Question

find two consecutive odd integer such that two fifth of the smaller exceeds two ninth of the greater by 4

Answer 1

Let the smaller consecutive odd integer be x.

Then the larger number will be x + 2.

Given that 2/5th of the smaller exceeds 2/9th of the greater by 4.

 2/5 * x = 2/9 (x+2) + 4

9 * 2x = 5 * 2(x+2) + 4 * 45

18x = 10(x+2) + 4 * 45

18x = 10x + 20 + 180

18x - 10x = 200

8x = 200

x = 25.

If one number is 25 then the other number is 25 +2 = 27.

The numbers are 25 and 27.


Hope this helps!

Answer 2

Let the first odd number be x.
The second odd number is x+2 as they are consecutive.

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

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

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

18-10/45 x = 40/9

8/45 x = 40/9

x = 40/9* 45/8

x = 25

Therefore The two consecutive odd number are 25,27