Question

find two consecutive odd integers such that 2/5th of the smaller exceeds two-ninth of the greater by 4 ? solve in step form

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.

Answer 2

What is the answer

Answer:25,27

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

2x/5=2x+40/9

NOW WE WILL DO CROSS MULTIPLICATION

9(2x)=5(2x+40)

18x=10x+200

18x-10x=200

8x=200

Now we will divide by 4

2x=50

Now we will divide by 2

x=25

x=25

x+2=25+2=27