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12 Nov 2024, 03:51
Kudos
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Maximum: 7
Minimum: 4
In a class of 30 students, each student reads exactly one newspaper, with no more than nine students reading the same newspaper. Furthermore, each newspaper is read by a different number of students.
Select for Maximum the highest possible number of newspapers that could have been read by the class, and select for Minimum the lowest possible number of newspapers that could have been read by the class. Make only two selections, one in each column.
Select for Maximum the highest possible number of newspapers that could have been read by the class, and select for Minimum the lowest possible number of newspapers that could have been read by the class. Make only two selections, one in each column.
| Maximum | Minimum | |
| 3 | ||
| 4 | ||
| 6 | ||
| 7 | ||
| 8 | ||
| 30 |
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Official Answer
Maximum: 7
Minimum: 4
12 Nov 2024, 03:51
Kudos
Bookmarks
Official Solution:
To maximize the number of different newspapers read by the class, we need to minimize the number of students reading each newspaper. Let 1 student read one newspaper, 2 students read another, 3 students read a third newspaper, and so on. This gives the sum: \(1 + 2 + 3 + 4 + 5 + 6 + 7 + ...\). Adding these, we see that 7 newspapers would account for a total of 28 students. We cannot exceed 7 newspapers because doing so would require more than 30 students. However, we can adjust the total from 28 to 30 by increasing the last group, reading the seventh kind of newspaper, from 7 students to 9 students. Thus, the distribution becomes \(1 + 2 + 3 + 4 + 5 + 6 + 9 = 30\). Therefore, the maximum number of newspapers that could have been read by the class is 7.
To minimize the number of different newspapers read by the class, we need to maximize the number of students reading each newspaper. Since no more than nine students can read the same newspaper, the first group of 9 students could read one newspaper, the next group of 8 could read another, the third group of 7 could read a third newspaper, and the fourth group of 6 could read a fourth newspaper. Adding these groups gives us \(9 + 8 + 7 + 6 = 30\). Therefore, the minimum number of newspapers that could have been read by the class is 4.
Correct answer:
Maximum "7"
Minimum "4"
Bunuel
To maximize the number of different newspapers read by the class, we need to minimize the number of students reading each newspaper. Let 1 student read one newspaper, 2 students read another, 3 students read a third newspaper, and so on. This gives the sum: \(1 + 2 + 3 + 4 + 5 + 6 + 7 + ...\). Adding these, we see that 7 newspapers would account for a total of 28 students. We cannot exceed 7 newspapers because doing so would require more than 30 students. However, we can adjust the total from 28 to 30 by increasing the last group, reading the seventh kind of newspaper, from 7 students to 9 students. Thus, the distribution becomes \(1 + 2 + 3 + 4 + 5 + 6 + 9 = 30\). Therefore, the maximum number of newspapers that could have been read by the class is 7.
To minimize the number of different newspapers read by the class, we need to maximize the number of students reading each newspaper. Since no more than nine students can read the same newspaper, the first group of 9 students could read one newspaper, the next group of 8 could read another, the third group of 7 could read a third newspaper, and the fourth group of 6 could read a fourth newspaper. Adding these groups gives us \(9 + 8 + 7 + 6 = 30\). Therefore, the minimum number of newspapers that could have been read by the class is 4.
Correct answer:
Maximum "7"
Minimum "4"
