43.2k views
3 votes
Amy has $1,000 in a savings account at the beginning of the fall. She wants to have at least $500 in the account by the end of the fall. She withdraws $100 a week for living expenses. Write an inequality for the number of weeks Amy can withdraw money, and solve.

A) 1000 - 100w ≤ 500; w ≥ 6
B) 1000 + 100w ≤ 500; w ≤ 5
C) 1000 + 100w ≥ 500; w ≥ 6
D) 1000 - 100w ≥ 500; w ≤ 5

User Tato
by
4.6k points

1 Answer

2 votes

Answer:

D.
1000 - 100w \geq 500; w \leq 5

Explanation:

Given:

Initial amount in the bank = $1000

Money withdrawn each week = $100

Final amount should be at least $500.

Now, let the number of weeks the money is withdrawn be 'w'.

Therefore,

Money withdrawn in 'w' weeks =
\textrm{Money withdrawn each week}* w

Total Money withdrawn in 'w' weeks =
100w

Now, final amount after 'w' weeks is equal to the difference between initial amount and total withdrawal amount. Therefore,

Final amount = Initial amount - Total withdrawal amount

Final amount =
1000 - 100w

Now, final amount must be greater than or equal to $500. So,


\textrm{Final amount}\geq500\\\\1000-100w\geq500

Therefore, the inequality that represents the inequality for the number of weeks Amy can withdraw money is:


1000-100w\geq500

Now, let us solve for 'w'.

Adding -500 and 100w both sides, we get:


1000-500-100w+100w\geq500-500+100w\\\\500\geq100w\\\\\textrm{The above inequality is reversed when taking 100w on the left side}\\\\100w\leq500\\\\w\leq(500)/(100)\\\\\therefore w\leq5

Therefore, the correct option is (D).

User Cmbaxter
by
5.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.