Number of non-negative integer solutions of equations 6 x+4 y+z=200 and x+y+z=100 is

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4-2.Quadratic Equations and Inequations

hard

The number of non-negative integer solutions of the equations $6 x+4 y+z=200$ and $x+y+z=100$ is

$3$

$5$

$7$

Infinite

(KVPY-2019)

Solution

(c)

Given equations

$6 x+4 y+z=200$ and $x+y+z=100$

By Eqs.$(i)$ and $(ii)$, we get

$5 x+3 y=100$

For non-negative integer solutions, when

$x=2$, then $y=30$

$x=5$, then $y=25$

$x=8$, then $y=20$

$x=11$, then $y=15$

$x=14$, then $y=10$

$x=17$, then $y=5$

and $x=20$, then $y=0$

In every case $z=100-(x+y) > 0$

So, total number of non-negative integral solutions are $7.$

Standard 11

Mathematics

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(JEE MAIN-2021)

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Solution

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