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10 Jul 2025, 15:15
Kudos
Bookmarks
Bunuel
In the first step, we bring the numbers to an standardised format.
11/60
8/60
14/60
18/60
9/60
Now we can look at the questions.
1) Alone insufficient. In theory, the distribution could be completely random and all the other books (except no matter what additional one) can be on the other shelf.
So we cant verify, that we need necessarily T3 to make up the balance.
2) Alone sufficient. In this case (and the question does already imply a case like that), we have two informations given. First: We know, that at least T1 and T5 are at the top. Second: The bottom makes up at least 50%. That is possible with T3 + T4, or all Three-Book-Combinations, except T1 + T2 and T5. As we know, that T1 and T5 are at the top, we need three books at least, to reach a percentage above 50%.
Therefore the result is B)
10 Jul 2025, 15:42
Kudos
Bookmarks
We need to find if T3 is on lower shelf. we know that 1/2 of total pages are on lower shelf.
we can rewrite the fractions to match denominator.
T1 = 11/60
T2 = 8/60
T3 = 14/60
T4 = 18/60
T5 = 9/60
(1) T2 and T4 have been placed on the lower shelf. =>
so T2+T4 = (8+18)/60 = 26/60
if T3 is in lower shelf = (26 +14)/60 = 2/3 which is greaeter than 1/2
if T3 is not in lower and some other book is there lets say T5 = (26+9)/40 = 35/40 = 7/8 which is again greater than 1/2.
So this alone does not tell if T3 is on lower shelf. Not Suffcient
(2) T1 and T5 have been placed on the upper shelf. =>
So T1+T5 = (11+9)/60 = 1/3
if only T1 and T5 are in upper then lower will be 1-1/3 = 2/3 which is greater than 1/2
if lets say T3 is in upper then (11+9+14)/60 = 34/60 then lower will be 26/60 => 0.43 < 1/2 so there is no way T3 is upper it needs to be in lower to satisfy the condition. this is Sufficient
Hence Ans B
we can rewrite the fractions to match denominator.
T1 = 11/60
T2 = 8/60
T3 = 14/60
T4 = 18/60
T5 = 9/60
(1) T2 and T4 have been placed on the lower shelf. =>
so T2+T4 = (8+18)/60 = 26/60
if T3 is in lower shelf = (26 +14)/60 = 2/3 which is greaeter than 1/2
if T3 is not in lower and some other book is there lets say T5 = (26+9)/40 = 35/40 = 7/8 which is again greater than 1/2.
So this alone does not tell if T3 is on lower shelf. Not Suffcient
(2) T1 and T5 have been placed on the upper shelf. =>
So T1+T5 = (11+9)/60 = 1/3
if only T1 and T5 are in upper then lower will be 1-1/3 = 2/3 which is greater than 1/2
if lets say T3 is in upper then (11+9+14)/60 = 34/60 then lower will be 26/60 => 0.43 < 1/2 so there is no way T3 is upper it needs to be in lower to satisfy the condition. this is Sufficient
Hence Ans B
10 Jul 2025, 16:45
Kudos
Bookmarks
Topic(s)- Least Common Denominator (LCD), Max/Min
Strategy- Max/Min Scenarios
Variable(s)- Textbook # = subscript "i"; Lower Shelf = "L"; Upper Shelf = "U"
0. Pre-Work
i) LCD of the provided fractions: 60
T1 = (11*1/60*1), T2 = (2*4)/(15*4), T3 = (7*2)/(30*2), T4 = (3*6)/(10*6), T5 = (3*3)/(20*3)
T1 = 11/60, T2 = 8/60, T3 = 14/60, T4 = 18/60, T5 = 9/60
ii) For a total of 60, L must be greater than (60)*(1/2) = 30
L > 30
Rephrase the Question: Is T3 on L?
(1) At least T2 + T4 = 8 + 18 = 26 on L
This implies at least 1 other book is on the shelf (L must be greater than 30)
i) T2 + T4 + T1 [Question: NO]
ii) T2 + T4 + T5 [Question: NO]
iii) T2 + T4 + T3 [Question: YES]
Conflicting scenarios [Eliminate A,D]
(2) At least T1 + T5 = 11 + 9 = 20 on U
Max/Min test: Could U hold T3 and still have 30 or fewer books?
i) U = 20 + (T3) = 20 + 14 = 34 > 30 [NO]
T3 MUST be on L
[Answer: B]
Strategy- Max/Min Scenarios
Variable(s)- Textbook # = subscript "i"; Lower Shelf = "L"; Upper Shelf = "U"
0. Pre-Work
i) LCD of the provided fractions: 60
T1 = (11*1/60*1), T2 = (2*4)/(15*4), T3 = (7*2)/(30*2), T4 = (3*6)/(10*6), T5 = (3*3)/(20*3)
T1 = 11/60, T2 = 8/60, T3 = 14/60, T4 = 18/60, T5 = 9/60
ii) For a total of 60, L must be greater than (60)*(1/2) = 30
L > 30
Rephrase the Question: Is T3 on L?
(1) At least T2 + T4 = 8 + 18 = 26 on L
This implies at least 1 other book is on the shelf (L must be greater than 30)
i) T2 + T4 + T1 [Question: NO]
ii) T2 + T4 + T5 [Question: NO]
iii) T2 + T4 + T3 [Question: YES]
Conflicting scenarios [Eliminate A,D]
(2) At least T1 + T5 = 11 + 9 = 20 on U
Max/Min test: Could U hold T3 and still have 30 or fewer books?
i) U = 20 + (T3) = 20 + 14 = 34 > 30 [NO]
T3 MUST be on L
[Answer: B]
