9. Five mark exercise problems - IV

1. Solve \(\frac{1}{3}(x + y - 5) = y - z = 2x - 11 = 9 - (x + 2z)\).

 

Answer:

 

The solution set is {6,2,1}{2,6,1}{1,2,6}.

 

(Note: Select the answer in the form of \(\{x,y,z\}\))

 

2. One hundred and fifty students are admitted to a school. They are distributed over three sections \(A\), \(B\) and \(C\). If \(6\) students are shifted from section \(A\) to section \(C\), the sections will have an equal number of students. If \(4\) times of students of \(C\) is equal to the number of students in section \(A\) by the number of students in section \(B\), find the number of students in the three sections.

 

Answer:

 

The number of students in section \(A\) is .

 

The number of students in section \(B\) is .

 

The number of students in section \(C\) is .

 

3. In a three-digit number, when the tens and the hundreds digit are interchanged, the new number is \(54\) more than three times the original number. If \(198\) is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as the tens digit exceeds the unit digit. Find the original number.

 

Answer:

 

The original number is .