Lets say Fri is lesst than 12 = 11moneyman wrote:Wats the fastest approach here ?
then from mon - thurs we will have sales < 11
so maximum sum can be 11+10+9+8+7+6=51
so Sat=90-51 = 39
from B we know that Sat < 39 so fri>11
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Source: Beat The GMAT — Data Sufficiency |
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netigen
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then from mon - thurs we will have sales < 11
so maximum sum can be 11+10+9+8+7+6=51
so Sat=90-51 = 39
from B we know that Sat < 39 so fri>11
Q says the store sold 90 copies over a period of 7 days with Sat had max sales and Friday the second most
Stmt 1 - On thursday the store sold 8 copies. So on Friday the store must sell more than 8 it can be 10 or it can be 12 or any other no to satisfy saturday being the highest. NOT SUFF
Stmt 2 - On sat the store sold 38 copies. No of copies sold from sunday to friday = 90-38 = 52. IMO, the best approach will be to see what happens if the store sells only 11 copies on Friday.
If we assume the store sold 11 copies on Friday no of copies sold from Sunday to Thursday = 52-11 = 41. Since, each day the no of copies sold is different that means the store would have sold, best case scenario, 10+9+8+7+6 = 40 copies. Means, the store did sell more than 11 copies on Friday. SUFF
ANS B
Stmt 1 - On thursday the store sold 8 copies. So on Friday the store must sell more than 8 it can be 10 or it can be 12 or any other no to satisfy saturday being the highest. NOT SUFF
Stmt 2 - On sat the store sold 38 copies. No of copies sold from sunday to friday = 90-38 = 52. IMO, the best approach will be to see what happens if the store sells only 11 copies on Friday.
If we assume the store sold 11 copies on Friday no of copies sold from Sunday to Thursday = 52-11 = 41. Since, each day the no of copies sold is different that means the store would have sold, best case scenario, 10+9+8+7+6 = 40 copies. Means, the store did sell more than 11 copies on Friday. SUFF
ANS B