himanshu nautiyal
CAT Preparation Community 10y ago
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himanshu nautiyal
CAT Preparation Community 10y ago
Pv Aaditya 10y ago
Hi Himanshu,
Let the number of students who pass in exactly one subject be a, exactly 2 be b, exactly 3 be c, exactly 4 be d and exactly 5 be e.
a+b+c+d+e = 50
a+2b+3c+4d+5e = 25+27+29+33+35 = 149
We have to get the maximum number of students who passed in at least three subjects, i.e. we have to maximize c+d+e.
From the 2 equations, we have, b+2c+3d+4e = 99.
So, b+2(c+d+e)+d+2e = 99. b =1 and c = 49 and d = e = a = 0 => c + d + e = 49 is the maximum value.
Please let us know which part is not clear. We will try to explain that in greater detail.
Prateek Singhal 8y ago
could you explain the second equation, a+2b+3c+4d+5e=149? How did you get this equation?
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