Problem Solving

This is not a question that the GMAT expects you to do in 2 minutes. I fully agree with Mitch - this is a question that most savvy test-takers (even those aiming for a 700+) should SKIP! The GMAT designs a certain number of questions to be too hard to reasonably solve, primarily for 2 reasons: 1) to bog down the stubborn students who insist on solving every problem, and 2) to make distinctions between 780 and 800 level test takers. (There have to be some questions at the crazy-high end of the spectrum!).

If you're not aiming for a 780 or above, this kind of question isn't worth your time. You can still score well above a 700 if you skip questions like this - in fact, you're probably more likely to get a 700+ if you do!

Remember - the GMAT is above all a decision-making test, and there will be plenty of times that the best decision you can make is to skip an excruciatingly hard question, to save time for other questions that you can get.

utkalnayak wrote:The following question from OG - Is this really doable in around 2 mins? Do we have any way other than how it is explained in the guide ?

"218. List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digit is odd is rounded down to the nearest integer; E is the sum of the resulting integers. If of the decimals in T have a tenths digit that is even, which of the following is a possible value of E - S ?

I. -16
II. 6
III. 10
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III"

Solution:

I, too, agree with everyone's sentiments about the difficulty of this question. If you see something like this on the exam and if you don't know how to solve it, or if you spend more than two minutes trying to solve it, you should probably skip it and focus your attention on other problems.

When reading through this question, notice that we are asked which of the following is a POSSIBLE value of E - S. This tells us that we will not be looking for a definite answer here.

We are given that list T has 30 decimals, and that the sum of this list is S.

Next we are given that each decimal whose tenths digit is even is rounded up to the nearest integer and each decimal whose tenths digit is odd is rounded down to the nearest integer.

We are next given that E is the sum of these resulting integers.

Finally, we are given that 1/3 of the decimals in list T have a tenths digit that is even. This this means that 2/3 have a tenths digit that is odd. This means we have 10 decimals with an even tenths digit and 20 decimals with an odd tenths digit.

This is very helpful because we are going to use all this information to create a RANGE of values. We will calculate both the maximum value of E - S and the minimum value of E - S.

Another way to say this is that we want the maximum value of the sum of our estimated value in list T minus the sum of the actual values in list T and also the minimum value of the sum of the estimate values in list T minus the sum of the actual values in list T. Thus, we need to determine the largest estimated values and the smallest estimated values for the decimals in list T.

To do this, let's go back to some given information:

Each decimal whose tenths digit is even is rounded up to the nearest integer.

Each decimal whose tenths digit is odd is rounded down to the nearest integer.

Let's first compute the maximum estimated values for the decimals in list T. To get this, we want our 10 decimals with an even tenths digit to be rounded UP as MUCH as possible and we want our 20 decimals with an odds tenths digit to be rounded DOWN as LITTLE as possible. Thus, we can use decimals of 1.2 (for the even tenths place) and 1.1 (for the odd tenths place). Let's first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.2) + 20(1.1) = 12 + 22 = 34

Now we can round up 1.2 to the nearest integer and round down 1.1. We see that 1.2 rounded up to the nearest integer is 2, and 1.1 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers. We can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the maximum value of E - S, in this case, is 40 - 34 = 6.

Let's next compute the minimum estimated values for the decimals in list T. To get this we want our 10 decimals with an even tenths digit to be rounded UP as LITTLE as possible and we want out 20 decimals with an odds tenths digit to be rounded DOWN as MUCH as possible. Thus, we can use decimals of 1.8 (for the even tenths place) and 1.9 (for the odd tenths place). Let's first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.8) + 20(1.9) = 18 + 38 = 56

Now we can round up 1.8 to the nearest integer and round down 1.9 to the nearest integer. 1.8 rounded up to the nearest integer is 2, and 1.9 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers, so we can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the minimum value of E - S, in this case, is 40 - 56 = -16.

Thus, the possible range of E - S is between -16 and 6, inclusive. We see that I and II fall within this range.

The answer is B