Saved Bookmarks
| 1. |
Find two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4. |
|
Answer» 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 + 4Taking 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 = 25Putting the value of x, we getx + 225 + 2 = 27So, the smaller integer is 25 and the greater odd integer is 27 |
|