Qualification > Math

math doubts [NEW]

<< < (27/29) > >>

vanibharutham:

--- Quote from: Z.J.110 on December 10, 2009, 03:35:12 pm ---Find the sum of the infinite series 1/103+1/106+1/109+... expressing ur answer as a fraction in its lowest terms.

Hence express the infinite recurring decimal 0.108108108...as a fraction in its lowest terms.

--- End quote ---

We find that r = 0.001, and a = 0.001

S to infinity = a / 1 - r

=====> 0.001 / 1 - 0.001
=> 1 / 999

Since

1/ 999 = 0.001001001001001 etc....
2/999 must equal 0.002002002002 etc

Hence 108/999 must equal 0.108108108108108

In its simplest form 108/999 = 4/37

:) 

zara:
thank you mate!

zara:
Charles borrow $6000 for a new car. Compound interest is charged on the loan at a rate of 2% per month. Charles has to pay off the loan with 24 equal monthly payments. Calculate the value of each monthly payment.

@d!_†oX!©:
I am not sure if it's right or not...i just gave it a try...please check the answer if you have one and reconfirm....
A = P ( 1 + r/100 )n
6000=P(1+2/100)24
thus, P= $3730.328928
and because the monthly payments are equal, thus each monthly payment= 3730.328928/24 = $155.43

 :) :)

zara:
if anyone uses Pure mathematics 1 by Hugh Neill & Douglas Quadling can u plz turn over to page 253 Ex.16D Q4.
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?

Navigation

[0] Message Index

[#] Next page

[*] Previous page