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store S selling books

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by tasshy06 » Tue Aug 17, 2010 4:44 am
Store S sold a total of 90 copies of a book during 7 days 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 copies on Sat. and and the 2nd greatest number of copies on Fri., did store S sell more than 11 copies on Friday?

(1) Last week store S sold 8 copies on Thurs.
(2) Last week store S sold 38 copies on Sat.
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Source: — Data Sufficiency |
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by gmatmachoman » Tue Aug 17, 2010 5:51 am
Pick B

St2 : Last week store S sold 38 copies on Sat.

Sun Mon Tue Wed thu Fri Sat

WCS 5 6 7 8 9 11 38 = 84 < 90

Since friday is to be second largest day for sales, it has to be greater than 11

BCS : 10 9 8 7 6 12 38 = 90

St 2 is suuficient to say that friday sales is more than 11 for the stated conditions to be valid

Coming to st 1

we can easily make the stated conditions invalid ; let see so,

WCS : 4,5,6,7,8,9,51 = 90 ( here friday sales is less than 11)

case 2: 5 6 7 8 9 17 38 = 90 ( Here friday sales greater than 11.....
)

Insufficient

pick B
Last edited by gmatmachoman on Tue Aug 17, 2010 6:17 am, edited 1 time in total.
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by Gurpinder » Tue Aug 17, 2010 6:00 am
Stmt 1: Insuff

All we know is that 8 copies were sold on Thursday. We know nothing about the rest of the 82 copies.

Stmt 2: Insuff

Again, now all we know is that the greatest number of copies sold were 38. We know nothing about the rest!

Together:

8 on Thurs
38 on Sat - LARGEST

So now we have 46-90=44 copies left and 5 days in the week left. The greatest copies sold in a day CANNOT exceed 38. And the copies sold on any 2 given days is to be different! And friday must have the 2nd largest copies sold!

So lets distribute:
fri --11
sun --10
mon --9
tues--8
wed--6

in this case, store S sold exactly 11 copies.

lets try a different distribution:
fri --10
sun --9
mon --8
tues--7
wed--10 <<<< doesent work.

so the only way this works is if we choose the last distribution.

Therefore the answer is (C). And with that we know that store S did not sell more copies than 11
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by sanju09 » Tue Aug 17, 2010 6:03 am
gmatmachoman wrote:Pick B

St2 : Last week store S sold 38 copies on Sat.

Sun Mon Tue Wed thu Fri Sat

WCS 5 6 7 8 9 11 38 = 84 < 90

Since friday is to be second largest day for sales, it has to be greater than 11

BCS : 5 6 7 8 9 17 38 = 90

St 2 is suuficient to say that friday sales is more than 11 for the stated conditions to be valid

Coming to st 1

we can easily make the stated conditions invalid ; let see so,

WCS : 4,5,6,7,8,9,51 = 90 ( here friday sales is less than 11)

case 2: 5 6 7 8 9 17 38 = 90 ( Here friday sales greater than 11.....
)

Insufficient

pick B
This is the pick of the day, if B is your pick, then explain Statement 2 first. I'm too with B.
The mind is everything. What you think you become. -Lord Buddha



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by Gurpinder » Tue Aug 17, 2010 6:18 am
whats the OA
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by sanju09 » Tue Aug 17, 2010 6:20 am
Gurpinder wrote:Stmt 1: Insuff

All we know is that 8 copies were sold on Thursday. We know nothing about the rest of the 82 copies.

Stmt 2: Insuff

Again, now all we know is that the greatest number of copies sold were 38. We know nothing about the rest!

Together:

8 on Thurs
38 on Sat - LARGEST

So now we have 46-90=44 copies left and 5 days in the week left. The greatest copies sold in a day CANNOT exceed 38. And the copies sold on any 2 given days is to be different! And friday must have the 2nd largest copies sold!

So lets distribute:
fri --11
sun --10
mon --9
tues--8
wed--6

in this case, store S sold exactly 11 copies.

lets try a different distribution:
fri --10
sun --9
mon --8
tues--7
wed--10 <<<< doesent work.

so the only way this works is if we choose the last distribution.

Therefore the answer is (C). And with that we know that store S did not sell more copies than 11
The most important part of this question is to assume five different non-negative integers each less than 11, which could add up to 90 - (38 + 11) = 41 or more. We cannot be repetitive at the same time. If we can find such a set of 5, then Friday sales could have been 11 too; but if we cannot find such a set of 5, then Friday sales got to be more than 11.

So let's try our max, take 10 + 9 + 8 + 7 + 6 = 40, just falling short of 41, so yes, Friday sales got to be more than 11.

[spoiler]B[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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by gmatmachoman » Tue Aug 17, 2010 6:22 am
sanju bhai,

I did explained my st2 .plz have a look.

I said Best case scenario will be like :

BCS : 10 9 8 7 6 12 38 = 90 ....


Friday will be more than 11 sales.. Bhai its my 2000 post..so all the credit goes to you when u "Ping a Thank button just infront of u"..remember yesterday Shewag was denied a century....but in my case i can make this 2000 post very much memorable....

@Sanju ..remember..ur 50 half ton thanks was from me..LOL!!
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by tasshy06 » Tue Aug 17, 2010 6:53 am
the answer is B

thank you for the responses!
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by sanju09 » Wed Aug 18, 2010 1:48 am
gmatmachoman wrote:sanju bhai,

I did explained my st2 .plz have a look.

I said Best case scenario will be like :

BCS : 10 9 8 7 6 12 38 = 90 ....


Friday will be more than 11 sales.. Bhai its my 2000 post..so all the credit goes to you when u "Ping a Thank button just infront of u"..remember yesterday Shewag was denied a century....but in my case i can make this 2000 post very much memorable....

@Sanju ..remember..ur 50 half ton thanks was from me..LOL!!
yeh Kaikeyi ki tarah mujhe mere vachan yaad mat dilaao Govind Moshai, lo de ditta Thank twaanu, 2000 mubaarkaa.n!!

No need to mention your favors mate, here comes one Thank for you, 2000 congrats!!
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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by gmatmachoman » Wed Aug 18, 2010 3:20 am
Sanju ,

hi...na na..i just said that it will be so nice when u ping one thanks for me...Small moments to feel good...

C ya...
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