Question

Sam initially invested $4,500 into a savings account that offers an interest rate of 3% each year. He wants to determine the number of years, x, for which the account will have less than or equal to $7,020.

Determine the solution set to the inequality that represents this situation.

Answer 1

The solution set to the given inequality : 0 < x ≤ 18.67

Explanation:

Principal = $4500

Simple Interest Rate = 3%

= 0.03

Time = x year

Amount in the account will be less than or equal to $7020

⇒ Amount ≥ Principal × (1 + Rate × Time)

⇒ 7020 ≥ 4500(1 + 0.03·x)

⇒ 7020 ≥ 4500 + 135·x

⇒ 2520 ≥ 135·x

⇒ 18.67 ≥ x

The solution set to the given inequality : 0 < x ≤ 18.67

Answer 2

Principal = $4,500

Rate of interest=3%

Time = x years

We have to find number of years such that

Amount ≤$7020

Amount ≤ Principal + Simple interest

7020≤4500+ S.I

S.I≥7020-4500

S.I≥$2520

Formula for simple interest

S.I=
`(P* R * T)/(100)`, where P=Principal, R=Rate of interest, T=time

2520≥
`(4500 * 3 * x)/(100)`

`(2520 * 100)/(4500 * 3) \leq x`

x≤
`(56)/(3)`

So, the solution set is (0,
`(56)/(3)`]