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

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?

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.
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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 there be x shelf so total initial number of books will be 10x and thus we need to fbd 10x

Also given that( 10x+10 )/12 is an integer lets say a

On solving we get 10(x+1)/12 =a or 5(x+1)/6= a now for a to be integer x+1 will have to be a multiple of 6

So possible values of x=5,11,17 and so on

St1 says 10x<96 or x< 9.6 where x is the no of initial books shelf. Now out of all the possible value of x only 5 satisfies the given equation so st1 is sufficient

St 2 says 10x>24 or x> 2.4 as we can see there are many values of x satisfying this inequality thus st2 is not sufficient

Ans is A
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Hello,

Just getting started with the GMAT prep!

Here's my first attempt.

From the stem, we get that LCM of 10 and 12 is 60. So, the number of books are anyone of 60, 120, 180, 240...etc

St1: Before Michael acquired the 10 additional books, he had fewer than 96 books.

Sufficient. This means he had 50 books. He cannot have 120-10 = 110 books because that is more than 96.

Down to A or D.

St2: Before Michael acquired the 10 additional books, he had more than 24 books.

Not sufficient. This means he could have had anyone of 50, 110, 170, 230..etc.

Answer (A).

Experts, please give feedback if my approach is right. Thanks!

Answer: A

The books should always remain in the multiples of 10.

Now it is important to note that after acquiring 10 additional books, the number of books should not only be a multiple of 10, but also of 12.

To solve, calculate the LCM of 10 and 12 -> 60, 120, 180...

Statement 1: Number of books is less than 96, which is sufficient to answer since we have only one LCM available less than 96 (i.e. 60)
Therefore he had 50 books earlier; after acquiring 10 more, he had 60.

Statement 2: Number of books is more than 24, which is not sufficient since we can have 120 or 180 or 240 ...so on.

Answer: A

Let X be the original number of books.
Since we can place 10 books on each shelf with no books left over, we know that X is a multiple of 10.

Let Y be the new number of books, once we add 10 books (i.e., Y = X+10).
Since we can now place 12 books on each shelf with no books left over, we know that Y is a multiple of 12.
Important: If X is a multiple of 10, we also know that X+10 is a multiple of 10. In other words, we know that Y is a multiple of 10.

So, we know that Y is a multiple of 10 and 12.

Since 60 is the least common multiple of 10 and 12, we can conclude that Y is a multiple of 60.

So, some possible values of Y are 60, 120, 180, 240, etc.
This means that some possible values of X are 50, 110, 170, 230, etc.

The target question asks us to find the value of X

Statement 1: X < 96
So, the original number of books must be 50.
SUFFICIENT

Statement 2: X > 24
So, the original number of books could be 50, 110, 170, etc
NOT SUFFICIENT

Hence A

Bunuel i have a doubt on this quesstion esp. Part B 10 additional books....equally distributed with no left over indicates 5 rows...earlier 10 books equally distributed....so 50....answer should be D .....whatz wrong with my logic here ?

I don't understand what you mean but the second statement is not sufficient because there are infinitely many values of x are possible. For example, 110, 170, 230, ...
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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. We can eliminate answer choices B, C, and E.

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 how we analyzed 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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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.

Situation 1-
Let us suppose earlier there were x no. of books
since 10 books are arranged in the book shelf and no book is left out
====> x is a multiple of 10

Situation 2-
10 additional books is acquired which is then arranged in a row. 12 books are arranged in a row.
=======> x+10 is a multiple of 12

When we connect situation 1 and situation 2-
since no. of shelf has not increased and no of books has increased by 10 or 2 books per row
=====> No. of row * 2=10
=====> No. of row=5

Thus we can calculate the value of x, which is
x= no. of row* no. of books in row
x= 5* 10=50

My point is that there is some issue in the problem statement as it gives the answer without the help of any statement.
Earlier thread provided reasoning about statement 2 that value can be 50, 110, 170 etc. But any value except 50 can not provide the correct answer.
Let us try the process with no. of books 110
In this case No. of rows will be = no. of books / no. of books per row= 110/10=11
After addition of 10 books, no. of books will be= 120/12=10
since the no. of row has to be the same, hence 110 can not be the right answer.

please correct me if i am wrong

kurien01 I had the same idea when I looked at it initially:

10+10*rows = 12 *rows
10 = 2*rows
5 = rows

The thing is, you can't assume the number of shelves are equal so you can't assign them the same variable. The problem even says that he puts it in a NEW bookcase.

We have x books and they are put in 10*p (p=some int number of shelves) with 0 remainder
x books / # shelves = 10p

Then we add 10 books, so x+10 is now arranged in 12*q (q=another int number of shelves) with 0 remainder.
x books + 10 / # NEW shelves = 12q

To equate, we add 10 to the number 10*p .... 10*p+10 = 12*q (number of books*p shelves + 10 = number of books*q shelves)

Let's say p=5, q=5 ... then 50+10 = 60. So, originally he had 50 books.

Now let's say p=11 and q=10... then 110+10 = 120. So, originally he had 110 books.

Both are perfectly viable until we get to the statements. (1) x<96 means only 50 as the original value works since the next value will be 110. (2) means any value will work (50, 110, 170 ... any multiple of 60 - 10 will be his original amount)

R(N/10)=0;

R(N+10/12)=0;

N=?

10 20 30 40 50 60
12 24 36 48 60

50 60---P1

110 120--P2

170 180--P3


1)N<96---Sufficient
2) N>24 ---Can be 50, 110, etc --NS

Hence A

I echo the same logic .....
Bunuel,
Please help (in context with the logic given here....
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Let's say the total number of books is N. We know the number of shelves is equal (I mean I guess it makes sense from the question stem?)
We have,
\(N/10=(N+10)/12\)
\(12N=10N+100\)
\(N=50\)

And hence I itself is sufficient?

Can someone validate my approach?

I don't understand how the initial amount of books could be anything other than 50?


10 +2 = 12
10 +2 = 12
10 +2 = 12
10 +2 = 12
10 +2 = 12

50 +10 = 60


???

statement 1 says Before Michael acquired the 10 additional books he had fewer than 96 books.
In this scenario why do we consider 12 (LCM approach). As the statement says before scenario when we did not put the books in the new bookcase with 12 books.

I took it as multiple of 10 in the before scenario and came up with E. Brunel or anybody please tell me whats wrong with my approach here?

@Bunnuel VeritasKarishma plz help read my last post

Why cannot the # of books Michael had before he acquired the 10 additional books be 110, or 170, or 230, ...?

For example, if Michael had 110 books before he acquired the 10 additional books, he arranged 110 books in a bookcase with 10 books on each of the 11 shelves. After he acquired the 10 additional books, he arranged 120 books in NEW bookcase with 12 books on each of the 10 shelves.
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Not sure I understand your question. What value other than 50 satisfy the first statement?
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Bunuel

According to statement 1 in the before scenario when he did not put the 12 books in the shelf he had 10 books in each row.
So according to statement 1 which has fewer than 96 books. this should be teh multiple of 10 like values could be 90,80,70,60 etc..