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

[plum]

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!

[Jeff]

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

[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

[Borcho]

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

[Pathik]

It took me more than 4 min to solve this.

Is there any faster way to solve this problem?

Thanks
Pathik

[Captain]

I am sorry for the partial solution - I would solve

(2) Last week Store S sold 38 copies of the book on Saturday.

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).

[RonPurewal]

It took me more than 4 min to solve this.

Is there any faster way to solve this problem?

Thanks
Pathik

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.

[tobias-m.schulz]

My understanding of this question was a little bit different due to the
"it sold different numbers of copies on any two of the days"-statement

I understood that any two days in a row have different values.

So it could be that the values are e.g.
9; 8; 9; 8; (Fr-)10; (Sa-)38; 8
or
9; 8; 9; 8; (Fr-)12; (Sa-)38; 6

thus making E the right answer...

If the "any two of THE days" refered only to Sat and Fri, then the other days could even have equal values.

Did I misinterpreted The Q? I am really confused by the "any TWO of THE days"
if it would be "sold sold differnt number of copies on any day", then I would understand it

[RonPurewal]

My understanding of this question was a little bit different due to the
"it sold different numbers of copies on any two of the days"-statement

I understood that any two days in a row have different values.

So it could be that the values are e.g.
9; 8; 9; 8; (Fr-)10; (Sa-)38; 8
or
9; 8; 9; 8; (Fr-)12; (Sa-)38; 6

thus making E the right answer...

If the "any two of THE days" refered only to Sat and Fri, then the other days could even have equal values.

Did I misinterpreted The Q? I am really confused by the "any TWO of THE days"
if it would be "sold sold differnt number of copies on any day", then I would understand it

yes, you misinterpreted the question.

the way the problem is written is the only literally correct way in which to write it: namely, if you select any two days, then the numbers of sales will be different. that's exactly what the problem says.

your suggestion -- "sold different number of copies on any day" -- doesn't make literal sense, since that would require different numbers of copies to be sold on one day.

the mistake of which you are guilty here is mistaking "any two days" for "any two CONSECUTIVE days".
if the problem meant consecutive days, the problem would have to say "consecutive days".

[rajnath99]

There is a great Thursdays with Ron Study Hall where Ron explains this problem. I apologize, but I do not have the exact week/date of the session. Replay is available on vimeo.

[RonPurewal]

There is a great Thursdays with Ron Study Hall where Ron explains this problem. I apologize, but I do not have the exact week/date of the session. Replay is available on vimeo.

here:
http://www.manhattangmat.com/thursdays-with-ron.cfm
the session date is NOVEMBER 4, 2010

after a popular outcry, we've added the Elluminate recordings back to the site, so those are now available in addition to the vimeo recordings.

[shelco07]

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.

(1) For each let's see if we can find solutions for copies sold on Friday where value is 10 or 12. If we can find a solution for both, the data provided is N/S.

(i) Find solution where Thurs = 8, Friday = 10, Sat = K, K>10, and no values are the same.

(M) + (T) + (W) + (TH) + (F) + (S) + (Su) = 90

(1) + (2) + (3) + (8) + (10) + ( K ) + (4) = 90 - 28 - K

K must be > 10, thurs = 8, and all days sell unique numbers of books. Therefore we have a solution which satisfies all requirements.

(ii) Find solution where Thurs = 8, Friday = 12, Sat = K, K>12, and no values are the same.

(M) + (T) + (W) + (TH) + (F) + (S) + (Su) = 90

(1) + (2) + (3) + (8) + (12) + ( K ) + (4) = 90 - 30 - K

Therefore K must be > 12, thurs = 8, and all days sell unique numbers of books. This also satisfies all requirements.

Since we can find a solution for but books sales on fri being both greater and less than 11, this information is N/S.

ELIMINATE A, D.


(2) We will use the same approach. For each let's see if we can find solutions for copies sold on Friday where value is 11 or 12. If we can find a solution for both, the data provided is N/S. If only one works than we are good to go.

(i) Find solution where Sat = 38, 11 < Fri < 38, all other values are less than friday, and each day sells a unique number of books:

(M) + (T) + (W) + (TH) + (F) + (S) + (Su) = 90

(M) + (T) + (W) + (TH) + (12) + (38) + (Su) = 90

This leaves us with 90-38-12 = 40 to spread between 5 days, where none can be the same.
40/5 = 8. Therefore we can use (8-2) (8-1) (8) (8+1) and (8+2)

(6) + (7) + (8) + (9) + (12) + (38) + (10) = 90

- Good to go.


(ii) Find solution where sat = 38, Fri = 11, all other values less than friday.

(M) + (T) + (W) + (TH) + (11) + (38) + (Su) = 90

This leaves us with 90-11-38 = 41
10+9+8+7+6 = 40 - Therefore not possible as 10, 9, 8, 7 and 6 are the greatest unique values all less than 11 that can be used.

Since we can not find a solution when Friday = 11, then we know that friday must be > 11.

Information Sufficient.

Ans: B

[RonPurewal]

shelco, that is ok.

random question: why'd you focus on the values 10 and 12 for the first statement, but then 11 and 12 for the second statement? that's an interesting shift in strategy -- you'd probably have an easier time, conceptually speaking, if you just looked for the same cases in both statements.

[ccamankulor]

Store sold a total of 90 copies during seven days and it sold different numbers of copies on any two of the days.

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.

1. Assuming 8 is the value of the third greatest number of copies,
the least value of the Friday and Saturday sales combined is
where the number of copies sold in the third to 6th greatest is
8+7+6+5.

74 copies would be sold over Friday and Saturday.

The saturday sale, as the greatest, is greater than the mean
74/2=37

Values for the friday sale, range from 9 to (74-38=36)

2. 38 copies sold on Saturday, 52 on the remaining days

n2 + R(n) = D, where

n2 = second greatest number of copies sold
R(n) = an approximating function
where n is the integer value of the 3rd greatest
number of copies and the numbers of copies are
consecutively summed from 3rd greatest to least.

D = the sum of the 2nd greatest, 3rd greatest....to the least.

Minimum value of n2
occurs at the maximum value of R(n)

At D = 52,
the greatest value for R(n) is 40 (R(10) = 10+9+8+7+6)

Least value for n2 is 12

*Note R2 is only a tuning function that allows approximation of R(n). Actual specified minimum of n2 depends on the "probability" spaces between the integer values. Consider the following counter. To the left is the probable value of n2 and to the right are indexed R(n) values.

12 [40] [10 9 8 7 6] -- or --12 [40] [11 10 9 8 5]

-1 +1

11 [41] [10 cannot move up]
[no currency/space built up at 12--10 9 8 7 6][diff. is 1]

Compare: gmat-prep-1-ds-six-countries-in-a-certain-region-sent-t2797.html

[RonPurewal]

ccamankulor, your contribution is appreciated. however, i believe that, by including such things as "tuning function", "probability spaces", and "r(n) values", you are unnecessarily complicating the issue.
after all, the solution to this problem doesn't require anything beyond simple arithmetic:
post10828.html#p10828

that doesn't necessarily mean that the problem is easy, but we don't want people thinking it's too much more complicated than it actually is.