Find a 2 decimal place number between 0 and 1 such that the sum of its digits is 13 and such that when the digits are reversed the number is decreased by 0.45.

Question

 I understand there is a standard formula for figuring these out, but I don't really understand why.

I think it's something along the lines of 0.1t +0.01h =13 and 0.01t + 0.1h = 12.55.

But I'm having trouble understanding why this is and applying it to other problems, especially fractions.

Answer 1

The first statement tells us the the sum of the tenths digit (t) and the hundredths digit (h) is 13.  The equation for this is:  t + h = 13

The second statement tells us that reversing the digits decreases the number by .45 this means that .th is .45 greater than .ht.  To get .th, we need to multiply t by .1 and h by .01.  From the second statement, we would get the equation:  .1t + .01h = .1h + .01t + .45

To solve this, I would solve the first equation for one of the variables (I will solve for t) and substitute into the second equation.

t + h = 13

t = 13 - h

.1t + .01h = .1h + .01t + .45

.1(13 - h) + .01h = .1h + .01(13 - h) + .45

1.3 - .1h + .01h = .1h + .13 - .01h + .45

1.3 -.09h = .09h + .58

1.3 - .58 = .09h + .09h

.72 = .18h

h = 4

Now that we have one digit, we can substitute the number into the first equation to find the other digit

t + h = 13

t + 4 = 13

t = 9