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Difficult Math Problem #121 - Interest

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Difficult Math Problem #121 - Interest

by 800guy » Fri Apr 20, 2007 8:44 am
A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum?

A) 38%
B) 42%
C) 19%
D) 40%
E) 50%

from diff math problem doc, oa coming when ppl respond with explantions, thanks!
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Re: Difficult Math Problem #121 - Interest

by Shaki » Sat Apr 21, 2007 7:53 am
1000*(1+x)^3=1190
x=0.06
so, the answer is ~6%

I can't see it among the choices though :roll:

I think the choices you have shown are too big...
na etot raz poluchitsya!
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by rajeshvellanki » Sat Apr 21, 2007 8:10 am
my bet is on Answer B: here is the explanation for it:

A=P(1+(r/n))^(nt)

Where
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times the interest is compounded per year
t = number of years
A = amount after time t

==> p=1000, r=r(for now), n=1, t(3)=3 and t(6)=6

first lets find A(6)

==> A(6)=p(1+(r/1))^6 ==> A(6) = P(1+r)^6 ----- eq1

A(3)=p(1+r)^3

A(3)=principa+compund interest==> P(1+r)^3 = P + .19P

==> (1+r)^3 = 1.19

so (1+r)^6 = ((1+r)^3)((1+r)^3) = 1.19x1.19 = 1.4161 ==> bases are same so poweres can be added.

ok now subsittue the vale of (1+r)^6 from above in eq1

A(6)= 1000 (1.4161) = 1416.1

% increase from initial value = ((1416.1-1000)/ 1000)*100= 41.6% = ~42%

So answer is B
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Re: Difficult Math Problem #121 - Interest

by jayhawk2001 » Sat Apr 21, 2007 4:18 pm
800guy wrote:A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum?

A) 38%
B) 42%
C) 19%
D) 40%
E) 50%

from diff math problem doc, oa coming when ppl respond with explantions, thanks!
Using compound interest formula: A = P(1 + r/100)^n

1190 = 1000 (1 + r/100)^3
(1+r/100)^3 = 119/100

We are asked to find

1000 * (1+r/100)^6 = 1000 * 119/100 * 119/100 = 1416

416 is roughly 42% of the original capital 1000.
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by Cybermusings » Sun Apr 22, 2007 3:55 am
19% of the original sum = 1000*19/100 = 190
So total amount of CI = 190 after 3 years on 1000
So 1190 = 1000[1 + (r/100)]^3
1190/1000 = [1 + (r/100)]^3
We want to know 1000[1 + (r/100)]^6 which is {[1 + (r/100)]^3}^2
= Substituting [1 + (r/100)]^3 with 1190/1000
= 1000*1190/1000*1190/1000
=1416.1
Amount = 1416.1
Hence I for 6 years = 1416.1-1000=416.1
416.1/1000*100 = 41.61 = 42%
Whats the OA
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OA

by 800guy » Mon Apr 23, 2007 9:22 am
OA:

assume, interest = r
so after 3 years total money = 1000*(1+r)^3 = 1000*1.19
(1+r)^3 = 1.19
so after 6 years total money = 1000*(1+r)^6 = 1000*1.19^2 = 1000*1.42
so percentage of interest is 42%
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