sukriteez Wrote:Hey Ron,
Reopening this thread yet again.. is the answer to the question then E?
no, it's (c). that part was just a proof that the second statement by itself is insufficient.
if you have the two statements together:
* it's definitely possible to have exactly 6 hardcover books. (you can establish this by testing cases -- but you can also realize that this is the minimum allowed value, so it MUST work. if it were impossible to have 10 hardcover books under these restrictions, then
no combination of books would work; they don't write vacuous problems that don't have solutions.)
* if you tried to have 7 hardcover books, that's $175 worth of hardcover books. since you also have to have more than 10 (so 11 or more) paperback books, that's at least 11 x 8 = $88 worth of paperback books. so, in that case, your total is at least 175 + 88 = $263, violating the condition that the total be no more than $260.
therefore, it's impossible to have 7 (and thus,
a fortiori, also impossible to have more than 7) hardcover books. so, with both statements together, there must be exactly 6 hardcover books.
[edited. the original post contained some arithmetic mistakes; i've corrected those mistakes, and simply deleted the posts pointing the mistakes out so as to minimize potential confusion
--ron]