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01 Jul 2026, 19:00
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E
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Difficulty:
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29% (02:31) correct
71%
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wrong
based on 83
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At the end of Monday, a bookstore display had only hardcover books and paperback books, and 7/12 of the books on the display were hardcovers. On Tuesday, the store added hardcover books and paperback books to the display and removed none of the books that were already there. After the additions, was the fraction of paperback books on the display less than it had been at the end of Monday?
(1) Less than 1/2 of the books added on Tuesday were paperbacks.
(2) After the additions on Tuesday, the total number of books on the display was 1/3 greater than it had been at the end of Monday.
(1) Less than 1/2 of the books added on Tuesday were paperbacks.
(2) After the additions on Tuesday, the total number of books on the display was 1/3 greater than it had been at the end of Monday.
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M47-224
05 Jul 2026, 07:00
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GMAT Club Official Solution:
At the end of Monday, a bookstore display had only hardcover books and paperback books, and 7/12 of the books on the display were hardcovers. On Tuesday, the store added hardcover books and paperback books to the display and removed none of the books that were already there. After the additions, was the fraction of paperback books on the display less than it had been at the end of Monday?
At the end of Monday, 5/12 of the books were paperbacks. The paperback fraction will decrease only if the fraction of paperbacks among the books added on Tuesday is less than 5/12.
(1) Less than 1/2 of the books added on Tuesday were paperbacks.
This is not enough, because less than 1/2 could be less than 5/12 or greater than 5/12.
Not sufficient.
(2) After the additions on Tuesday, the total number of books on the display was 1/3 greater than it had been at the end of Monday.
This tells us how many books were added relative to the Monday total, but it does not tell us what fraction of the added books were paperbacks.
Not sufficient.
(1)+(2) Let the Monday total be 12 books. Then there were 5 paperbacks. Statement (2) says 4 books were added. By statement (1), fewer than 2 of those 4 added books were paperbacks, so at most 1 paperback was added. Then the Tuesday paperback fraction is at most:
6/16 = 3/8
This is less than 5/12, so the answer is YES.
But let the Monday total be 36 books. Then there were 15 paperbacks. Statement (2) says 12 books were added. By statement (1), fewer than 6 of those 12 added books were paperbacks, so 5 paperbacks could have been added. Then the Tuesday paperback fraction is:
20/48 = 5/12
This is not less than the Monday fraction, so the answer is NO.
So, even together, the statements do not give a definite answer.
Not sufficient.
Answer: E.
At the end of Monday, a bookstore display had only hardcover books and paperback books, and 7/12 of the books on the display were hardcovers. On Tuesday, the store added hardcover books and paperback books to the display and removed none of the books that were already there. After the additions, was the fraction of paperback books on the display less than it had been at the end of Monday?
At the end of Monday, 5/12 of the books were paperbacks. The paperback fraction will decrease only if the fraction of paperbacks among the books added on Tuesday is less than 5/12.
(1) Less than 1/2 of the books added on Tuesday were paperbacks.
This is not enough, because less than 1/2 could be less than 5/12 or greater than 5/12.
Not sufficient.
(2) After the additions on Tuesday, the total number of books on the display was 1/3 greater than it had been at the end of Monday.
This tells us how many books were added relative to the Monday total, but it does not tell us what fraction of the added books were paperbacks.
Not sufficient.
(1)+(2) Let the Monday total be 12 books. Then there were 5 paperbacks. Statement (2) says 4 books were added. By statement (1), fewer than 2 of those 4 added books were paperbacks, so at most 1 paperback was added. Then the Tuesday paperback fraction is at most:
6/16 = 3/8
This is less than 5/12, so the answer is YES.
But let the Monday total be 36 books. Then there were 15 paperbacks. Statement (2) says 12 books were added. By statement (1), fewer than 6 of those 12 added books were paperbacks, so 5 paperbacks could have been added. Then the Tuesday paperback fraction is:
20/48 = 5/12
This is not less than the Monday fraction, so the answer is NO.
So, even together, the statements do not give a definite answer.
Not sufficient.
Answer: E.
General Discussion
01 Jul 2026, 23:32
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say monday total books = 60x
hardcover= 35x
paperback = 25x
tuesday, added hardcover = m
paperback = n
no books were removed.
so total book on tuesday = 60x+m+n
hardcover= 35x+m
paperback= 25x+ n
find: 25x+n/ 60x+m+n < 25x/60x
25x+n / 60x+m+n < 5/12
300x+12n < 300x+5m+5n
12n < 5m+5n
7n< 5m
this is what we need to find.
S-1
total books added on tuesday = m+n
paperback = n
paperback < (1/2) (m+n)
n < (1/2) (m+n)
2n < m+n
n<m
here with different values we get both yes and no for our inequality.
so not sufficient.
S-2
total books added on tuesday= m+n
60x+m+n = (4/3) (60x)
60x+m+n = 80x
m+n = 20x
now, we can try two extreme cases here.
C-1 m = 19x, n= x
in this case 7n< 5m holds true.
C-s, m=x, n=19x
in this case 7n< 5m doesnt hold true.
not sufficient
S- 1&2
we know n<m and m+n = 20x
again we can try two cases.
n= 5, m= 15, then 7n< 5m holds true.
n=9, m=11, then 7n< 5m doesnt hold true.
choice E
hardcover= 35x
paperback = 25x
tuesday, added hardcover = m
paperback = n
no books were removed.
so total book on tuesday = 60x+m+n
hardcover= 35x+m
paperback= 25x+ n
find: 25x+n/ 60x+m+n < 25x/60x
25x+n / 60x+m+n < 5/12
300x+12n < 300x+5m+5n
12n < 5m+5n
7n< 5m
this is what we need to find.
S-1
total books added on tuesday = m+n
paperback = n
paperback < (1/2) (m+n)
n < (1/2) (m+n)
2n < m+n
n<m
here with different values we get both yes and no for our inequality.
so not sufficient.
S-2
total books added on tuesday= m+n
60x+m+n = (4/3) (60x)
60x+m+n = 80x
m+n = 20x
now, we can try two extreme cases here.
C-1 m = 19x, n= x
in this case 7n< 5m holds true.
C-s, m=x, n=19x
in this case 7n< 5m doesnt hold true.
not sufficient
S- 1&2
we know n<m and m+n = 20x
again we can try two cases.
n= 5, m= 15, then 7n< 5m holds true.
n=9, m=11, then 7n< 5m doesnt hold true.
choice E
Bunuel
