find two consecutive odd integers are added such that two fifth of the smaller exceeds two ninth of the greater by 4
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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, ATP,
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.
Hope it helps u....
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