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Lana invested $2500 in an account at the beginning of the year 1990 an

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Lana invested $2500 in an account at the beginning of the year 1990 an  [#permalink]

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New post 02 Dec 2019, 01:15
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Lana invested $2500 in an account at the beginning of the year 1990 and withdrew all of the money in the account at the end of the year 2000. In even-numbered years, the value of the account increased by 21%. In odd-numbered years, the value of the account increased by 10%. Which of the following expressions is equal to the value of the account in dollars at the end of the year 2000?


A. \(2500(1.1)^{17}\)

B. \(2500(1.155)^{11}\)

C. \(2500(1.1)^6(1.21)^5\)

D. \(2500(1.31)^{11}\)

E. \(2500(1.155)^{10}\)


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Re: Lana invested $2500 in an account at the beginning of the year 1990 an  [#permalink]

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New post 02 Dec 2019, 06:57
Bunuel wrote:
Lana invested $2500 in an account at the beginning of the year 1990 and withdrew all of the money in the account at the end of the year 2000. In even-numbered years, the value of the account increased by 21%. In odd-numbered years, the value of the account increased by 10%. Which of the following expressions is equal to the value of the account in dollars at the end of the year 2000?


A. \(2500(1.1)^{17}\)

B. \(2500(1.155)^{11}\)

C. \(2500(1.1)^6(1.21)^5\)

D. \(2500(1.31)^{11}\)

E. \(2500(1.155)^{10}\)


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Explanation:
=2500 x (1.1)^5 x (1.21)^6
=2500 x (1.1)^5 x (1.1)^12
=2500 x (1.1)^ {5+12}
=2500 x (1.1)^17

IMO-A
Easy Question!!!!

:please Please Give kudos, if you find my explanation good enough :please
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Re: Lana invested $2500 in an account at the beginning of the year 1990 an  [#permalink]

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New post 02 Dec 2019, 11:49
Since the amount was invested at the beginning of 1990 and withdrawn at the end of 2000,
we have a total of (2000-1990)+1 years i.e. 11 years (1990 included), out of which 5 are odd and 6 are even.
Now for every odd amount increases (1+10/100) = 1.1 times
Now for every even amount increases (1+21/100) = 1.21 or 1.1^2 times

Thus Total amount at end of 2000:
=2500*(1.1)^5 *(1.21)^6
=2500*(1.1)^5 *(1.1^2)^6
=2500(1.1)^17

Answer: A
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Re: Lana invested $2500 in an account at the beginning of the year 1990 an   [#permalink] 02 Dec 2019, 11:49
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Lana invested $2500 in an account at the beginning of the year 1990 an

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