Is the median of a set of five distinct positive integers {p, q, r, s, t} greater than 8?

✓ Correct Answer: Option C

(1) Mean is 10. Sum = 50. Smallest possible integers are 1,2,3,4. 1+2+3+4=10. Fifth could be 40. {1,2,3,4,40}, median is 3 (not >8). Or {8,9,10,11,12}, median is 10 (>8). Not sufficient. (2) Largest is 15. {1,2,3,4,15}, median is 3 (not >8). Or {10,11,12,13,15}, median is 12 (>8). Not sufficient. (1)+(2): Sum=50, largest=15. Remaining sum for 4 integers = 35. For median to be >8, the middle integer must be >8. If the integers are ordered x1<x2<x3<x4<x5=15, then x3 is the median. To make x3 <= 8, try to maximize x1, x2, x4. Smallest set with x3=8 is {6,7,8,14,15}. Sum = 50. Median is 8 (not >8). If median is 9, {7,8,9,11,15}, sum is 50. Median is 9 (>8). Since both cases are possible for sum 50, and max is 15, let's re-evaluate. If median is 8, (x1+x2+8+x4+15)=50. x1+x2+x4=27. x1<x2<8<x4<15. Smallest possible x1,x2 is 1,2. Then x4=24, but x4 must be <15. This implies x3 cannot be 8 if the other constraints are met. Let's try to make median > 8. x3 > 8. Smallest x3=9. So x1,x2<9, x4<15. Smallest possible set for sum=50, largest=15, and x3=9. E.g., {5,7,9,14,15}. Sum=50. Median is 9, which is >8. This works. Now try to make median <= 8. Smallest possible x3 is 1,2,3... Let x3=8. Then x1<x2<8<x4<15. And x1+x2+x4+15+8=50 => x1+x2+x4=27. x1 and x2 must be smaller than 8. x4 must be between 8 and 15. Smallest possible x1,x2 are 1,2. Then x4=24. This violates x4<15. So, x3 cannot be 8. Therefore, the median must be greater than 8. Sufficient.