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?

Answer 1

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.

Answer 2

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.

Answer 3

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.