Data Interpretation Actual Question in XAT 2022
Q1-3. Instructions : Read the following scenario and answer the THREE questions that follow.
The enrolment of students (in 1000s) at each of the five universities named — MPU, JSU, LTU, PKU and TRU — during each of the eight years from 2014 to 2021 is represented in the following chart. The names of these universities are not shown in the chart, Stead they are labelled Unit 1, Unit 2, Unit 3, Unit 4 and Unit 5.
However, these four pieces of information are available:
W: The magnitudes of TRU's and MPU's net change in enrolment between 2014 and 2021 are the closest among any two universities.
X: LTU had the same enrolment in consecutive years at least twice between 2014 and 2021.
Y: The increase in JSU's enrolment from 2015 to 2019 is about 50% of TRU's total enrolment in 2020.
Z: The enrolment in one of LTU and PKU had a steady decline between 2014 and 2021, while the enrolment in the other had no decline between any two consecutive years in the same period
1. Which of the five universities can Univ 4 possibly be?
A Either TRU or MPU
B Either MPU or PKU
C Only PKU
D Only TRU
E Only MPU
EXPLANATION
D
Using condition W :
The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities
for TRU and MPU.
Going by the color of the lines the net change for different universities is:
Univ 1: 0.7
Univ 2: 4.4
Univ 3: 0.1
Univ 4: 0.2
Univ 5: 3.3
The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)
Using condition X :
The university LTU must have the same enrollment in consecutive years at least twice :
LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.
Using condition Y :
The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment
in 2020.
Considering :
TRU = Univ 4
The enrollment is 5.
TRU = Univ 3
The enrollment is 0.7
For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of
TRU.
Hence TRU = Univ 4.
Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5
The only possible case is JSU = Univ 2.
MPU = Univ 3.
LTU = Univ 1
PKU = Univ 5.
2. Which Univ’s enrolment was around twice that of LTU in 2014?
A Only JSU’s
B Only PKU’s
C Either PKU’s or TRU’s
D Either JSU’s or MPU’s
E Only MPU’s
EXPLANATION
B
Using condition W :
The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities
for TRU and MPU.
Going by the color of the lines the net change for different universities is:
Univ 1: 0.7
Univ 2: 4.4
Univ 3: 0.1
Univ 4: 0.2
Univ 5: 3.3
The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)
Using condition X :
The university LTU must have the same enrollment in consecutive years at least twice :
LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.
Using condition Y :
The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment
in 2020.
Considering :
TRU = Univ 4
The enrollment is 5.
TRU = Univ 3
The enrollment is 0.7
For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of
TRU.
Hence TRU = Univ 4.
Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5
The only possible case is JSU = Univ 2.
MPU = Univ 3.
LTU = Univ 1
PKU = Univ 5.
SInce LTU is univ 1 the university with an enrollment twice that of LTU = 2*(3.2)
= 6.4.
Hence the only possible case close by is Univ 5 (PKU)
58.C
Using condition W :
The magnitudes of the net change in enrollment between 2014 and 2021 is closest among any two universities
for TRU and MPU.
Going by the color of the lines the net change for different universities is:
Univ 1: 0.7
Univ 2: 4.4
Univ 3: 0.1
Univ 4: 0.2
Univ 5: 3.3
The closest among these are: Univ 3 and Univ 4. They can possibly be : (TRU/MPU)
Using condition X :
The university LTU must have the same enrollment in consecutive years at least twice :
LTU can either be Univ 3 or Univ 1 but since Univ 3 must be among TRU and MPU. LTU is university 1.
Using condition Y :
The increase in the enrollment of JSU between the years 2015 and 2019 is 50 percent of TRU's total enrollment
in 2020.
Considering :
TRU = Univ 4
The enrollment is 5.
TRU = Univ 3
The enrollment is 0.7
For TRU as Univ 3, there is no university whose increase in enrollment between 2015 and 2019 is 50 percent of
TRU.
Hence TRU = Univ 4.
Since the increase in enrollment for JSU is half of TRU. The increase must be half of 5 = 2.5
The only possible case is JSU = Univ 2.
MPU = Univ 3.
LTU = Univ 1
PKU = Univ 5.
All the universities can be uniquely determined without using the condition Z.
3. Which amongst the pieces of information mentioned below, if removed, will not prevent us from uniquely identifying the five universities?
A Either X or Y
B Y
C Z
D NONE, since all four pieces of information are necessary to uniquely identify the five universities.
E X
EXPLANATION
Q4-6. Instructions: Read the following scenario and answer the THREE questions that follow.
A pencil maker ships pencils in boxes of size 50, 100 and 200. Due to packaging issues, some pencils break. About the 20 boxes he has supplied to a shop, the following information is available:
- Box no. 1 through 6 have 50 pencils, Box no. 7 through 16 have 100 pencils and Box no. 17 through 20 have
200 pencils. - No box has less than 5% or more than 20% broken pencils.
Following is the frequency table of the number of broken pencils for the twenty boxes:
4. Which of the following can possibly be the sequence of the number of broken pencils in Boxes 7-16?
A 6,7,9, 11,15,19,20,20,20,29
B 5,6,6,6,11,15,15,20,20,20
C 7,7,7,7,11,15,15,19,20,20
D 7,7,9,9,11,13,15,19,20,20
E 5,7,7,7,9,11,15,20,20,20
EXPLANATION
C
Boxes 7 to 16 contain a total of 100 pencils each. The minimum number of broken pencils the box can hold is 5
percent of the total pencils and a maximum of 20 percent of the total pencils.
5 percent of 100 = 5 and 20 percent of 100 = 20 pencils.
Hence the number of broken pencils must be in the range of 5 to 20.
The frequency of the different number of broken pencils is :
5 - 1
6 - 2
7 - 4
9 - 3
11 - 1
15 - 2
19 - 1
20 - 3
29 - 1
31 - 1
33 - 1
The boxes cannot contain 29, 31, 33 to be the number of broken pencils because they are beyond 20 percent.
Since boxes 1- 6 can contain only between 2.5 to 10 pencils. The remaining boxes which include broken pencils
of numbers less than 10 must be a part of 7 - 16. Because boxes 17 - 20 cannot contain broken pencils of numbers less than 10.
Hence 7 - 16 must have 4 boxes that contain less than 10 broken pencils.
Going through the options :
Option A fails because this includes only 3 boxes with less than 10 pencils.
Option B fails because we only have 2 boxes with 6 broken pencils but this includes 3.
Option D fails because it does not include a box of 15 and a box of 20 pencils which can only be a part of boxes
with 100 or boxes with 200 pencils. Since boxes 17 - 20 can include only one among 15 or 20 because 29, 31, 33 are a part of this group. Hence this case fails.
Option E fails because this includes 5 boxes with broken pencils less than 10 but this is not possible because this must exactly contain 4 boxes with less than 10 pencils.
Option C is a feasible case containing :
1-6 ( 5, 6, 6, 9, 9, 9)
7 - 16 ( 7, 7, 7, 7, 11, 15, 15, 19, 20, 20 )
17 - 20 (20, 29, 31, 33)
5. Which of the following cannot be inferred conclusively from the given information?
A No box numbered 1-6 has more broken pencils than any box numbered 17-20.
B A box with the highest percentage of broken pencils has 100 pencils.
C Four among the boxes numbered 7 to 16 have less than 10 broken pencils.
D Exactly three boxes have 20% broken pencils.
E Three among the boxes numbered 17 to 20 have 29, 31 and 33 broken pencils in someorder.
EXPLANATION
D
Going by the given options :
Option A: The boxes numbered 1- 6 have a capacity of 50 pencils. The maximum number of broken pencils they can contain is 20 percent of 20 pencils = 10 pencils. For boxes numbered 17 to 20 they must contain a minimum of 5 percent which is equivalent to 5 percent of 200. Hence this is true.
Option B: Boxes with broken pencils of 29, 31, 33 must be a part of 17 - 20. There must be one box containing broken pencils in the range of 10 to 20. There are three boxes in total containing exactly 20 pencils. A maximum of only one of the three can be a part of 17 - 20. The remaining must be a part of boxes 7- 16 because they cannot be a part of 1-6. Hence at least one box among 7 - 16 contains 20 percent of broken pencils which is the highest.
Option C: There are a total of ten boxes with less than 10 broken pencils. They can either be a part of boxes 1-6 or 7 - 16. Since boxes 1-6 can only take broken pencils with less than 10 in number. Hence of the 10 six must be a part of 1-6 and the remaining 4 must be a part of 7 - 16.
Option D: The only possibility for containing 20 percent of the broken pencils is only possible for 20 broken pencils which is 20 percent of 100. There must be at least 2 boxes in the range of 7 - 16 which contain 20 broken pencils which is equal to 20 percent. The third box can either be a part of (7-16) to (17-20). If this
belongs to 17 - 20 then the case is not possible and hence cannot be concluded.
Option E: Boxes containing 29, 31, and 33 broken pencils must be a part of boxes 17 -20 because they are higher than the 20 percent range of boxes (1-6) and (7-16). Hence this can be concluded.
6. Suppose that additionally it is known that the number of broken pencils in Boxes 17-20 are in increasing order. Which among the following additional information, if true, is not sufficient to uniquely know the number of defective pencils in each of the boxes numbered 17-20?
A Boxes no. 7-16 contains a total of 124 defective pencils.
B Boxes no. 17-20 contain a total of 108 defective pencils.
C Boxes no. 11-16 contain a total of 101 defective pencils.
D Box no. 17 contains more defective pencils than any box from among boxes no. 1-14.
E Boxes no. 7-16 contain a total of 133 defective pencils.
EXPLANATION
E
Going by the options :
Option A: Boxes 7- 16 contain a total of 124 pencils. Boxes (1-6) has 6 boxes with broken pencils which can be included from :
(5, 6, 6, 7, 7, 7, 7, 9, 9, 9). The minimum possible sum of the 6 pencils is : (5+6+6+7) = 24 and the maximum possible sum is (7+9+9+9) = 34.
Boxes 7 - 16 contains all the boxes with broken pencils except one among the boxes with broken pencils among 11 - 20 and hence ;
This can contain : (11+15+15+19+20+20) or (15+15+19+20+20+20) or (11+15+15+20+20+20) or (11+15+19+20+20+20 ) =100/109/101/105.
The only possible case to contain 124 pencils is by considering the case : (24+100) = (5, 6,6,7, 11+15+15+19+20+20) = 124 .
Hence box 17 - 20 must contain (20, 29, 31, 33).
Q7-9. Instructions: Read the following scenario and answer the THREE questions that follow.
An examination had ten multiple choice questions; labelled Q1 to Q10 respectively. Each question had four answer options — A, B, C and D — of which one and only one was the correct answer. For each correct answer, the candidate obtained 1 mark. There were no negative marks for wrong answers. The answers chosen by six candidates named Om, Pavan, Qadir, Rakesh, Simranjeet and Tracey to each of the ten questions and the total marks obtained by each of them are shown in the table.
7. What is the correct answer for Q5?
A Not possible to determine uniquely
B A
C B
D C
E D
EXPLANATION
D
The correct answer for Q5 is C.
8. For which of these questions is D the correct answer?
A Both Q1 and Q9
B Both Q1 and Q8
C Q8
D Q1
E Q9
EXPLANATION
C
Q6 and Q8 have D as the correct answer.
9. Which of these questions witnessed the least number of the students answering correctly?
A Both Q3 and Q4
B Q4
C Q5
D Q10
E Q2
EXPLANATION
B
Of the different possibilities are ( 1, 6, 7), (1, 6, 9), (6, 7, 9), (1, 7, 9).
But the cases 6, 7, 9 fails because if he answered all three of them correctly then he must have answered 1 wrongly but since if 1 is answered wrong then all of 1, 6, 7, 9 are answered correctly by Simarjeet when Simarjeet can answer actually a total of 3 questions only correctly hence this case fails.
Similarly if considered the cases ( 1, 6, 7) and (1, 6, 9) as the questions which were answered correctly. The cases fail because if they are answered correctly we cannot possibly have Pavan answering 5 questions correctly.
Hence the only possibility is he must have answered the questions :
( 1, 7, 9) correctly.
Hence the correct answers are :
Now drawing the table based on the answers marked by them :
Simarjeet must have answered 1 question of the Q2, Q4, and Q6 correctly. Qadir must have answered the remaining two of them correctly.
If Simarjeet answered Q4 correctly and Q2, Q6 wrong the answers for Q2, Q4 and Q6 will be: B, B, and D
But if Q2 and Q4 are answered as B then Om must have scored 3 marks instead of2 and hence the case fails.
If Simarjeet answered Q6 correctly and Q2, Q4 wrong the answers for Q2, Q4 and Q6 will be: A, C, and C.
But Pavan cannot score 5 marks.
Hence Simarjeet must have answered Q2 correctly and answers for Q2, Q4, and Q6 will be :
(B, C, D).
Q4 was answered wrong by 5 members.
Q10-12. Instructions: Read the following scenario and answer the THREE questions that follow.
The given candlestick chart depicts the prices of a particular stock over 10 consecutive days. A candlestick comprises of a rectangular box pieced by a line. The top and bottom ends of the line respectively indicate the maximum and minimum prices of the stock on that day, while the horizontal edges of the rectangle correspond to the stock's opening and closing prices. If the rectangle is white, the opening price is lower than the closing price, but if the rectangle is black, then it is the other way around.
Using the above information, answer the questions that follow:
Q10. Which day saw the maximum percentage increase in the stock price at closing from the opening?
A Day 10
B Day 2
C Day 1
D Day 6
E Day 7
EXPLANATION
C
Going by the cases in the option for the five days :
Day 1 : (Opening price, Closing Price ) : (2365, 2395)
Day 2 : (Opening price, Closing Price ) : ( 2395, 2425 )
Day 6 : (Opening price, Closing Price ) : ( Closing price is lower than the opening price )
Day 7 : (Opening price, Closing Price ) : ( Closing price is lower than the opening price )
Day 10 : (Opening price, Closing Price ) : (2277.5, 2292.5)
The percentage increase for day 1 :
The percentage increase for day 2 :
The percentage increase for day 10 :
Hence Day 1 is the highest
Q11. What is the highest magnitude of change over two consecutive days (for example, Day 1 → Day 3 or Day 5 → Day 7), in the maximum price touched by the stock during the 10 day period ( choose the closest amongst the options given)?
A 60
B 70
C 80
D 50
E 40
EXPLANATION
B
Among the given days the magnitude of change in difference of the maximum price in an interval of two days is:
Day 1 - Day 3 : ( 2415, 2440) : 25
Day 2 - Day 4 : ( 2432.5, 2455) : 22.5
Day 3 - Day 5 : ( 2440, 2415) = 25
Day 4 - Day 6 : ( 2455, 2400) = 55
Day 5 - Day 7 : ( 2415, 2367.5) = 47.5
Day 6 - Day 8 : (2400, 2330) = 70
Day 7 - Day 9 : ( 2367.5, 2330) = 52.5
Day 8 - Day 10 : (2330, 2332.5) = 2.5
The maximum difference among the possible cases is : 70
Q12. On which day is the ratio of the maximum price to the opening price, the highest across the ten days?
A Day 3
B Day 4
C Day 10
D Day 1
E Day 9
EXPLANATION
C
Going by considering the given options :
The ratio is given by :
Day 3 : (Maximum price, Opening price ) : ( 2440, 2405) = the ratio = 1.014
Day 4 : (Maximum price, Opening price ) : (2455, 2432.5) = ratio = 1.009
Day 10 : (Maximum price, Opening price ) : ( 2330, 2292.5) = 1.016
Day 1 : (Maximum price, Opening price ) : ( 2415, 2395) = 1.008
Day 9 : (Maximum price, Opening price ) : ( 2330, 2297.5) = 1.014
Day 10 has the maximum ratio