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number of books

by clock60 » Sun Dec 12, 2010 11:41 am
hi guys please help with this one

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.
oa is B
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by shovan85 » Sun Dec 12, 2010 11:54 am
clock60 wrote:hi guys please help with this one

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.
oa is B
Did not find any better idea of solving it but kind of hit and trial.

(1) If you know about Thursday its of no use as Saturday is the highest and it can take any number.
Insufficient

(2) Now Saturday 38 Remaining 90 - 38 = 52 in rest 6 days
Friday is the highest in the rest 6 days.

Now minimize the condition

Say from Monday to Thursday 0 1 2 3 4 (Total = 10) sold as all days the sell is distinct. Remained 52 - 10 = 42 much greater than 11.

Now maximize the condition

52/6 = 8.66
In order to maximize take 8.66 as the median then the sequence will be 6 7 8 (Median = 8.66) 9 10 11
Now 6+7+8+9+10+11 = 51 Thus 1 book is remeining which ought to go in the count of Friday.
So in friday sell = 11+1 = 12.

In either cases the sell on Friday is greater than 11.
Sufficient
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by clock60 » Sun Dec 12, 2010 12:31 pm
hi shovan
thank you for quick reply, but have some perhaps silly queries
mean here 52/6=8,66. but in your example 6,7,8,9,10,11, mean=median=17/2=8,5 or it does not matter?
can you clarify a little bit more how to make set of integers that when summing gives max sum?
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by shovan85 » Sun Dec 12, 2010 12:46 pm
clock60 wrote:hi shovan
thank you for quick reply, but have some perhaps silly queries
mean here 52/6=8,66. but in your example 6,7,8,9,10,11, mean=median=17/2=8,5 or it does not matter?
can you clarify a little bit more how to make set of integers that when summing gives max sum?
It does not matter here. If you see the sum is comin 51 for the sequence which should be 52 (is not it?). This is the reason I consider 8.66 as the median.

Though 8.66 is the mean the word "Median" I used here just to EMPHASIZE that In a sequence of Consecutive Integer the Mean is equal to Median which resides exactly in the middle of the sequence.

For getting the maximum value: in this case we know the last one the MAXIMUM out of all here and the reason I took consecutive as I wanted all numbers to be distinct.

Say u know 11 is the maximum then how can you get the maximum values of other 5 integers just subtract 1 from the previous ones. This is the logic I applied here.

One more thing the other 5 values are not necessarily in sequence.

I must agree that concrete math cannot solve this but little bit intuition is required.
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by clock60 » Sun Dec 12, 2010 12:59 pm
it seems to me that i start to get the solution
if we have sum for the total 6 days=52 and we need to estimate whether friday is>than 11, let me consider that friday is less than 11 say 10 then possible set will looks like
5,6,7,8,9,10. with the sum=45. not possible as we need 52 and all numbers must be distinct
friday=11. and posible set
6,7,8,9,10,11. with the sum=51, again not possible we need 52
we need one more 1 to obtain 52 and as friday is the greatest. so no other days will be greater than friday, it will equal 11+1=12
and possible set
6,7,8,9,10,12 summing =52
Got it
thanks Shovan!!!
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