Answer to Question #86084 in Algebra for RAKESH DEY

Question

3 major projects P1, P2, P3 are being funded by 3 voluntary agencies V1,V2,V3. V1,V2,V3 are willing to pay Rs. 8000, Rs. 4000, Rs. 2000, respectively per person on the project P1; Rs. 4000, 3000, 4000 respectively per person on P2; and Rs. 3000, 5000, 8000 respectively on the project P3. Further, the amount that V1,V2,V3 have kept aside for paying people on these projects is Rs. 217000, 142000 and 132000 respectively. How many people should each project employ so that the total money available is utilised?

Answer 1

Denote the numbers of executors as x₁ for P₁ project, x₂ for P₂ project and x₃ for P₃ project.

Then we obtain the equations set:

$\begin{cases} 8000x_1+4000x_2+3000x_3=217000 \\ 4000x_1+3000x_2+5000x_3=142000 \\ 2000x_1+4000x_2+8000x_3=132000 \end{cases}$

Let's multiply the second equation by 2 and the third one by 4:

$\begin{cases} 8000x_1+4000x_2+3000x_3=217000 \\ 8000x_1+6000x_2+10000x_3=284000 \\ 8000x_1+16000x_2+32000x_3=528000 \end{cases}$

Subtract the first equation from the second and third ones:

$\begin{cases} 8000 x_1+4000x_2+3000x_3=217000 \\ 2000x_2+7000x_3=67000 \\ 12000x_2+29000x_3=311000 \end{cases}$

Rearrange the second and third equations:

$\begin{cases} 8000 x_1+4000x_2+3000x_3=217000 \\ 12000x_2+29000x_3=311000 \\ 2000x_2+7000x_3=67000 \end{cases}$

Multiply the third equation by 6:

$\begin{cases} 8000 x_1+4000x_2+3000x_3=217000 \\ 12000x_2+29000x_3=311000 \\ 12000x_2+42000x_3=402000 \end{cases}$

Subtract the second equation from the third one:

$\begin{cases} 8000 x_1+4000x_2+3000x_3=217000 \\ 12000x_2+29000x_3=311000 \\ 13000x_3=91000 \end{cases}$

$x_3 = \frac{91000}{13000} =7$

$12000x_2+29000 \cdot 7=311000$

$12000x_2=108000$

$x_2 = \frac{108000}{12000} =9$

$8000 x_1+4000 \cdot 9+3000 \cdot 7=217000$

$8000 x_1=160000$

$x_1 = \frac{160000}{8000} =20$

Answer: project P₁ should employ 20 peoples, project P₂ should employ 9 peoples and project P₃ should employ 7 peoples.