My Math Forum How they combine two C I equations in single step?

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May 23rd, 2018, 01:57 AM   #11
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Quote:
 Originally Posted by weirddave In your original post, part of your maths is 1 + 10/100, another part is 1 + 20/100
Thank you

But I understood how you got this values: 30000,1.10, 1.20 etc.

But I didn't understand how this equation obtained in my solution: 30000 [1+10/100]^2 x [1+20/100] - 30000 ?

May 25th, 2018, 10:22 AM   #12
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I thought that I had explained that.

Quote:
 Originally Posted by Ganesh Ujwal Naveen lends 30000 to Ravi for 2 years at 20% p.a on compound interest on condition that the interest is compounded half yearly for the first year
20% annually is 10% half yearly. 3000 compounded half yearly for one year (2 half years) is a total (both initial amount and interest) of $\displaystyle 3000[1+ 10/100]^2$.
+
Quote:
 Originally Posted by Ganesh Ujwal and compounded annually for the second year What will be the total interest earned by Naveen at the end of two years?
For the second year, we multiply (1 + 20/100) by the amount at the beginning of the year. That amount is $\displaystyle 3000[1+ 10/100]^2$, so the total amount at the end of the second year is $\displaystyle 3000[1+ 10/100]^2(1+ 20/100)$.

But the question asked for "total interest earned" for those two year. That is the total amount at the end of the second year minus the original amount: $\displaystyle 3000[1+ 10/100]^2(1+ 20/100)- 3000$.

Last edited by skipjack; May 25th, 2018 at 02:03 PM.

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