Bunuel
If \(a≤b≤c≤d\) is \(a≥20\)?
(1) \(a+b+c+d=170\)
(2) \(5a≥2d\)
Consider statement 1. If a=b=c=d then 170/4=42.5. We can easily determine that a,b,c,d can take 41,42,43,44 since 41+42+43+44=170 and a is greater than or equal to 20.
But, these are not the only values these variables can take. a can be 10, b=c=53 and d=54 so that 10+53+53+54=170. For the latter, a≤b≤c≤d holds true but a is less than 20.
Therefore, statement 1 is insufficient.
Consider statement 2. When a=41, b=42, c=43 and d=44 then 5a=205 and 2d=88. 205≥88 and a≥20.
But if a=10, b=c=53 and d=54 then 5a=50 and 2d=50. Here, 50≤104 and a≤20.
Therefore, statement 2 is insufficient too.
Combine 1 and 2. If 5a≥2d then the minimum value a can hold is 20 and d can hold 50 so that 100≥100.
Now we can say a=20, b=c=d=50. This is sufficient. Option (C) is correct.