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Compounded n times

by rahulvsd » Sat Nov 05, 2011 5:18 am
An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at $900 in the third year (after interest has compounded twice), and $1080 in the fourth year (after interest has compounded three times). What is the original amount?

A.540
B.$600
C.$625
D.$750
E.$800

OA: C
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by neelgandham » Sat Nov 05, 2011 5:24 am
Say X is the principal amount and R = Rate of Interest

Then
((1+(R/100))^2)*X = 900
((1+(R/100))^3)*X = 1080

=> (1+(R/100))= 1080/900

((1+(R/100))^2)*X = 900 => ((1080/900)^2)*X = 900 -> X = 625 !
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by vaibhavgupta » Sat Nov 05, 2011 8:34 am
rahulvsd wrote:An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at $900 in the third year (after interest has compounded twice), and $1080 in the fourth year (after interest has compounded three times). What is the original amount?

A.540
B.$600
C.$625
D.$750
E.$800

OA: C
Compound Interest is not much different than Simple Interest. It is just after every year we need to add interest in principal. Hence 180 is interest on 900. which means rate is 5%

So using that .. 625 will lead to interest of 125 at end of year 1
Year 2 : 750 is P and interest= 150.
hence , P=750+150 =900

[spoiler]Thus, C it is!!!![/spoiler] :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by mathewmithun » Sun May 12, 2013 11:10 pm
vaibhavgupta wrote:
rahulvsd wrote:An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at $900 in the third year (after interest has compounded twice), and $1080 in the fourth year (after interest has compounded three times). What is the original amount?

A.540
B.$600
C.$625
D.$750
E.$800

OA: C
Compound Interest is not much different than Simple Interest. It is just after every year we need to add interest in principal. Hence 180 is interest on 900. which means rate is 5%

So using that .. 625 will lead to interest of 125 at end of year 1
Year 2 : 750 is P and interest= 150.
hence , P=750+150 =900

[spoiler]Thus, C it is!!!![/spoiler] :)

Vaibhav, I am not sure what went wrong but this is how I carried out the calculation:

interest on 900 is 180, hence interest rate is 20%

P(120/100)^3=900 or P(6/5)^3=900 hence P=900*5/6*5/6*5/6=520 approx.

Please help me understand where I could have gone wrong.
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by Atekihcan » Mon May 13, 2013 2:59 am
mathewmithun wrote:P(120/100)^3=900
...
Please help me understand where I could have gone wrong.
read the problem again : "It is valued at $900 in the third year (after interest has compounded twice)

So, P*(120/100)² = 900
--> P*(6/5)² = 900
--> P = 900*25/36 = 25*25 = 625

Answer : C
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