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13 Jan 2025, 09:15
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x: 16
y: 3
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:28) correct
37%
(02:48)
wrong
based on 419
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A teacher distributes a batch of 56 cookies among \(x\) students such that each student receives exactly \(y\) cookies, where \(y > 0\), with some cookies left over. If these leftover cookies were redistributed, all except eight students would receive one additional cookie.
Select for x the possible value of \(x\), and select for y the possible value of \(y\) that would be jointly consistent with the given information. Make only two selections, one in each column.
Select for x the possible value of \(x\), and select for y the possible value of \(y\) that would be jointly consistent with the given information. Make only two selections, one in each column.
| x | y | |
| 2 | ||
| 3 | ||
| 4 | ||
| 15 | ||
| 16 | ||
| 32 |
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Official Answer
x: 16
y: 3
20 Jan 2025, 15:15
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Bunuel
Official Solution:
The number of cookies initially distributed was \(xy\).
Since redistributing the leftover cookies would give all except 8 students an additional cookie, the number of leftover cookies must have been \(x - 8\). This also indicates that there were more than 8 students in the group, so \(x > 8\).
Thus, we get \(xy + (x - 8) = 56\), which simplifies to \(x(y + 1) = 64\). This implies that \(x\) must be a factor of 64. The factors of 64 greater than 8 are 16, 32, and 64.
If \(x = 16\), then \(y = 3\).
If \(x = 32\), then \(y = 1\).
If \(x = 64\), then \(y = 0\), which is not possible since we are given that \(y > 0\).
Therefore, the possible pairs \((x, y)\) are \((16, 3)\) and \((32, 1)\). Among these pairs, only \((16, 3)\) is included in the options.
Correct answer:
x "16"
y "3"
Attachment:
GMAT-Club-Forum-s6cy86hd.png [ 8.55 KiB | Viewed 1568 times ]
General Discussion
13 Jan 2025, 11:07
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1. When x students received y cookies, we can write this as \(# \ of \ cookies \ given = # \ of \ students * # \ of \ cookies \ for \ each \ student = x * y\).
2. Now, if the rest of the cookies were given, x - 8 students received 1 cookie, which can be written as \(# \ of \ cookies \ given = # \ of \ students * # \ of \ cookies \ for \ each \ student = (x - 8) * 1\). Remember that x must then be larger than 8.
3. The total number of given cookies is 56, so let's add the two equations together. \(56 = x * y + (x - 8) * 1 = xy + x - 8 \rightarrow 64 = xy + x = x(y+1)\).
4. The options for x are: 15, 16, 32. The options for y + 1 are: 3, 4, 5, 16, 17, 33. The only pair that has a product equal to 64 is 16 and 4.
5. Our answer will be: x - 16 and y - 3.
2. Now, if the rest of the cookies were given, x - 8 students received 1 cookie, which can be written as \(# \ of \ cookies \ given = # \ of \ students * # \ of \ cookies \ for \ each \ student = (x - 8) * 1\). Remember that x must then be larger than 8.
3. The total number of given cookies is 56, so let's add the two equations together. \(56 = x * y + (x - 8) * 1 = xy + x - 8 \rightarrow 64 = xy + x = x(y+1)\).
4. The options for x are: 15, 16, 32. The options for y + 1 are: 3, 4, 5, 16, 17, 33. The only pair that has a product equal to 64 is 16 and 4.
5. Our answer will be: x - 16 and y - 3.
