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A financial adviser was showing a client the value,
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11 Dec 2023, 18:33
11 Dec 2023, 18:33
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A financial adviser was showing a client the value, rounded to the nearest cent, of an initial investment of $100.00 after 5, 10, 15, 20, 25, and 30 years, under the assumption that the value increases by r% per year for some positive constant r. The adviser correctly gave $130.01 for the value after 5 years. However, the adviser inadvertently made two transcription errors, and as a result, two of the remaining values shown to the client were incorrect.
Assuming the options provided are the remaining five amounts shown to the client, select for First error the lesser of the two incorrectly transcribed values and select for Second error the greater of the two incorrectly transcribed values. Make only two selections, one in each column.
Assuming the options provided are the remaining five amounts shown to the client, select for First error the lesser of the two incorrectly transcribed values and select for Second error the greater of the two incorrectly transcribed values. Make only two selections, one in each column.
| First error | Second error | |
| $160.02 | ||
| $219.75 | ||
| $285.70 | ||
| $317.43 | ||
| $482.90 |
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Official Answer
First error: $160.02
Second error: $317.43
Re: A financial adviser was showing a client the value,
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12 Dec 2023, 06:13
12 Dec 2023, 06:13
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The first error is $160.02 and the second error is $317.43.
Since the amount is $130.01 after 5 years, r = 30.01%
--> After 10 years = 130.01 * 1.3001 = $169.03 (incorrect)
--> After 15 years = 169.03 * 1.3001 = $219.75
--> After 20 years = 219.75 * 1.3001 = $285.70
--> After 25 years = 285.70 * 1.3001 = $371.44 (incorrect)
--> After 30 years = 371.44 * 1.3001 = $482.90
Since the amount is $130.01 after 5 years, r = 30.01%
--> After 10 years = 130.01 * 1.3001 = $169.03 (incorrect)
--> After 15 years = 169.03 * 1.3001 = $219.75
--> After 20 years = 219.75 * 1.3001 = $285.70
--> After 25 years = 285.70 * 1.3001 = $371.44 (incorrect)
--> After 30 years = 371.44 * 1.3001 = $482.90
Re: A financial adviser was showing a client the value,
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24 Dec 2023, 22:25
24 Dec 2023, 22:25
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Expert Reply
apoorvmht wrote:
A financial adviser was showing a client the value, rounded to the nearest cent, of an initial investment of $100.00 after 5, 10, 15, 20, 25, and 30 years, under the assumption that the value increases by r% per year for some positive constant r. The adviser correctly gave $130.01 for the value after 5 years. However, the adviser inadvertently made two transcription errors, and as a result, two of the remaining values shown to the client were incorrect.
Assuming the options provided are the remaining five amounts shown to the client, select for First error the lesser of the two incorrectly transcribed values and select for Second error the greater of the two incorrectly transcribed values. Make only two selections, one in each column.
Assuming the options provided are the remaining five amounts shown to the client, select for First error the lesser of the two incorrectly transcribed values and select for Second error the greater of the two incorrectly transcribed values. Make only two selections, one in each column.
So, we are looking at compound interest at r% per year.
Finding r here would be cumbersome, So, let us take rate of interest as R for a time period of 5 years as we are talking of amount after every five years.
So, if time period is 5 years, the R is \(\frac{130.01-100}{100}*100\) or 30.01%
10 years: After another 5 years, amount will become \(130.01*(1+\frac{30.01}{100})=130.01*1.3001~169\neq 160\)
15 years: After another 5 years, amount will become \(169*(1+\frac{30.01}{100})=169*1.3001~219\)
20 years: After another 5 years, amount will become \(219*(1+\frac{30.01}{100})=219*1.3001~285\)
25 years: After another 5 years, amount will become \(285*(1+\frac{30.01}{100})=285*1.3001~371\neq 317\)
30 years: After another 5 years, amount will become \(371*(1+\frac{30.01}{100})=371*1.3001~482\)
The values are 160 and 317
Mathematically
CI formula \(P(1+\frac{r}{100})^t=Amount\)
\(100(1+\frac{r}{100})^5=130.01\)
\((1+\frac{r}{100})^5=1.3001\)
After 10 years: \(100(1+\frac{r}{100})^10\)
\(100((1+\frac{r}{100})^5)^2=130.01*(1+\frac{r}{100})^5=130.01*1.3001\)
And so on
