|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
26 May 2017, 04:47
Kudos
Bookmarks
If 12 children purchased 34 candies and the range of the number of candies that they purchased is 2, which of the following can be the number of children who purchased 3 candies?
I. 2 II. 5 III. 11
A. I only
B. II only
C. I and II
D. I and III
E. I, II and III
I. 2 II. 5 III. 11
A. I only
B. II only
C. I and II
D. I and III
E. I, II and III
26 May 2017, 04:48
Kudos
Bookmarks
Official Solution:
If 12 children purchased 34 candies and the range of the number of candies that they purchased is 2, which of the following can be the number of children who purchased 3 candies?
I. 2 II. 5 III. 11
A. I only
B. II only
C. I and II
D. I and III
E. I, II and III
12 children bought 34 candies, and the rage of the candies purchased is 2, so the number of purchased candies should be 1, 2, 3 or 2, 3, 4.
Case 1) The case where 1, 2, 3 candies were purchased:
If you suppose that the numbers of children who purchased 1, 2, 3 candies are a, b, and c, respectively, you get \(a+b+c = 12\) and \(a+2b+3c = 34\). If two equations are taken away, from \(b+2c = 22\), only \((a,b,c) = (1,0,11)\) is possible. In addition, the number of children who purchased 3 candies is c, so c=11 is possible.
Case 2) The case where 2, 3, 4 candies were purchased:
If you suppose that the number of children who purchased 2, 3, 4 candies are x, y, and z, respectively, you get \(x+y+z = 12\) and \(2x+3y+4z = 34\). If you multiply the first equation \((x+y+z = 12)\) by 2, you get \(2x+2y+2z = 24\), and if you subtract the 2 equations, you get \(y+2z = 10\). Then, \((x,y,z) = (7,0,5), (6,2,4), (5,4,3), (4,6,2), (3, 8, 1)\) is possible. At this time, the number of children who purchased 3 candies is y, so \(y = 2\) is possible, but from \(y = 5\), you get \(z = 2. 5\), and since x, y, z must be integers (number of children), \(y = 5\) is impossible.
Therefore, from the options, 2 and 11 are possible. Therefore, I and III is the answer.
Answer: D
If 12 children purchased 34 candies and the range of the number of candies that they purchased is 2, which of the following can be the number of children who purchased 3 candies?
I. 2 II. 5 III. 11
A. I only
B. II only
C. I and II
D. I and III
E. I, II and III
12 children bought 34 candies, and the rage of the candies purchased is 2, so the number of purchased candies should be 1, 2, 3 or 2, 3, 4.
Case 1) The case where 1, 2, 3 candies were purchased:
If you suppose that the numbers of children who purchased 1, 2, 3 candies are a, b, and c, respectively, you get \(a+b+c = 12\) and \(a+2b+3c = 34\). If two equations are taken away, from \(b+2c = 22\), only \((a,b,c) = (1,0,11)\) is possible. In addition, the number of children who purchased 3 candies is c, so c=11 is possible.
Case 2) The case where 2, 3, 4 candies were purchased:
If you suppose that the number of children who purchased 2, 3, 4 candies are x, y, and z, respectively, you get \(x+y+z = 12\) and \(2x+3y+4z = 34\). If you multiply the first equation \((x+y+z = 12)\) by 2, you get \(2x+2y+2z = 24\), and if you subtract the 2 equations, you get \(y+2z = 10\). Then, \((x,y,z) = (7,0,5), (6,2,4), (5,4,3), (4,6,2), (3, 8, 1)\) is possible. At this time, the number of children who purchased 3 candies is y, so \(y = 2\) is possible, but from \(y = 5\), you get \(z = 2. 5\), and since x, y, z must be integers (number of children), \(y = 5\) is impossible.
Therefore, from the options, 2 and 11 are possible. Therefore, I and III is the answer.
Answer: D
Moderator:
