Question

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others using matrix category. Apart from these values namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

Answer 1

Hint: Find the equations as per the given conditions. The equation so formed are solved to find the value of x, y and z. The equations are:
\[\begin{align}
  & ~x+y+z=12\text{ } \\
 & 3\left( y+z \right)+2x=33 \\
 & \text{ }x+z=2y \\
\end{align}\]
Solve all the three equations.

Complete step-by-step answer:
In the question, it is given that the management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Also, three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, then we have to find the number of each awardees.
So, here we will need to find the equations. From the first condition that the sum of all the awardees is 12 , we get the equation as; \[x+y+z=12\text{ }\].
Next, from the condition that three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33, we get the equation as: \[3\left( y+z \right)+2x=33\]
Next, from the condition that the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, we get the equation as: \[x+z=2y\]. So all the three equations are:
\[\begin{align}
  & ~x+y+z=12\text{ } \\
 & 3\left( y+z \right)+2x=33 \\
 & \text{ }x+z=2y \\
\end{align}\]

So, now from the equation \[x+y+z=12\text{ }\], we will have \[x=12-y-z\], which we will substitute in the equations:
\[3\left( y+z \right)+2x=33\] and \[\text{ }x+z=2y\]
We will get:
\[\left[ \begin{matrix}
   3\left( y+z \right)+2\left( 12-y-z \right)=33 \\
   12-y-z+z=2y \\
\end{matrix} \right]\]
On simplifying we get:
\[\left[ \begin{matrix}
   y+z+24=33 \\
   12-y=2y \\
\end{matrix} \right]\]
Now here we have:
\[\begin{align}
  & \Rightarrow 12-y=2y \\
 & \Rightarrow y=4 \\
\end{align}\]
Now substituting \[y=4\], in the equation \[y+z+24=33\], we get:
\[\begin{align}
  & \Rightarrow z+28=33 \\
 & \Rightarrow z=5 \\
\end{align}\]
Now substituting \[y=4\]and \[z=5\]in the equation \[x=12-y-z\], we will get:
\[\begin{align}
  & \Rightarrow x=12-y-z \\
 & \Rightarrow x=12-4-5 \\
 & \Rightarrow x=3 \\
\end{align}\]
So, we get the value of \[x=3\], \[y=4\]and \[z=5\]. Hence, the number of awardees from honest is 3, number of awardees for helping others is 4 and number of awardees for supervising the workers is 5.
Now, one more category that can be included in the award category can be the quality work.

Note: When we are finding the second equation, then we have to be careful that we need to add \[3\left( y+z \right)\] with twice the number of x, and which will be then equal to 33. We can also use the elimination method to solve the given set of equations. Next, we can also verify the results if found are correct or not using the LHS=RHS method for each equation.