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02 Dec 2019, 01:15
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Difficulty:

45% (medium)

Question Stats:

37% (01:43) correct 63% (02:32) wrong based on 19 sessions

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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. $$2500(1.155)^{10}$$ Are You Up For the Challenge: 700 Level Questions _________________ Senior Manager Joined: 16 Feb 2015 Posts: 252 Location: United States Concentration: Finance, Operations Schools: INSEAD, ISB Re: Lana invested$2500 in an account at the beginning of the year 1990 an  [#permalink]

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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. $$2500(1.155)^{10}$$ Are You Up For the Challenge: 700 Level Questions 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 Give kudos, if you find my explanation good enough Manager Joined: 03 Nov 2019 Posts: 53 Re: Lana invested$2500 in an account at the beginning of the year 1990 an  [#permalink]

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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Since the amount was invested at the beginning of 1990 and withdrawn at the end of 2000,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
Re: Lana invested $2500 in an account at the beginning of the year 1990 an [#permalink] 02 Dec 2019, 11:49 Display posts from previous: Sort by # Lana invested$2500 in an account at the beginning of the year 1990 an

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