Answer to Question #111301 in Algebra for Ankit

Question

3 types of people go to a marriage party.
Type 1 people: 1 man eats in 5 plates. Type 2 people: 1 man eats in 1 plates. Type 3 people: 10 men eat in 1 plate.

You are given that exactly 100 people go to the party and there are exactly 100 plates. So you have to calculate how many people of type 1, type 2 and type 3 will go to the party. Whereby the condition is all the 100 plates should be consumed by 100 people. And no plate will remaining after the party.

Answer 1

Answer:(0, 100, 0); (9, 51, 40); (18, 2, 80)

$Let\;x\;-\;number\;of\;type\;1\;people,\;\\y\;-\;type\;2,\;10z\;-\;type3.\\x,y,z\geq0\\It's\;obvious\;that\;number\;of\;type\;3\;people\\divides\;by\;10,\;because\;there\;should\;be\\whole\;number\;of\;plates.\\x\;people\;-\;10x\;plates\\y\;people\;-\;y\;plates\;\\10z\;people\;-\;z\;plates\\x+y+10z\;=\;100\;=\;10x+y+z\\Substract\;(x+y+z)\;from\;both\;sides.\\9z=4x\Rightarrow z\vdots4\\10z\leq x+y+10z\leq100\Rightarrow z\leq10\\\left.\begin{array}{r}z\leq10\\z\vdots4\end{array}\right\}\Rightarrow z=0,\;4,\;8\\(1)\;z=0\Rightarrow4x=9\times0=0\Rightarrow x=0\\y=100\\(0,\;100,\;0)\\(2)\;z=4\Rightarrow4x=9\times4=36\Rightarrow x=9\\y=100-9-4\times10=51\\(9,\;51,\;40)\\(3)\;z=8\Rightarrow4x=9\times8=72\Rightarrow x=18\\y=100-18-8\times10=2\\(18,\;2,\;80)$