Expert: Sherman D. - 3/17/2009

Question
Use the numbers 1, 2, 3, 4, 5, and 6 to write a subtracion problem with two mixed numbers that have a difference of 4 13/20.

Answer
4 13/20 = 93/20

we can easily deduce that the denominators will be 4 and 5 because 20 would be their LCD.

x/4 - y/5 = 93/20

5x - 4y = 93
or
4y - 5x = 93, if the mixed numbers were switched

5x = 4y + 93

this leaves us with 1, 2, 3, and 6

5(21) = 4y + 93, y = 3

even though 21/4 - 3/5 = 93/20

those aren't 2 mixed numbers

5(23) = 4y + 93, y = 11/2
5(26) = 4y + 93, y = 37/4
5(32) = 4y + 93, y = 67/4
5(31) = 4y + 93, y = 31/2
5(36) = 4y + 93, y = 87/4
5(62) = 4y + 93, y = 217/4
5(63) = 4y + 93, y = 111/2
5(61) = 4y + 93, y = 21/2

4(26) = 5y + 93, y = 11/5
4(32) = 5y + 93, y = 7

that one doesn't work because its not one of our choices

4(36) = 5y + 93, y = 51/5
4(31) = 5y + 93, y = 31/5
4(63) = 5y + 93, y = 159/5
4(61) = 5y + 93, y = 151/5
4(62) = 5y + 93, y = 155/5 or 31

(62/5) - (31/4)

well as you can see it took me awhile to find the right ones that follows the question

ANS :
(62/5) - (31/4) = (93/20)
or
12 2/5 - 7 3/4 = 4 13/20

however if you want it in a b/c - d e/f = 4 13/20 form whereas a, b, c, d, e, and f stand for the numbers 1-6, i recommend checking with answers.yahoo.com because it took me a while just to get it to improper fraction.