A) 15
B) 17
C) 19
D) Data inadequate
Correct Answer: C
Solution :
[c] Let 3 consecutive odd no. be x, x+2 and x+4 ATQ \[\frac{x+x+2+x+4}{3}=12+\frac{1}{3}x\] \[\frac{3x+6}{3}-\,\frac{x}{3}=12\] \[=2x+6=36x=\frac{36-\,6}{2}=15\] Last no=15+4=19You need to login to perform this action.
You will be redirected in
3 sec