Is this answer right????

Question

Suppose that you invest $10,000 for 5 years.

* If the compound interest rate is doubled, does the amount accumulated also double at the end of 5 years? Defend your answer with an actual illustration


It will Not be doubled after 5 years.
Say you get an interest rate of 10%.

Then you would have (1.1 for 100% + 10%):

1 10,000 * 1.1 = 10,100
2 10,100 * 1.1 = 11,110
3 11,110 * 1.1 = 12,221
4 12,221 * 1.1 = 13,443.10
5. 13,443.10 * 1.1 = 14,787.41

So your interest is $14,784.41 - $10,000 = $4,787.41.

Now say you get an interest rate of 20%.

Then you would have (1.2 for 100% + 20%):

1 10,000 * 1.2 = 12,000
2 12,000 * 1.2 = 14,400
3 14,400 * 1.2 = 17,280
4 17,280 * 1.2 = 20,736
5. 20,736 * 1.2 = 24,883.20

So your interest is $24,883.20 - $10,000 = $14,883.20

That's almost three times as much interest as when the rate was 10%.

Answer 1

In the chart below I am showing the balance at 5% and 10% by year and the ratio between the two. Notice that the ratio grows each year. The higher interest rate gets further ahead in both absolute and percentage terms.

Year 1.05 1.10 Ratio
0 10,000.00 10,000.00 1.000
1 10,500.00 11,000.00 1.048
2 11,025.00 12,100.00 1.098
3 11,576.25 13,310.00 1.150
4 12,155.06 14,641.00 1.205
5 12,762.82 16,105.10 1.262
6 13,400.96 17,715.61 1.322
7 14,071.00 19,487.17 1.385
8 14,774.55 21,435.89 1.451
9 15,513.28 23,579.48 1.520
10 16,288.95 25,937.42 1.592

Sorry the columns are not lined up better but I think you can see the progression.

Answer 2

Looks ok to me, although a few of the computations may be off, the idea is right.. The reason why your amount earned would more than double is because the interest on each period is computed on the previous amount, as you showed. So it's not a simple linear relationship. And it's good to see someone learning to do their math instead of wanting other people to do it for them. Good luck on your test!

Answer 3

Looks good, yes. So you've proved that
1.2 ^ 5 - 1> 2 (1.1 ^5 - 1)

which is not surprising because (for x >0, n integer > 1)
(1 + 2x) ^ n - 1 = 1 + 2nx + ... (2x) ^ n - 1
>
2 ((1 + x)^n - 1) = 2 (1 + nx + .... x ^ n - 1)

Answer 4

No. You have to realize that compounding do not double if you double the rate. Only in simple compounding do that occur.

Answer 5

Sorry. You goofed where you said, 10,000*1.1 = 10,100, when it equals 11,000 instead.