Question

Find two consecutive odd numbers such that the sum of one-fifth of the smaller and four-sevenths of the larger is equal to fifty-nine.

Answer 1

Do you know how to set this type of a problem up? I will explain it to you!!! In order to better understand the consecutive odd number thing, think of the first two consecutive odd numbers you know: 1 and 3. If you start at 1, what do you do to get to 3? You add 2, right? So if you call 1 "x", then you have to add 2 to it to get to 3. So 3 would be x + 2. Those are not the numbers, I'm just using those as examples to start setting up this problem. So the two unknowns are "x" and "x + 2". one-fifth of the smaller number would translate to "1/5x" and 4/7 of the larger would translate to "4/7(x + 2)". Those are the 2 numbers now. We have to add them together to get 59. So the equation looks like this:\[\frac{ 1 }{ 5 }x+\frac{ 4 }{ 7 }(x+2)=59\] Distribute the 4/7 into the parenthesis to get 4/7x + 8/7. At this point, the easiest thing to do would be to rid yourself of the denominators by multiplying everything by the least common multiple of each, which would be 35. \[(35)\frac{ 1 }{ 5 }x+(35)\frac{ 4 }{ 7 }x+(35)\frac{ 8 }{ 7 }=(35)59\]35 divided by 5 is 7, so the first term is 7x. 35 divided by 7 is 5, and 5 times 4 is 20, so the second term is 20x. 35 divided by 7 is 5, and 5 times 8 is 40, so the third term is 40, all equal to 35 times 59 which is 2065.\[7x+20x+40=2065\]Combine your x terms to get 27x and subtract the 40 from both sides to get 2025. Now you have a much simpler equation:\[27x=2025\]Divide by 27 to solve for x. x = 75. So if x = 75, that's the smaller number, remember? The larger number was x + 2, or 75 + 2 which is 77. This seems reasonable even without checking our answer for accuracy, because they are both odd consecutive numbers. How nice that worked out!