore i = \frac{r}{100} = \frac{9}{100} = 0.09 \)
\( \therefore P = \frac{C}{i} [1-(1+i)^{-n}] \)
\( = \frac{36,000}{0.09} [1-(1+0.09)^{-3}] \)
\( = \frac{36,000 \times 100}{9} (1-0.7722) \)
\( = 4,00,000 \times 0.2278 \)
\( = \text{Rs. } 91,120 \) In simple words: To find the present value of an annuity, we determine the periodic interest rate and the total number of periods. Then, we apply the present value formula for an ordinary annuity to calculate the single sum that, if invested today, would yield the same future annuity payments.

🎯 Exam Tip: Differentiate between accumulated (future) value and present value formulas. For present value, the exponent for `(1+i)` is negative. Ensure your calculations for `(1+i)^-n` are correct.

Question 6. Find the present value of ordinary annuity of Rs. 63,000 p.a. for 4 years at 14% p.a. compounded annually.

Answer: Solution:
\( \therefore \) C = Rs. 63,000
\( \therefore \) n = 4 years
\( \therefore \) r = 14% p.a.
\( \therefore i = \frac{r}{100} = \frac{14}{100} = 0.14 \)
\( \therefore P = \frac{C}{i} [1-(1+i)^{-n}] \)
\( \therefore P = \frac{63,000}{0.14} [1-(1+0.14)^{-4}] \)
\( \therefore P = \frac{63,000 \times 100}{14} [1-(1.14)^{-4}] \)
\( = 4,50,000[1 - 0.5921] \)
\( = 4,50,000 \times 0.4079 \)
\( = \text{Rs. } 1,83,555 \) In simple words: The present value of this ordinary annuity is calculated by first establishing the periodic interest rate and the total number of payment periods. Then, using the specific formula for the present value of an ordinary annuity, we find the equivalent lump sum value today.

🎯 Exam Tip: Pay close attention to the exponent in the present value formula `(1+i)^-n`. A common error is omitting the negative sign or miscalculating the term `1 - (1+i)^-n`.

Question 7. A lady plans to save for her daughter's marriage. She wishes to accumulate a sum of Rs. 4,64,100 at the end of 4 years. What amount should she invest every year if she get an interest of 10%p.a. compounded annually?

Answer: Solution:
\( \therefore \) A = Rs. 4,64,100
\( \therefore \) n = 4 years
\( \therefore \) r = 10% p.a.
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore 4,64,100 = \frac{C}{0.1} [(1+0.1)^4 -1] \)
\( \therefore 46,410 = C[1.4641 - 1] \)
\( \therefore 46,410 = C \times 0.4641 \)
\( \therefore \frac{46,410}{0.4641} = C \)
\( \therefore \) C = Rs. 1,00,000 In simple words: To find the required annual investment, we rearrange the future value of an annuity formula. We use the desired accumulated amount, the number of years, and the interest rate to solve for the constant payment (C) needed each year.

🎯 Exam Tip: When solving for the annuity payment 'C', correctly isolate 'C' after substituting the given values. Algebraic accuracy is crucial to avoid calculation errors.

Question 8. A person wants to create a fund of Rs. 6,96,150 after 4 years at the time of his retirement. He decides to invest a fixed amount at the end of every year in a bank that offers him interest of 10% p.a. compounded annually. What amount should he invest every year?

Answer: Solution:
\( \therefore \) A = Rs. 6,96,150
\( \therefore \) n = 4 years
\( \therefore \) r = 10% p.a
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore 6,96,150 = \frac{C}{0.1} [(1+0.1)^4 -1] \)
\( \therefore 69,615 = C[1.4641 - 1] \)
\( \therefore 69,615 = C \times 0.4641 \)
\( \therefore \frac{69,615}{0.4641} = C \)
\( \therefore \) C = Rs. 1,50,000 In simple words: This problem requires us to determine the constant annual investment needed to reach a specific future sum. By utilizing the annuity accumulation formula and known values for the target amount, investment duration, and interest rate, we can calculate the required yearly payment.

🎯 Exam Tip: Clearly distinguish between problems asking for accumulated value versus those asking for the annuity payment. The same formula is used, but the unknown variable differs.

Question 9. Find the rate of interest compounded annually if an annuity immediate at Rs. 20,000 per year amounts to Rs. 2,60,000 in 3 years.

Answer: Solution:
\( \therefore \) C = Rs. 20,000
\( \therefore \) A = Rs. 2,60,000
\( \therefore \) n = 3 years
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore 2,60,000 = \frac{20,000}{i} [(1+i)^3 -1] \)
\( \therefore \frac{2,60,000i}{20,000} = 1 + 3i + 3i^2 + i^3 - 1 \)
\( \therefore 13i = 3i + 3 i^2 + i^3 \)
\( \therefore 13i = i(3 + 3i + i^2) \)
\( \therefore 13 = 3 + 3i + i^2 \)
\( \therefore i^2 + 3i + 3 - 13 = 0 \)
\( \therefore i^2 + 3i - 10 = 0 \)
\( \therefore (i + 5) (i - 2) = 0 \)
\( \therefore i + 5 = 0 \text{ or } i - 2 = 0 \)
\( \therefore i = -5 \text{ or } i = 2 \)
\( \therefore \) Rate of interest cannot be negative
\( \therefore \) i = 2 is accepted
\( \therefore \frac{r}{100} = 2 \)
\( \therefore \) r = 200% p.a. In simple words: This problem involves working backward from the accumulated value of an annuity to find the interest rate. By substituting the known values into the annuity formula and solving the resulting cubic equation, we determine the periodic interest rate.

🎯 Exam Tip: Solving for 'i' often involves algebraic manipulation and sometimes factoring quadratic or higher-order equations. Remember that interest rates cannot be negative in practical scenarios.

Question 10. Find the number of years for which an annuity of Rs. 500 is paid at the end of every years, if the accumulated amount works out to be Rs. 1,655 when interest is compounded annually at 10% p.a.

Answer: Solution:
\( \therefore \) C = 500
\( \therefore \) A = 1,655
\( \therefore \) r = 10% p.a.
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore 1,655 = \frac{500}{0.1} [(1+0.1)^n -1] \)
\( \therefore \frac{1,655 \times 0.1}{500} = (1.1)^n -1 \)
\( \therefore 0.331 = (1.1)^n -1 \)
\( \therefore 0.331 + 1 = (1.1)^n \)
\( \therefore 1.331 = (1.1)^n \)
\( \therefore (1.1)^3 = (1.1)^n \)
\( \therefore \) n = 3 years In simple words: To find the number of years (n), we use the annuity accumulation formula and solve for the exponent. After substituting the known values and simplifying, we equate the base numbers and determine 'n'.

🎯 Exam Tip: When solving for 'n', it usually involves logarithms. If the base values can be easily matched, as in this case `(1.1)^3 = (1.1)^n`, then direct comparison is quicker and often expected.

Question 11. Find the accumulated value of annuity due of Rs. 1,000 p.a. for 3 years at 10% p.a. compounded annually.

Answer: Solution:
\( \therefore \) C = Rs. 1,000
\( \therefore \) n = 3 years
\( \therefore \) r = 10% p.a.
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore A' = \frac{C(1+i)}{i} [(1+i)^n -1] \)
\( \therefore A' = \frac{1,000(1+0.1)}{0.1} [(1+0.1)^3 -1] \)
\( \therefore A' = 10,000 \times 1.1[(1.1)^3 - 1] \)
\( \therefore A' = 11,000 [1.331 - 1] \)
\( \therefore A' = 11,000 \times 0.331 \)
\( \therefore A' = \text{Rs. } 3,641 \) In simple words: This calculates the future value of an annuity due, where payments are made at the beginning of each period. The formula for an annuity due is a slight modification of the ordinary annuity formula, factoring in the extra interest earned on the first payment.

🎯 Exam Tip: Remember the distinction between an ordinary annuity (payments at the end) and an annuity due (payments at the beginning). The formula for an annuity due includes an additional `(1+i)` factor in the numerator.

Question 12. A person plans to put Rs. 400 at the beginning of each year for 2 years in a deposit that gives interest at 2% p.a. compounded annually. Find the amount that will be accumulated at the end of 2 years.

Answer: Solution:
\( \therefore \) C = Rs. 400
\( \therefore \) r = 2% p.a.
\( \therefore i = \frac{r}{100} = \frac{2}{100} = 0.02 \)
\( \therefore \) n = 2 years
\( \therefore A' = \frac{C(1+i)}{i} [(1+i)^n -1] \)
\( \therefore A' = \frac{400(1+0.02)}{0.02} [(1+0.02)^2 -1] \)
\( = \frac{40,000(1.02)}{2}[(1.02)^2 -1] \)
\( = 20,000 (1.02) (1.0404 - 1) \)
\( = 20,400 [0.0404] \)
\( = \text{Rs. } 824.16 \) In simple words: This problem involves finding the accumulated value of an annuity due, as payments are made at the beginning of each year. We apply the specific annuity due formula, adjusting for the beginning-of-period payments to determine the total accumulation.

🎯 Exam Tip: Always identify if payments are made at the beginning (annuity due) or end (ordinary annuity) of the period, as this dictates which formula to use for accumulation. The `(1+i)` factor is crucial for annuity due.

Question 13. Find the present value of an annuity due of Rs. 600 to be paid quarterly at 32% p.a. compounded quarterly.

Answer: Solution:
\( \therefore \) C = Rs. 600
\( \therefore \) n = 1 year
\( \therefore \) But invested every quarterly
\( \therefore \) n = 1 \( \times \) 4 = 4
\( \therefore \) r = 32% p.a. compounded quarterly
\( \therefore r = \frac{32}{4} = 8\% \)
\( \therefore i = \frac{r}{100} = \frac{8}{100} = 0.08 \)
\( \therefore P' = \frac{C(1+i)}{i} [1-(1+i)^{-n}] \)
\( \therefore P' = \frac{600(1+0.08)}{0.08} [1-(1+0.08)^{-4}] \)
\( = 7,500(1.08) [1 - 0.7350] \)
\( = 8,100 [0.2650] \)
\( = \text{Rs. } 2,146.5 \) In simple words: We calculate the present value of an annuity due with quarterly payments by adjusting the 'n' and 'r' for quarterly compounding. The formula for present value of annuity due includes an `(1+i)` multiplier to account for the payments being made at the start of each period.

🎯 Exam Tip: For annuity due calculations, especially with compounding periods other than annual, remember to adjust 'n' and 'i' correctly. The `(1+i)` factor in the present value of annuity due formula is critical.

Question 14. An annuity immediate is to be paid for some years at 12% p.a. The present value of the annuity is Rs. 10,000 and the accumulated value is Rs. 20,000. Find the amount of each annuity payment.

Answer: Solution:
\( \therefore \) r = 12% p.a.
\( \therefore i = \frac{r}{100} = \frac{12}{100} = 0.12 \)
\( \therefore \) P = Rs. 10,000
\( \therefore \) A = Rs. 20,000
\( \therefore \frac{1}{P} - \frac{1}{A} = \frac{i}{C} \)
\( \therefore \frac{1}{10,000} - \frac{1}{20,000} = \frac{0.12}{C} \)
\( \therefore \frac{2-1}{20,000} = \frac{0.12}{C} \)
\( \therefore \frac{1}{20,000} = \frac{0.12}{C} \)
\( \therefore \) C = 0.12 \( \times \) 20,000
\( \therefore \) C = Rs. 2,400 In simple words: This problem connects the present value, accumulated value, and the periodic interest rate to find the annuity payment. A key relationship between these values allows us to calculate the constant payment (C) directly.

🎯 Exam Tip: Understand the relationship `1/P - 1/A = i/C` as it efficiently links present value, accumulated value, and annuity payment, often simplifying calculations when multiple variables are given.

Question 15. For an annuity immediate paid for 3 years with interest compounded at 10% p.a. the present value is Rs. 24,000. What will be the accumulated value after 3 years?

Answer: Solution:
\( \therefore \) n = 3 years
\( \therefore \) P = Rs. 24,000
\( \therefore \) r = 10% p.a.
\( \therefore i = \frac{r}{100} = \frac{10}{100} = 0.1 \)
\( \therefore A = P(1 + i)^n \)
\( \therefore A = 24,000 [1 + 0.1]^3 \)
\( \therefore A = 24,000 \times (1.1)^3 \)
\( \therefore A = 24,000 \times 1.331 \)
\( \therefore A = 31,944 \) In simple words: To find the accumulated value from a given present value, we use the compound interest formula. The present value acts as the principal amount, which is compounded over the specified number of periods at the given interest rate.

🎯 Exam Tip: When both present value and accumulated value (or one and you need to find the other) are involved over the same period and interest rate, use the simple compounding formula: `A = P(1+i)^n`.

Question 16. A person sets up a sinking fund in order to have Rs. 1,00,000 after 10 years. What amount should be deposited bi-annually in the account that pays him 5% p.a. compounded semi-annually?

Answer: Solution:
\( \therefore \) A = Rs. 1,00,000
\( \therefore \) n = 10 years
But, invested half yearly
\( \therefore \) n = 10 \( \times \) 2 = 20
\( \therefore \) r = 5% p.a. compounded half yearly
\( \therefore r = \frac{5}{2} = 2.5\% \)
\( \therefore i = \frac{2.5}{100} = 0.025 \)
\( \therefore A = \frac{C}{i} [(1+i)^n -1] \)
\( \therefore 1,00,000 = \frac{C}{0.025} [(1 + 0.025)^{20} -1] \)
\( \therefore 1,00,000 \times 0.025 = C [(1.025)^{20} - 1] \)
\( \therefore 2,500 = C[1.675 - 1] \)
\( \therefore 2,500 = C \times 0.675 \)
\( \therefore \frac{2,500}{0.675} = C \)
\( \therefore \) C = Rs. 3,703.70 In simple words: To reach a future target amount (sinking fund), we need to determine the constant bi-annual deposits. We adjust the total number of periods and the interest rate for semi-annual compounding and then solve the annuity accumulation formula for the periodic payment (C).

🎯 Exam Tip: Sinking fund problems are applications of future value of an annuity. Accurately adjusting 'n' and 'i' for the compounding frequency (e.g., bi-annually means n times 2, r divided by 2) is paramount for correct calculation.