An insurance company provides coverage for a certain dental procedure

For first 1200$. Patient pays 20% = 20/100*1200 = 240 $
For remaining = 490 - 240 = 250 $, his contribution = 50%

Let x be the cost above 1200 $
--> 50% of x = 250
--> 50/100*x = 250
--> x = 250*2 = 500 $

So, Total cost of procedure = 1200 + 500 = 1700 $

IMO Option E

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Let n = the total cost of the procedure. Since the insurance company pays 80% of the first $1200, then the patient pays 20% of the first $1200. The patient must also pay for 50% of the cost above $1200, which is (n - 1200). Thus, we have:

0.2(1200) + 0.5(n - 1200) = 490

240 + 0.5n - 600 = 490

0.5n - 360 = 490

0.5n = 850

n = 1,700

Alternate Solution:

Looking at the answer choices, we know the procedure costs at least 1,200$. On the first 1,200$, the patient paid 1200 x 0.2 = 240$ and thus, 490 - 240 = 250$ was the amount paid for the excess of 1,200$. Since the company and the patient pays the same amount after the first 1,200$, the operation costs 250 x 2 = 500$ more than 1,200$; i.e. 1200 + 500 = 1,700$.

Answer: E