Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven 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.

The answer is B.

Thanks!

Plum -

Let's look at (1): If the Store sold 8 books on Thursday this implies that it has sold somewhere between (0+1+2+3+8) = 14 to (4+5+6+7+8) = 30 books through Thursday. This means that the store has to sell between 76 and 60 books total on Friday and Saturday. You can split this up in any number of ways between the two days without violating the constraints in the problem (e.g 9, 51 25, 35 , etc). So (1) is insufficient by itself which rules out A&D.

Let's look at (2): If the store sells 38 copies of the book on Saturday, then that means it has to sell 52 books on the proceeding 6 days. If F is the number of books sold on Friday, then the lowest possible value of X is given by F+(F-1)+(f-2)+(f-3)+(f-4)+(f-5) = 52. So 6F - 15 = 52, 6F = 67 , F = 11.16. Since books are integers, the lowest possible value of F is 12, making (2) sufficient.

The lowest possible value of F when 38 books are sold on Saturday is given by S,M,T,W,T,F,S 6,7,8,9,10,12,38.


/Jeff

Please disregard my answer above - sent it a bit late at night. I'll try again:

From fact (1) the store sells 8 books on Thursday. All fact (1) tells you is that the store sold more than 8 books on both Friday and Saturday. For example, it might sell 1 book on sunday, two on Monday, 3 on Tuesday, 4 on Wed, 9 on Friday and 63 on Saturday. Or it could sell 1 on Sunday, 2 on Monday, 3 on Tuesday, 4 on Wed, 8 on Thursday, 30 on Friday and 42 on Saturday. It is clearly insufficient by itself to answer whether the store sold more than 11 copies on Friday. We can rule out answer choice A&D.

Let's consider (2), the fact that the store sold 38 copies on Saturday. This means that it must sell 52 books on the first six days of the week. Is there anyway to do this subject to the constraints in the problem statement? There are two approaches to this that will work, plugging in number and algebraic:

a) Plug in numbers. It's pretty easy to see that there are values greater than 11 books sold on Friday that will work. For example, 30 books on Friday, 8 books on Thursday, 6 books on Wed, 3 books on Tuesday, 4 books on Monday and 1 on Sunday. What about when the books sold on Friday are 11 or less? Well, start with 11 books on Friday. Then because of the constraints in the problem statement, the most books sold on Thursday would be 10. Similarly the most books sold on Wed would be 9, Tuesday 8, Monday 7, and Sunday 6. Summing all these up only gets you to 51 books sold Monday - Friday, so there is no way of getting to 90 books sold for the week when you sell 38 on Saturday and only 11 or fewer on Friday. So (2) is sufficient by itself and the answer is b.

b) You can also approach this algebraically. If F is the number of books sold on Friday, then the smallest acceptable value of F is given by the equation F+(F-1)+(F-2)+(F-3)+(F-4)+(F-5) = 52. This gives a value of F = 11 1/6. So it's not possible to satisfy the constraints with a value of F equal to 11 or less.

/Jeff

Does someone have a good (that is quick) approach to min/max DS problems of this sort? I agree with the solution above but can one really get this done in 2 min? Borcho

It took me more than 4 min to solve this.

Is there any faster way to solve this problem?

Thanks
Pathik

I am sorry for the partial solution - I would solve as -
If 38 are sold on the highest day the the rest 6 days sold 52 copies.
ie. each day on average sold 8.66 copies.
Since we know Friday is 2nd highest and all days are unique. I might have something like this
6, 7, 8, (8.66), 9, 10, 11+. Note that since ave is 8.66 so 11 will not do it has to be 11+. (I am using the property median = average for consecutive nos,)
Note that this is the best case. There is no other way to reduce 11+ (try it, remember all nos are unique).

the solution posted by 'captain' here is the real deal - the fastest way to solve this problem. in a nutshell, here's the method: consider the lowest and highest possible values in the problem.

to minimize the number of copies sold on friday, you should try to make the sales for all days except saturday (which is fixed) as even as possible. (you can't make friday 0 copies, because friday has to be the second greatest sales day.)
you have 90 - 38 = 52 copies to distribute between six days.
the closest you can get together is consecutive integers; if you try to sell only 11 copies on friday, the best you can do on the other days is 10, 9, 8, 7, 6. that's a total of 51 copies on those days - not good enough. you must sell one extra copy; but because your numbers are currently consecutive integers, you are forced to increase friday. therefore, you must sell at least 12 copies on friday.

to maximize the number of copies sold on friday, you'd make the other days' sales 0, 1, 2, 3, and 4 copies - but note that you don't have to do this, because we've already established that the minimum possibility is greater than 11 copies (so we don't particularly care what the maximum is).

this sort of reasoning shouldn't take four minutes, provided you're used to it. just realize that if you want to make the greatest number in a group of different numbers as small as possible (or make the smallest one as big as possible), then consecutive integers are the way to go.