Let us see this with the help of an example:Īnswer: The rule here is that the first two letters have a gap of one letter between them e.g.

You no longer have to worry about the meaning or the use or even the way the words are pronounced, rather you can just focus on the patterns. But what if the words don’t mean anything? What if they are just a collection of letters? That is actually easier as it takes a lot of rules off the table. In the above section, we saw words could be arranged on the basis of certain properties. As opposed to the bird which doesn’t rhyme and is thus the odd word out. The first three rhyme so they form a set. In the above example, all the options seem to be related. It could be the number of letters, the pronunciation, or any property of the words etc. The sole purpose is to identify a contrasting property. Thus the arrow has a remarkable difference from the others and hence is the odd word out.Įxample 2: Select the word that doesn’t belong:Īnswer: If you think linearly, you will find this example rather difficult. They must be used very close to the target. What can be done with an axe or a dagger that can’t be done with the others? Well, an arrow can be used from a distance. Try and think of a difference that they have. Let us see our first example:Ī) Arrow B) Axe C) Knife D) Dagger E) SwordĪnswer: They are all weapons. Here we will mainly focus on those types of categories that are relevant with respect to the reasoning ability tests. There are several rules or methods that can be used to group words together and thus there are several types of questions that can be asked in the Odd Words category.

I leave them as an exercise to you.Here we will try and get acquainted with the several types of problems that a candidate is expected to come across while appearing for many graduate levels banking exams like IBPS PO, SBI PO, and other relevant exams. Note that there are other ways to accomplish this. Finally, str.join joins each word back into a single string. yield transforms the function into a generator, yielding one word at a time from the loop. Use the index to determine whether the word is an odd or even word.

enumerate transforms the list into a list of (index, word) tuples. What f does is split the string into a list of words. 'WHAT si THE gnidraob TIME ni BANGALORE noitats' > ' '.join(f('what is the boarding time in bangalore station')) join the transformed words into a single string.for each word, check if it is an even word.Define a little function to do this for you.

Here's one simple way of tackling your problem. Instead of slicing every alternate word and performing one kind of operation at a time, iterate over each one in turn, handling it as appropriate.

Different words for odd