List of top Mathematics Questions on Applications of Compound Interest Formula

A person invested Rs.15,000 in a mutual fund; it became Rs.25,000. If CAGR is 8.88%, then the number of years \( n \) is: [Use \( -\log 1.667 = 0.2219 \); \( \log 1.089 = 0.0370 \)]
A TV set costing Rs.55,000 has a useful life of 8 years. If annual depreciation is Rs.5,000, then the scrap value by straight line method is:
For any square matrix A, A - A\(^T\) is always:
Shyam takes a loan of ₹5,00,000 with 8% annual interest rate for 10 years. The value of EMI under flat rate system is :
A vehicle whose cost is ₹7,00,000 will depreciate to scrap value of ₹1,50,000 in 5 years. Using linear method of depreciation, the book value of the vehicle at the end of the third year is :
The money needed to invest now, so as to get ₹7500 at the beginning of each month forever (starting from the current month) if the money is worth 9% per annuum compounded monthly is.
A machine costing ₹ one lakh depreciates at constant rate 10%. Estimated useful life of machine is 8 years.
Match List I with List II
List IList II
A.Total depreciation in 2nd and 3rd year isI.₹81,000
B.Value of machine after one year isII.₹17,100
C.Value of machine after 2 year isIII.₹43050
D.Scrap value of machine is :
given (1.1)3 - 2.144 & (0.9)3 - 0.4305 		IV.₹90,000
Choose the correct answer from the options given below :