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Author | Topic: Mathematics problem |
---|---|

happy123@2014-02-02 06:31:05 |
Hi Do you know how to solve this problem: 1.0605 = (1+(0.0589/m))power m m = ? Thank you. |

aLEx78@2014-02-02 17:37:08 |
m=12 means a monthly compounding frequency rate |

happy123@2014-02-03 12:47:58 |
Hi Thank you for your reply. Is it possible for you to show me the working instead of just the answer. Thank you. |

happy123@2014-02-03 13:05:41 |
Another mathematics problem is: 100,000= 35,000 (1 + (5%/12))power 12N N=? Thank you. If there is an alpha in the power, I do not know how to turn the power into normal equation to get to the answer. |

muna@2014-04-30 18:37:12 |
Hi Mr Happy! In doing both level I and level II, I have not come across such a problem. Your first question puzzled me for a while until I realised, as alex poited out, that 'm' refers to the number of months. (just means the rate calculated is compounded monthly) Theoretically, the equation can be reduced to a quadratic - i have done it but takes time to explain. Rest assured you won't be required to do this, but I suppose you misinterpreted the question. Your last question leads me to beleive the following simple algrbra will cover quite a few of your problems... If K = Y^p, where ^ = power of. then Y = K^(1/p) eg. 9 = 3^2, 3 = 9^(1/2). Cheers mate. Muna. 4 = 2^2 |