Tuesday, July 14th, 2020
Logical reasoning questions based on ranking involve ordering people/objects based on certain characteristics like height/weight/score etc. In these types of questions, you are supposed to place people/objects in a decreasing or increasing order based on certain attributes. There is not one common method to solve ranking problems and strategy should vary with questions. But there are certain key-steps that can be applied while approaching such questions. They are:
Following examples would help illustrate the required key-skills and acquaint you with them.
Aarti, Bharati, Chandni, Divya, Esha and Flora are running in a 100m race.
A) Aarti beat only Bharati and Flora
B) Esha was beaten only by Divya
C) Bharati did not finish last in the race
D) The bronze medallist finished the race in 55 seconds
E) Bharati finished the race in 63 seconds
Which of the following could be Aarti’s time?
a) 40
b) 51
c) 61
d) 64
The gold medallist beat Chandni by 20 seconds. Which of the following can be Esha’s time?
a) 31
b) 41
c) 57
d) 60
This problem can be considered similar to arranging 6 girls in a line from left (1st position) to right (last place)
We know that as Esha was beaten only by Divya, Divya must be in 1st place followed by Esha in 2nd place:
D E
Aarti beat only Bharati and Flora and hence must be 3rd last in the race:
D E A __
As Bharati and Flora are last 2 and Bharati is not last in the race
D E _ A B F
Therefore Chandni must be in 3rd place. Hence, the order is D E C A B F.
The bronze medalist Chandni’s time is 55s and Bharati’s is 63s. As Aarti finishes between these two runners, her time can be anywhere between these 2 values. Thus, the answer is 61 s.
If Divya beats Chandni by 20s, her time will be 35s and Esha’s time will be between 35 and 55 seconds. Hence, the correct answer is option B – 41 seconds.
Let’s see one medium difficulty example now.
A, B, C, D, E, F and G are seven persons who travel to office every day by a particular train which stops at five stations – I, II, III, IV and V respectively after it leaves base station. Three among them get in the train at the base station. D gets down at the next station at which F gets down. B does not get down either with A or E. G alone gets in at station III and gets down with C after one station. A travels between only two stations and gets down at station V. None of them gets in at station II. C gets in with F but does not get in with either B or D. E gets in with two others and gets down alone after D. B and D work in the same office and they get down together at station III. None of them gets down at station I.
At which station does E get down?
(a) II
(b) III
(c) IV
(d) Data inadequate
At which station do C and F get in?
(a) I
(b) II
(c) III
(d) Data inadequate
At which of the following stations do B and D get in?
(a) I
(b) Base station
(c) III
(d) Data inadequate.
After how many stations does E get down?
(a) One
(b) Two
(c)Three
(d) Four
E gets down after how many stations at which F gets down?
(a) One
(b) Two
(c) Three
(d) Four
 First imagine the structure of the table to fill the clues. After going through the clues, we can draw a table with stations of getting in and getting out corresponding to each person as below.
Person | Get-in Station | Get-out station |
A | IV | V |
B | Base or I | III |
C | Base or I | V |
D | Base or I | III |
E | Base | IV |
F | Base or I | II |
G | III | V |
Now, let’s go through each question.
(c) E gets in with two others and gets down alone after D and we know that D get down at station 3, so option c.
(d) Since we know that G gets in at station 3 alone and there was nobody who gets in from station 2, therefore D may get in from base station or station 1.C and F also can get in either at Base station or station I . So, any of the pair (B,D and C,F) can get in at base (with E) or station I.
(d) we know that B and D work in the same office and they get down together at station III. It means they cannot get in at station 3 so they also may get in from Base station or Station 1 .
(c) B and D work in the same office and they get down together at station III .We know that E gets in with two others and gets down alone after D. So E gets in at base station and gets down at station IV after three stations. So, E gets down after three stations (I, II and III).
(b) B and D work in the same office and they get down together at station III.We know that D gets down at the next station at which F gets down and E gets in with two others and gets down alone after D. From this it is clear that E gets down after two stations at which F gets down.
When number of variables involved in a question increase, its difficulty level also increases. In that case, it is better to draw a cross-tabular grid which can accommodate clues common to 2 or more than 2 variables. Let’s see an example.
Five sisters all have their birthday in a different month (February, March, June, July and December) and each on a different day of the week(Sunday, Monday, Wednesday, Friday and Saturday). Using the clues below, determine the month and day of the week on which each sister’s birthday falls.
We will draw the above grid and start to fill it according to clues provided. When you find a right match, draw a black dot in the corresponding square. When a combination is false, draw a cross in the corresponding square. Now, let’s start filling the grid.
1st clue: The first clue states that Paula was born in March so put a black dot in the square for “Paula” and “March” as shown. Since Paula was born in March, none of the other possibilities are valid, so the rest of the row and column can be marked with an ‘x’ as shown. The clue also states that Paula’s birthday is not on Saturday, thus “Saturday” for “Paula” can also be crossed out as shown. The clue also states that Abigail’s birthday is not on Wednesday or Friday so you can mark Wednesday and Friday for Abigail with an ‘x’ in the right section of the grid as shown.
2ndClue :Brenda and Mary didn’t have a birthday on Monday. Also, Brenda and Mary could not have been born in February since the girl whose birthday is Monday was born earlier in the year. So these squares can be crossed out. Also, in the bottom part of the grid, the Monday birthday cannot happen in July or December because both Brenda and Mary had birthdays after the girl whose birthday is on Monday so these two squares can also be crossed out.
3rdClue: Tara was not born in February and so that square is crossed out. Now the only empty square in that row is Abigail. By elimination therefore, Abigail was born in February. Put a black dot in that square and cross out the rest of the horizontal row. Moreover, since you know from clue 1 that Abigail’s birthday is not on a Wednesday or a Friday and you know she was born in February, you can cross out Wednesday and Friday in the lower block of the grid for February. Also, from the second part of Clue 3, Tara’s birthday is on the weekend, so Monday, Wednesday, and Friday are crossed out for Tara in the right section of the grid.
4th Clue: Mary wasn’t born in December so cross it out the corresponding square. Since her birthday isn’t on a weekday, you can also cross out Wednesday and Friday for Mary in the right-hand section of the grid. From Clue 2 you know that Tara’s birthday was on a weekend too. So in the right section of the grid, Saturday and Sunday can be crossed out for Abigail, Brenda, and Paula. This leaves only Monday for Abigail to put a black dot in this square and cross out the rest of the column for Monday. From the last part of Clue 4, in the bottom section of the grid, put a dot for June and Sunday, crossing out the rest of the column and row respectively as shown. Also, in the upper grid, cross out Brenda for June because you know her birthday wasn’t on the weekend.
5thClue :Cross out December for Tara since she couldn’t have been born last as she was born before Brenda. This leaves only December for Brenda so put a black dot in that square and cross out the rest of the row. From the next part of the clue, you know that Mary wasn’t born in July so by elimination Tara was born in July. Put a black dot in that square and cross out the rest of the row. Also put a black dot in the square for Mary and June since it’s the last square open. You’ve now determined all of the months in which everyone was born. Also, since we know that June goes with Sunday and Mary was born in June, her birthday is on Sunday so mark a dot in the corresponding section of the grid. This then leaves Tara for Saturday so mark a dot there too. Lastly, Brenda’s birthday wasn’t on Friday. So by elimination, Brenda’s birthday was on Wednesday and Paula’s birthday was on Friday.
So, from the final grid, we can see the birth-day and birth-month corresponding to each girl. In questions like these, one does not need to go through the clues sequence-wise. It is possible that some clues given at the end of the sequence can provide insights about the previous clues and help to fill the grid. Just start filling the grid with the most straight-forward and direct clues and empty squares can be filled later based on the deduction from available clues.
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All questions from CAT Exam Logical Reasoning and Data Interpretation
Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.
Logical Reasoning – Set 2: There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE).
Logical Reasoning – Set 3: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata by a flight operated by Cheap Air.
Logical Reasoning – Set 4: tea taster was assigned to rate teas from six different locations – Munnar, Wayanad, Ooty, Darjeeling, Assam and Himachal.
Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).
Data Interpretation – Set 1: Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET).
Data Interpretation – Set 2: An old woman had the following assets: (a) Rs. 70 lakh in bank deposits
Data Interpretation – Set 3: Simple Happiness index (SHI) of a country is computed on the basis of three parameters: social support (S), freedom to life choices (F) and corruption perception (C).
Data Interpretation – Set 4: Funky Pizzaria was required to supply pizzas to three different parties.
Data Interpretation – Set 5: There were seven elective courses – El to E7 – running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven.
This post was contributed by Praveen Bharadwaj. We are looking for contributions for our blog (paid). If you wish to write for us, please send an email to [email protected]
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How to prepare for LRDI (Logical Reasoning and Data Interpretation) for the CAT Exam?
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