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06 Jan 2022, 09:45
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D
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Difficulty:
45%
(medium)
Question Stats:
70% (02:38) correct
30%
(02:32)
wrong
based on 223
sessions
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There were 2 classes A& B, with a total of 45 students in them. Students of both the classes were asked to go on a school trip but only 1/5th of the students from class A and 2/3rd of the students from class B agreed to go. What is the minimum possible number of students who did not agree to go on the trip?
A. 8
B. 16
C. 17
D. 22
E. 23
A. 8
B. 16
C. 17
D. 22
E. 23
06 Jan 2022, 11:55
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Few inferences from the question
1) The number of students in Class A are in multiples of 5
2) The number of students in Class B are in multiples of 3
The combined total of students from both the class is 45.
Therefore the possible combination of students, with above points
Class A = 15 | Class B = 30
Class A = 30 | Class B = 15
It is give that 1 out of 5 students in class A agreed to go, so 4 out of 5 students did not agree to go
Also,
2 out of 3 students in class B agreed to go, so 1 out of 3 students did not agree to go
Class A = 15 | Class B = 30
Number of students in class A who did not agree to go = \(\frac{4}{5} * 15\) = 12
Number of students in class B who did not agree to go = \(\frac{1}{3} * 30\) = 10
Total = 22
Class A = 30 | Class B = 15
Number of students in class A who did not agree to go = \(\frac{4}{5} * 30\) = 24
Number of students in class B who did not agree to go = \(\frac{1}{3} * 15\) = 5
Total = 29
We want the minimum number , hence IMO Option D
1) The number of students in Class A are in multiples of 5
2) The number of students in Class B are in multiples of 3
The combined total of students from both the class is 45.
Therefore the possible combination of students, with above points
Class A = 15 | Class B = 30
Class A = 30 | Class B = 15
It is give that 1 out of 5 students in class A agreed to go, so 4 out of 5 students did not agree to go
Also,
2 out of 3 students in class B agreed to go, so 1 out of 3 students did not agree to go
Class A = 15 | Class B = 30
Number of students in class A who did not agree to go = \(\frac{4}{5} * 15\) = 12
Number of students in class B who did not agree to go = \(\frac{1}{3} * 30\) = 10
Total = 22
Class A = 30 | Class B = 15
Number of students in class A who did not agree to go = \(\frac{4}{5} * 30\) = 24
Number of students in class B who did not agree to go = \(\frac{1}{3} * 15\) = 5
Total = 29
We want the minimum number , hence IMO Option D
07 Jan 2022, 07:14
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To find the minimum number of students who might not attend, we need to maximize the number who could attend.
This will be maximized when the greatest number of students are allocated to class B.
However, the number of students allocated to each class have to be divisible by 15.
The greatest number allocated to class B that doesn't leave 0 for class A is 30.
2/3*30= 20
1/5*15= 3
Maximum students that could attend: 23
Therefore minimum students who might not attend is
45-23= 22
Posted from my mobile device
This will be maximized when the greatest number of students are allocated to class B.
However, the number of students allocated to each class have to be divisible by 15.
The greatest number allocated to class B that doesn't leave 0 for class A is 30.
2/3*30= 20
1/5*15= 3
Maximum students that could attend: 23
Therefore minimum students who might not attend is
45-23= 22
Posted from my mobile device
