I found this question in one of the practice tests.

Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each. If Juan bought more than 10 paperback books, how many hardcover books did he buy?

(1) The total cost of the hardcover books that Juan bought was atleast $150
(2) The total cost of all the books that Juan bought was less than $260

I thought answer would be (e) but turned out the ans is (c)

From the above 2 conditions, I get a range for the number of hardcover books.

From Stem : Price Paperback = $ 8
Price Hardcover = $ 25 , Number paperback >10 , P = { 11, 12, ....}
P and H are integers

St1>
25* Number Hardcover >=150
Number Hardcover > =6 Then , H = { 6, 7, ....}

Since question requires a unique value, INSUFFICIENT

St2> 8P + 25H < 260
Since Greatest Common Factor of { 8, 25, 260} is different from 1,
More than one solution , INSUFFICIENT

From St1 and St2 >

While P and H can only be integers

Plug the minimmum values for H and P
at 8P + 25H < 260,

Eg. ( P, H ) at 8P + 25H < 260,
( 11, 6 ) , 238 < 260, VALID
( 11, 7 ) 263 < 260, WRONG
( 12, 7 ) 271 < 260, WRONG

From this point, all other combinations will NOT comply the inequality.
Then , there is a unique value for H = 6

answer = C

Hey

From Stem : Price Paperback = $ 8
Price Hardcover = $ 25 , [b]Number paperback >10 , P = { 11, 12, ....}[/b]

:}

Nice work luc2r4. I just wanted to emphasize a point that you clearly picked up, because it's a pretty common issue. You wrote "while P and H can only be integers." Right, the OP came up with a range of values because he missed what we call a "hidden constraint," namely that in this question the variables must represent integers.

Check out pages 19-21 of Manhattan GMAT's Word Translations guide for more on this issue.

Reopening an old thread:

I did not quite understand the below statement:

St2> 8P + 25H < 260
Since Greatest Common Factor of { 8, 25, 260} is different from 1,
More than one solution

Isn't the GCD of 8 (2*2*2), 25 (5*5) and 260 (13*2*2*5) = 1

Moreover, could someone elaborate on the rule relating GCD and linear equations?

yeah, i'm not following that one, either. if this poster could come back and elaborate a bit, that would be a good thing.
also, whatever this poster is doing, it doesn't seem like a very good idea, since it doesn't take into account the additional restriction given in the problem statement (i.e., there must be more than 10 paperback books).

this is another one of those instances where people are going to the ends of the earth to come up with theory-based solutions for things that are much more easily investigated by simply TESTING CASES.
you are not expected to know advanced number theory on this exam, and, in the vast majority of these "integer solution" problems, such theory will make the problem much more difficult. in general, for problems that are restricted to integer solutions, TESTING CASES is the way to go unless there is a very obvious solution.

for the statement in question:
there must be more than 10 paperback books. so let's just start with 11 paperback books.
that's a total of $88 on paperback books.
subtract $88 from $260 --> we must have spent less than $172 on hardcover books.
this would allow any number of hardcover books up to 6, so the statement is insufficient.
that's it.

for the statement in question:
there must be more than 10 paperback books. so let's just start with 11 paperback books.
that's a total of $88 on paperback books.
subtract $88 from $260 --> we must have spent less than $172 on hardcover books.
this would allow any number of hardcover books up to 6, so the statement is insufficient.
that's it.[/quote]


Hey Ron,
Reopening this thread yet again.. is the answer to the question then E?

Hey Ron,
Reopening this thread yet again.. is the answer to the question then E? [/quote]

See the first post.Its C and not 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]

Thanks a ton!
Missed replying..

hmm? "missed replying"?
i don't understand -- please explain. thanks

Please point out the flaw in my approach. I followed the MGMT inequalities rules in handling this question..

let the quantity of hardcover books be x

Statement 1
25x is greater than or equal to 150
therefore x is greater than or equal to 6
Insufficient.

Statement 2

8(GT10)+25x < 260
Therefore GT 80 +25x<260
25x<LT180
x< 7.2

Statement 1 and 2 (x must be an integer)

x is less than 7.2 and greater than or equal to 6,

therefore x can be 6 or 7.

E..I can't wrap around my head how this approach can be wrong?

I guess u are missing the trick when it says that no. of Paperback is more than 10. What u are doing it taking 10 and then substracting.

The answer provided that Paperback equals 6 makes no sense, it says he bought more than 10 paperbacks.

yep, be sure to read the problem statement carefully. if there are any extra restrictions on the numbers in the problem, be sure to write them down on your paper, just as a reminder.