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Sam invests a principal of $10,000, which earns interest ove
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30 Jul 2018, 10:34
30 Jul 2018, 10:34
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Question Stats:
70% (01:27) correct
29% (01:10) wrong based on 133 sessions
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Sam invests a principal of $10,000, which earns interest over a period of years.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The final value of the investment after 2 years at 8% interest, compounded annually |
The final value of the investment after 4 years at 4% interest, compounded annually |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: Sam invests a principal of $10,000, which earns interest ove
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12 Aug 2018, 07:10
12 Aug 2018, 07:10
1
Expert Reply
Explanation
Solve this problem by applying the compound interest formula:
\(V=P(1+\frac{r}{100})^t\)
Since the principal P is the same in both cases, leave it out and just compare the rest:
Quantity A: \((1+\frac{r}{100})^t=(1+\frac{8}{100})^2= (1.08)^2 = 1.08 × 1.08 = 1.1664\)
Quantity B: \((1+\frac{r}{100})^t=(1+\frac{4}{100})^4= (1.04)^4 = 1.04×1.04×1.04×1.04 ≈ 1.1699\)
Quantity B is greater.
Solve this problem by applying the compound interest formula:
\(V=P(1+\frac{r}{100})^t\)
Since the principal P is the same in both cases, leave it out and just compare the rest:
Quantity A: \((1+\frac{r}{100})^t=(1+\frac{8}{100})^2= (1.08)^2 = 1.08 × 1.08 = 1.1664\)
Quantity B: \((1+\frac{r}{100})^t=(1+\frac{4}{100})^4= (1.04)^4 = 1.04×1.04×1.04×1.04 ≈ 1.1699\)
Quantity B is greater.
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Sam invests a principal of $10,000, which earns interest ove
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06 Feb 2021, 23:29
06 Feb 2021, 23:29
1
Amount = P \((1 + \frac{r}{100n})^t\)
Col. A: r = 8, t=2
= 10000 \((1 + \frac{8}{100})^2\)
= 108 x 108
=\(108^2\)
Col. B: r = 4, t = 4
= 10000 \((1 + \frac{4}{100})^4\)
= \((\frac{104}{10})^4\)
= \(10.4^4\) = \(108.16^2\)
We can see, A < B
Hence, option B
Col. A: r = 8, t=2
= 10000 \((1 + \frac{8}{100})^2\)
= 108 x 108
=\(108^2\)
Col. B: r = 4, t = 4
= 10000 \((1 + \frac{4}{100})^4\)
= \((\frac{104}{10})^4\)
= \(10.4^4\) = \(108.16^2\)
We can see, A < B
Hence, option B
