Question

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

Answer 1

Let the consecutive odd integers be x and x+2.
Then according to the question,
2x/5 - 2/9(x+2)=4.
2x/5 -2x/9 - 4/9 = 4.
Taking lcm.
(18x-10x-20)/45=4.
8x-20= 45* 4.
8x-20= 180.
8x= 180+20=200.
So x = 200/8.
Therefore x= 25.
x+2= 25+2= 27.
25 and 27 are the two consecutive odd integers.
Hope it helps.

Answer 2

Answer:

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.