Juan bought some paperback books that cost $8 each and some

Juan bought some paperback books that cost $8 each and some hard cover 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 hardcover books that Juan bought was atleast 150$ He bought more than 6 hardcover books.
NOT SUFF
2 The total cost of all the books that Juan bought was less than $260
He spent at least $88 on paperback books, and thus less than 260 - 88 = $172 on hardcover books. This tells us he bought at most 6 hardover books

Together, we know he bought exactly 6 hardcover books. SUFF

I agree with 1

but in stmt 2 : How do we get 88 ? Juan bought atleast 10 paperback books so it has to 80 ...Next the total cost is less than 260 ..Lets say total cost is 100 ..

Then 100 - 80 = 20 ...So juan bought 0 hard cover books...

Combined do we get the answer still 6 here ???

As pointed out by esbee above, Juan bought more than 10 paperback books so he must have bought at least 11 paperbacks. This means he must have spent at least $88 on the paperbacks.

Statement 2 alone is not sufficient. All it tells us is that he spent less than $260. This means he spent less than $260 - $88 = $172 on hardcover books. He could have spent $88 + $50 = $138 OR $98 + $100 = $198 OR some other combination. We do not know how many hardcover books he bought but we do know that he could not have bought more than 6 since for 7 (or more) hardcover books, he would need $25*7 = $175 (or more). But he spent less than $172 on hardcover books. So he bought at most 6 hardcover books.

Using both statements together, statement 1 says that he bought at least 6 hardcover books and statement 2 says he bought at most 6 hardcover books. What does this mean? This means he must have bought exactly 6 hardcover books.
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HI CAN SOMEONE PLEASE EXPLAIN WHY ADDING INEQUALITIES DOES NOT WORK IN THIS QUESTION

let the quantity of hardcover books be H and paperback P

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

Statement 2

8(P)+25(H) < 260 AND
10< P

MULTIPLYING BY 8 AND ADDING IF WE SOLVE

H is less than 7.2 insuff

Statement 1 and 2 (H must be an integer)

and greater than or equal to 6,

therefore H can be 6 or 7.

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

Because the inequality 10 < P does not include all the information you have available. You know that P cannot be 10.5 i.e. a decimal. It must be an integer since it represents the number of paperbacks. So you might want to use 11<= P to get a tighter value for H.

8P + 25H < 260
88 <= 8P

When you add, you get 25H < 172 i.e. H < 6.9

All in all, algebra will always give you the correct result; but it may not be sufficient if you don't use reasoning as well. You got that H is less than 7.2 which is correct but it was not enough.
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Statement 1: Juan spent at least $150 on Hardcovers. So that means he bought at least 6 harcovers. We have no upside limit though.

Insufficient.

Statement 2: Juan spent less than $260 total on books. This tells us at most that Juan spent $259 - $88 (minimum amount possible spent on paperbacks) = $171 maximum on hardcovers. 171 divided by 25 is 6 and change. So round down. The maximum number of harcovers that could be purchased is 6. This statement could also be true with any number of hardcovers below six being purchased. For example, 0 hardcovers + 12 paperbacks = $96, which is less than $260.

Insufficient.

Together:

Looking at Statement 1, we know that at least 6 hardcovers had to be purchased. Looking at Statement 2, we know that no more than 6 hardcovers could be purchased. So Juan must have purchased 6 hardcovers.

Hi Karishma,

These exact type of problems give me a lot of trouble but I'm definitely getting better at them. I just get overwhelmed with information despite making charts etc and before I know it, it's 3+ minutes in and i'm stuck! What a complete waste of time.

Do you have a post that tackles on how to deal with these problems, perhaps algebraically.

Looking at the solution posted above, I can see how you came to the actual answer. That's never my problem. I can always FOLLOW the work, but for some reason, I can't start it myself. How do I fix this?

For this problem, I created a table but it ended up in shambles. Would really appreciate your help here.

Statement 1/2 insufficient solo. I'm always skeptical of C so I guessed E at the end of 3 minutes, like I said, horrible waste of time. My equations were as such.

Statement 1: (x=# of hc books)(25) < 150
Statement 2: (x)(25) + (11)(8)<260

I tried to see what value will make statement 2 valid. I noticed that 8(whatever number) has to end in a 0 to make this valid and it has to be greater than 88, so I had 120 and 160.

That made x only valid if the other value was 160, therefore, x had to 4. This is completely off than the explanation above. Where did I go wrong?

I have discussed this here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/09 ... ct-answer/
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1) 25h>=150
Sol : h>=150/25 => h>=6
it can be from 6 to +infinite. not a definite answer. so, NOT SUFFICIENT ALONE.

2) 8 p + 25 h < 260
sol: its been mentioned that p>10.

lets say that she bought only one hardcover, h=1, makes p=29

but "p" can be any number from 11 to 29 (price of p=8*11=88 to 8*29=232) : (lets say that she bought only one hardcover, h=1, makes p=29)
depending upon "p", value of "h" can vary from 1->6 (since p>10, so if p=11, h=6)
so this again is NOT SUFFICIENT ALONE.

NOW LETS TAKE BOTH statements 1 and 2:

1) h>=6
2)1<= h <=6

so h=6." C " is the answer

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


To find the GCF of two numbers:
1) List the prime factors of each number.
2) Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1.
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We are given that Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each, and that he bought more than 10 paperback books. If we let p = the number of paperback books Juan bought and h = the number of hardcover books, the total cost of the paperback books is 8p, the total cost of the hardcover books is 25h, and the total cost of all books is 8p + 25h.

We need to determine the value of h, the number of hardcover books purchased by Juan.

Statement One Alone:
The total cost of the hardcover books that Juan bought was at least $150.

Using the information in statement one, we can create the following inequality:

25h ≥ 150

h ≥ 6

Thus, at least 6 hardcover books were purchased. However, we cannot determine the exact number of hardcover books purchased by Juan. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The total cost of all the books that Juan bought was less than $260.

Using the information in statement two, we can create the following inequality:

8p + 25h ≤ 260

We still do not have enough information to determine how many hardcover books were purchased by Juan.

Statements One and Two Together:

From the given information as well as our two statements, we know that p > 10, h ≥ 6, and that 8p + 25h ≤ 260.

Since p > 10, Juan, at minimum, purchased 11 paperback books. If we use 11 for p in the inequality 8p + 25h ≤ 260, then we have:

88 + 25h ≤ 260

25h ≤ 172

h ≤ 6 22/25

Recall that h ≥ 6, and since h has to be a whole number, Juan must have purchased 6 hardcover books.

Answer: C
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Dear IanStewart,

Do you recommend using algebraic approach in solving inequality for whole number items?

Let's #paperback books = p, and #hardcover books = h.
p > 10
h = ?

Statement 1: 25h ≥ 150 ---> h ≥ 6
Statement 2:

260 > 8p + 25h

Since
p > 10 *** Here's the trap I've fallen. It should be p ≥ 11***
8p > 80
260 > 8p + 25h > 80 + 25h
260 > 80 + 25h
7.2 > h

When combining both statements, I've got h = 6 or 7. (Hence, E -- most popular choice, btw)
The trap is very subtle.

It very much depends on the question. Sometimes algebra is truly necessary, but any time a conceptual solution is available, it will be faster than an algebraic one.

In this question, there's no reason at all to spend time doing algebra. Each Statement is clearly insufficient alone, and using both, Juan spent at least $88 on paperbacks, and bought at least 6 hardcovers for $150. That already accounts for $238, and adding one more $25 hardcover would put him over $260, so he must have bought 6 hardcovers.
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Info from stem : Juan spent at least $88 on paperback books.

1) Juan bought at least 7 hardcover books. That means, she spent at least $175 on hardcover books. Not sufficient

2) Juan spent at most 266 - 88 = $178 on hardcover books. She can bought any number of H.B from 1 to 7. Not sufficient.

Together, Juan bought 7 hardcover books. Sufficient.

C is the answer.

Hi Karishma please correct me on my logic from statement one we get that number of hard books bought is > or = 6

Now statement two says 8p+25h<260
Now least number of paperbacks Juan could have bought is 11 this implies least amount of money he could have spent on paperback is 88 this maximum he could have spent on hardback would be less than 260-78= 172

25h<172 thus h< 172/25 this h< 6.8 as number of books bought has to be integer maximum number of books bought has less than or equal to 6

Together 1 and 2

Statement 1 hard books h =or> 6
Statement 2 less then or equal to 6

Thus 6

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