IPMAT 2020 - Quantitative Aptitude Question 25
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IPMAT 2020 - Quantitative Aptitude
Sum of marks of student = 64 × 6
= 384
a + b + c + d + e + f = 384
a + b + c + d + e + = 314
since we need to maximize the difference between second highest & second lowest so we need to assign as minimum values the highest & second highest value
Minimum value for smallest number = 40
all values should be distinct
40 + 41 + 42 + d + e = 314
123 + d + e + = 314
d + e =191
To maximize both assign almost equal values so
d = 190/2
e = 96
So difference = 95 – 41
= 54