for non-native speakers, note rounded to the nearest tens' not tenths In set P, 5 integers are EVEN and 15 are ODD. The integers can be odd or even (excluding 0 as non-positive). In set R 5 are rounded DOWN and 15 rounded UP. question- which could be Sum(numbers in R) - Sum(numbers in P) So the fixed terms is rounding and assigned around the even/the odd. If the numbers are rounded to the nearest tens' their digits will be 0. We may have different scenarios how the numbers are rounded to the tens, but for all the numbers in set R the digits will be ZERO and their numbers will be added/summed with the digit ZERO too.. Thus, we have some value for the Sum(numbers in P) minus some value with the digit 0, and there are total 20 integers in set P. Case I) -25 is possible if set P contains 17 even and 3 odd numbers, 30-Sum(set P)=-25, Sum(set P)=55 -> 17*2 + 3*7 (set P contains seventeen 2s, and three 7s, in total 20 numbers in P, three odd numbers) could be Case II) 75 is not quite possible as set P contains 10 even and 10 odd numbers, 100-Sum(set P)=75, Sum(set P)=35 -> 10*2 + 7*1 + ... Case III) 145 is possible if set P contains 3 even and 17 odd numbers, 170-Sum(set P)=145, Sum(set P)=25 -> 3*2 + 16*1 + 3 (set P contains three 2s, sixteen 1s, and 3, in total 20 numbers in P, sevetnee odd numbers) could be Cases possible: I and III only correct answer e

Sorry Pemdas, I think otherwise. There is no need to consider 17 odd or 3 even or vice -versa. The question itself says 1/4 of the numbers are even i.e 5 Even and 15 odd. Odd numbers are rounded up and even are rounded down. So for odd number valus increased and for even it decreases. Minimum unit digit for odd = 1, Maximum unit digit = 9 Minimum unit digit for Even = 0, Maximum unit digit = 8 Let's consider extreme cases: For R-P to be minimum: Odd has to be minimum and even has to be maximum. So on rounding 5 even digits sum decreased by 8*5 = 40 So on rounding 15 odd digits sum decreased by 1*15 = 15 Overall change = 15 - 40 = -25 For R-P to be maximum: Odd has to be maximum and even has to be minimum. So on rounding 5 even digits sum decreased by 0*5 = 0 So on rounding 15 odd digits sum decreased by 9*15 = 135 Overall change = 135 - 0 = 135 So -25 <= R -P <= 135 Ans D.