# GMAT Quant: The Roman Numeral Problem

The GMAT Quant generally throws up a few problems that designed to act as speed-breakers during the course of the 75-minute Quantitative section. Not surprisingly, these questions are what are usually referred to as the Roman Numeral problems — information followed by III statements, with the question asking you identify the statements that could be true, must be true or is trueDepending upon the the question stem — could be or must be — you need to follow a specific approach to nail these questions without wasting much time. But one still has to proceed with the knowledge that these problems will take a tad longer to solve than the others since the equivalent of since almost three questions is built into one question.

## Roman Numeral Problem – MUST BE TRUE

Whenever the problem is a MUST BE TRUE question then you should know that the question itself will have enough information for you to be able to solve or develop a framework to select or reject statements.

Hence, roman numeral questions involving MUST BE TRUE framework must not be solved by jumping directly into the statements. These are questions that need you to some pre-work before your jump into the statements. Let us look at a few sample GMAT PS questions.

```QUESTION # 1

If k and t are integers, and k² – t² is an odd integer, which of the following
must be even?

I.   k + t + 2
II.  k² + 2kt + t²
III. k² + t²

(1) None
(2) I
(3) II
(4) III
(5) I, II & III```

Since it is a MUST BE TRUE question you need to do a bit of pre-work before you evaluate each of the the statements.

t² is odd, means one of the two terms is even and one of terms is odd, making one of k and t even and the other odd. If both are odd or both are even, then the difference cannot be odd.

If we now proceed to the statements,

I. is odd since odd + even will be odd and odd + 2 will be odd,
II. is odd since k²  + will be odd and  2kt will be even, the sum of the two will be odd
III. is odd since  k²  + will be odd

making option (1) the right choice.

```QUESTION # 2

If n and y are positive integers and 450y = n³, which of the following is an integer?

I.   y / 3 * 2² * 5
II.  y / 3² * 2 * 5
III. y / 3 * 2 * 5²

(1) None
(2) I
(3) II
(4) III
(5) I, II & III```

This question is equivalent to the previous one, since both IS TRUE and MUST BE TRUE are the same and hence some pre-work is rquired before you exmaine the statements.

450 can be factorised as 2*3²*5²*y; if n is an integer and the cube root of2*3²*5²*y, then the powers of 2,3 and 5 have to be alteast 3 (2³*3³*5³) or multiples of three.

So y has to make up for the remaining powers and hence should have atleast two powers of 2,  one of 3 and one of 5 or  in other words y should be atleast 2²*3*5.

From this it is clear that both II and III do not divide y since they have an extra power of 3 and 5 each. They could be true since y can have higher powers but need not be true since it is not a must, making option (2) the only choice.

## Roman Numeral Problem – COULD BE TRUE

The other type of question stem is the could be true type that will definitely require you to move into the statements at an earlier stage but it is crucial that you approach the questions with a clear strategy. Let us take a GMAT question to understand the implications.

```QUESTION # 3

If x is positive, which of the following could be the correct ordering of
1/x, 2x, and x²?
￼
I.   x² < 2x  < 1/x
II.  x² < 1/x < 2x
II.  2x < x²  < 1/x

(1) None
(2) I
(3) II
(4) I  & II
(5) I, II & III```

In such questions you have to first go by substituting values but there is a way to go about it.

If they have mentioned positive, then substitute numbers not only greater than 1 but also all fractions, vlaus between 0 and 1; on most questions, the catch lies in the fractions more importantly fractions that lie close to 1!

So on all such questions be sure to check the inequalities using values close to 1 such as 9/10, 8.5/10 and 8/10; it is on this range that statement that you think could not be true starts becoming true.

Statement 1 fails for positive values greater than 1 but becomes true for any fraction or value lying between 0 and 1 and hence could be true.

Statement II should again be a fraction since x² < 1/x but at the same time 1/x < 2x.Our normal tendency whenever we think fractions is to think 1/2, 1/3 or 1/4. If you take 1/2  you will get 1/x = 2x and then you will move to checking 1/3 and 1/4 and find that 1/x > 2x. Since it is not satisfying these three values, you will assume that II cannot be true. What if you take a value close to 1 such as 9/10, you will quiclly see that 1/x will be 10/9 and 2x will be 18/10 making 1/x < 2x. So, always check for fractions close to the 0 and close to 1.

Statement III will fail the test of substitution, making option (4) the right one.

But is there is a better way or rather faster method?

Evaluate the inequalities two at a time, if you get inconsistent results then the statment cannot be true.

If we take Statement III and evaluate 2 inequalities at a time.

• 2x < x² or x² > 2x,  means x > 2
• x² < 1/x or x³ < 1, means x < 1

Since x cannot be both greater than 2 and less than 1 this statement cannot be true, you can try the sam method on the first two stahemtn and you will find that they yield consistent results.

```QUESTION # 4

If 0 < r < 1 < s < 2 , which of the following must be less than 1?

I.   r/s
II.  rs
III. s – r

(1) I
(2) II
(3) III
(4) I  & II
(5) I & III```

True, this question is a MUST BE TRUE question, and thus should have been in the earlier section but whenever the question involves an inequality it becomes a question that needs subsititution.

When you are asked to test a MUST BE TRUE situation that involes an inequality, you should look for cases to falsify the statements or look for cases where the inequality will not hold and where is an inequality least likely to hold? At the edges.

Statement I is true since r < s , r/s has to be less than 1!

Statement II is where the real test begins and to do this with clarity you need to plugin the inequality at the edges.

r lies between 0 & 1 and s between 1 & 2 and you need to check whether there are values for whihc the statement, rs < 1, is falsified, or in other words check if rs can be greater than 1.

To check that you need to take the maximum possible values for r and s — for close to 1 andfor s close to 2, say 9/10 and 18/10, for which rs becomes greater than 1, making the statement false.

Statement III can also be tested in the same format — by checking if  s – r can be greater that 1. For that you need to take the largest possible value for s, close to 2, and smallest possible value for r, close to 0, the difference will be greater than 1 and hence this statement also need not be true, making option (1) the right option.

This post was a result of a mail sent by one of the readers of this blog. So do let me know if any specific GMAT topic, question type or area is giving you a bit of a bother and I will work a post around it.

1. dvs28 says

If the square root of p2 is an integer greater than 1, which of the following must be true?

I. p2 has an odd number of positive factors

II. p2 can be expressed as the product of an even number of positive prime factors

III. p has an even number of positive factors

A.I

B.II

C.III

D.I and II

E.II and III

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2. I only. Every perfect square has an odd number of factors since if P is a^x.b^y.c^z then P^2 will be a^2x.b^2y.c^2z and number of factors will be (2x+1)(2y+1)(2z+1) which will be odd. Hence option A, the rest need not be true.

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3. Sameer Kamath says

How many liters of water must be evaporated from 50 liters of a 3-percent sugar solution to get a 10-percent solution?

A) 35
B) 33 1/3
C) 27
D) 16 2/3
E) 15
Would love to know your approach in this problem sir. I was a part of the Bootcamp workshop and i thoroughly enjoyed it. So i came back home and started solving sums with your ‘non-swayamvar” approach especially for mixture problems like this one. I could’nt get the answer for this so am posting it here.

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4. Sameer Kamath says

So i worked a bit more on this and came up with the solution. Could you please check the approach?
I used the weighted averages approach you used in class. 50 litres/3%(a) , X litres of 0%(b) sugar(because we are adding pure water) to get 10%(c). So (50/X)= (0-10)/(10-3). Hence X=35!

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