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Michael arranged all his books in a bookcase with 10 books [#permalink]
10 Dec 2010, 15:10

1

This post was BOOKMARKED

00:00

A

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E

Difficulty:

65% (hard)

Question Stats:

61% (02:21) correct
39% (01:30) wrong based on 106 sessions

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.

Re: Quant Review V.2 DS #82: Michael's Book Case [#permalink]
10 Dec 2010, 15:45

3

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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.

Michael arranged all his books in a bookcase with 10 books [#permalink]
19 Jul 2011, 08:58

2

This post received KUDOS

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? _________________

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. _________________

He could also have 110 books initially.. so arranged in 11 shelves each having 10 books

lets assume this...

agdimple333 wrote:

when 10 additional books are acquired.. 120 books can be arranged in 10 shelves each having 12 books...

I dont think you can just change numbers like that.

Say he had 110 books initially, arranged in 11 book shelves. When +10 books are put in the SAME book shelf you CANNOT have 12 books per shelf. The only way to distribute 10 books in 2's is if there were 5 shelves. _________________

He could also have 110 books initially.. so arranged in 11 shelves each having 10 books

lets assume this...

agdimple333 wrote:

when 10 additional books are acquired.. 120 books can be arranged in 10 shelves each having 12 books...

I dont think you can just change numbers like that.

Say he had 110 books initially, arranged in 11 book shelves. When +10 books are put in the SAME book shelf you CANNOT have 12 books per shelf. The only way to distribute 10 books in 2's is if there were 5 shelves.

There is NO Assumption. I almost first thought of 50 books... but when i saw the source being official guide, i am sure there could not be any flaw in the question.

for your above statements, please read the question again, it says he arranges in NEW BOOK CASE. no where its mentioned in the question that # of book shelves in both books cases are same. i just gave another example of 110 to clarify the doubt... but if you want to use algebra..

lets say he originally had x books.. he arranged 10 books on each shelf, so x is multiple of 10. later he has x+10 books, he arranged 12 books in each shelf... so x+10 is multiple of 12.

statement 1 - he has less than 96 books.. so x could be 10,20, 30, 40, 50, 60.. then but x+10 could be only 60.. (since it is multiple of 12) x+10 = 60 x = 50 sufficient.

statement 2 - he has more than 24 books..

so x could be 30, 40, 50, 60, 70, 80, 90, 100, 110, 120.... 180, 190 and so on... x+10 could be now 60, 120, 180... so x could be 50, 110, 170 not sufficient.

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.

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.

Re: Quant Review V.2 DS #82: Michael's Book Case [#permalink]
15 Aug 2011, 10:02

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? _________________

Re: Quant Review V.2 DS #82: Michael's Book Case [#permalink]
15 Aug 2011, 10:08

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

Re: Quant Review V.2 DS #82: Michael's Book Case [#permalink]
15 Aug 2011, 18:18

what if there are left overs..let's say this:

Michael arranged all his books in a bookcase with 10 books on each shelf with 2 left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf with 3 leftovers.

Can someone help me understand how to solve such a question?

Re: Quant Review V.2 DS #82: Michael's Book Case [#permalink]
16 Aug 2011, 11:19

DeeptiM wrote:

what if there are left overs..let's say this:

Michael arranged all his books in a bookcase with 10 books on each shelf with 2 left over. After Michael acquired 10 additional books, he arranged all his books in a new bookcase with 12 books on each shelf with 3 leftovers.

Can someone help me understand how to solve such a question?

Michael arranged all his books in a bookcase with 10 books on each shelf with 2 left over. => Number of books, N1 = 10M+2 ; M is an integer he arranged all his books in a new bookcase with 12 books on each shelf with 3 leftovers. => Number of books, N2 = 12K+3 ; K is an integer N2 comes After Michael acquired 10 additional books in N1 => N1+10 = N2 => 10M+2+10 = 12K+3 => (10M+9)/12 = K

We need to see, if we can find one and only value of M so that (10M+9) is divisible by 12

Statement I : Before Michael acquired the 10 additional books, he had fewer than 96 books

N1<96 => 10M+2<96 => M < 9.4 => so possible values of M = {1,2,3,4,5,6,7,8,9}

none of these possible values satisfy the condition (10M+9) is divisible by /12

So Not - sufficient

Statement II : (Before Michael acquired the 10 additional books, he had fewer than 96 books)

N1>24 => 10M+2>24 => M > 2.2 => so possible values of M = {3,4,5,6,7,8,9......Infinite set}

This is an Infinite set of possible values,

So Not - sufficient either

Combining both of these also doesn't give us anything which could give us any possible value of M that could satisfy the required condition.

So answer of this modified question would be E _________________

Michael arranged all his books... [#permalink]
13 Nov 2012, 22:28

Michael arranged all his books in a bookcase with 10 books on each shelf and no books leftover. 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.

In my mind, you don't even need the options. The answer is that he had 50 books before he added 10 new books. The OA does not agree with me. Why can't you automatically determine that the answer is 50? What am I missing?

Re: Michael arranged all his books... [#permalink]
13 Nov 2012, 22:36

Expert's post

egiles wrote:

Michael arranged all his books in a bookcase with 10 books on each shelf and no books leftover. 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.

In my mind, you don't even need the options. The answer is that he had 50 books before he added 10 new books. The OA does not agree with me. Why can't you automatically determine that the answer is 50? What am I missing?

Its a case of LCM. Initially Michael arranged the books in a shelf of 10 with no left overs. MULTIPLE OF 10 After Michael acquired 10 additional books, he arranged those in a shelf of 12 with no left overs. A MULTIPLE OF 12 Therefore the question is asking the desired LCM of 10 and 12 i.e. 60, 120, 180 etc Statement 1)Before Michael acquired the 10 additional books, he had fewer than 96 books. The only LCM of 10 and 12 under 96 is 60. Hence we have sufficient information to answer the question. Statement 2)Before Michael acquired the 10 additional books, he had more than 24 books. More than 24 could be anything in thousands, millions which are an LCM of 10 and 12. Clearly Insufficient Hence A _________________

Re: Michael arranged all his books... [#permalink]
13 Nov 2012, 23:31

1

This post received KUDOS

egiles wrote:

Michael arranged all his books in a bookcase with 10 books on each shelf and no books leftover. 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.

In my mind, you don't even need the options. The answer is that he had 50 books before he added 10 new books. The OA does not agree with me. Why can't you automatically determine that the answer is 50? What am I missing?

We have 10x + 10 = 12y i.e 5(x+1) = 6y

Note that we are not told that he arranges the books in the same number of shelves. Hence the number of shelves before the purchase and no. of shelves after the purchase are unknown.We can have several integer values of x giving integer values for y. Eg: x = 5, y = 5, or x = 11, y = 10 and many more. But you can see that the question statement alone is not sufficient. We need a statement limiting the values that x and hence y can take.

1) Only x = 5 and y = 5 satisfy this condition. The next lowest values of x can only be 11 for which 10x will exceed 96. Sufficient.

2)Insufficient. x can be 5,11 or even several other numbers greater.

Answer is hence A.

Kudos Please... If my post helped. _________________

Did you find this post helpful?... Please let me know through the Kudos button.

Re: Michael arranged all his books... [#permalink]
14 Nov 2012, 01:46

Expert's post

egiles wrote:

Michael arranged all his books in a bookcase with 10 books on each shelf and no books leftover. 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.

In my mind, you don't even need the options. The answer is that he had 50 books before he added 10 new books. The OA does not agree with me. Why can't you automatically determine that the answer is 50? What am I missing?

Merging similar topics. Please refer to the solutions above. _________________

Re: Michael arranged all his books in a bookcase with 10 books [#permalink]
26 Feb 2013, 01:32

not easy at all, want to follow to remember this question. _________________

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Re: Michael arranged all his books in a bookcase with 10 books [#permalink]
24 Sep 2014, 19:46

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