3. 9% compounded quarterly is equal to The semi-annual effective rate is: A) 9.10%. B) 9.31%. C) 9.20%.

4. solution... Suppose you pay interest @ 9 % compounded quarterly, it means you are paying 9/4% every quarter and interest is added in the principal every time. Suppose you start with 100. After 1 st quarter, 100 will become : (100+2.25) = 102.25 after 2 nd quarter = (102.25 *1.0225) = 104.55 thus on semi-annual basis we are getting interest of 4.55 or 9.1% p.a. Semi annually compounding Answer

5. What is the annual effective rate, if there is continous compounding rate of 9% ? Formula of rate = e^(rate) - 1 =(2.71828)^(.09) -1 = rate is 9.417% answer

6. I invest Rs. 1000 every year from today onwards for 10 years. Rate of interest is 8%. Calculate future value of annuity ? Formula = amt (((1+rate)^n – 1) / rate ) =1000(((1.08^10)-1)/.08) =14486.56 but here I am starting my investment from today, so this first instalment, the interest will also be added : 14486.56 * 1.08 =15645.48 answer

7. What is log? Here we use shortcuts to exponentials for speedy calculations. logarithm of a number to a given base is the power or exponent to which the base must be raised in order to produce that number. For example, the logarithm of 1000 to base 10 is 3 it can be written as : log 10 1000 = 3 usually we use the base of 10, so we can write : log 1000 = 3

8. What is log 10 100000? = 5 answer

9. What is natural log ? Here the base is e, which is 2.71828 (not 10, which we had discussed earlier).

10. What is log 2 8? = 3 (here base is 2, and we know that 2^3 = 8)

11. What is log 1982 1? Answer = 0 We know that in any base b, the logarithm of 1 is 0.

12. What is log b b^x = ? Answer = x it is a rule that : log b b^x = x

13. What is : 5 = log 2 n ( what is n?) 32 (answer ) 2^5 = 32

14. What is log 10 .001 = -3

15. Find log of 1728 to the base 2*sqrt(3) ? Let us take factors of 1728 : 2*2*2*2*2*2*3*3*3 thus we know that 2*sqrt(3) ^ 6 = 1728 so answer is 6.

16. Change the base of log 200 to 2? = log 2 200 / log 2 10

17. Given: log 2 = .3010 log 3 = .4771 log 5 = .6990, what is log 6? We know that log 6 = log 2 * 3 we know that log 2*3 = .3010+.4771 =.7781

18. Given: log 2 = .3010, log 3 = .4771 Evaluate log 18. Log 2 * log 3*log 3 =.3010+.4771+.4771 =1.2552

19. Given: log 6 = .7781 Use the laws of logarithms to evaluate log 600 =log 6 * 100 = .7781 + 2 = 2.7781 answer