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02 Dec 2019, 01:15
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A
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Difficulty:
95%
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Question Stats:
48% (02:27) correct
52%
(02:22)
wrong
based on 122
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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
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
02 Dec 2019, 06:57
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Bunuel
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 General Discussion
Ansh777
Joined: 03 Nov 2019
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Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Posts: 53
02 Dec 2019, 11:49
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Since the amount was invested at the beginning of 1990 and withdrawn at the end of 2000,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
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
