Problems based on Numbers Test 6

Let the three consecutive multiples of 3 are 3x, 3(x + 1) and 3(x + 2)
then, 3x + 3(x + 1) + 3(x + 2) = 72
⇒ 9x + 9 = 72
⇒9x = 63
⇒ x = 7
Hence, largest number = 3(7 + 2) = 3 × 9 = 27