Answer:
Choice D is correct.
Explanation:
We have given that
If y varies directly as x and z,
y ∝ xz
y = kxz eq(1)
where k is constant of variation.
As given that y = 4 when x = 6 and z = 1
4 = k(6)(1)
4 = k(6)
4 = 6k
k = 4/6
k = 2/3
Putting the value of k in eq(1), we have
y = 2/3xz
Now, we have to find the value of y when x = 7 and z = 4
y = 2/3(7)(4)
y = 56/3
Hence, Choice D is correct.