BTGmoderatorDC wrote: ↑Sun Jul 05, 2020 4:55 pm
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
OA
B
Source: Princeton Review
No. of pizzas baked on Saturday = 3/5 of 125 = 75;
No. of pizzas baked on Sunday = 3/5 of (125 – 75) = 30;
Total no. of pizzas baked on two days = 75 + 30 = 105;
No. of remaining pizzas = 125 – 105 = 20
=> No. of pizzas baked on Monday, Tuesday, Wednesday, Thursday, and Friday = 20
Since we have to maximize the no. of pizzas baked on Wednesday, let's assume that George baked only 1 pizza on Friday; thus, 2 on Thursday, x on Wednesday, (x + 1) on Tuesday, and (x + 2) on Monday
=> 20 = 1 + 2 + x + (x + 1) + (x + 2)
3x = 14
x = 4.67;
However, x = 4.67 is not possible since x must be a positive integer. Looking at 3x =14, note that the nearest number that is divisible by 3 and closest to 14 is 12. Thus, x = 12/3 = 4. This way, on Thursday and Friday, together, George baked (x + 1) + (x + 2) + 2 = (2x + 5) pizzas
x = 4
Correct answer:
B
Hope this helps!
-Jay
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