CAT 2003 Slot 2 — QA Question 28

5x + 19y = 64 where x, y ∈ I

This means that the values of x have an interval of 19 between each other and the values of y will have an interval of 5 between each other.

Now, there are 2 possible cases; y could either be positive or negative:

Case 1:
When y = 1, then x = 9
When y = 6, then x = −10
When y = 11, then x = −29 and so on
You will notice that the values of y are in intervals of 5 and that of x are in intervals of 19.
Generally speaking, when y is positive, we will get integral values of x when y’s unit’s digit is either 1 or 6.

Case 2:
When y = −4, then x = 28
When y = −9, then x = 47
Again, the values of y are in intervals of 5 and that of x are in intervals of 19.
That is, when y is negative, we will get integral values of x when y’s unit’s digit is either 4 or 9.

Now, let’s evaluate the options:

Option 1: “no solution for x < 300 and y < 0” is False.

∵ According to Case 2, we should get integral values of x when y is −4, −9 or −14 and so on.

Option 2: “no solution for x > 250 and y > –100” is False.

According to Case 2, we should get integral values of x when y is −99, −94, −74 or −69 etc.
Now, when y = −74, x = 294
∴ A solution exists.

Option 3: “a solution for 250 < x < 300” is True.

∵ y = −74, x = 294 is a possible solution

Option 4: “a solution for –59 < y < –56” is False.

∵ From Case 2, when y is negative, we will get integral values of x only when y’s unit’s digit is either 4 or 9.

Hence, option (c).