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Michael arranged all his books in a bookcase with 10 books on each she

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

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New post 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.

2- he could have 50, 110, etc. not sufficient.

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

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

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New post 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

Thus A is the correct answer.
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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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

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New post 05 Dec 2015, 13:09
1
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.
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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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

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New post 17 Jan 2016, 10:07
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.


No you are not. Please show HOW you can get the answer from the stem only.
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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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New post 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

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

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New post 17 Jan 2016, 10:52
robu wrote:
robu wrote:
Bunuel wrote:

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

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

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New post 17 Jan 2016, 22:46
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.
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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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

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New post 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.

Answer: A.


Clear answer. Thanks!

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

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New post 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

1) because 10k< 96, k=5

2> not sufficient
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Re: Michael arranged all his books in a bookcase with 10 books on each she [#permalink]

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New post 27 Nov 2017, 19:19
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.


We are given that Michael arranged all his books in a bookcase with 10 books on each shelf and no books left over, and that after he acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf and no books left over. We need to determine how many books Michael had before he acquired 10 additional books.

Using the given information, we can determine that Michael originally had a total number of books that was a multiple of 10, and after he acquired 10 new books, he had a total number of books that was a multiple of 12.

Statement One Alone:

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

The information in statement one is sufficient to answer the question. Since we know the original number of books was a multiple of 10, the number of books could have been the following:

10, 20, 30, 40, 50, 60, 70, 80, or 90.

Using the above numbers, he could have had the following number of books after acquiring 10 more:

20, 30, 40, 50, 60, 70, 80, 90, or 100.

Remember, after acquiring the 10 new books, the total number of books was a multiple of 12. Of the numbers above, only 60 is a multiple of 12. Thus, Michael originally had 50 books. Statement one is sufficient to answer the question.

Statement Two Alone:

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

We can analyze statement two in a similar way to that of statement one. Michael could have originally had any one of the following numbers of books:

30, 40, 50, 60, 70, 80, 90, 100, 110, 120, and so on.

Using the above numbers, he could have had any one of the following numbers of books after acquiring 10 more:

40, 50, 60, 70, 80, 90, 100, 110, 120, 130, and so on.

Once again, remember that after acquiring the 10 new books, the total number of books was a multiple of 12. Of the numbers above, 60 and 120 are both multiples of 12. Thus, Michael could have originally had 50 or 110 books. Statement two is not sufficient to answer the question.

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

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New post 28 Nov 2017, 01:15
eragotte wrote:
fluke wrote:
eragotte wrote:
I am not understanding something here.

X has to be divisible by 10, and x+10 has to be divisible by 12...

To me this only occurs when x=50, x+10 = 60

If x =100, 110 is not divisible by 12?


Maybe my brain is just broken atm?


Yeah, even I overlooked the surreptitious adjective "new" when I was solving it.

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?



for whatever reason this response made my brain start functioning again. thanks haha


Right but if x=110, then 120 is divisible by 12. Thus (2) is insufficient.
Re: Michael arranged all his books in a bookcase with 10 books on each she   [#permalink] 28 Nov 2017, 01:15

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