Browse · harp Print → jmc prealgebra junior Problem (1+11+21+31+41)+(9+19+29+39+49)= (A) 150 (B) 199 (C) 200 (D) 249 (E) 250 Solution — click to reveal We make use of the associative and commutative properties of addition to rearrange the sum as (1+49)+(11+39)+(21+29)+(31+19)+(41+9)=50+50+50+50+50=250⟹250 Final answer E ← Previous problem Next problem →