Guest Posted January 2, 2010 Report Share Posted January 2, 2010 I have X apples, I gave 1/2 to you and you gave me 6/9 of yours. In the end, you had X-4 apples and I X-1 apples. How many apples do you and I have? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 what is answer? this is imposible beacuse: i solve equ: http://www.wolframalpha.com/input/?i=3*x-12*y%3D18%2C15*x-6*y%3D72 and the answer is possitive! Quote Link to comment Share on other sites More sharing options...
0 DudleyDude Posted January 2, 2010 Report Share Posted January 2, 2010 I don't really think this is a word riddle since it's a straight up algebra problem and I'm not sure why Vahid insists it is impossible when there is clearly an answer. In the end, the "I' in the puzzle has 13 and the "you" has 10 (in other words, X in the puzzle is 14) This is assuming that the "you" calculated 2/3 of their apples based on their initial count. I found an answer for if they calculated the 2/3 after receiving the extra apples but it did not come out with nice, integral answers so I discounted that case. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 what is answer? this is imposible beacuse: i solve equ: http://www.wolframalpha.com/input/?i=3*x-12*y%3D18%2C15*x-6*y%3D72 and the answer is possitive! i solve it anfter see the answer! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 And what is good in this "puzzle"? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 Why is everyone complaining? This was posted in Math Puzzles, not word riddles. Why is it surprising that it is all math? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 assuming "you" doesnt have any apples to begin with, both "you" and "I" have the same amount of apples that "I" started with. that means that the number of apples "you" has and "I" has add up to the total number of apples "I" had in the beginning x-1+x-4=x (the number of apples "I" had and "you" had equal the original amount of apples) 2x-5=x (commutative property/adding like terms) 2x=x+5 x=5 from this, you can then find out what each had. hope it helped! seventh grade algebra 1! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2010 Report Share Posted January 3, 2010 assuming "you" doesnt have any apples to begin with, both "you" and "I" have the same amount of apples that "I" started with. that means that the number of apples "you" has and "I" has add up to the total number of apples "I" had in the beginning x-1+x-4=x (the number of apples "I" had and "you" had equal the original amount of apples) 2x-5=x (commutative property/adding like terms) 2x=x+5 x=5 from this, you can then find out what each had. hope it helped! seventh grade algebra 1! you cannot assume "you" had no apples to start with. I am with Dudley Dude on this answer. In the end, you have x-4 apples and I have x-1 apples. The total number of apples in the problem is therefore 2*x - 5 In the beginning, I had x apples, meaning you had the remaining 2*x-5 - x = x-5 apples. I give you half of my original amount or x/2 and you give me 2/3 of your original amount or 2/3(x-5) My final amount is therefore x - x/2 + 2/3(x-5) = 7/6 * x -10/3 We know from the problem that I end up with x-1 so 7*x/6 -10/3 = x-1 Solving for x gives x/6 = 7/3 or x = 14 Therefore I started with 14, and you started with 9. I gave you 7 and you gave me 6. I ended with 13 and you ended with 10. This matches every aspect of the problem's requirements. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2010 Report Share Posted January 3, 2010 I have X apples, I gave 1/2 to you and you gave me 6/9 of yours. In the end, you had X-4 apples and I X-1 apples. How many apples do you and I have? I have 10 apples and you have 6 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2010 Report Share Posted January 3, 2010 x = "I"s apples to start, y = "your" apples to start, therefore x-1 = x - (x/2) + (2y/3) which eventually settles down to 4y = 3x - 6. The other side of the equation is y - (2y/3) + (x/2) = x - 4 which boils down to 2y = 3x - 24. If you then do some simple substitution, you get y=9 and x=14... x (starting number for "I") was 14, gave 7 (half), received 6, now has 13 (x-1 or 14-1); y "you" started with 9, have 6 (two thirds) but received 7 so are not on 10 (x-4... 14-4). Cool? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2010 Report Share Posted January 3, 2010 yes mmiguel1 i understand what you mean, caught my mistake Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2010 Report Share Posted January 3, 2010 1/2X + 6/9X=(X-1)+(X-5) Works out : I have 6 and you have 9 Is this correct? Looks algebraically correct to me. Quote Link to comment Share on other sites More sharing options...
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I have X apples, I gave 1/2 to you and you gave me 6/9 of yours. In the end, you had X-4 apples and I X-1 apples. How many apples do you and I have?
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