Browse · harp Print → jmc prealgebra junior Problem 2(81+83+85+87+89+91+93+95+97+99)= (A) 1600 (B) 1650 (C) 1700 (D) 1750 (E) 1800 Solution — click to reveal Find that (81+83+85+87+89+91+93+95+97+99)=5⋅180 Which gives us 2(5⋅180)=10⋅180=1800 Thus 1800 is the correct answer Final answer E ← Previous problem Next problem →