Question

A sum of money is invested at 12% compounded quarterly. About how long will it take for the amount of money to double?

Compound interest formula:mc009-1.jpg

t = years since initial deposit
n = number of times compounded per year
r = annual interest rate (as a decimal)
P = initial (principal) investment
V(t) = value of investment after t years
5.9 years
6.1 years
23.4 years
24.5 years

Answer 1

Answer:

Here, the rate percent is divided by 4 and the number of years is multiplied by 4.

Therefore, CI = A - P = P{(1 + r4100r4100)4n4n - 1}

Note:

A = P(1 + r4100r4100)4n4n is the relation among the four quantities P, r, n and A.

Given any three of these, the fourth can be found from this formula.

CI = A - P = P{(1 + r4100r4100)4n4n - 1} is the relation among the four quantities P, r, n and CI.

Given any three of these, the fourth can be found from this formula.

Word problems on compound interest when interest is compounded quarterly:

1. Find the compound interest when $1,25,000 is invested for 9 months at 8% per annum, compounded quarterly.

Solution:

Here, P = principal amount (the initial amount) = $ 1,25,000

Rate of interest (r) = 8 % per annum

Number of years the amount is deposited or borrowed for (n) = 912912 year = 3434 year.

Therefore,

The amount of money accumulated after n years (A) = P(1 + r4100r4100)4n4n

= $ 1,25,000 (1 + 8410084100)4∙344∙34

= $ 1,25,000 (1 + 21002100)33

= $ 1,25,000 (1 + 150150)33

= $ 1,25,000 × (51505150)33

= $ 1,25,000 × 51505150 × 51505150 × 51505150

= $ 1,32,651

Therefore, compound interest $ (1,32,651 - 1,25,000) = $ 7,651.

2. Find the compound interest on $10,000 if Ron took loan from a bank for 1 year at 8 % per annum, compounded quarterly.

Solution:

Here, P = principal amount (the initial amount) = $ 10,000

Rate of interest (r) = 8 % per annum

Number of years the amount is deposited or borrowed for (n) = 1 year

Using the compound interest when interest is compounded quarterly formula, we have that

A = P(1 + r4100r4100)4n4n

= $ 10,000 (1 + 8410084100)4∙14∙1

= $ 10,000 (1 + 21002100)44

= $ 10,000 (1 + 150150)44

= $ 10,000 × (51505150)44

= $ 10,000 × 51505150 × 51505150 × 51505150 × 51505150

= $ 10824.3216

= $ 10824.32 (Approx.)

Therefore, compound interest $ (10824.32 - $ 10,000) = $ 824.32

3. Find the amount and the compound interest on $ 1,00,000 compounded quarterly for 9 months at the rate of 4% per annum.

Solution:

Here, P = principal amount (the initial amount) = $ 1,00,000

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (n) = 912912 year = 3434 year.

Therefore,

The amount of money accumulated after n years (A) = P(1 + r4100r4100)4n4n

= $ 1,00,000 (1 + 4410044100)4∙344∙34

= $ 1,00,000 (1 + 11001100)33

= $ 1,00,000 × (101100101100)33

= $ 1,00,000 × 101100101100 × 101100101100 × 101100101100

= $ 103030.10

Therefore, the required amount = $ 103030.10 and compound interest $ ($ 103030.10 - $ 1,00,000) = $ 3030.10

4. If $1,500.00 is invested at a compound interest rate 4.3% per annum compounded quarterly for 72 months, find the compound interest.

Solution:

Here, P = principal amount (the initial amount) = $1,500.00

Rate of interest (r) = 4.3 % per annum

Number of years the amount is deposited or borrowed for (n) = 72127212 years = 6 years.

A = amount of money accumulated after n years

Using the compound interest when interest is compounded quarterly formula, we have that

A = P(1 + r4100r4100)4n4n

= $1,500.00 (1 + 4.341004.34100)4∙64∙6

= $1,500.00 (1 + 1.0751001.075100)2424

= $1,500.00 × (1 + 0.01075)2424

= $1,500.00 × (1.01075)2424

= $ 1938.83682213

= $ 1938.84 (Approx.)

Therefore, the compound interest after 6 years is approximately $ (1,938.84 - 1,500.00) = $ 438.84.

Answer 2

Therefore the amount becomes double after 5.9 years.(Option-a )

Given:

The sum of the amount is compounded quarterly.

Rate of interest = 12%

t = years since the initial deposit

n = number of times compounded per year = 4

r = annual interest rate (as a decimal)

P = initial (principal) investment

V(t) = value of the investment after t years = 2A

To Find:

Time taken to double the investment.

Solution:

The given question can be solved as shown below.

As the amount is compounded quarterly, every quarter is 3 months and 1 year has 4 such quarters.

Now, V(t) = P( 1 + (r/n )/100 $)^n^t$

⇒ 2P = P ( 1 + ( 12/4 )/100 $)^4^t$

⇒ 2 = (1.03 $)^4^t$

Taking ln on both sides,

⇒ ln 2 = ln ( 1.03 $)^4^t$

⇒ ln 2 = 4t ln 1.03

⇒ t = ln 2 / (4 ln 1.03 ) = 0.693/( 4 × 0.0295 ) = 5.86 ≈ 5.9 years

Therefore the amount becomes double after 5.9 years.

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