Another approach

[user44]

For this, if s = 0, it would mean that for List 1 it is 2+5+t, but divided by 4 to get the mean; versus for List 2 where it is 2+5+t, but divided by 3 to get the mean. Which means that the average for List 1 will be smaller since its the same numerator but divided by a larger denominator. As such, in order for both means to be the same (i.e., balance out the larger denominator), something positive needs to be added on to 2+5+t (i.e., s needs to be > 0). Hence answer is A

is my approach correct or where am I wrong

[Sunny]

I did it this way. if we are only equating the two means, we reach to only one value of s, which is greater than 0. but what if we take a negative value of s, can the two means still remain equal? the answer is yes because we have an unknown variable, t, which can also be negative. if s=-1; mean are equal as long as t=-10. you can test with whatever value you like.