George baked a total of 125 pizzas for 7 straight days, beginning on

Question

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

Skywalker18 · Answer

Number of pizzas George baked on Saturday = (3/5)*125 = 75
Number of pizzas George baked on Sunday = (3/5)*(125-75)=(3/5)*50 = 30
We now have 20 pizzas which has to distributed among the remaining 5 days - Monday- Friday
Average pizzas per day for last 5 days = 20/5 = 4
Since each successive day George baked fewer pizzas than the previous day , to maximize number of pizzas 
baked on Wednesday , we need to bake the least number of pizzas possible on Monday and Tuesday .
Number of pizzas baked on Monday = 7 
Number of pizzas baked on Tuesday = 6
Max number of pizzas baked on Wednesday = 5
number of pizzas baked on Thursday = 2
number of pizzas baked on Friday = 0

Answer C

P.S-If there was restriction that atleast one pizza was baked on each day then max number of pizzas baked on Wednesday would have been =4 
Pizzas baked from Monday - Friday would have been 6,5,4,3,2

EMPOWERgmatRichC · Answer

Hi All,

This question could have been written in a slightly clearer way (The intent is that a total number of pizzas is meant to be 125 over the course of 7 days; at least one pizza is cooked on each day).

3/5 of the 125 pizzas cooked on Saturday = 75 pizzas
3/5 of the remaining pizzas on Sunday = 30 pizzas

We're left with 20 pizzas for the remaining 5 days. The prompt tells us that each day has FEWER pizzas than the day before it, so we can't have duplicate numbers. We're asked for the maximum number of pizzas that could have been made on Wednesday (re: on the 5th day). To maximize THAT number, we have to minimize all of the other numbers, while still following all of the other restrictions that the prompt described.

_ _ _ _ _

We'll start by making the 6th and 7th days 2 pizzas and 1 pizza, respectively...

_ _ _ 2 1

This leaves us with 17 pizzas for the remaining 3 days....we could have....

10 4 3 2 1 with 3 on Wednesday, but that's NOT the maximum
8 5 4 2 1 here we have 4 on Wednesday.

_ _ 5 2 1 is NOT possible though, since we would have 12 pizzas for the remaining 2 days, but no way to satisfy the other restrictions.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

voraniket · Answer

Why isn't 75 30 8 6 5 1 0 possible? It satisfies all the conditions of the question. The question doesn't say that george cannot bake a non-zero pizza any day.

chetan2u · Answer

Hi,
if the 7th day there have been none pizzas cooked, the Q would hav esaid that "George baked a total of 125 pizzas for  6  straight days, beginning on Saturday"...
we have 20 for remaining 5 days.. 
for the middle term, wednesday here, to be max, the higher term should be as close as possible to middle term and the lower terms the smallest possible..
so last two days.. 2,1
remaining three ddays 20-3=17..
as all are different, the middle should be closest to average..
17/3 nearly 6,,
if we take three terms then as 7,6,5, total =18, which is one shorter..
that shortage has to be made up from the lowest term here, ..
so wed , max can be 4..
B

Fdambro294 · Answer

Sat ----> (3/5) * 125 = 75 baked

Sun-----> (125 - 75) * (3/5) = 30 baked

Pizzas Remaining to be Baked:

125 Total - (75) - (30) = 20 Remaining Pizzas

"if each day he baked fewer pizzas than the previous day"

we want to MAXIMIZE the # on Wednesday

however, we are given the Condition that:

Mon > Tues > Wed > Thurs. > Fri

in other words, given the 5 Remaining days, if we listed the Baked Pizzas in DECREASING Order ------> Wednesday = MEDIAN

We want to make Thursday and Friday as low as we can: Let Friday = 1 pizza and Let Thursday = 2 Pizzas

leaves us with 17 pizzas remaining to distribute between: Monday > Tuesday > Wednesday

Given that Monday and Tuesday must be GREATER than Wednesday, Minimizing Monday and Tuesday will only hurt the amount that can be baked on Wednesday

If we were to try 6 pizzas on Wednesday:

Thursday would at MIN have to be = 7
and Friday would at MIN have to be = 8

8 + 7 + 6 = 21 --------> BUT we only have 17 pizzas left to distribute

If were to try 5 pizzas on Wednesday:

Thursday would at MIN have to be = 6
Friday would at MIN have to be = 7

7 + 6 + 5 = 18, which is GREATER THAN > 17 remaining pizzas we have

Lowering Thursday or Friday would only have the effect of DECREASING Wednesday even more. Thus, the remaining 1 pizza MUST come out of Wednesday and we are left with:

Monday = 7
Tuesday = 6

Wednesday = 4 = MAXIMUM that can be made on Wednesday

Thursday = 2

Friday = 1
________________________

20 Remaining Pizzas

-B- 4 is the MAX that can be made on Wednesday

chetan2u · Answer

So 125 pizzas over 7 days starting on Saturday.
To get maximum on Wednesday( 5th day),  we should minimize on thursday and friday  => 2 and 1 = 3 pizzas on last 2 days
1st day and 2nd day,  the number of pizzas is fixed  =>  \frac{3}{5}*125=75 , and  \frac{3}{5}*(125-75)=30 .....105 pizzas on first two days.
Therefore we have 125-105-3=17

Now,  we have to maximize the SMALLEST value, as we looking for 5th day out of 3rd, 4th and 5th day. 
So take average and spread the number around it with minimum difference between them..
Thus 4th day becomes 17/3 or 6 pizzas, but 3rd day has to be one more, so 6+1=7.
5th day = 17-7-6=4

B

GMATGuruNY · Answer

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.

B

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