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Class 12 Applied Maths Chapter 12 Solutions
Perpetuity, Sinking Funds and EMI
EXERCISE- 12.2
Q.1 A company establishes a sinking fund to provide for payment of ₹250000 debt, maturing in Contribution to the fund are to be made at the end of each year. Find the amount of 4 years. each annual deposit if the interest is 18% per annum.
Ans. A = Rs.250000, n = 4, R = ?, i = 18% = 0.18
A = R [(1 + i)^n – 1]/i
250000 = R [(1 + 0.18)^4 – 1]/0.18
250000 x 0.18 = R [(1.18)^4 – 1]
45000.00 = R (1.939 – 1)
45000 = R x 0.939
R = 45000/0.939
= (45000 x 100)/939
= 47923.53
Q.2 A person has set up a sinking fund so that he can accumulate ₹100000 in 10 years for his children’s higher education. How much amount should he deposit every six months if interest is 5% per annum compounded semi-annually?
Ans. A = Rs.100000, n = 10 = 10×2 = 20, R = ?, i = 5/2 = 2.5% = 0.025
A = R [(1 + i)^n – 1]/i
100000 = R [(1 + 0.025)^20 – 1]/0.025
2500 = R (1.637 – 1)
2500 = R x 0.637
R = 2500/0.637
= (2500 x 100)/637
= 3924.64
Q.3 A firm anticipates a capital expenditure of Rs.50000 for a new equipment in 5 years. How much should be deposited quarterly in a sinking fund carrying 12% per annum compounded quarterly to provide for the purchase?
Ans. A = Rs.50000, n = 5 x 4 = 20, R = ?, i = 12/4 = 3% = 0.03
A = R [(1 + i)^n – 1]/i
50000 = R [(1 + 0.03)^20 – 1]/0.03
50000 x 0.03 = R [(1.03)^20 – 1]
1500 = R (1.803 – 1)
1500 = R x 0.803
R = 1500/0.803
= (1500 x 100)/803
= 1867.99
Q.4 A machine costs a company Rs.525000 and its effective life is estimated to be 20 years. A sinking fund is created for replacing the machine at the end of its lifetime when its scrap realizes a sum of Rs25000 only. Calculate what amount should be provided every year out of profits for the sinking fund if it accumulates an interest of 5% per annum.
Ans. A = 525000 – 25000 = 500000
n = 20, R = ?, i = 5% = 0.05
A = R [(1 + i)^n – 1]/i
500000 = R [(1 + 0.05)^20 – 1]/0.05
500000 x 0.05 = R [(1.05)^20 – 1]
25000 = R (2.6533 – 1)
25000 = R x 1.6533
R = 25000/1.6533
= Rs.15121.27
Q.5 A machine being used by a company is estimated to have a life of 15 years. At that time a new machine would cost Rs.75000 and the scrap of the old machine would yield Rs.9600 only. A sinking fund is created for replacing the machine at the end of its life. What sum should be retained by the company at the end of every year to accumulate at 6% per annum?
Ans. A = 75000 – 9600 = Rs.65400
n = 15, R = ?, i = 6% = 0.06
A = R [(1 + i)^n – 1]/i
65400 = R [(1 + 0.06)^15 – 1]/0.06
65400 x 0.06 = R [(1.06)^15 – 1]
3924 = R (2.396 – 1)
3924 = R x 1.396
R = 3924/1.396
= Rs.2810.89
Q.6 A company sets aside a sum of Rs.1000 at the end of each year in a sinking fund so that at the 10 that the cost of machinery remains constant at the end of 10 years and that money earns 5% per annum compound interest, find the cost of the machinery.
Ans. A = Rs.1000, n = 10, R = ?, i = 5% = 0.05
A = R [(1 + i)^n – 1]/i
= 1000 [(1 + 0.05)^10 – 1]/0.05
= (1000 x 100)/5 [1.629 – 1]
= 20000 x 0.629
= Rs.12580
Q.7 10 years ago, Mr Mehra set-up a sinking fund to save for his daughter’s higher studies. At the end of 10 years, he received an amount of Rs.10,21,760. What amount did he put in the sinking fund at the end of every 6 months for the tenure, which paid him 5% p.a. compounded semi-annually? [Use (1.025)20 = 1.6386]
Ans. A = Rs.1021760, n = 10 x 2 = 20, R = ?, i = 5/2 = 2.5% = 0.025
A = R [(1 + i)^n – 1]/i
1021760 = R [(1 + 0.025)^20 – 1]/0.025
1021760 x 0.025 = R [(1.025)^20 – 1]
25544.000 = R (1.6386 – 1)
25544 = R x 0.6386
R = 25544/0.6386
= Rs.40000
FAQ’s related to Class 12 Applied Maths Chapter 12 on Perpetuity, Sinking Funds and EMI:
Q.1 What is a perpetuity?
Ans. A perpetuity is a type of annuity that pays a fixed amount of money indefinitely. Unlike typical annuities that end after a set period, perpetuities have no end date.
Q.2 What is a sinking fund?
Ans. A sinking fund is a fund established to save money over time for a future expense or to pay off a debt. Regular contributions are made to this fund, which earns interest, to accumulate enough to meet the intended obligation.
Q.3 What are the practical applications of sinking funds?
Ans. Sinking funds are commonly used by companies to pay off debt, like bonds, to replace assets, or to fund large future expenses. Individuals may use sinking funds to save for goals like buying a car or paying for a weddi
Q.4 What is EMI?
Ans. EMI stands for Equated Monthly Installment. It is a fixed payment amount made by a borrower to a lender at a specified date each month. EMIs are used to pay off both interest and principal over a specific period.
Q.5 What is an amortization schedule?
Ans. An amortization schedule is a table detailing each periodic payment on a loan (typically a mortgage), showing the amount of principal and interest that comprise each payment until the loan is paid off at the end of its term.
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 12
In Class 12 Applied Maths chapter 12, 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, Class 12 Applied Maths Chapter 12 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 12, we delve deep into advanced mathematical concepts that are crucial for understanding.
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