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Question 1 of 5
If you deposit $ 3,000 $ 3,000 into an account earning 7 % interest compounded weekly, how much money will be in the account after 4 years?
Round your answer to two decimal places
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Label Known Values
Principal Value = P = $ 3,000
Interest Rate = r = 7 % per annum = 0.07 (as a decimal )
Time Elapsed = t = 4 years
Number of times compounded per year = n = 52 (Weekly )
Solve for the Amount ( A ) , using the formula
A
=
P ( 1 + r n ) n t
Compound Interest Formula
=
3,000 ( 1 + 0.07 52 ) 52 × 4
Substitute known values
=
( 3,000 ) ( 1 + 0.00134615384 ) 208
=
( 3,000 ) ( 1.3228807 )
Simplify
=
3,968.64
Round to two decimal places
Question 2 of 5
An amount of $ 375 invested for 2 years grows to $ 431.35 . The amount is compounded annually. What is the interest rate?
Round your answer to two decimal places
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Label Known Values
Amount Earned= A = $ 431.35
Principal = P = $ 375
Number of Years = n = 2 years
Solve for the Interest Rate ( r )
A
=
P ( 1 + r ) n
Compound Interest Formula
431.35
=
375 ( 1 + r ) 2
Substitute known values
431.35 375
=
( 1 + r ) 2
Divide both sides by 375
1.15
=
( 1 + r ) 2
1.0725
=
1 + r
Get square root of both sides
0.0725
=
r
Subtract 1 from both sides
r
=
0.0725
r
=
0.0725 × 100 %
Multiply by 100 %
r
=
7.25 %
Question 3 of 5
How much money would you need to deposit today at an interest rate of 7.5 % compounded annually to have $ 65,000 in the account after 8 years?
Round your answer to two decimal places
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Label Known Values
Amount Earned= A = $ 65,000
Rate = r = 0.075
Number of Years = n = 8 years
Solve for the Principal Amount ( P )
A
=
P ( 1 + r ) n
Compound Interest Formula
65,000
=
P ( 1 + 0.075 ) 8
Substitute known values
65,000
=
P ( 1.075 ) 8
Evaluate
65,000 1.075 8
=
P
Divide both sides by 1.075 8
65,000 1.783477826
=
P
36,445.65
=
P
P
=
36,445.65
Question 4 of 5
How much money would you need to deposit today at an interest rate of 6 % compounded annually to have $ 30,000 in the account after 5 years?
Round your answer to two decimal places
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Label Known Values
Amount Earned= A = $ 30,000
Rate = r = 0.06
Number of years = n = 5 years
Solve for the Principal Amount
A
=
P ( 1 + r ) n
Compound Interest Formula
30,000
=
P ( 1 + 0.06 ) 5
Substitute known values
30,000
=
P ( 1.06 ) 5
Evaluate
30,000 1.06 5
=
P
Divide both sides by 1.06 5
30,000 1.338225578
=
P
22,417.75
=
P
P
=
22,417.75
Question 5 of 5
Two banks receive an investment of $ 20 000 each with an interest rate of 7 % per annum. If one bank uses simple interest and the other bank uses compound interest, what will be the difference in the amount (A ) after a period of 3 years?
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Label Known Values
Principal Value = P = $ 20,000
Interest Rate = r = 7 % per annum = 0.07 as a decimal
Time Elapsed = t = 3 years
Number of times compounded per year = n = 1 (Annually)
To understand the concept Simple and Compound interest, let us first try to solve this problem without any formulas.
Simple Interest:
The principal value where the interest is applied stays the same every period.
20 000 + ( 0.07 × 20 000 )
=
20 000 + 1400
=
21 400
21 400 + ( 0.07 × 20 000 )
=
21 400 + 1400
=
22 800
22 800 + ( 0.07 × 20 000 )
=
22 800 + 1400
=
24 200
Compound Interest:
The principal value where the interest is applied compounds every period.
20 000 + ( 0.07 × 20 000 )
=
20 000 + 1400
=
21 400
21 400 + ( 0.07 × 21 400 )
=
21 400 + 1498
=
22 898
22 898 + ( 0.07 × 22 898 )
=
22 898 + 1602.86
=
24 500.86
Therefore, the difference between the two amounts is 24 500.86 - 24 200 = 300.86
This time, let’s see more of how Simple and Compound Interest work by using their formulas to solve the problem.
I
=
P r t
=
20 000 × 0.07 × 3
=
4200
This is the interest earned over the 3 years, not the final amount.
A (Final amount)
=
P + I
=
20 000 + 4200
=
24 200
A
=
P ( 1 + r n ) n t
=
20 000 ( 1 + ( 0.07 ) ( 1 ) ) ( 1 ) ( 3 )
=
20 000 ( 1 + 0.07 ) 3
=
20 000 ( 1.07 3 )
=
20 000 × 1.225043
=
24 500.86
We can confirm again, through the formulas, that the difference between the two amounts is 24 500.86 - 24 200 = 300.86