56.3k views
2 votes
George wants to buy a clarinet that cost 450 he puts 360 into the savings account that pays 1% per month using trial and improvement workout the smallest number of a whole month that George will have to wait to have enough money in this account to buy the clarinet

User WebBrother
by
7.4k points

2 Answers

3 votes

Answer:

7 whole months

Explanation:

User Lea Verou
by
7.4k points
3 votes

Answer:

7 whole months

Explanation:

To determine the smallest number of whole months George will have to wait to have enough money in his savings account to buy the clarinet, we can set up an equation and use trial and improvement.

Let M be the number of months George has to wait, and A be the amount he will have in the account after M months.

The monthly interest rate is 1%, which is equivalent to 0.01 as a decimal. So, each month, his savings account balance will increase by 1%:

A = 360 * (1 + 0.01)^M

We want to find the smallest whole number of months M when the balance A is greater than or equal to $450 (the cost of the clarinet):

360 * (1 + 0.01)^M ≥ 450

Now, you can use trial and improvement to find the smallest whole number of months that satisfies this equation:

1. Start with M = 1:

A = 360 * (1 + 0.01)^1 = 360 * 1.01 = $363.60 (not enough)

2. Try M = 2:

A = 360 * (1 + 0.01)^2 ≈ 360 * 1.0201 ≈ $367.24 (not enough)

3. Try M = 3:

A = 360 * (1 + 0.01)^3 ≈ 360 * 1.030301 ≈ $370.91 (not enough)

4. Try M = 4:

A = 360 * (1 + 0.01)^4 ≈ 360 * 1.04060401 ≈ $374.62 (not enough)

5. Try M = 5:

A = 360 * (1 + 0.01)^5 ≈ 360 * 1.0510100501 ≈ $378.37 (not enough)

Continue this process until you find the smallest M for which A is greater than or equal to $450. You will find that M = 7:

A = 360 * (1 + 0.01)^7 ≈ 360 * 1.0714576279 ≈ $386.92 (enough)

So, George will have to wait for 7 whole months to have enough money in his savings account to buy the clarinet.

User Fedoranimus
by
7.0k points