Process of elimination is only next to number plugging in popularity as a strategy for solving Quant questions on the GMAT. I am not a fan of either method. Yes, they are useful sometimes, and even necessary in some questions but for most questions, I like to use logic/reasoning.
That said, there is a set of questions in which we should think of these strategies. Number plugging is very useful when you have one or two variables in the options. Algebra can be time consuming in these cases because of equation manipulation required.
Similarly, some questions beg you to use the process of elimination. Their question stem goes something like ”which of the following options can be the value of x?”, “which of the following options cannot be the sum of a and b?” etc. These questions are framed like this because often they have multiple solutions. x could possibly take many different values but the options would have only one of them. So it makes sense to check which values x can take from the options. Let’s look at one such instance of a tricky question where process of elimination can be very useful.
A list of numbers has six positive integers. Three of those integers are known – 4, 5 and 24 and three of those are unknown – x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).
Which of the following CANNOT be the value of any one of the unknowns?
Solution: The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.
Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.
Now we know that the mean = 10 and median = 7.5
Method 1: Algebra/Logic
Let’s try to solve the question algebraically/logically first.
There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60
4 + 5 + 24 + x + y + z = 60
x + y + z = 27
Median of the list = 7.5
So sum of third and fourth elements must be 7.5 * 2 = 15
There are two cases possible:
Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.
The sum of the third and fourth elements of the list is 15 so
15 + z = 27
z = 12
So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be
5 and 10
6 and 9
7 and 8
x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.
Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.
If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.
The sum of the other two unknown numbers would be 27 – 10 = 17.
One of them must be less than 5 and the other greater than 10. So possible options are
4 and 13
3 and 14
2 and 15
1 and 16
x, y and z could take various values but none of them could be 11
Method 2: Process of Elimination
Let’s now try to look at the process of elimination here and see if we can find an easier way.
The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.
Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.
Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.
From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.
Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!