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A total of 100 customers purchased books at a certain bookstore last week. If these customers purchased a total of 200 books, how many of the customers purchased only 1 book each? (1) None of the customers purchased more than 3 books. (2) 20 of the customers purchased only 2 books each.
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I don't get how the OA can be C. Together with both statements, we don't know how many customers buy exactly one book or how many bought exactly three books. Shouldn't the answer be E? Can someone explain?
I don't get how the OA can be C. Together with both statements, we don't know how many customers buy exactly one book or how many bought exactly three books. Shouldn't the answer be E? Can someone explain?
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read my equation for statement 1, the equation resolves to x = z. The number of people who purchased 1 book (x) is equal to the number of people who purchased 3 books (z). Together, they are 80, so 40 purchased 3 books and 40 purchased 1 book.
Stmt 1 and 2: if 20 people bought 40 books then remaining 80 people bought a total of 160 books. But these 80 people bought either 1 or 3 books (since number for 2 is already deducted).
Thus, if number of people who bought 1 book is n then n*1 + (80-n)*3 = 160
I don't get how the OA can be C. Together with both statements, we don't know how many customers buy exactly one book or how many bought exactly three books. Shouldn't the answer be E? Can someone explain?
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we don't know how many customers buy exactly one book or how many bought exactly three books
>>>yes we can find the information
Given - # of customers - 100 (1) None of the customers purchased more than 3 books. ->customers can purchase 1or 2 or 3 books ->INSUFF (2) 20 of the customers purchased only 2 books each. ->This tells us how 40 books were sold.What about theremaining 160 books ->INSUFF
Combining 1 and 2
Let m - nos of customer who bought 1 book we know 20 customers bought 2 book
Therfeore 100 -(m+20) bought 3 books
m*1 + 20*2 + [100 -(m+20) ]*3 = 200
Now you can solve for m.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.