X,Y,Z share N68 so that for every N1 that Z gets, Y gets N2 and for every N3 that Y gets, X gets N4, how much does Y get?
Let's simplify the given information:
- For every N1 that Z gets, Y gets N2. This can be written as Y = N2/N1 * Z.
- For every N3 that Y gets, X gets N4. This can be written as X = N4/N3 * Y.
Now, let's substitute the first equation into the second equation to get X in terms of Z:
X = N4/N3 * (N2/N1 * Z)
Simplifying this expression:
X = (N2*N4/N1*N3) * Z
We know that X+Y+Z = N68, so let's substitute X and Y into this equation:
(N2*N4/N1*N3) * Z + Y + Z = N68
Simplifying this expression:
Y = N68 - Z * (N2*N4/N1*N3 + 1)
We are not given any values for N1, N2, N3, or N4, so we cannot solve for Y numerically. However, we can see that Y depends on the values of Z and the ratios between N1, N2, N3, and N4.
From the ratios, we have
Z/Y = N1/N2
X/Y = N4/N3
X+Y+Z = 68
dividing by Y, we have
X/Y + 1 + Z/Y = 68/Y
N4/N3 + 1 + N1/N2 = 68/Y
Y = 68/(N4/N3 + 1 + N1/N2)
Now you just need to find values for N1,N2,N3,N4 so that Y is an integer.
Thank you for providing the additional steps! Yes, if we know the values of N1, N2, N3, and N4, we can solve for Y. For example, let's say N1 = 2, N2 = 3, N3 = 4, and N4 = 5.
Then,
Z/Y = 2/3 -> Z = (2/3)Y
X/Y = 5/4 -> X = (5/4)Y
X+Y+Z = 68 -> (5/4)Y + Y + (2/3)Y = 68 -> Y = 36
Therefore, Y gets N36.
To determine how much Y gets, we need to understand the given ratios in the problem.
According to the problem, for every N1 that Z gets, Y gets N2. This can be expressed as:
Y = (N2 / N1) * Z
Similarly, for every N3 that Y gets, X gets N4. This can be expressed as:
X = (N4 / N3) * Y
Now, let's find the value of Y.
Since we know that X, Y, and Z share N68, we can express this as:
X + Y + Z = N68
Substituting the value of X from the second equation, we get:
((N4 / N3) * Y) + Y + Z = N68
Combining like terms:
Y * (1 + (N4 / N3)) + Z = N68
Now, let's simplify:
Y * ((N3 + N4) / N3) + Z = N68
To isolate Y, subtract Z from both sides:
Y * ((N3 + N4) / N3) = N68 - Z
Finally, divide both sides by ((N3 + N4) / N3) to solve for Y:
Y = (N68 - Z) * (N3 / (N3 + N4))
Thus, Y gets (N68 - Z) * (N3 / (N3 + N4)).