Yes the correct answer should be C. Because, lets assume he bought x paperback books, and y hardcover books. then your equation becomes :

8X+25Y = total money he spent. (we do not have info about this).
lets say he bought 11 paperback books ( i.e. more than 10 books) so the equation becomes 8x11+25Y= ?

Ok now when we check out (1) it says that cost for hardcover was at least $150, which does not help us out to solve the equation above. So (A) and (D) are out.

(2) Gives info about the total amount he spent was less than $250, which is again not sufficient to solve the equation as we may have any no. of vaules.

When we combine both equations then for certian we know the the total amount he spent was less than 260 and the $$ amt spent on hardcover book was less then 150. So we know the limits and hence it is solvable. ( you do not have to solve the equation) ........
I hope this is the way to solve this question any suggestions??

ruchisharma wrote: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 at least $150.
2.)The total cost of all books that Juan bought was less than $260.

How is C the correct answer?

8P + 25H = total amount spent.
P>10
H=??
1.) The total cost of the hardcover books that Juan bought was at least $150.
25H>=150
No exact info of H is available.
hence INSUFF

2.)The total cost of all books that Juan bought was less than $260.
8P + 25H<260
Still no definite info about H.
Hence INSUFF

Combining:
8P+25H<260
25H >=150
Hence 8P <=110
P<=13 [taking integer value]
Now it is given P>10
Hence P can be 11 or 12 or 13
When P is 11,
25H<172
H=6 [Integer value]
When P=12
25H<164
H=6 [Integer Value]
When P=13
25H<156
H=6 [Integer Value]

Hence for all the three case, H is only 6.

pick C.

I believe this is difficult one. Difficult to calculate and check in the actual test pressure.

@kentandoshi, I still dont see the logic in which you mentioned that we can solve the number for the hardcover book based on the information that it limits within the limit of 150 and 260

8P + 25H < 260

And we know 25H >= 150

So we know the total cost is between 150 and 260

And we can do increment 150, 175, 200, 225, 250 for the hardcover cost and deduce what the paperback cost..

Anyway, the other person's solution makes sense, but there is way too complicated. There is no way I can ever do this within the time limit

I was wondering if there is a simpler solution

ruchisharma wrote: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 at least $150.
2.)The total cost of all books that Juan bought was less than $260.

How is C the correct answer?

No of paperback books = p
No of hardcover books = h

8p + 25h

p>10

h=?



1.) The total cost of the hardcover books that Juan bought was at least $150.
25h >= 150
h>=6

NOT SUFFICIENT

2.)The total cost of all books that Juan bought was less than $260.
8p + 25h < 260
p=11 h<7
p=12 h<7

NOT SUFFICIENT

Combining the two statements

From 1 we get, h>=6
From 2 we get, h<7
hence h= 6

Option C