Question

A construction company will be fined for each day it is late completing its current project. The daily fine will be $4000 for the first day and will increase by $1000 each day. Based on its budget, the company can only afford $60,000 in total fines. What is the maximum number of days it can be late?
a.
10 days
c.
12 days
b.
8 days
d.
5 days

Answer 1

For the 1st day, the company will pay $4000.
For the 2nd day, the company will pay $4000 + 1($1000)
For the 3rd day, the company will pay $4000 + 2($1000)
For the nth day, the company will pay $4000 + (n -1)($1000)

Let nth day be the day where the company's fine exceed $60000.
The total amount of fines they will have paid by then will be:
(4000 + 4000 + (n - 1)(1000))(n / 2)

Setting up an inequality, we will have:
(4000 + 4000 + (n - 1)(1000))(n / 2) ≥ 60000
(8000 + 1000n - 1000)n ≥ 120000
(8 + n - 1)n ≥ 120
8n + n^2 - n ≥ 120
n^2 + 7n - 120 ≥ 0
(n - 8)(n + 15) ≥ 0
n ≤ -15 or n ≥ 8

The number of days cannot possibly be negative, thus n can only be 8.
Hence, the maximum numbers of days is 8 days.

Answer 2

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