From my experience so far, the most important thing in DS questions is to take each statement separately. It also helps to pick the "easier" or "simpler" statement first, if you can identify one.

This is how I approach such questions. It may or may not work well for you so I can't guarantee anything. I pick "good" variables and simplify the information given in the question.

The question you have mentioned can be written as:

p = number of paperbacks
h = number of hardcovers
t = total cost
t = 8p + 25h
p > 10
h= ?

Lets approach statement 2 first. It says that the total cost was less than $260. Is this and the information given in the question enough to calculate the number of hardcover books bought? Think about it. All we know are the prices of hardcover and paperback books and the fact that Juan bought more than 10 paperbacks. He could have bought 13 paperbacks and 6 hardcovers or he could have bought 10-32(260/8) paperbacks and no hardcover. Hence, statement 2 is insufficient.

Lets look at statement 1 now. The english->math translation we did can help us here. It says that the total cost of hardcover books is atleast $150.

25h>=150
h>=6

Remember that right now (when we're assessing statement 1), statement 2 and any information in it doesn't exist for us! So we know that p>10 and h>=6, but as we have no clue about how much Juan spent, we don't know how many books he may have bought. He could have bought a million hardcovers and a billion paperbacks for all we know! Statement 1 is insufficient.

NOW, lets consider information in both the statements, together!
h>=6
p>10 (p>=11)
25h+8p<260

Lets say h=7 (minimum possible value for h is 6), 25*7=175. That leaves us $85, not enough money to buy 11 paperbacks. Hence, h has to be less than 7 for p to be more than 10.

The only other possibility left is h=6. When h=6, 25*6=150 leaving $110 for paperbacks (enough to buy more than 10 paperbacks).

The statement taken alone were insufficient but taken together solved our problem!

The answer is C!