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

Answer:

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=$\frac{P\times R \times T}{100}$, where P=Principal, R=Rate of interest, T=time

2520≥$\frac{4500 \times 3 \times x}{100}$

$\frac{2520 \times 100}{4500 \times 3} \leq x$

x≤$\frac{56}{3}$

So, the solution set is (0,$\frac{56}{3}$]

Answer 2

Answer:    

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

Step-by-step 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