x+y+z=19 1 < x < 5, 4 < y <8, 10 < z < 14 How many outcomes are possible for integral values of x,

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15 Nov 2021 inequalities

x+y+z=19
1 < x < 5, 4 < y <8, 10 < z < 14
How many outcomes are possible for integral values of x, y and z

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) none of these

Solution

according to the given conditions
possible values of x, y and z are
x = 2, 3, 4
y = 5, 6, 7
z = 11, 12, 13
so possible combinations are
(2, 5, 12), (2, 6, 11), (3, 5, 11)

Correct Option: 3 Best Solution (2)