On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any

Question

1 month ago

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On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any

additional deposits or withdrawals since then, and if the interest rate has stayed the same, in what year did his balance hit $2300, according to the rule of 72?

Mathematics

2 answers:

zzz [12.3K] · Answer 1 · 2026-03-01T19:07:08+03:00

zzz [12.3K]1 month ago

7 0

According to the rule of 72,
72/rate=time
72÷9.6=7.5 years

An alternative method for resolution using the main formula
2300=1150(1+0.096/4)^4t
Isolate t
t=((log(2,300÷1,150)÷log(1+(0.096÷4))÷4))=7.31 years

I hope this is helpful:-)

Svet_ta [12.7K] · Answer 2 · 2026-03-01T19:07:56+03:00

Answer: 7.5 years.

Step-by-step explanation:

In this scenario, the initial amount = $1150

The sum after accruing compound interest becomes = $2300

Since, $2300 is twice the initial amount of $1150.

Following the rule of 72, an investment doubles if the result of the annual rate multiplied by time (in years) equals 72. This rule provides an approximate value for the interest rate or duration.

In this case, the annual interest rate is 9.6%.

Let's determine how long it takes for the $1150 to double to $2300 in t years.

Then, applying the equation, 9.6 × t = 72

⇒ t = 72/9.6 = 7.5

Therefore, the initial amount of $1150 doubles in about 7.5 years.

Verification: How to calculate the years using the compound interest formula.

$2300=1150(1+\frac{9.6}{100})^{t}$

$2=(1+\frac{9.6}{100})^{t}$

$2^{1/t}=(1+\frac{9.6}{100})$

Thus, t = 7.562 years.