In today’s post, we will discuss some smart formats in which we can assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.
Here are some examples:
An even number: 2a
Logic: It must be a multiple of 2.
An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.
Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).
Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.
Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.
A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.
Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.
Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.
Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:
The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a
Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)
The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b
Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.
We are given that the sum 8a is equal to the sum 6b.
8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.
With this in mind, possible solutions for a and b are:
a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
We are also given that the middle term of the even numbers is greater than 101 and less than 200.
So 101 < 2b < 200, i.e. 50.5 < b < 100.
B must be an integer, hence, 51 ≤ b ≤ 99.
Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96
The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12
For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences.
Therefore, the answer to our question is A.
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