Question

Cameron opens a checking account and a savings account at his local bank. He deposits $10,000 into his checking account and $2,000 into his savings account. He will withdraw $1,200 from his checking account each year. His savings account earns 8% interest each year. In how many years will both accounts have the same balance?

Answer 1

Let x be the number of needed years.

1. He deposits $10,000 into his checking account and he will withdraw $1,200 from his checking account each year. Then after x years he will have $10,000-$1,200x in his checking account.

2. He deposits $2,000 into his savings account and his savings account earns 8% interest each year, then after x years he will have
`\$2,000\cdot (1.08)^x.`

3. Equate these amounts of money:

`10,000-1,200x=2,000\cdot (1.08)^x.`

4. Solve this equation:

• divide the equation by 400:

`25-3x=5\cdot (1.08)^x;`

• plot graphs of the function
  `y=25-3x` and
  `y=5\cdot (1.08)^x` (see attached diagram);
• find the common point of these two graphs: (6.663,5.01).

Conclusion: he needs nearly 6.663 years.

Answer 2

The correct answer is:
5.9 years.

Step-by-step explanation:
Let x be the number of years. For the checking account, our expression would be 10000-1200x, since he withdraws $1200 from his account each year, and he begins with $10,000.

The formula for simple interest is:
I=prt
where:
p is the principal, r is the rate and t is the time

using our information, we have
p=2000,
r=8%=8/100=0.08, and
t=x.

This gives us 2000(0.08)(x).
However, we must add the $2000 to this (interest is added on to the principal); this gives us:
2000+2000(0.08)(x).

Setting these two equal, we have:
10000-1200x = 2000+2000(0.08)(x).

Simplifying the right hand side, we have:
10000-1200x=2000+160x.

Add 1200x to both sides:
10000-1200x+1200x=2000+160x+2000x;
10000=2000+1360x.

Subtract 2000 from both sides:
10000-2000=2000+1360x-2000;
8000=1360x.

Divide both sides by 1360:

`(8000)/(1360) = (1360x)/(1360)`;
5.9=x.