Odd One Out Or Classification -

Odd One Out Or Classification - Short-cut Tricks And Examples

You must have in your mind that what does classification mean. In fact, in classification we take out an element out of some given elements and the element to be taken out is different from the rest of the elements in terms of common properties, shapes, sizes, types, nature, colours, traits etc. In this way the rest of the elements form a group and the element that has been taken out is not the member of that group as this single element does not possesses the common quality to be possessed by rest of the elements. For example, if we compare the elements like, lion, cow, tiger, panther, bear and wolf then we find that this is a group of animals. How do we classify them? To understand this let us see the presentation given below:

Here, if we want to separate out one animal then definitely that animal will be cow because cow is the only animal in the group which is a domestic animal. Rest of the animals (Lion, Tiger, Panther, Bear and Wolf) are wild animals. Hence rest of the animals (Lion, Tiger, Panther, Bear & Wolf) form a group of wild animals separating out the domestic animal (Cow).

Similarly, out of 6 letters A, M, N, U, P & Q, we will take out A and form a group of 5 letters M, N, U, P & Q because out of given six letters only A is a vowel while rest of the letters form a group of consonants.

Types of Classification

Type 1. Word Classification

In this type of classification, different objects are classified on the basis of common features / properties – names, places, uses, situations, origin, etc.

Example 1: Four of the following five-are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. Work : Leisure
2. Day : Night
3. Expedite : Procrastinate
4. Frequently : Always
5. Happy : Unhappy

Solution. (4): All others are the antonym of each other.

Example 2: Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. March
2. January
3. July
4. June
5. May

Solution. (4): All other months have 31 days.

Type 2. Alphabet Classification

In this type, alphabet are classified in a group using a particular method or rule.

Rules or methods used for such classification are often simple and hence can easily be understood.

Example 3: Four of the following five-are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. BY
2. LO
3. EW
4. GT
5. SH

Solution. (3): All others letter pairs are the antonym of each other.

Type 3. Number Classification

In this type, numbers are classification in a group using a particular method or rule. Rules or methods used for such classification may be based on mathematical operations.

Example 4: Four of the following five-are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. 25—5
2. 16—4
3. 144—12
4. 64—7
5. 36—6

Solution. (4): In all other number pairs, the first number is the square of second number.

25 = 5², 16 = 4², 144 = 12², 64 ≠ 7², 36 = 6²

Example 5: Four of the following five-are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. 28
2. 42
3. 35
4. 21
5. 65

Solution. (5): All other numbers are divisible by 7 while 65 is not divisible by 7.

Type 4. Number and Letter Classification

Example 6: Four of the following five-are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. 25—E
2. 16—D
3. 144—L
4. 64—G
5. 36—F

Solution. (4): In all other number-letter pairs, the first number is the square of the position of second number.

25 = (E→5²), 16 = (D→4²), 144 = (L→12²), 64 ≠ (G→7²), 36 = (F→6²)

Type 5. Miscellaneous Classification

In this type of classification, any rule other than described above can be used for classification or grouping. Questions on such pattern do not necessarily use the alphabets and words. Here the numerics and other mathematical symbols can also be used.

Example 7: In each of the following five options each has a combination of three words group. In which, four groups are alike in a certain way and so form a group. Which is the one that does not belong to that group?

1. Driver, passenger, vehicle
2. Chair, table, bench
3. Ship, boat, pilot
4. Apple, orange, winter
5. Mango, flower, orchard

Solution. (2): Chair, table and bench belong to a category of furniture.

Solved Examples

Example 8. Which one is different from the rest three?

1. Door
2. Gate
3. Table
4. Window

Solution. (3): All the rest are the parts of a building.

Example 9. In this question, there is four words with the letters jumbled up. Three of them are alike. Find the odd one out.

1. CIRE
2. NAIR
3. LOUDSC
4. RNUTHDE

Solution. (1): By arranging the letters of NAIR, LOUDSC and RNUTHDE we get RAIN, CLOUDS and THUNDER respectively which are all related with one other except CIRE i.e., RICE.

Example 10. Which one is different from the rest three?

1. NMLK
2. RQPO
3. UTSR
4. WXUV

Solution. (4): In all the other options, the letters are in reverse order of alphabet.

Example 11. Which one letter group differs from the other three?

1. WRONG
2. GREEN
3. WHITE
4. RIGHT

Solution. (2): In other options, no letter is repeated.

Example 12. Three of the following are alike in a certain way and form a group. Find the odd one out.

1. Bird
2. Insect
3. Aeroplane
4. Kite

Solution. (2): All except the insect fly in the sky.

Example 13. Find out the odd one out.

1. 28
2. 14
3. 49
4. 64

Solution. (4): Except 64, all the rest number 28, 14 and 49 are divisible by 7 while 64 is not divisible by 7. Therefore 64 is different from the rest.

 

 

Ranking And Time Sequence

Ranking And Time Sequence - Short-cut Tricks And Examples

Dear Reader, below are five simple types of ‘ranking and time sequence’ problems. You will find detailed solutions with each of the problems

As far as ranking problems are concerned, you have to remember one simple formula.

Total number of people = Rank of a person from START + Rank of person from the END – 1

Let we start with our tutorial. At the end of the tutorial, you will find a short practice test. Please do take the test so that you can be double sure that you have understood well.

Type I: Finding Rank From the Start (or the End)

In type 1, you will know the rank of a person from either the start (or the end). Using that data, you have to find the rank of that person from the end (or the start). Below example will help you.

Example Question 1: Reena is 10 ranks ahead of Priya in a class of 40. If Priya`s rank is 20th from the last, what is Reena`s rank from the start?
Answer: 31th

Solution:
Priya’s rank from the last is 20. Reena is 10 ranks ahead. Therefore, Reena’s rank from the last = Rank of Reena from the last = 20 – 10 = 10th
Now you have to apply the formula that you saw in the introduction.
Total number of students = Rank of Reena from the start + Rank of Reena from the end – 1
40 = Rank of Reena from the start + 10 – 1
Rank of Reena from the start = (40 – 10) +1
= 31th

Type II: Finding Total Number of People in a Sequence

In type 2, you will find ranks of a person from the start and the end. You have to find the total number of people. Let us see an example.

Example Question 2: Venkatesh ranks 8th from the top and 24th from the bottom in the class. How many students are there in total?
Answer: 31

Solution:
Total number of Students = Rank of Venkatesh from the top + Rank of Venkatesh from the bottom – 1
= [8 + 24] – 1= 31

Type III: Interchanging Positions

In this type, positions of the people in a sequence will be interchanged. You have to solve these problems after processing the data given.

Example Question 3: In a row of girls, Uma is 10th from the left and Meena is 20th from the right. If they interchange their positions, Uma becomes 15th from the left. How many girls are there in the row?
Answer: 34

Solution:
After interchange, rank of Uma from the left = 15
But, before interchange, Meena was 20th from the right. After interchange, Uma would have occupied the same position of Meena’s earlier spot. Therefore,
Present rank of Uma from the right = 20
Now you know the current ranks of Uma from left as well as right. Therefore,
Total number of girls = Uma’s rank from the left + Uma’s rank from the right – 1
= (15+20)-1
=34.

Type IV: Intervention in a Frequent Event

Though the heading of this type looks complicated, this is one of the easiest. There will be intervention at certain time of a frequent event. Based on that time data you have to solve the problem logically.

Let us see an example.

Example Question 4: A bus leaves to Chennai from Bangalore every 30 minutes. A passenger inquired about the next bus to Bangalore, and he was informed that the bus left 15 minutes before, and the next bus will be at 5.00 pm. Find at what time the passenger had enquired?
Answer: 4.45 pm

Solution:
For every 30 minutes, there is a bus. The next bus will be at 5.00 pm, so the last bus must have left at 4.30 pm.
The informer said that the bus had left 15 minutes before his inquiry. So the time of inquiry is 4.30 + 0.15 = 4.45 pm.

Type V: Day of Week Based on Frequency

This type is very easy just like the previous one. You can answer this with little effort. (You may not expect questions simple in your exam. However, this will be a first step in understanding more difficult questions.)

Example Question 5: Gita went to the temple five days ago. If she goes to the temple every Friday, then what day of the week is today?
Answer: Wednesday

Solution:
She went to temple five days ago i.e., Friday.
Five days from Friday is Wednesday.