Bunuel wrote:
George baked a total of 125 pizzas for 7 straight days, beginning on Saturday. He baked 3/5 of the pizzas the first day, and 3/5 of the remaining pizzas the second day. If each successive day he baked fewer pizzas than the previous day, what is the maximum number of pizzas he could have baked on Wednesday?
A. 3
B. 4
C. 5
D. 7
E. 10
Pies baked on Saturday \(= \frac{3}{5} * 125 = 75\)
Remaining pies to be baked \(= 125-75 = 50\)
Pies baked on Sunday \(= \frac{3}{5} * 50 = 30\)
Remaining pies to be baked \(= 50-30 = 20\)
We can PLUG IN THE ANSWERS, which represent the maximum number that can be baked on Wednesday.
Since the correct answer must represent the greatest possible number, start with the largest answer choice.
The following constraints must be satisfied:
Each successive day fewer pizzas are baked than on the previous day.
Since pies are baked for 7 straight days, at least 1 pie must be baked on each day Monday through Friday.
E: W=10
Here, the least possible case for Monday through Wednesday is as follows:
M=12, T=11, W=10
Total = 12+11+10 = 33
Not viable, since the total exceeds 20.
D: W=7
Here, the least possible case for Monday through Wednesday is as follows:
M=9, T=8, W=7
Total = 9+8+7 = 24
Not viable, since the total exceeds 20.
C: W=5
Here, the least possible case for Monday through Friday is as follows:
M=7, T=6, W=5, TH=2, FR=1
Total = 7+6+5+2+1 = 21
Not viable, since the total exceeds 20.
The total in C is only one greater than the required value.
Implication:
The next smallest answer choice must be correct.
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