Class 12 CBSE Applied Maths Perpetuity, Sinking Funds, Bonds & EMI Exercise 12.4
Class 12 CBSE Applied Maths aims to develop an understanding of basic
mathematical and statistical tools and their
applications in the field of commerce (business/ finance/economics) and social
sciences. Topics covered in Class 12th Applied Maths includes : Numbers, Quantification and
Numerical Applications, Algebra, Calculus, Probability Distributions , Inferential Statistics, Index
Numbers and Time-based data , Financial Mathematics , Linear Programming.
Please Select
Q1.
Without expanding, find the values of :
Solution :
Q2.
Without expanding, find the values of :
Solution :
Q3.
Without expanding, find the values of :
Solution :
Q4. Mrs. Dubey borrowed Rs 500000 from a bank to purchase a car and decided to repay by monthly installments in 5 years. The bank charges interest at 8% p.a. compounded monthly. Calculate the EMI. (Given (1.0067)60 = 1.4928)
Solution :
Q5. Utkarsh purchased a laptop worth ₹80000. He paid 20000 as cash down and balance in equal monthly installments in 2 years. If bank charges 9% p.a. compounded monthly. Calculate the EMI. (Given (1.0075)24 = 1.1964)
Solution :
Q6. Mr. Bharti wishes to purchase a flat for ₹6000000 with a down payment of ₹1000000 and balance in equal monthly payments for 20 years. If bank charges 7.5% p.a. compounded monthly, calculate the EMI. (Given (1.00625)240 = 4.4608)
Solution :
Q7. Mr. Dharmendra Patel purchased a motorcycle of Rs 150000 with Rs 25000 down payment. He wishes to repay balance in equal monthly payments in 4 years. If bank charges 9% p.a. compounded monthly, calculate the EMI. (Given (1.0075)48 = 1.4314)
Solution :
Q8.
A person amortizes a loan of ₹150000 for a new home by obtaining a 10 year mortgage at the
rate of 12% compounded monthly. Find
(i) EMI.
(ii) Total interest paid (Given a120/0.01 = 69.6891)
Solution :
Q9.
A loan of Rs 400000 at the interest rate of 6.75% p.a. compounded monthly is to be amortized
by equal payments at the end of each month for 10 years. Find
(i) the size of each monthly payment.
(ii) the principal outstanding at the beginning of 61st month.
(iii) the interest paid in 61st payment.
(iv) the principal contained in 61st payment.
(v) total interest paid.
(Given (1.005625)12
120 1.9603, (1.005625)60 = 1.4001)
Solution :
Q10.
A person amortizes a loan of Rs 1500000 for renovation of his house by 8 years mortgage at the
rate of 12% p.a. compounded monthly. Find
(i) the equated monthly installment.
(ii) the principal outstanding at the beginning of 40th month.
(iii) the interest paid in 40th payment.
(iv) the principal contained in 40th payment.
(v) total interest paid.
(Given (1.01)96 = 2.5993, (1.01)57 = 1.7633)
Solution :
