When numbers are in proportion to each and in reduced form, then at some point they had to have a number in common that would then reduce to 3/4.
Consider this number x. So, before it was reduced, we know that the proportion looked like this: 3x/4x and we also know that the sum is 36. Therefore, 3x + 4x =36
Thus, 7x = 36 and x = 36/7
Putting this back into the proportion of 3/4, you have 3(36/7) and 4(36/7), which means the two numbers are:
108/7 and 144/7.
This process is used quite a bit in Geometry in the study of proportional figures and angles.
The two numbers will not be whole numbers.
Questions of this type can be solved using the following general steps.
If two numbers add x and y are such that:
x + y = N and
x/y = a/b
x = N*[a/(a + b)]
y = N*[b/(a + b)]
In the above question:
a = 3, b = 4, and N = 36
x = 36*3/(3 + 4) = 36*3/7 = 108/7
y = 36*4/(3 + 4) = 36*4/7 = 114/7
x + y = N = 108/7 + 144/7 = (108 +144)/7 = 252/7 = 36
We assume that one number is x and the other is 3x/4 so that their ratio is 3/4.
Their algebraic sum = x+(3x/4) =7x/4 which is equal to 36 by hypothesis.
Therefore, 7x/4 = 36 or
x=36*4/7 = 144/7 and the other numver,3x/4 = (3/4)*144/7 =108/7
The numbers are 144/7 and 108/7.
Aliter: 1 is distrinbeted in the format m/n or p:q implies,
p/(p+q)) and q/(p+q). So if N is to be distributed in this prportion, then Np/((p+q) and Nq/(p+q) are the required results. Putting p=1 and q= 3/4 you get:36*1/(1+3/4) = 36*4/7 and 36*(3/4)/(1+3/4) = 36*3/7 are the required results.