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Here we provide you with Applied Maths Chapter 14, to help you gain a comprehensive understanding of the chapter and its concepts.
Applied Maths Chapter 14 Solutions
Compound Interest And Annuity
EXERCISE – 14.1
Q.1 At the beginning of each month, Rs.500 is deposited into a savings account that pays 12% per annum compounded monthly. Find the balance in the account at the end of 6 years (use log table).
Ans. i = 12/12 = 1 1% = 1% 0.01,
n = 6 x 12 = 72 72
A = R (1 + i) [(1 + i)^n – 1/i]
= 500 (1 + 0.01) [(1 + 0.01)^72 – 1/0.01]
= 500 x 1.01/0.01 [(1.01)^72 – 1]
= 50500 [(1.01)^72 – 1]
[let, x = (1.01)^72
taking logs on both sides
log x = 72 log (1.01)
log x = 72 x 0.0043
log x = 0.3096
x = antilog (0.3096)
x = 2.040]
= 50500 (2.040 – 1)
= 50500 x 1.040
= Rs.52520
Q.2 What is the present value of an annuity due of Rs.1000 payable quarterly for 6 years, if the money is worth 8% per annum compounded quarterly? (Use annuity table)
Ans. R = Rs.1000, n = 6 x 4 = 24,
i =8/4 = 2%
P = R (1 + a n-1/i)
= 1000 (1 + a 23/0.02)
= 1000 (1 + 18.2922)
= 1000 x 19.2922
= Rs.19292.20
Q.3 What is the present value of an annuity due of Rs.1000 payable quarterly for 6 years, if the money is worth 8% per annum compounded quarterly? (Use annuity table)
Ans. A = Rs.1200, n = 5 x 4 = 20,
i =8/4 = 2% 0.02
A = R (S n+1/I – 1)
1200 = R (S 21/0.02 – 1)
1200 = R (25.7833 – 1)
1200 = R x 24.7833
R = 122/24.7833
R = Rs.484.19702
= Rs.484.20
Q.4 Find the present value of a sequence of annual payments of Rs.1500 each, the first being made at the end of the 7th year and the last being made at the end of the 16th year, if money is worth 7% per annum compounded annually. (Use annuity table)
Ans. R = 1500, m = 6, n = 10,
i = 0.07
P = R[a m+n/i – a m/i]
P = 1500 [a 16/0.07 – a 6/0.07]
= 1500 (9.4467 – 4.7665)
= 1500 x 4.6802
= Rs.7020.30
Q.5 Find the amount of deferred annuity of Rs.3200 payable at the end of every six months for 10 years, the first being paid at the end of 2 years. The money is worth 15% per annum compounded semi-annually. (Use annuity table)
Ans. R = Rs.3200, n = 10 x 2 = 20,
i = 15/2 = 7 1/2%
A = R (S n/i)
= 3200 x S 20/7 1/2%
= 3200 x 43.3047
= Rs.138575.04
Q.6 Sheela purchased a washing machine paying Rs.5000 down and promising to pay Rs.200 every month for 3 years, the first being made at the end of the first year. Find the cash price of the washing machine, assuming that the money is worth 9% per annum compounded monthly. (Use annuity table)
Ans. Downpayment = Rs.5000, R = Rs.200, n = 36, m = 1 x 12 -1 = 11
m + n = 11 + 36 = 47, i = 9/12 = 3/4%
P = R[a m+n/i – a m/i]
P = 200 [a 47/3/4% – a 11/3/4%]
= 200 (39.4862 – 10.5207)
= 200 x 28.9655
= Rs.5793.100
Cost of Machine = 5793.10 + 500
= Rs.10793
FAQ’s related to Applied Maths Chapter 14 on compound interest and annuity:
Q.1 What is compound interest, and how does it differ from simple interest?
Ans. Compound interest is the interest calculated on the initial principal and also on the accumulated interest from previous periods. Unlike simple interest, which is calculated only on the principal amount, compound interest takes into account the interest earned over time, resulting in exponential growth of the investment.
Q.2 What are some common mistakes to avoid when solving compound interest and annuity problems?
Ans. Common mistakes include misinterpreting the problem statement, using incorrect formulas, neglecting to convert interest rates to decimal form, and misunderstanding the frequency of compounding or payment periods. It’s essential to carefully read the problem and double-check calculations for accuracy.
Q.3 How do I calculate compound interest for a given principal, interest rate, and time period?
Ans. The formula to calculate compound interest is A = P(1 + r/n)^(nt), where A is the future value, P is the principal amount, r is the annual interest rate (in decimal form), n is the number of times interest is compounded per year, and t is the time in years.
These are few Frequently Asked Questions relating to Applied Maths Chapter 14
In Applied Maths chapter 14, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 14 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 14, we delve deep into advanced mathematical concepts that are crucial for understanding.
