KVPY 2019 SA Exam Previous Papers

Question : 12 of 80

Marks: +1, -0

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

Solution:

Subtracting the given equations we get

5x + 3y = 100

x = 20 -{3y}/5

⇒ y is multiple of 5, let y = 5k
x = 20 – 3k
∴ k = 0, 1, 2, ……6
Hence numbers of solutions are 7

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