examPAL DS Forum Expert David - Ask Me Anything about GMAT DS

Hello David,

Can you Just suggest me upon how to prepare for DS from scratch?And resources which would prove to be helpful. Everything I need to do for acing the DS section.

Posted from my mobile device

Hello Shrinidhi,

Thanks for the question!

First off, we made a video on exactly that topic, see it here

In words, the first thing you need to do is realize that there is a very large difference between 'regular multiple choice' PS questions and DS questions.
In PS, you are asked a question and need to calculate the answer to the question.
In DS, you only need to figure out if a certain statement gives you enough information so that you could potentially calculate the answer should you really want to.

For example, if a PS question were to ask you "3^10 = ?" you would actually need to calculate 3 to the power of 10 and look for the relevant answer choice.
If a DS question were to ask you "x^10 = ?" then all you would need to do is look for a statement that tells you the value of x (which would then let you calculate x^10, if you really wanted to).
So "x = 3" would be sufficient as would be "2x + 5 = 11" as would be "x^3 = 27". Each of these allows us to calculate the value of x, which then allows us to calculate the value of x^10. Note that we didn't actually have to complete the calculations! All we had to do was realize that the above statements allow us to calculate the answer, should we want to. This focus on 'sufficiency' and not on 'actually calculating' is the most important thing to realize when going into DS. Once you've got this down, your life becomes much easier and your scores much higher.

The answer choices also reflect this 'information-focus':
Select (A) when the first statement on its own gives you enough information, and the second does not
Select (B) when that the second statement on its own is enough, and the first is not
Select (C) when you must combine both statements to have enough information
Select (D) when each of the statements on its own gives enough information
Select (E) when, even when you combine both statements together, you still don't have enough information

Other than that, much of the routine 'learning the material' work is the same. You need to know the concepts behind, for example, number properties, rate and work, powers and roots, linear equations and so on... You need to learn common GMAT traps, such as missing a negative option or forgetting about zero.

In particular, you need to develop cognitive flexibility - the ability to choose the right tool for the right question. Is it better to just count the number of variables and equations and see if you have enough equations to solve (a Logical approach)? Should you maybe try out a few numbers to help make things concrete (an Alternative approach)? Maybe just dive straight into algebra and simplify (a Precise approach)?

Additionally, you need to learn from your mistakes. Did you miss a negative or positive option? Did you get confused with prime factoring? Every mistake you make should be listed in an error log and you should try 'fixing' your mistake before moving on to the next question. You can either build up this error log on your own or you can let someone else do it for you. On our platform, we use an AI to choose, based on each individual student's preferred solution tools and common mistake reasons, the set of questions and solution approaches that are best for them. If you like, give us a go.

If you're having difficulty with a specific question and would like my input on it, please don't hesitate to ask!

(ex 1) A certain store, books are sold. Books are hard cover or soft cover and hard cover books sold $10 each and soft cover books sold $6. Is the number of hard cover books sold greater than that of soft cover books sold?
1) The average price sold of total books is $9
2) The number of hard cover books sold is 100

(integer) If m and n are positive integers, what is the greatest common divisor of m and n?

1)m=n+1
2)m∗n is divisible by 2

Hey indu1954,

Sorry for only getting to this now, it slipped through my inbox...

Where is this exercise from? At any rate, it is a relatively simple word problem, and the easiest way to approach such problems is usually to first write them out as equations. (Which lets you 'get concrete' instead of getting lost in mental logic)

Specifically, we can label h = 10 and s = 6 for the prices and numh,nums as the respective number of books. Then we need to find out if numh > nums is true.

(1) tells us that (10*numh + 6*nums)/(numh + nums) = 9. We can either do some algebra to get to numh = 3nums and therefore numh > nums or we can we the concept of weighted average - if we had equal number of hard-cover vs. soft-cover books the average would be the exact middle - 8. Since it is 9, we have more of the expensive books, which are hard-cover.
Either way, this is sufficient.

(2) So we know that numh = 100 but have no way to link nums to an exact number.
This is insufficient.

Hey indu1954,

The moment you see a question which has the word 'divisor' in it, your immediate thought should be "ooh this is about number properties or factorizations!"
In particular, the GCD is the largest integer that divides both m and n, so we'll either factorize both m and n to find the largest common factor (the Precise approach), or look for some number-theoretic rule that allows us to infer what these factors are without calculations (the Logical approach).

(1) This tells us that the GCD must be 1, which is to say the numbers have no common divisors larger than 1. An easy way to see this is to just try a few examples, a more theoretical (Logical) way inolves using rules of divisors:
Let's label the GCD of m and n as x. Since x divides both m and n, it must also divide m - n (do you understand why?). Then we can write m - n = x*(an integer). Then the equation in the stmt simplifes to x*(an integer) = m - n = 1. This implies that x must divide 1. The only such integer is, of course, 1.
Then (1) is sufficient.

(2) The only thing this tells us is that at least one of m,n is even. With no other information, this cannot be sufficient.

(A) is our answer.

Hi DavidTutorexamPAL,

Can you provide an alternative explanation to the below DS problem from one of the examPAL quizes?

Q. a and b are integers. [x] is an integer less than or equal to x. Is \([\frac{a}{b}]\)≥ 1?
(1) \(ab = 64\)
(2) \(a = b^2\)

Here is the official explanation:
Since \([\frac{a}{b}]\) is defined as some integer either equal to or less than \(\frac{a}{b}\), no possible value of (a,b) can ever make \([\frac{a}{b}]\) necessarily greater than any integer. This means a definitive answer to the question stem, if there is sufficient information, can only be ‘NO!’ – when \([\frac{a}{b}]\) < 1. This would be the case if b is greater than a. (1) gives us no such information, and (2) gives us the opposite: the greater integer b is, integer a becomes even greater. Thus, combining the two we’ll still get a > b. Therefore, (E) is correct.

And here is my thought on this question:
Combining (1) and (2), we can get b = 4 and a = 16. So, \(\frac{a}{b} = \frac{16}{4} = 4\) and hence \([\frac{a}{b}] = [4] = 4\).
So, the answer to the above DS question should be C.
Kindly let me know where I am wrong.

Hey tarunanandani
This is a confusing question, because it deals with a function. these can be really tricky sometimes...
read the instructions again: [x] is an integer less than or equal to x. This means that if \(\frac{a}{b} = \frac{16}{4} = 4\), then \([\frac{a}{b}]\) equals... something less or equal to 4 - but we don't know what! it could be 3, 1, -100... we don't know! hence, the answer is insufficient.

A certain box has only a total of 7 red balls and green balls. If two balls are selected randomly from the box and one ball at a time with replacement, what is the number of red balls?

1) The probability that two balls selected are green balls is (4/7)(4/7)
2) The probability that two balls selected are not green balls is (3/7)(3/7)

So first off, the current wording is a bit unclear; a better wording could be: "A box contains exactly 7 balls, all of which are either red or green. If two balls are selected from the box randomly and with replacement, such that one ball is chosen each time, how many red balls are in the box?"

For (1), the only way to get a probability of \((\frac{4}{7})^2\) is if there were 4 green balls (as the equation implies that you have 4 choices for the first, then replace it and have 4 for the second), implying that there were 3 red balls.
Similarly, stmt (2) says there are 3 red balls.

Then (D) is our answer.

Were you interested in this specific question or in more general concepts relating to probability? As the question itself is a bit basic...

A history class has recently finished their second test. A quarter of the class passed both tests. What percentage of those that passed the first test also passed the second?

S1) A third of the class passed the first test.
S2) Half of the class passed the second test.

Hi David,
/ (statistics) A new set is made by 2 numbers are removed from {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. What is the standard deviation of new set?
1) The new set has 10 as a median
2) The old set and new set have the same average (arithmetic mean)

Hi David,
statistics) 5 points v, w, x, y, and z are on the number line. Is the median of them equal to their average (arithmetic mean)?
1) v<w<x<y<z
2) z-y=y-x=x-w=w-v

Hey paulbro4,
First, apologies for the delay in response.
If you want to be sure that I see a message, please tag me (with '@' and my full screenname) or send a PM

This is a standard 'sets' question: you are given 2 overlapping sets ('passed first test' and 'passed second test') and asked something numerical about them.
In this case, you need to calculate 'passed both tests' / 'passed the first test' and are given only that 'passed both tests = 25%.
So to answer we need to know the % that passed the first test.
(1) gives you this but (2) does not, so (A) is the answer.

Hey Sandeepanisha,

To calculate the std, we either need to know exactly which two numbers changed, or need to information that tells us that the 'spread' of the data changed. The first is much easier to do and is most likely our answer. Let's see which statements give us the information we need, a Logical approach.

(1) All we know is that the sum of the two middle numbers in the new set is 20. So we don't know which 2 numbers were removed and have no information on the new 'spread' of data.
Insufficient

(2) This tells us that the two removed numbers are equidistant from the old average. Again -- not the information we need as there are many ways to do this.
Insufficient.

Combined: Similarly to the above, there are many ways to do this: by removing 1,19 or 3,17, or any other 'symmetrical' pair.

(E) is our answer.

Hi Sandeepanisha,

If we're unsure how to address this formally, we can always pick easy numbers, then try to change them so as to get the opposite answer.
This is an Alternative approach.

(1) If our points are 1,2,3,4,5 then the median equals the average, (=3) giving a YES. So we now need to change this to get a NO. If we don't change the middle number but change one of the other numbers, then the median won't change but the average will. For example if our points are 0,2,3,4,5 then the median (3) does not equal the average (which is now less than what it was before because we have decreased one of the numbers). Then we have found a set which gives NO.
Insufficient.

(2) Picking two random numbers for z and y, say z = 1 and y = 0, we can then see that x = -1, w = -2 and v = -3. So once again we have a sequence of 5 consecutive numbers giving a YES. We'll try to look for a different choice of numbers which gives a NO. If we pick say z = 3 and y = 0, we'll get 3,0,-3,-6,-9 which is still a 'symmetrical set' so gives a YES. Hopefully, having gone over the math for 2 examples, we can see that the constraints of (2) will always give us a 'symmetrical set' of numbers, and therefore (2) is always sufficient.

(B) is the answer.

(statistics) 5 points v, w, x, y, and z are on the number line. Is the median of
them equal to their average (arithmetic mean)?
1) v<w<x<y<z
2) z-y=y-x=x-w=w-v