Question

find two consecutive odd numbers such that two fifth of the smaller exceed two-ninths of the greater by 4

Answer 1

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 2

Let the numbers be x and x+2

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

On solving you will get

8x/45=40/9

x=25

x+2=27

Numbers are 25 and 27