Instructions :Read the below information carefully and answer the foll...
It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
We have to maximize the number of students who chose both Finance and Operations as their specialization. We know that the number of students who chose Marketing will be greater than the number of students who opted for Finance and Operations as their specialization. This can be done as shown in the following Venn diagram.
Thus, the maximum number of students who opted for Finance and Operations but not Marketing = 48
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Instructions :Read the below information carefully and answer the foll...
Understanding the problem
The problem states that there are 100 students in a class who have opted for one or more of the three specializations: Finance, Operations, and Marketing. The given information suggests that the number of students who opted for Marketing is more than those who opted for Finance, and the number of students who opted for Finance is more than those who opted for Operations. Additionally, at least one student opted for all three specializations.
We need to find the maximum number of students who opted for Finance and Operations but not Marketing.
Approach
To solve this problem, we need to use the given information and logic to find the maximum number of students who opted for Finance and Operations but not Marketing.
Let's break down the given information step by step and see how we can use it to solve the problem.
Step 1: Number of students who opted for Marketing is more than Finance
Let's assume that the number of students who opted for Marketing is 'M', and the number of students who opted for Finance is 'F'. As per the given information, we have:
M > F
Step 2: Number of students who opted for Finance is more than Operations
Let's assume that the number of students who opted for Operations is 'O'. As per the given information, we have:
F > O
Step 3: Number of students who opted for exactly two specializations is more than those who opted for all three specializations
Let's assume that the number of students who opted for exactly two specializations is 'T', and the number of students who opted for all three specializations is 'A'. As per the given information, we have:
T > A
Step 4: At least one student opted for all three specializations
As per the given information, we know that at least one student opted for all three specializations. Therefore:
A >= 1
Step 5: Finding the maximum number of students who opted for Finance and Operations but not Marketing
We need to find the maximum number of students who opted for Finance and Operations but not Marketing. Let's assume that the number of students who opted for both Finance and Operations is 'X'. Therefore:
X = F + O - A
Here, we subtract 'A' because the students who opted for all three specializations would have been counted twice (once for Finance and once for Operations). We subtract 'M' from 'X' to get the maximum number of students who opted for Finance and Operations but not Marketing:
X - M = F + O - A - M
We need to maximize this value. Let's rearrange the above equation by substituting the values we derived from the given information:
X - M = F + O - A - M
X - M = F + O - 1 - T
X = F + O + M - T - 1
To find the maximum value of 'X', we need to minimize the value of 'T'. As per the given information, we know that 'T' is greater than the number of students who opted for all three specializations. Therefore:
T >= A
We can substitute this value in the above equation to get:
X = F + O + M - A - 1
To maximize 'X', we need to minimize 'A'. As per the given information, we know that at least one student opted for all three specializations. Therefore:
A >= 1
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