Question

Find 3 consecutive odd integers such that 10 times the first is 59 more than the third?

Answer 1

Answer: The required three consecutive odd integers are 7, 9 , 11.

Explanation:

Let x , x+2 , x+4 be the three consecutive odd integers.

As per given ,

10(x) = 59+x+4

⇒10x= 59+x+4

⇒ 9x= 63

⇒ x= 7 [ divide both sides by 7]

Hence, the required three consecutive odd integers are 7, 9 , 11.