jmc
prealgebra junior
Problem
If 991+993+995+997+999=5000−N, then N=
(A)
5
(B)
10
(C)
15
(D)
20
(E)
25
Solution — click to reveal
991+993+995+997+999=5000−N⇒(1000−9)+(1000−7)+(1000−5)+(1000−3)+(1000−1)=5000−N⇒5×1000−(1+3+5+7+9)=5000−N⇒5000−25=5000−N⇒N=25→25.