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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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12 Nov 2015, 04:10

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

I don't understand we can not use both statements? i mean x = number of shelf total books = 10x total with additional books = 10 + 10x = 12x => x = 5 number old books = 5x = 50

what do you think ? what iam missing? what is the problem ?

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

I don't understand we can not use both statements? i mean x = number of shelf total books = 10x total with additional books = 10 + 10x = 12x => x = 5 number old books = 5x = 50

what do you think ? what iam missing? what is the problem ?

When the first statement is sufficient alone and the second is not, the answer is A. Also, you cannot use information given in one statement when evaluating another.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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12 Nov 2015, 07:37

to arrange the books on 12book-shelf without left overs, the new number of books must be a multiple of 12

multiple of 12: 12-24-36-48-60-72-84-96... 1 - tells us that he had fewer than 96 books, the only way he could have had before so that to arrange the books without leftovers is 50 Sufficient.

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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28 Nov 2015, 05:11

Sudhanshuacharya wrote:

fluke wrote:

386390 wrote:

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 book

My question is, You can figure out that he had 50 books just by reading the question itself. After he got 10 additional book, each shelf had 12 books. Meaning each shelf had +2. 2(x)=10 =5

Hence 5 shelves. Therefore originally, he had 10 books in each shelf = 50 books.

Whats the point of the statements? Or am I missing something?

Brilliant!!! Even I overlooked that.

IMO: This OG question is flawed. Your reasoning is perfect.

Fluke i don't think the question is flawed. The question says the books are equally divided by 10 and again by 12 after 10 more books are added. Hence lets check the factor of 10 and 12.

Factor 10: 2,5 Factor 12: 2,2,3 Factor N (number of books) = 2,2,3,5,???

Hence the number of after addition of 10 books can be a multiple of 60. So it can be 60, 120 ,180,etc with initial no. of books as 50,110,170,etc...

Hence from statement 1: no. of books is less than 96. Hence sufficient as we have only one value Statement 2: no of books greater than 24 so we can have more than one value hence insufficient.

So A.

This is so helpful, thank you. I had exactly the same question about what the point was when you can find x=5 just by reading the question stem. Now I got it.

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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03 Dec 2015, 10:15

Lets say first bookshelf has n1 racks wherein 10 books are arranged in each shelf=10 n1 assume second bookshelf has n2 racks wherein 12 books are arranged in each shelf=12 n2 we are told 10 n1 + 10 = 12 n2 therefore n2= (10n1 +10)/12 for integer values of n2; values of n1 are 5, 11,..... If n1= 5 there were 50 books If n1= 11 there were 110 books as given in (i) if books are to be less than 96 then n1 can have one definite value that is 5 and 50 books initially.....gives definite answer thus sufficient in (ii) if books are to be more than 24; n1 can have infinite possible values and no definite answer van be established

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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03 Dec 2015, 20:48

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Bunuel! I am missing one point in the last portion of your second statement.

" ...as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12". Isn't that the statement has to be x+10 to be a multiple of12".....120, 180, 240.....

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

We get 10n+10=12m(n: number of bookcases before, m: number of bookcases after acquirement) if we modify the question a little. If we divide both sides by 2, from 5n+5=6m, (n,m)=(5,5),(11,10), and we get a set of possible numbers as (before,after)=(50,60),(110,120),(170,180),... There are 2 variables (n,m) and one equation (10n+10=12m), whereas 2 equations are given by the 2 conditions, which makes (D) a likely answer. For condition 1, (50,60) is an unique set of answer, making the condition sufficient. For condition 2, (50,60),(110,120)... is not unique, making the condition insufficient and the answer (A).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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17 Jan 2016, 09:47

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Sir I don't think we need any of the above two statements B/c I can get the answer from the question stem only. Am I correct? Please clarify.

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Sir I don't think we need any of the above two statements B/c I can get the answer from the question stem only. Am I correct? Please clarify.

No you are not. Please show HOW you can get the answer from the stem only.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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17 Jan 2016, 10:44

robu wrote:

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Sir I don't think we need any of the above two statements B/c I can get the answer from the question stem only. Am I correct? Please clarify.

here is my analysis:

let No. of shelf=N total books before addition =10N

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Sir I don't think we need any of the above two statements B/c I can get the answer from the question stem only. Am I correct? Please clarify.

here is my analysis:

let No. of shelf=N total books before addition =10N

after addition Books =10N+10

10N+10=12N N=5

hence before addition, books= 10x5=50.

please correct if i m wrong /missing something.

thanks in advance.

You should read the questions more carefully: Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

We cannot say whether the bookcases have the same number of shelves.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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17 Jan 2016, 11:19

robu wrote:

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Answer: A.

Sir I don't think we need any of the above two statements B/c I can get the answer from the question stem only. Am I correct? Please clarify.

yes sir you are very correct. there will be huge penalty for carelessness. thanks for the advice.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

When you modify the original condition and the question, they become 10n+10=12m(Michael arranged 10 books initially and after acquiring 10 additional books, he arranged 12 books). Then, (10n,12m)=(50,60), (110,120), (170,180), ..... There are 2 variables(n,m) and 1 equation(10n+10=12m), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), only (10n,12m)=(50,60) is derived, which is unique and sufficient. For 2), (10n,12m)=(50,60),(110,120),... are derived, which is not unique and not sufficient. Therefore, the answer is A.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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03 Jun 2017, 21:31

tonebeeze wrote:

Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

The goal is to find the original number of books that Michael owned. We know that when stored in one bookcase, the number of books was a multiple of 10 and in the new bookcase with 10 new books it was a multiple of 12. So the total number of books must be a multiple of 60 (LCM of 10 and 12).

Statement 1) B < 96

Multiples of 60: 60 120 180 ...

Only one of these values is < 96. Sufficient.

Statement 2) B > 24.

Multiples of 60: 60 120 180

Insufficient. An infinite number of multiples of 60 are > 24.

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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08 Sep 2017, 22:54

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12: 110, 170, 230, ... Not sufficient.

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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12 Oct 2017, 08:32

Bunuel wrote:

tonebeeze wrote:

Hey all,

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12:110, 170, 230, ... Not sufficient.

Answer: A.

Hi Bunuel - i assumed the number of shelves was constant...why is that not a fair assumption to make ?

In S2.... you said 110 is a possibility for X

if 110 --- thats 10 books each with 11 shelves 120 ----- thats 12 books each with 10 shelves

How can you assume the number of shelves between the book cases is different ? i assumed it would be the same and hence got this answer wrong ..

I got this problem correct, but I spent way too much time on it. I was looking for an alternate method of solving this problem.

Thanks!

82. Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

Let the # of books Michal had before he acquired the 10 additional books be x. Then x must be multiple of 10 as Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. For the same reason x+10 must be a multiple of 12 (basically x+10 must be multiple of 60).

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books --> x<96, so x can be 10, 20, 30, ... 90. From these possible values of x there is only one for which x+10 is a multiple of 12: x=50 --> x+10=60. Sufficient.

(2) Before Michael acquired the 10 additional books, he had more than 24 books --> x>24, so x can be 30, 40, 50, ... But as the upper limit of x is not limited there will be infinitely many values of x possible so that x+12 to be a multiple of 12:110, 170, 230, ... Not sufficient.

Answer: A.

Hi Bunuel - i assumed the number of shelves was constant...why is that not a fair assumption to make ?

In S2.... you said 110 is a possibility for X

if 110 --- thats 10 books each with 11 shelves 120 ----- thats 12 books each with 10 shelves

How can you assume the number of shelves between the book cases is different ? i assumed it would be the same and hence got this answer wrong ..

I'd suggest to read the whole discussion before posting a question.
_________________

Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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31 Oct 2017, 08:33

tonebeeze wrote:

Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. How many books did Michael have before he acquired the 10 additional books?

(1) Before Michael acquired the 10 additional books, he had fewer than 96 books.

(2) Before Michael acquired the 10 additional books, he had more than 24 books.

before having more 10 book, Mi has 10 k book after having 10 books, Mi has 10k+10=10(k+1) to make 10(k+1)/12= interger, 5(k+1)/6 must be interger, so k+1 must be divided by 6 this means k=5, 11