Question

Find the two consecutive odd numbers such that two-fifths of the smaller exceeds two-ninths
of the greater by 4.​

Answer 1

Answer:

Let x be the smaller odd integer and (x + 2) be the greater odd integer respectively.

2/5th of the smaller odd integer exceeds 2/9th of the greater odd integer by 4.

So, according to the question.

2x/5 = 2/9*(x + 2) + 4

⇒ 2x/5 = (2x + 4)/9 + 4

Taking L.C.M. of the denominators of the right side, we get.

2x/5 = (2x + 4 + 36)/9

Now, cross multiplying, we get.

⇒ (2x*9) = 5*(2x + 40)

⇒ 18x = 10x + 200

⇒ 18x - 10x = 200

⇒ 8x = 200

⇒ x = 200/8

⇒ x = 25

Putting the value of x, we get

x + 2

25 + 2 = 27

So, the smaller integer is 25 and the greater odd integer is 27

Answer 2

Answer:

Let x be the smaller odd integer and (x + 2) be the greater odd integer respectively.

2/5th of the smaller odd integer exceeds 2/9th of the greater odd integer by 4.

So, according to the question.

2x/5 = 2/9*(x + 2) + 4

⇒ 2x/5 = (2x + 4)/9 + 4

Taking L.C.M. of the denominators of the right side, we get.

2x/5 = (2x + 4 + 36)/9

Now, cross multiplying, we get.

⇒ (2x*9) = 5*(2x + 40)

⇒ 18x = 10x + 200

⇒ 18x - 10x = 200

⇒ 8x = 200

⇒ x = 200/8

⇒ x = 25

Putting the value of x, we get

x + 2

25 + 2 = 27

So, the smaller integer is 25 and the greater odd integer is 27