In an engineering college of 10,000 students, 1,500 like neither their core branches nor other branches. The number of students who like their core branches is 1/4th of the number of students who like other branches. The number of students who like both their core and other branches is 500.
The number of students who like their core branches is
Engineering College Core Branches Likes Calculation
This section details the calculation to find the number of students favouring their core engineering branches.
Given Data Summary
- Total students: \( 10,000 \)
- Students liking neither branch: \( 1,500 \)
- Relation: Students liking core branches = \( \frac{1}{4} \) \times Students liking other branches
- Students liking both core and other branches: \( 500 \)
Calculation Steps
-
Determine the number of students who like at least one type of branch (core or other).
Students liking at least one = Total students - Students liking neither
\( \text{At least one} = 10,000 - 1,500 = 8,500 \)
-
Apply the principle of inclusion-exclusion: \( |C \cup O| = |C| + |O| - |C \cap O| \).
Let \( |C| \) be students liking core branches and \( |O| \) be students liking other branches.
We have \( |C \cup O| = 8,500 \) and \( |C \cap O| = 500 \).
\( 8,500 = |C| + |O| - 500 \)
This simplifies to: \( |C| + |O| = 9,000 \)
-
Use the given condition: \( |C| = \frac{1}{4} |O| \).
Substitute this into the equation from the previous step:
\( (\frac{1}{4} |O|) + |O| = 9,000 \)
\( \frac{5}{4} |O| = 9,000 \)
-
Solve for \( |O| \) (number of students liking other branches):
\( |O| = 9,000 \times \frac{4}{5} \)
\( |O| = 7,200 \)
-
Calculate \( |C| \) (number of students liking core branches):
\( |C| = \frac{1}{4} \times |O| = \frac{1}{4} \times 7,200 \)
\( |C| = 1,800 \)
Result
The number of students who like their core branches is 1,800.
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Statement 1: Columns of matrix $A$ are linearly independent
Statement 2: Inverse of matrix $A$ exists
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