A restaurant with an all-you-can-eat buffet served 42 customers on Friday and 56 customers on Saturday. If the price per head on Saturday is ₱200 more than the price for Friday, how much is Friday’s price if the restaurant earned ₱60,200 for the two days?1 pointA. ₱500B. ₱600C. ₱700D. ₱800
Question
A restaurant with an all-you-can-eat buffet served 42 customers on Friday and 56 customers on Saturday. If the price per head on Saturday is ₱200 more than the price for Friday, how much is Friday’s price if the restaurant earned ₱60,200 for the two days?1 pointA. ₱500B. ₱600C. ₱700D. ₱800
Solution
Let's denote the price on Friday as F and the price on Saturday as S.
From the problem, we know that:
- The restaurant served 42 customers on Friday and 56 customers on Saturday.
- The price on Saturday is ₱200 more than the price on Friday, so S = F + ₱200.
- The total earnings for the two days is ₱60,200.
We can set up the following equations based on these facts:
42F + 56S = ₱60,200 (This is the total earnings for the two days)
S = F + ₱200 (This is the price difference between Saturday and Friday)
We can substitute the second equation into the first to get:
42F + 56(F + ₱200) = ₱60,200
Solving this equation will give us the price on Friday (F).
Let's solve it:
42F + 56F + ₱11,200 = ₱60,200
98F + ₱11,200 = ₱60,200
98F = ₱60,200 - ₱11,200
98F = ₱49,000
F = ₱49,000 / 98
F = ₱500
So, the price on Friday is ₱500. The answer is A. ₱500.
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