Elimination method is a powerful technique of solving multiple choice questions.

Even without being able to solve many a question in the test, test takers can successfully circumvent the difficulty in solving many of multiple choice questions with the help of elimination method.

Let us check the use and power of elimination method with the following multiple choice question.

Question:

A is a set of odd numbers from 1 to 100. B is a set of two even integers 2 and 4. The elements of set A and the elements of set B are multiplied and written in another set C. Which of the following is the number of elements in the set C?

A. 50

B. 100

C. 140

D. 150

E. 200

Let us now check, how helpful the elimination method is in solving multiple choice questions with the above example.

Given, A = {1, 3, 5,...,99} B = {2, 4}. So, C = ?

Now, Set A contains 50 odd numbers and set B, 2 positive even integers.

Multiply each of 2 and 4 with all of the 50 numbers in set A, to give 100 products.

So, set C can have 100 products:

*50 from 2, and another 50 from 4. *

But the question is, are the two groups of products all *different?*

How to ascertain that the products of 2 and 4 in set B with the elements of set A are all ** different** from each other?

Only if different, then set C can be said to contain 100 elements.

If not, then set C will not have 100 elements, and instead it will only contain fewer than 100 elements.

Exactly during such difficulty, the *elimination method** *proves a powerful technique in solving multiple choice questions easily and quickly.

First, take a close look at the answer choices given below the above multiple choice question.

Answer options C, D and E each contain numbers greater than 100.

Each of these is impossible.

The number of elements in set C can be ** at most100**; it cannot exceed 100.

So, options C, D and E are gone; they are said to be *eliminated.*

Now, options A and B remain.

Look at option A. It contains 50.

A serious implication of option A-50 is that all of the products of 2 and 4 with the odd numbers of set A are *common, which is too far-fetched. *

*It appears more improbable than the answer options C, D and E. *

So, option B having 100 must be *somehow* right.

Whatever may be the exact reason, which is not necessary to know while taking the test, the option B should indeed be the correct answer.

That’s it.

The correct answer for the above multiple choice question is therefore B.

This is how*elimination method** *zeroes us on the right answer without even being able to solve the question.

many of the ** multiple choice questions** in the test can be solved using elimination method inspite of not having the necessary knowledge or the requisite skill.

By the way, what’s the exact reason. It is:

In the above question, common products will arise if set A contains even integers.

For example, consider number 12.

12 is a product of 2 and 6, and also of 4 and 3.

i.e., 2 × 6 = 12, and 4 × 3 = 12.

So, number 12 will be a common product of 2 and 4 provided there is 6 in set A. But set A contains only odd integers. Therefore, common product do not exist in the question above.

Therefore, all of the above products are different. So, set C will contain 100 elements-products.

Consider the following multiple choice question.

**Question: **

When an experiment is done, the probability of four events that are likely to happen are *p, q, r and s*. Which of the following answer choice contains the probability values of the four events?

A. p = 0.2, q = 0.3, r = 0.4, s = 0.2

B. p = 0.3, q = 0.1, r = 0.1, s = 0.4

C. p = 0.1, q = 0.5, r = 0.2, s = 0.3

D. p = 0.4, q = 0.2, r = 0.3, s = 0.1

E. p = 0.3, q = 0.4, r = 0.3, s = 0.1

Use the ** elimination method **to prove your answer is choice D.

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