A) 30
B) 66
C) 33
D) 27
Correct Answer: C
Solution :
Let three consecutive multiples of 3 be 3x, 3x+3 and 3x+6. According to questions. \[3x+(3x+3)+(3x+6)=90\] \[\Rightarrow \] \[3x+3x+3+3x+6=90\] \[\Rightarrow \] \[\,9x=90-9\Rightarrow x=\frac{81}{9}=9\] Largest number \[=3\times 9+6=33\]You need to login to perform this action.
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