Find two consecutive odd integers such that two-fifth of the smaller exceeds two- ninth of the greater by 4.
Answers
Answer:
Let the both smaller and greater consecutive odd integers numbers be x & (x+2)
If Two-fifth of smaller number(x) is greater than Two-ninth of greater number(x+2) by 4 then the equation will be-
2/5 × x = 2/9(x+2) + 4
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9 8x/45 = 40/9
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9 8x/45 = 40/9 8x×9 = 45×40
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9 8x/45 = 40/9 8x×9 = 45×4072x = 1800
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9 8x/45 = 40/9 8x×9 = 45×4072x = 1800x = 1800/72
2/5 × x = 2/9(x+2) + 42x/5 = 2x/9 + 4/9 + 4 2x/5 - 2x/9 = 4/9 + 4 18x-10x/45 = 4+36/9 8x/45 = 40/9 8x×9 = 45×4072x = 1800x = 1800/72 x = 25
Smaller no. = x = 25
Smaller no. = x = 25Greater number = (x+2) = 25+2 = 27