Store S sold a total of 90 copies of a certain book during

Store S sold a total of 90 copies of a certain book during the seven days last week and it sold different numbers of copies on any 2 days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Firday, did Store S sell more than 11 copies on Friday?

Given: (Any of other five days different sales)<Fri<Sat(max) and Sun+Mon+Tue+Wed+Thu+Fri+Sat=90.

Question: is Fri>11?

(1) Last week Store S sold 8 copies on Thursday --> Sun+Mon+Tue+Wed+8+Fri+Sat=90 --> if the sales before Friday were: 0+1+2+3+8(Thu)=14 then Fri+Sat=90-14=76, so Store S could have sold less than 11 copies on Friday (for example if Fri+Sat=10+66=76) as well as more than 11 copies (for example if Fri+Sat=12+64=76). Not sufficient.

(2) Last week Store S sold 38 copies on Saturday --> Sun+Mon+Tue+Wed+Thu+Fri+38=90 --> Sun+Mon+Tue+Wed+Thu+Fri=52. Now, if Store S didn't sell more than 11 copies on Friday, then max possible copies sold before Saturday would be Sun+Mon+Tue+Wed+Thu+Fri=6+7+8+9+10+11=51<52 --> so, Store S must have been sold more than 11 copies on Friday. Sufficient.

Answer: B.

Hope it's clear.
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Very good question indeed, I have to accept it took me a little longer than it should have.
But let's tackle this one

So total of 90 books, 7 days to sell and no # is equal in any two days. This last constraint is very important because we are going to start elaborating from here.
Ok, then we are asked if the copies sold on Friday are more than 11. And we are also told that Saturday sold the most followed by Friday

So Statement 1 says that 8 copies were sold on Thursday. We don't really know where Thursday stands in regards to its "position" from the other days. So we have a remainder of 82 books to distribute among the other 6 days with not too many restrictions. Friday could very well be >11 or less than 11.

This statement is Insuff

Now, statement 2 says that 38 copies were sold on Saturday.
So one key thing here, is that we are asked if S>11. So let's try extreme cases.

Could there be more than 11 copies sold on Friday? Well, our only constraint is that it will have to be 37 or less, so yes. Now, on the other hand can the number be not more than 11, lets say exactly 11 for example.
OK, so let's see. If we sell 11 books on Friday, then we'll have a remainder of 41 book (90 - 49). And we will have 5 more days. But then these numbers for the number of books sold will have some constraints. First all of them <11, then all different. So in a best case scenario we could have 6+7+8+9+10 = 40 which is not enough to reach the total of 41.

Therefore, our answer is yes. It will have to be more than 11 books

Hope it helps
Kudos if you like! I'm trying hard to get those GMAT Club tests!
Show some love!
Cheers
J

Let me correct my earlier post:

1: S+M+TU+W+Th+F+SAT = 90
S+M+TU+W+ 8 +F+SAT = 90

2 possibilities:

I: S+M+TU+W+ 8 +F+SAT = 90
1+2+3+4+8+F+63 = 90
F = 9........No.

II: S+M+TU+W+ 8 +F+SAT = 90
S+M+TU+W+ 8 +F+SAT = 90
1+2+3+4+8+F+40 = 90
F = 32.... Yes.
NSF.

2: S+M+TU+W+Th+F+SAT = 90
S+M+TU+W+TH+F+38 = 90
S+M+TU+W+TH+F = 52
(n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) = 52

5n + 21 = 52
n = 31/5 = 6.2

Since 6.2 cannot be a possible no. of copies, n has to be 7. Then no. of copies sold on friday is (n+6), which equal to 13. suff.

B.

Store S sold a total of 90 copies of a certain book during the 7 days of last week, and it sold different number of copies on any two of the days. If for the 7 days store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did store S sell more than 11 copies on Friday?

1. Last week store S sold 8 copies of the book on Thursday

2. Last week store S sold 38 copies of the book on Saturday


This Q has been discussed multiple times. My question is "which statement in the Question tells that each day they sold different number of books"

It says in the question stem that they didn't sell the same number of books on any two of the days. This means that they didn't sell same number of books on any day.

As per the question M +Tu +W +Th + Fr + Sa +Su =90 and M,Tu,W,Th,Su < F< Sa

1) If Thursday sold 8 copies
For values of Th=8,W=7,Tu=6,M=5,Su=4,F=10,Sa=50 (Fr <11)
For values of Th=8,W=7,Tu=6,M=5,Su=4,F=12,Sa=48 (Fr >11). Hence not sufficient.
2) If Saturday sold 38 copies
M +Tu +W +Th + Su + Fr = 52
Since the second highest number of books were sold on Friday and no two days sold the same number of books.
If Friday sold <=11 books The maximum value of M +Tu +W +Th + Su is 10+9+8+7+6
which makes max value of M +Tu +W +Th + Su + Fr = 51 (< 52). So Friday should have definitely sold more than 11 books. Hence sufficient.

So answer is B

Hi Bunnel,
I interpreted the question that except friday and saturday the number of books sold on all the other days were same. Since it is given the number of books sold were different on any 2 days?. Shouldnt the question be on all the days the number of books sold were different?.. Please clarify....

No, your interpretation is not correct, we are told that "(store) sold different numbers of copies on any 2 days", so there are no 2 days on which it sold the same number of copies --> store sold different number of copies on each day.
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Hello Bunnel,
The explanation is very clear..Just would like to ask one Q..
It is mentioned: Store S sold a total of 90 copies of a certain book during the seven days last week and it sold different numbers of copies on any 2 days
and to explain this you have taken 0,1,2,3,4 to explain... But I am just wondering can we take 0 in this case because he has sold some copies and its different from that of the other days..I think we should start from 1,2,3,4,5.
(The answer does not depend on this query but I am just asking out of curiousity)

I think the store could have sold zero copies on any day (except Friday and Saturday). Why not? Nothing in the stem indicates that the # of sold copies were more than zero.
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OK..
Thanks for clarifying...

Store S sold a total of 90 copies of a certain book during the seven days last week and it sold different numbers of copies on any 2 days. If for the seven days Store S sold the greatest number of copies on Saturday and the second greatest number of copies on Firday, did Store S sell more than 11 copies on Friday?

1) Last week Store S sold 8 copies on Thursday
2) Last week Store S sold 38 copies on Saturday

Lets produce the contradiction.

1) The store sold 11 copies on Friday. The answer is NO
8 (Thursday) + 11 (Friday) + 71 copies on rest of the days (with highest copies on Sat)

The store sold 12 copies on Friday. The answer is YES
8 (Thursday) + 12 (Friday) + 70 copies on rest of the days (with highest copies on Sat)

Insufficient.

2) Lets assume the contradiction. The store 11 copies on Friday. Copies sold on Thursday = 38. hence the remaining copies are 90 - 38 = 52. Maximizing the number of copies sold - lets assume
11 (Friday) + 10 (Thu) + 9 (Wed) + 8 (Tue) + 7(Mon) + 6 (Sun) copies are sold on the rest of the six days. Total = 51 copies. Not high enough - we are short by 1 copy. The contradiction is wrong - the store indeed sold more than 11 copies on Friday. Proved.

Sufficient. Hence B

Thanks "jimmy86" to help us dig out this interesting question for others to take a look at. Kudos+1.

Sol:
1) Last week Store S sold 8 copies on Thursday.

	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
Copies Sold(Friday>11)	1	2	3	4	8	12	90-24=66
Copies Sold(Friday<=11)	1	2	3	4	8	11	90-23=67

We did not violate any condition mentioned in the stem:
Every day the store sold different number of copies.
Greatest number of copies were sold on Saturday.
Second greatest number of copies were sold on Friday.

Order in which the rows were filled:
Row1: Thursday(8), Friday(12), Sunday(1), Monday(2), Tuesday(3), Wednesday(4), Saturday(90-(Sum of rest))
Row2: Thursday(8), Friday(11), Sunday(1), Monday(2), Tuesday(3), Wednesday(4), Saturday(90-(Sum of rest))

We proved that with 8 copies sold on the Thursday, the store could have sold 11 or more on the Friday.
Not Sufficient.

2) Last week Store S sold 38 copies on Saturday

	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
Copies Sold(Friday>11)	1	2	3	4	5	90-53=37	38
Copies Sold(Friday<=11)	X	X	X	X	X	11	38

Row1: We showed one of the possible combinations of copies sold when the store sold more than 11 copies on Friday.
Order: Saturday(38), Sunday(1), Monday(2), Tuesday(3), Wednesday(4), Thursday(5), Friday(90-Sum of rest)

Row2: Saturday=38 and Friday=11
The store must sell (90-38-11)=(90-49)=41 copies in order to satisfy the stem.

We cannot possibly assign a number bigger than 11 for the rest of the days because Friday needs to be second greatest. Saturday is standing greatest with 38 and Friday is the second greatest with 11.
The maximum number of copies the store could sell on other days, given it sold 11 copies on Friday, would be 10+9+8+7+6=40, which is 1 short of the required number.
This proves that there were indeed more than 11 copies that were sold on the Friday. In fact, the minimum number of copies that could be sold on Friday with these conditions would be "12".

Sufficient.

Ans:"B"

(1) NS. This tells us we must sell atleast 9 books on Friday, but not necessarily 11, because the excess demand can all be taken by saturday since there is no restriction.

(2) Check the maximums

IF we set friday to 11 and everything else to the most we possibly can, we have to check if we can still get 90 of total volume

M 6 T 7 W 8 T 9 F 11 S 38 S 10 = total of 89, and we can't increase anything but friday, because if we increase another day, two days will have the same number of sales, or will have more sales that friday, which is not allowed. -> sufficient

A : INSUFFICIENT. There is no information about Friday or Saturday which are the main constraints of this question.

B: SUFFICIENT.

Sat---Fri---Thu---Wed---Tue---Mon---Sun
38----11---10-----9------8-----7------6----

This above configuration indicates the minimum possible value for each day given the constraint:
1) Different copies for each day
2) Saturday > Friday (# of copies)
3) Saturday = 38 copies
4) Friday = 11 copies?

When we add this possibility up we get 89. Since 89 < 90. We know that the value for Friday MUST be greater than 11.

1st attempt I got this wrong,, but eventually dod it rigt here's how.

I is insufficient
II - sale from sunday to Fri is 52. now, since no to values are same, the value of friday is maximised when sale on the other days are consecutive integers
so, let sale on sunday = x, monday = x+1 etc.. therefore , 5x+4 + friday = 52.
friday = 48-5x. At x=7, the value on friday is 13, at x=8, the value is 8. but if x=8, friday cannot be the day with second largest sale.. therefore x>7 and friday sales is atleast 13.. hence suff.

Statement 1 is obviously insufficient
Statement 2, let Friday be x. To obtain the least value of x, the other five days should be, x-1, x-2, x-3, x-4, x-5
So, 38+x+x-1+x-2+x-3+x-4+x-5=90
6x=67
x=67/6>11

For statement (2).

Suppose on Friday, store sold 11 copies.

Given: 38 copies on Saturday

So, copies on other days: 90 - (38+11) = 41

Max copies that can be sold on other days = 10,9,8,7,6.

But 10+9+8+7+6 = 40

So, this doesn't add up to 41. Hence, this configuration is not possible at all. It has to be more than 11 on Friday.

GMATNinja, what may be the most efficient approach to solve constraint questions under 2 min?