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Juan bought some paperback books that cost $8 each and some [#permalink]

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09 Jul 2008, 22:04

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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 hardcover books that Juan bought was at least $150 (2) The total cost of all books that Juan bought was less than $260

Re: Juan bought some books ,Gmat prep question [#permalink]

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10 Jul 2008, 00:27

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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

Re: Juan bought some books ,Gmat prep question [#permalink]

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18 Jul 2011, 02:31

kevincan wrote:

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...

Re: Juan bought some books ,Gmat prep question [#permalink]

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18 Jul 2011, 11:45

Siddhans, the question states that Juan bought "more than 10 paper books" and not atleast 10. Also didnt quite get how you arrived that the total cost is 100 . Its told that the total cost is less than 260, given your way of working the total cost could also equal to 259, which leads to the right answer

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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Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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29 Jun 2013, 06:05

Great discussion just to point out when combined it can't be 7 books because the cost will be 263 so that's above the the required condition.
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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.
_________________

Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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11 Sep 2013, 21:54

VeritasPrepKarishma wrote:

GMAT98 wrote:

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.

ok..i got it. I should have applied hidden constraint. Thanks for prompt reply.

Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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18 Feb 2014, 18:06

Statement 1 alone only tells us that H>=6.

Statement 2 tells us that the total cost that is 25H + 8P<260, which together with P>=11 is not sufficient.

Both statements together tell us that H has to be 6. Given that P is the minimum possible value = 11, if H were 7 then total cost >260 and violate statement 2. Therefore C stands.

Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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18 May 2014, 10:40

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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.

Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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30 Aug 2014, 10:22

VeritasPrepKarishma wrote:

siddhans wrote:

kevincan wrote:

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.

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?

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?

Juan bought some paperback books that cost $8 each and some [#permalink]

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19 Sep 2014, 02:54

neeraj.kaushal 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 hardcover books that Juan bought was at least $150 (2) The total cost of all books that Juan bought was less than $260

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.

Re: Juan bought some paperback books that cost $8 each and some [#permalink]

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