Question

Find two consecutive odd integers such that 2/5th of the smaller exceeds two-ninth of the greater by 4 ?

Answer 1

Let the two numbers be x and x+2


Given, 2x/5 = 2/9( x+2) + 4

2x/5 = 2x/9 + 4/9 + 4

2x/5 - 2x/9 = 40/9

(18x - 10x)/45 = 40/9

8x = 200

x = 200/8

    = 25.

The number will be 25 + 2 = 27.

Answer 2

Solution:-

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.