Concept: The question tests on Inequalities and their application

Question data tells us that x, y and z are positive numbers. Quite often, a lot of test takers assume this to be positive integers and end up falling for trap answers. x, y and z could be fractions too.

In this question, plugging in numbers is a good way to prove/disprove the individual statements, while using a property is a good way of combining them.

From statement I alone, x<2z<y.

If x = 1, z = 2 and y = 5, 1<4<5. Is z between x and y? YES.

If x = 1, z = 0.75 and y = 2, 1<1.5<2. Is z between x and y? NO.

Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, 2x<z<2y.

If x = 0.5, z = 2 and y = 1.5, 1<2<3. Is z between x and y? NO.

If x = 1, z = 3 and y = 5, 2<3<10. Is z between x and y? YES.

Statement II alone is insufficient. Answer option B can be eliminated. Possible answer options are C or E.

Combining statements I and II, we have the following:

From statement I, x<2z<y; from statement II, 2x<z<2y.

Since the inequality signs are the same, the two inequalities can be added. Adding the inequalities, we have,

3x<3z<3y.

Dividing all terms of the inequality by 3, we have x<z<y.

Is z between x and y? YES.

The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.