Date: 4/14/96 at 23:18:30 From: Anonymous Subject: Math puzzle Could you please show me how to do this problem. The two digits in the numerator of a fraction are reversed in its denominator. If 1 is subtracted from both the numerator and the denominator, the value of the resulting fraction is 1/2. The fraction whose numerator is the difference and whose denominator is the sum of the units and tens digits equals 2/5. Find the original fraction. I tried using 10t + U divided by 10u + t to represent the reversed numbers, and (10t+u)-1 divided by (10u+t)-1 = 1/2 to represent the second clue, and t-u divided by t+u = 2/5 to represent the third clue,but it doesn't work.
Date: 4/26/96 at 7:50:24 From: Doctor Steven Subject: Re: Math puzzle Okay we'll search for this answer but we'll do it systematically. First let's look to see if we can find some conditions we want. Say a and b are the digits. We want (ab - 1)/(ba-1) = 1/2. So (ba-1) must be even. This says a is odd. We also want ba > ab, so ab/ba is less than one, so b > a. Now we want our denominators ba and (ba-1) to be two digit numbers. If a >= 5, then our numerator is > 50, so the denominator would be three digits. So a < 5. This says a is odd, a < 5, and b > a. We also want b almost 2*a but this is inexact. So a can be 1, or a can be 3. If a = 1 then b had better be 2, or else equation 1 wouldn't hold, so check this: (ab - 1) = 11 and (ba - 1) = 20, almost 1/2 but not quite, so a does not equal 1. So a = 3. Since the numerator is in the thirties we know b = 6, or b = 7. If b = 6 our denominator would have to be in the 70's, so b cannot equal 6, so b = 7. So a = 3, and b = 7 if they equal anything at all. Check this. (ab - 1) = 36 and (ba - 1) = 72, so (ab - 1)/(ba - 1) = 1/2. This is the answer. Check to see if (their difference/their sum) = 2/5. (7 - 3)/(7 + 3) = 4/10 = 2/5 And the original fraction is: 37/73 Sort of a neat puzzle. The trick was you didn't want to use systems of equations to solve it, but rather think about the restrictions on a and b. -Doctor Steven, The Math Forum
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