Get Answers to all your Questions

header-bg qa

A school wants to award its students for regularity and hardwork with a total cash award of Rs 6,000. If three times the award money for hardwork added to that given for regularity amounts to Rs 11,000, represent the above situation algebraically and find the award money for each value, using matrix method. Suggest two more values, which the school must include for award.

 

 

 

 
 
 
 
 

Answers (1)

Let the award for regularity be Rs x and for hard work be Rs y
\therefore x+y= 6000 and
x+3y= 11000
\Rightarrow \begin{bmatrix} 1 &2 \\ 1& 3 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 6000 \\ 11000 \end{bmatrix}\, or\, AX= B
\therefore X= A^{-1}B\, as\, \left | A \right |= 2\neq 0
= \begin{bmatrix} x\\ y \end{bmatrix}= \frac{1}{2}\begin{bmatrix} 3 &-1 \\ -1& 1 \end{bmatrix}\begin{bmatrix} 6000\\ 11000 \end{bmatrix}
\begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 3500\\ 2500 \end{bmatrix}\therefore x= Rs\, 3500,y= Rs\, 2500
Any 2 values like obedience, respect for elders

Posted by

Ravindra Pindel

View full answer