slope = -3/5
and y intercept = 30
so eqn of line
3x+5y = 150
y = (150-3x)/5
for y to be integer x has to be multiple of 5
x varies from 0 to 50
so 11 possible values of x: 0 5 10 15 . . . . . 45
Whats the OA?
T-52:13
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Indradeep
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It is 11 points because any point on the line (x,y) is in the ration of 5:3
Thus x = (5/3)y which is only possible if y = multiple of 3 from 0->30 including both, i.e. 11 points.
Thus D
Thus x = (5/3)y which is only possible if y = multiple of 3 from 0->30 including both, i.e. 11 points.
Thus D













