Question

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

Answer 1

Answer:

Step-by-step explanation:

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

please mark as a brainliest

Answer 2

Answer:

25 and 27

Step-by-step explanation:

Let the numbers be x and x+2

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

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

2x/5 - 2x/9 - 4/9 = 4

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

8x/45 = 40/9

8x = (40 x 45 )/9

8x = 200

x =200/8

x= 25

Therefore the numbers are (x) and (x+2) = 25 and 27