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Bunuel
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$3,000 is the initial amount placed in an account and the interest 16 Nov 2022, 09:21
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55% (hard)

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68% (02:42) correct 32% (02:31) wrong based on 168 sessions

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$3,000 is the initial amount placed in an account and the interest compounds monthly, and the total value is $3,090 at the end of the first month. At the end of the second month, what fraction of the total interest of the two months is the interest of the second month?

A. 100/203
B. 1/2
C. 102/203
D. 103/203
E. 105/203
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D
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$3,000 is the initial amount placed in an account and the interest 25 Nov 2022, 04:22
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Bunuel
$3,000 is the initial amount placed in an account and the interest compounds monthly, and the total value is $3,090 at the end of the first month. At the end of the second month, what fraction of the total interest of the two months is the interest of the second month?

A. 100/203
B. 1/2
C. 102/203
D. 103/203
E. 105/203
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Interest rate \(= \frac{90}{3000}*100 =3\%\)

Principal for second month \(3090\)

Interest for the second month \(\frac{3}{100}*3090 =92.7\)

Total interest \(=90+92.7 = 182.7\)

Required fraction \(= \frac{92.7}{182.7}= \frac{309}{609}=\frac{103}{203}\)

Ans D

Hope it's clear.
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$3,000 is the initial amount placed in an account and the interest 11 Mar 2024, 08:18
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Please explain any easy method to solve this problem. Thanks
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NamDuong
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$3,000 is the initial amount placed in an account and the interest 01 Apr 2026, 11:35
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Gmat860sanskar
Please explain any easy method to solve this problem. Thanks
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I believe it's best to remember this formula: A=P(1+r/n)^(nt). Then you can apply it to all other problems related to exponential growth without thinking much.

  • [color=#000000][b][color=#000000][size=100]P : principal (initial amount)[/size][/color][/color][/b]
  • [color=#000000][b][color=#000000][size=100]r : annual interest rate (decimal)[/size][/color][/color][/b]
  • [color=#000000][b][color=#000000][size=100]n : the frequency of compounds per year [/size][/color][/color][/b]
  • t : total length of time the money stays in the account (in years)

Total amount received in the first month: 3000(1+r/12)^(12*1/12)=3090 -> r=0.36
Total amount received in the second month: 3000(1+0.36/12)^(12*2/12)=3182.7
Total interest earned in two months: 3182.7-3000=182.7
Interest earned in 2nd month: 182.7-90=92.7

-> 92.7/182.7=103/205

Btw do you have any tips to simplify such type of functions quickly? Please share~~
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Bunuel
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