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Given, If more than 1/2 of the total pages that comprise the five books are found on the lower shelf

Target Question ---- > is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
T2+T4 = 2/15+3/10 = 13/30 , which is less than 1/2. Therefore one more book can be added to lower shelf.
Note that adding either T1 or T3 will push the probability above 1/2. Hence Insufficient.

(2) T1 and T5 have been placed on the upper shelf.
T1+T5 = 11/60+3/20 = 20/60, which is less than 1/2. Now to keep this probability below 1/2 any book which is placed on upper shelf must not push the probability above 1/2.
Note that when adding T3 the probability goes beyond 1/2. Hence T3 cannot be placed on upper shelf and only on lower shelf. Sufficient.

Correct Choice is thus B.
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let total books be 60
so
T1 = 11
T2= 8
T3 = 14
T4= 18
T5=9

we are given that If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, i.e. lower shelf has more than 30 books

target is T3 book in lower shelf
#1
T2 and T4 have been placed on the lower shelf.
8+18 =26 books are on lower shelf since we need to have more than 30 books so any of T1, T3,T4 can be present in lower shelf
insufficient

#2

T1 and T5 have been placed on the upper shelf.
we need to have T3 in lower shelf as without it more than 30 books will not be possible as per given condition
as T2 & T4 will make up to 26 books only so T3 has to be there in lower shelf
sufficient

OPTION B is correct
Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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First make the fractions denominator same for easy comparison.
T1=11/60
T2=8/60
T3=14/60
T4=18/60
T5=9/60

In option 1The
total pages of T2 and T4 is 26/90, but the lower shelf needs to be more than 30 pages out of 90,and any of T1,T3,T5 can be picked for this condition
Hence, this option is not sufficient

In option 2
Total pages of T1 and T5 in top shelf is 20 pages. Then, when we checkthe lower shelf.If we add T2 and T6 its still only 26,so T3 should be added. Whatever case we take,T3 will be needed in the lower shelf to meet the criteria where pages>30
Hence this option is sufficient

The answer is B.
Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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Converting everything to common bases we have:
T1: 11/60
T2: 8/60
T3: 14/60
T4: 18/60
T5: 9/60

More than 1/2 of the total pages are on the lower shelf, hence less than half of the total pages would be in upper shelf.

We have to find, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
T2+ T4 = 8/60 + 18/60 = 26/60 which is less than half, to make it more than half, lower shelf could have T1/T3/T5 or any combination of these, hence Not Sufficient

(2) T1 and T5 have been placed on the upper shelf.
T1+ T5 = 11/60 + 9/60 = 20/60
We can't have T3 or T4 in the upper shelf as adding either one of them will make the fraction go above 1/2:
If we add T3 to upper shelf, fraction becomes 34/60> 1/2
If we add T4, fraction becomes 38/60>1/2

Hence T3 will be there in the lower shelf. Sufficient

Ans B
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Now that the table with the fractions is provided, we can solve this GMAT Data Sufficiency problem definitively.

The table gives the following fractions of the total pages:

T1: 11/60

T2: 2/15

T3: 7/30

T4: 3/10

T5: 3/20

First, let's convert all fractions to a common denominator (60) to make comparisons easier:

T1: 11/60

T2: 2/15 = (2 * 4) / (15 * 4) = 8/60

T3: 7/30 = (7 * 2) / (30 * 2) = 14/60

T4: 3/10 = (3 * 6) / (10 * 6) = 18/60

T5: 3/20 = (3 * 3) / (20 * 3) = 9/60

Let's check the sum of these fractions to ensure it's 1:
11/60 + 8/60 + 14/60 + 18/60 + 9/60 = (11 + 8 + 14 + 18 + 9) / 60 = 60/60 = 1.0. Correct.

The condition given in the stem: More than 1/2 of the total pages are found on the lower shelf.
1/2 in terms of 60ths is 30/60. So, the sum of fractions on the lower shelf must be >30/60.

We need to determine: Is T3 (14/60) found on the lower shelf?

Statement (1): T2 and T4 have been placed on the lower shelf.
Sum of pages for T2 and T4 = T2 + T4 = 8/60 + 18/60 = 26/60.
This sum (26/60) is NOT greater than 30/60.
This means that other books must also be on the lower shelf for the condition to be met.
The remaining books are T1 (11/60), T3 (14/60), and T5 (9/60).
Let's see if T3 must be on the lower shelf.

If T3 is added: 26/60 + 14/60 = 40/60. This is >30/60. (T3 is on lower shelf)

If T1 is added instead of T3: 26/60 + 11/60 = 37/60. This is >30/60. (T3 is NOT on lower shelf)

If T5 is added instead of T3: 26/60 + 9/60 = 35/60. This is >30/60. (T3 is NOT on lower shelf)
Since we can satisfy the condition without T3 being on the lower shelf (e.g., if T1 and T2 and T4 were on lower shelf, or T5 and T2 and T4 were on lower shelf), Statement (1) alone is not sufficient.

Statement (2): T1 and T5 have been placed on the upper shelf.
This means T1 (11/60) and T5 (9/60) are not on the lower shelf.
The books that could be on the lower shelf are T2 (8/60), T3 (14/60), and T4 (18/60).
Let SL be the sum of fractions on the lower shelf.
We know SL >30/60.
Let's test combinations of T2, T3, T4:

T2 + T4 = 8/60 + 18/60 = 26/60. (Not > 30/60)

T2 + T3 = 8/60 + 14/60 = 22/60. (Not > 30/60)

T3 + T4 = 14/60 + 18/60 = 32/60. (This IS > 30/60)

In this case, T3 is on the lower shelf. (YES)

What if only T2 and T4 are on the lower shelf? Sum is 26/60, which is not > 30/60. So, T2 and T4 alone are not sufficient to meet the condition.

What if only T2 and T3 are on the lower shelf? Sum is 22/60, not > 30/60.

What if only T3 and T4 are on the lower shelf? Sum is 32/60, which is > 30/60. In this scenario, T3 is on the lower shelf.

What if T2, T3, and T4 are all on the lower shelf? Sum = 8/60 + 14/60 + 18/60 = 40/60. This is >30/60. In this scenario, T3 is on the lower shelf.

Since we found a scenario (T3 + T4 on lower shelf) where T3 is on the lower shelf, and the condition (SL >30/60) is met, and there are no other combinations of T2, T3, T4 (excluding T1, T5) that satisfy the condition without T3, this statement is sufficient. Let's re-verify.

The maximum sum of pages possible on the lower shelf without T3 is T2 + T4 = 8/60 + 18/60 = 26/60.
Since 26/60 is NOT greater than 30/60, it is impossible to satisfy the condition (lower shelf > 30/60) if T3 is NOT on the lower shelf.
Therefore, T3 must be on the lower shelf for the total pages on the lower shelf to be greater than 30/60.
Statement (2) alone is sufficient.

Re-evaluating Statement (1) after the values are known:
T2 (8/60) and T4 (18/60) are on the lower shelf. Current sum = 26/60.
Remaining books: T1 (11/60), T3 (14/60), T5 (9/60).
We need the lower shelf total to be > 30/60.

Scenario 1: T3 is on the lower shelf. Sum = 26/60 + 14/60 = 40/60. This is > 30/60. (Answer: Yes, T3 is on lower shelf)

Scenario 2: T1 is on the lower shelf, T3 is not. Sum = 26/60 + 11/60 = 37/60. This is > 30/60. (Answer: No, T3 is not on lower shelf)
Since we get both a "Yes" and a "No" answer, Statement (1) is NOT sufficient.

Final Conclusion:
Statement (1) is not sufficient.
Statement (2) is sufficient.

The final answer is B
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Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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For the simplicity, we can modify the ratios by multiplying with 60. We know that More than 1/2 of the total pages that comprise the five books are found on the lower shelf

So pages in T1 : T2 : T3 : T4 : T5 are in 11 : 8 : 14 : 18 : 9.

Now we need to find the position of T3.

Stmt 1 : T2 and T4 have been placed on the lower shelf.

If T2 T4 are in lower shelf the pages are 26x, So the lower shelf has to take more than 1 book it can be T3, T1,T5 or any combination to meet the 50% requirement.

Hence, statement 1 is not sufficient.

Stmt 2 : T1 and T5 have been placed on the upper shelf.
If T1 & T5 are in uppershelf, Then the total count of pages is 20x, So If T3 is also in uppershelf it breaks the rule. Hence T3 has to be in the lower shelf.

Hence IMO B
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LCM(10,15,20,30,60)=60
T1=11, T2=8, T3=14, T4=18,T5=9..Pages...total=60 pages

1. T2 & T4 on lower shelf..T2+T4=26..so it can accommodate T3 or any other book..NOT SUFFICIENT
2. T1 & T5 on upper shelf.. T1+T5=20...So upper shelf can accommodate book <=9 pages..so T3 will be on lower shelf..SUFFICIENT

Ans B
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By analyzing the fractions, 5n math textbooks, T1 through T5, may have page counts of 11, 8, 14, 18, and 9 pages respectively, totaling 60 pages. The books are placed on two shelves, and more than half of the total pages (over 30 pages) are on the lower shelf. Statement (1) says T2 and T4 are on the lower shelf, contributing 8 + 18 = 26 pages. Since 26 is less than 30, at least one more book must be on the lower shelf to exceed half the total pages, but it could be T1, T3, or T5, so we cannot be sure if T3 is included, making statement (1) insufficient. Statement (2) says T1 and T5 are on the upper shelf, meaning T2, T3, and T4 must be on the lower shelf. Their combined pages are 8 + 14 + 18 = 40 pages, which is more than half, so T3 is definitely on the lower shelf. Therefore, statement (2) alone is sufficient to answer the question, while statement (1) alone is not.
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First, let's make the denominator of the fraction the same for the ease of the calculation

LCM of 60, 15, 30, 10,20=60

T1=11/60, T2= 8/60, T3= 14/60, T4=18/60, T5=9/60

So we have to find the sum of books on lower shelf that include T3 or not and fraction of lower shelf should be greater than either 1/2 or 30/60

Now let's evaluate statement 1

If T2 and T4 are on the lower shelf, their combined fraction of pages is: (T2) + (T4) = 8/60+18/60=26/60

Which is less than 1/2 or 30/60

Now lets consider if T3 is the part Then T2+T4+T3=40/60 so we can say T3 is the part of lower shelf
Now let's consider if T3 is not the part and T1 and T5 is the part so T1+T5+T2+T4=40/60 which is also greater than 30/60 hence we can say T3 is not part which is disputing with the above one so, statement one is not enough

Now Let's evaluate Statement 2

If T1 and T5 are on the upper shelf, their combined fraction of pages is: Fraction(T1) + Fraction(T5) = 11/60+9/60=20/60
Since there are only two shelves and five books, the remaining books must be on the lower shelf.
The remaining books are T2, T3, and T4 = 8/60+14/60+18/60=40/60 which is greater than 1/2 or 30/60

Hence Statement 2 is sufficient to answer so answer will be B




Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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T1T2T3T4T5
11/608/6014/6018/609/60


We know the bottom shelf is over 30/60.
Is T3 in the bottom shelf?

Condition one:
T2 and T4 are in the bottom shelf.

8/60 + 18/60 = 26/60
If we add any of T1, T3, or T5 to it, it'll be over 30/60. So we can't tell if T3 is in the bottom shelf for sure.
The conditions are insufficient.

Condition two:
Top shelf has T1 and T5.
11/60 + 9/60 = 20/60
Since we already know the bottom shelf is over 30/60,
the bottom could have either (T2, T3, T4) or just (T3, T4). Both work out to more than 30/60.

So the condition is sufficient
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Let's first convert all the fractions to a common denominator for easier calculations. We can use 60, since that is the LCM for all the denominators given:

T1 = 11/60
T2 = 8/60
T3 = 14/60
T4 = 18/60
T5 = 9/60

There are 2 shelves, upper and lower; and lower shelf contains >50% of all the pages of the 5 books. So, the lower shelf contains >1/2 or >30/60 pages

Statement 1:

This tells us that T2 and T4 are on the lower shelf.

Adding T2 and T4, we get, 26/60. If we were to put any other book doesn't matter which one (T1,T3, or T5) in the lower shelf, we will get >1/2 of all pages in that shelf.

So, insufficient.

Statement 2:

This tells us that T1 and T5 are on the upper shelf.

Let's see if we can put T3 on the upper shelf
  • The upper shelf with T1 and T5, already has 20/60 pages, if we put T3 there with them then we will have 34/60 pages in the upper shelf, which breaches our given statement. Since, lower shelf has to have >1/2 of all pages, we can say that T3 is on the lower shelf.
  • Sufficient.

Answer B.
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TextbookT1T2T3T4T5
Pages if total = 6011814189

More than 1/2 of the total pages that comprise the five books are found on the lower shelf
Using total =60, lower shelf has >30

is T3 found on the lower shelf?

S1
T2 and T4 have been placed on the lower shelf.
8+18=26
Lower shelf needs to have atleast 1 more book but it's not clear which one
Insufficient

S2
T1 and T5 have been placed on the upper shelf.
11+9=20
The only other book if at all that can be added here is T2, else the total pages would exceed 30
Regardless, T3 is on the lower shelf
Sufficient

Answer B


Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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Because T2 and T4 together account for less than half of the total pages. knowing that both of those are on the lower shelf doesn’t tell us whether T3 must also be there in order to push the lower‐shelf total over one‐half. The lower shelf could instead contain T5 (or T1) which by itself would raise the share above one‐half without ever placing T3 there. Thus statement (1) is not enough.
On the other hand, if T1 and T5 are both on the upper shelf, then all remaining books including T2, T3, and T4 must lie on the lower shelf. Since the lower shelf must exceed one‐half of the total pages, and the only books left to supply those pages are T2, T3, and T4, T3 unquestionably ends up on the lower shelf. Statement (2) alone is therefore sufficient. Thus B is correct.

Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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11/60+2/15+7/30+3/1+3/20=60/60
1/2 of 60=30
T3= 7/2*60=14 pages
T1=11 pages, T2=8, T4= 18 , T5= 9.

So , T3 is not found in lower shelf.
Hence, point C, both statemnts togther are sufficient, alone is not sufficient.
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Statement (1):
If T2 and T4 are at lower shelf , total pages in lower shelf = 13/30 < 1/2
There must have been another or more books on lower shelf.
But we cannot determine anything else, so this is not sufficient.

Statement (2):
If T1 and T5 are upper shelf, total pages in upper shelf = 1/3 < 1/2. --> confirms to logic
If T3 was added to upper shelf, total pages in upper shelf would be 17/30 > 1/2 and that would break the logic
So T3 has to be in lower shelf.
Its sufficient.

Answer B.
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Assume T1 = 11, T2 = 8, T3 = 14, T4 = 18, T5 = 9, a total of 60 pages. We know that >30 pages are on lower shelf. Question: is T3 (14 pages) on lower shelf?

(1) only: T2 and T4 have been placed on the lower shelf.
So 26 pages are on lower shelf. We don't know the answer

(2) T1 and T5 have been placed on the upper shelf.
20 pages are on upper shelf, leaving 40 pages. Looking at the remaining books, we know that at least T3 & T4 must be on the lower shelf so that more than half of the pages are on the lower shelf

Answer: B
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Dav2000
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Bunuel


Five mathematics textbooks, T1, T2, T3, T4, and T5, have been placed on two shelves, neither of which houses any other books. The table displays the number of pages in each book as a fraction of the total number of pages in the five books together. If more than 1/2 of the total pages that comprise the five books are found on the lower shelf, is T3 found on the lower shelf?

(1) T2 and T4 have been placed on the lower shelf.
(2) T1 and T5 have been placed on the upper shelf.


 


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Statement 1: T2 and T4 have total fraction as 2/15 + 3/10 = 13/30. Since we are given the lower shelf is more than half full so we need to check whether we add T1 or T5 to the lower shelf satisfies this condition.
So with T1 we get 13/30 + 11/60 = 37/60 which is greater than 1/2
and with T5 we get 13/30 + 3/20 = 35/60 which is greater than 1/2
and with T3 we get 13/30 + 7/30 = 20/30 which is less than 1/2. so Not Sufficient as any of three book combinations can be in lower shelf.

Statement 2: T1 and T5 have a 20/30 fraction of pages. Hence we can get T3 in the upper shelf but the fraction will be more than 1/2 which does not satisfy the condition of having more than 1/2 number of pages in lower shelf. So sufficient. Hence the answer is B Option.
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