Study Notes & Short Tricks: Order and Ranking and Direction
Dear Students,
Here are the study notes of Reasoning Ability based on Order and Ranking and Direction which will help you to ace your preparation.
Order and Ranking and direction is a very important chapter in reasoning section of any competitive exams.
There are usually 3 – 5 questions in any exam from the topic. So, we are here with detail concept of “Order and Ranking and Direction” which will help a student in boosting their marks in the exam.
Ranking Test: – In ranking Test, types of questions that comes in examination are : –
Total number of persons = [(Rank of a particular person from the left +Rank of same particular person from the right)–1]
‘left’ which is also called ‘top’
‘right’ which is also called ‘bottom’
Example : – Bikki ranks tenth from the top and seventh from the bottom in a certain exam
How many students are there in his class?
Solution : – Total number of students in the class
=[(Rank of Bikki from the top + Rank of Bikki from the bottom)–1]
= (10 + 7) –1
= 17–1
= 16
Example -2 : Shipra ranks fourth from the top and total number of students in the class are 35. Then find the rank of shipra from the bottom.
Solution : – As, one basic formula is discussed above
Total number of students = [(left + right) – 1]
So, rank of Shipra from the bottom
= 35 – 3 = 32nd from the bottom
Example -3 : Rahul is 15th from the left end in a row of a boys and Rohit is 4th to left of Rahul. Then find the position of Rahul from the left end.
Solutions: -
By keeping above table in your mind, you can solve questions of Ranking in less time.
Example -4 : Raj is elder than Rajesh but not as elder than Piyush. Mahesh is elder than Piyush but not as elder as Rohit. Rohit is younger than Vivek but not as Gaurav who is eldest among all ?
Solution : –
From the given information,
Piyush > Raj > Rajesh …..(i)
Rohit > Mahesh > Piyush .....(ii)
Vivek > Rohit > Gaurav ….(iii)
Combining (i) (ii) and (iii) we get vivek is eldest among all
Example -5 : – How many such pairs of letters are there in the word ‘LABOUR’ each of which has as many letters between them in the word (in both forward and backward direction) as they have between them in English alphabetical series.
Solution : –
Total number of word = 2
Example -6: If the letter of the word “ANIKET” are arranged alphabetically from left to right. How many letters will remain at the same position.
Solution: – “Word - “ANIKET”
Arranging alphabetically - AEIKNT
So, four alphabets will remain at the same position i.e. “A, I, K, T”
Example -7: If position of A from left side of a row is 10th and position of B from right side of a row is 15th and only 2 person is sitting in middle of A & B. Find the minimum number of persons that can be seated in this row?
Solution: – In the question it is asked to find minimum number of persons in a row then it is a always a case of overlapping ie given positions of persons from either sides overlap each other. For this type there is a short trick i.e.
Minimum no. of persons
= Sum of positions of person from both sides – person in middle of them –2
So, total number of persons = 10 + 15 – 2 –2
= 25 – 4
= 21
Direction Test: – Direction test are introduced to test the ‘sense of direction’ of the candidate. This test is for ascertain the final direction or distance between two points.
AC^2=AB^2+ BC^2
∴AC=√(AB^2+BC^2 )
Where, ∆ABC is right-angled triangle
Example -1 : Amit walked 2km west of his house and then turned south covering 4 km. Finally he moved 3 km towards east and then again 1 km west. How far is he from the initial position 2.
Solution : –
Amit starts from his house at A, mores 2 km west upto B, then 4 km to the south upto C, 3 km east upto D and finally 1 km west upto E, thus his distance from the initial position A will be AE.
And, AE = BC = 4 km.
Example -2: Ramesh walks 10 km towards north. From there he walks 6 km towards south. Then he walks 3 km towards east. How far and in which direction is he with reference to his starting point?
Solution: –
Ramesh walks 10 km from A to B. After that 6 km south from B to C. And 3 km towards east from C to D. Therefore, his distance from starting point A
AD=√(AC^2+CD^2 )
=√(4^2+3^2 )=5 km in North East direction.
As per today's scenario in the Banking exam even in other exams like SSC, it has become important that to solve a particular problem quickly you have to know the concept behind the question and also the way how to solve that problem quickly. Keeping in mind that thing we are providing here some important study notes and short tricks of various topics of Reasoning Ability so that a student can attempt more questions in less time (especially in Banking and SSC exams) and can make a higher probability of being selected. Best of hard work to all students. Keep using Bankersadda.
Here are the study notes of Reasoning Ability based on Order and Ranking and Direction which will help you to ace your preparation.
Order and Ranking and direction is a very important chapter in reasoning section of any competitive exams.
There are usually 3 – 5 questions in any exam from the topic. So, we are here with detail concept of “Order and Ranking and Direction” which will help a student in boosting their marks in the exam.
- First we will discuss about “Order and Ranking test”.
Ranking Test: – In ranking Test, types of questions that comes in examination are : –
- Rank of a particular person mentioned from left and right and the total number of persons are asked in the question.
- Sometimes, total number of persons are mentioned in the question and rank of a particular person from left are mentioned and rank from right is asked for that particular person.
- Sometimes, total number of persons are mentioned in the question and rank of a particular person from right are given and rank from left is asked for that particular person.
- There is one basic formula which will help you to solve questions of Ranking.
Total number of persons = [(Rank of a particular person from the left +Rank of same particular person from the right)–1]
- There are important terms regarding this topic are : –
‘left’ which is also called ‘top’
‘right’ which is also called ‘bottom’
- Let’s here a look on different types of examples which generally comes in exam : –
Example : – Bikki ranks tenth from the top and seventh from the bottom in a certain exam
How many students are there in his class?
Solution : – Total number of students in the class
=[(Rank of Bikki from the top + Rank of Bikki from the bottom)–1]
= (10 + 7) –1
= 17–1
= 16
Example -2 : Shipra ranks fourth from the top and total number of students in the class are 35. Then find the rank of shipra from the bottom.
Solution : – As, one basic formula is discussed above
Total number of students = [(left + right) – 1]
So, rank of Shipra from the bottom
= 35 – 3 = 32nd from the bottom
Example -3 : Rahul is 15th from the left end in a row of a boys and Rohit is 4th to left of Rahul. Then find the position of Rahul from the left end.
Solutions: -
By keeping above table in your mind, you can solve questions of Ranking in less time.
Example -4 : Raj is elder than Rajesh but not as elder than Piyush. Mahesh is elder than Piyush but not as elder as Rohit. Rohit is younger than Vivek but not as Gaurav who is eldest among all ?
Solution : –
From the given information,
Piyush > Raj > Rajesh …..(i)
Rohit > Mahesh > Piyush .....(ii)
Vivek > Rohit > Gaurav ….(iii)
Combining (i) (ii) and (iii) we get vivek is eldest among all
Example -5 : – How many such pairs of letters are there in the word ‘LABOUR’ each of which has as many letters between them in the word (in both forward and backward direction) as they have between them in English alphabetical series.
Solution : –
Total number of word = 2
- We can count from both forward and backward direction
Example -6: If the letter of the word “ANIKET” are arranged alphabetically from left to right. How many letters will remain at the same position.
Solution: – “Word - “ANIKET”
Arranging alphabetically - AEIKNT
So, four alphabets will remain at the same position i.e. “A, I, K, T”
Example -7: If position of A from left side of a row is 10th and position of B from right side of a row is 15th and only 2 person is sitting in middle of A & B. Find the minimum number of persons that can be seated in this row?
Solution: – In the question it is asked to find minimum number of persons in a row then it is a always a case of overlapping ie given positions of persons from either sides overlap each other. For this type there is a short trick i.e.
Minimum no. of persons
= Sum of positions of person from both sides – person in middle of them –2
So, total number of persons = 10 + 15 – 2 –2
= 25 – 4
= 21
Direction Test: – Direction test are introduced to test the ‘sense of direction’ of the candidate. This test is for ascertain the final direction or distance between two points.
- Some important points which should keep in mind while solving direction test :–
- Always try to use the direction planes as the reference for all the questions.
- Always mark the starting point and end point different from other points.
- Always be attentive while taking right and/or left turns.
- While solving direction test, you should keep in mind the below diagram as reference :–
- You should be aware of basic geometric rule, such as Pythagoras Theorem.
AC^2=AB^2+ BC^2
∴AC=√(AB^2+BC^2 )
Where, ∆ABC is right-angled triangle
Example -1 : Amit walked 2km west of his house and then turned south covering 4 km. Finally he moved 3 km towards east and then again 1 km west. How far is he from the initial position 2.
Solution : –
Amit starts from his house at A, mores 2 km west upto B, then 4 km to the south upto C, 3 km east upto D and finally 1 km west upto E, thus his distance from the initial position A will be AE.
And, AE = BC = 4 km.
Example -2: Ramesh walks 10 km towards north. From there he walks 6 km towards south. Then he walks 3 km towards east. How far and in which direction is he with reference to his starting point?
Solution: –
Ramesh walks 10 km from A to B. After that 6 km south from B to C. And 3 km towards east from C to D. Therefore, his distance from starting point A
AD=√(AC^2+CD^2 )
=√(4^2+3^2 )=5 km in North East direction.
- We have given our best to make you understand the concept of “Order and Ranking and Direction” with some different examples with their explanation. Hope you will be able to get things clear.
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