Answer

Correct Answer : c ) 34

It is given that,

x, y, and z are natural numbers. So, x, y, and z can be 1, 2, 3, ....

xy + yz = 19 ........(1)

yz + xz = 51 ..........(2)

Form eq 1,

y(x + z) = 19

Now, for the product of y and (x + z) to be 19, y and (x + z) need to be factors of 19.

19 has only two factors, i.e., 1 and 19

As x, y, and z are natural numbers, x + z cannot be 1

So, y = 1 and x + z = 19

From eq 2,

z(x + y) = 51

we already have y = 1

z(x + 1) = 51

x + 1 = 51/z

Now, similarly to above, factors of 51 = 1, 3, 17, 51

For the product of z and (x + 1) to be 51, z can be either 1, 3, 17 or 51

Case 1: z = 1, x + 1 = 51, x = 50, not possible as it will give x + z = 51, which is not true for eq. 1

Case 2: z = 3, x + 1 = 51, x = 16, possible case

Case 3: z = 17, x + 1 = 3, x = 2, possible case

Case 4: z = 51, x + 1 = 1, x = 0, not possible as x is natural number

So, x, y, and z can be 16, 1, and 3 or 2, 1, and 17 respectively

So, xyz can be 48 or 34.

Hence, the minimum value of xyz is 34.

Thus, option (c) is correct.