CAT 2019 LRDI Slot 2 | Previous Year CAT Paper
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Answer the following questions based on the information given below.
To compare the rainfall data, India Meteorological Department (IMD) calculated the Long Period Average (LPA) of rainfall during period June-August for each of the 16 states. The figure given below shows the actual rainfall (measured in mm) during June-August, 2019 and the percentage deviations from LPA of respective states in 2018. Each state along with its actual rainfall is presented in the figure.
If a ‘Heavy Monsoon State’ is defined as a state with actual rainfall from June-August, 2019 of 900 mm or more, then approximately what percentage of ‘Heavy Monsoon States’ have a negative deviation from respective LPAs in 2019?
- A.
57.14
- B.
14.29
- C.
75.00
- D.
42.86
Answer: Option D
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Explanation :
There are a total of 7 states which have actual rainfall greater than or equal to 900 mm.
Out of these only 3 have negative deviation. (Arunachal, Kerela and Meghalaya)
Required percentage = (3/7) × 100 ≈ 42.86%.
Hence, option (d).
Workspace:
If a ‘Low Monsoon State’ is defined as a state with actual rainfall from June-August, 2019 of 750 mm or less, then what is the median ‘deviation from LPA’ (as defined in the Y-axis of the figure) of ‘Low Monsoon States’?
- A.
10%
- B.
–30%
- C.
–10%
- D.
–20%
Answer: Option C
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Explanation :
There are 9 states whose actual rainfall is less than or equal to 750 mm.
The states arranged according to the deviation is: Gujarat (25%), Karnataka(20%), Rajasthan(15%), MP(10%), Assam(−10%), WB(−30%), Jharkhand(−35%), Delhi(−40%) and Manipur(−60%).
So the median deviation is of Assam (−10%).
Hence, option (c).
Workspace:
What is the average rainfall of all states that have actual rainfall of 600 mm or less in 2019 and have a negative deviation from LPA?
- A.
460 mm
- B.
500 mm
- C.
450 mm
- D.
367 mm
Answer: Option A
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Explanation :
There are five states who have actual rainfall of 600 mm or less in 2019 and have a negative deviation from LPA [Assam(600). WB(600), Jharkhand (400), Manipur (400) and Delhi (300).
Required average = (600 + 600 + 400 + 400 + 300)/5 = 460 mm.
Hence, option (a).
Workspace:
The LPA of a state for a year is defined as the average rainfall in the preceding 10 years considering the period of June-August. For example, LPA in 2018 is the average rainfall during 2009-2018 and LPA in 2019 is the average rainfall during 2010-2019. It is also observed that the actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009. The LPA of Gujarat in 2019 is closest to
- A.
505 mm
- B.
525 mm
- C.
490 mm
- D.
475 mm
Answer: Option C
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Explanation :
Actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009.
So, considering Gujarat, Actual rainfall in 2019 = (6/5) × Actual rainfall in 2009 = 600 mm (given).
∴ Actual rainfall in 2009 = 500 mm.
As deviation for Gujarat is 25%, so average rainfall for 2009 to 2018 is 600/1.25 = 480 mm.
So, total rainfall from 2009 to 2018 = 480 × 10 = 4800 mm.
LPA for 2019 = (Rainfall in 2010 to 2019)/10 = [(Rainfall in 2009 to 2018) − Rainfall in 2009 + Rainfall in 2019]/10
= (4800 − 500 + 600)/10 = 490 mm.
Hence, option (c).
Workspace:
Answer the following questions based on the information given below.
The first year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common endterm (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
- Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
- Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
- All questions prepared by a faculty member appeared consecutively in MT as well as ET
- Chetan prepared the third question in both MT and ET; and Esha prepared the eighth question in both.
- Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.
The second question in ET was prepared by:
- A.
Beti
- B.
Esha
- C.
Chetan
- D.
Dave
Answer: Option D
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Explanation :
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15-mark questions followed the 10-mark questions.
Therefore, Each of MT and ET had at least 4 + 3 + 2 = 9 questions. Also, these 9 questions were worth 80 marks. So, remaining 20 marks can be from (one 5-mark and one 15-mark questions OR two 10-mark questions OR two 5-mark questions and one 10-mark question OR four 5-mark questions).
From (ii), Annie prepared exactly one question (i.e., fifth question) for MT and minimum number of questions in MT = 1 + 5 × 5 = 11
From (iv) and (v), Fakir and Chetan prepared the first and the third questions respectively. Now from (ii) and (iii), Fakir and Chetan prepared the second and the fourth questions also.
Similarly it can be concluded that Dave prepared first two questions of ET and Chetan prepared the third and the fourth questions of ET.
Also, as each faculty member prepared the same number of questions and Chetan prepared 4 questions, each prepared 4 questions.
Total number of questions in ET and MT together = 4 × 6 = 24. As ET contained more questions than MT, MT had 11 questions and ET had 13 questions.
Therefore, Beti (6th and 7th), Esha (8th and 9th) and Dave (10th and 11th) prepared exactly two questions for MT. Also, first five questions were of 5 marks, then three 10-mark questions and in the last, three 15- mark questions. As ET has 13 questions and 9 among these contribute to 80 marks, we are left with the possibility that there were four 5-mark questions other than the 9 questions.
Annie prepared 3 questions for ET (i.e., the 5th, 6th and 7th). Esha prepared 8th and 9th questions. And Fakir prepared last two questions. Thus, we have
So, The second question in ET was prepared by Dave.
Hence, option (d).
Workspace:
How many 5‐mark questions were there in MT and ET combined?
- A.
13
- B.
10
- C.
12
- D.
Cannot be determined
Answer: Option A
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Explanation :
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15-mark questions followed the 10-mark questions.
Therefore, Each of MT and ET had at least 4 + 3 + 2 = 9 questions. Also, these 9 questions were worth 80 marks. So, remaining 20 marks can be from (one 5-mark and one 15-mark questions OR two 10-mark questions OR two 5-mark questions and one 10-mark question OR four 5-mark questions).
From (ii), Annie prepared exactly one question (i.e., fifth question) for MT and minimum number of questions in MT = 1 + 5 × 5 = 11
From (iv) and (v), Fakir and Chetan prepared the first and the third questions respectively. Now from (ii) and (iii), Fakir and Chetan prepared the second and the fourth questions also.
Similarly it can be concluded that Dave prepared first two questions of ET and Chetan prepared the third and the fourth questions of ET.
Also, as each faculty member prepared the same number of questions and Chetan prepared 4 questions, each prepared 4 questions.
Total number of questions in ET and MT together = 4 × 6 = 24. As ET contained more questions than MT, MT had 11 questions and ET had 13 questions.
Therefore, Beti (6th and 7th), Esha (8th and 9th) and Dave (10th and 11th) prepared exactly two questions for MT. Also, first five questions were of 5 marks, then three 10-mark questions and in the last, three 15- mark questions. As ET has 13 questions and 9 among these contribute to 80 marks, we are left with the possibility that there were four 5-mark questions other than the 9 questions.
Annie prepared 3 questions for ET (i.e., the 5th, 6th and 7th). Esha prepared 8th and 9th questions. And Fakir prepared last two questions. Thus, we have
There were eight 5-mark questions in ET and five 5-mark questions in MT.
Thus, there were 13 (5-mark) questions in MT and ET combined.
Hence, option (a).
Workspace:
Who prepared 15-mark questions for MT and ET?
- A.
Only Dave, Esha and Fakir
- B.
Only Beti, Dave, Esha and Fakir
- C.
Only Dave and Fakir
- D.
Only Esha and Fakir
Answer: Option A
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Explanation :
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15-mark questions followed the 10-mark questions.
Therefore, Each of MT and ET had at least 4 + 3 + 2 = 9 questions. Also, these 9 questions were worth 80 marks. So, remaining 20 marks can be from (one 5-mark and one 15-mark questions OR two 10-mark questions OR two 5-mark questions and one 10-mark question OR four 5-mark questions).
From (ii), Annie prepared exactly one question (i.e., fifth question) for MT and minimum number of questions in MT = 1 + 5 × 5 = 11
From (iv) and (v), Fakir and Chetan prepared the first and the third questions respectively. Now from (ii) and (iii), Fakir and Chetan prepared the second and the fourth questions also.
Similarly it can be concluded that Dave prepared first two questions of ET and Chetan prepared the third and the fourth questions of ET.
Also, as each faculty member prepared the same number of questions and Chetan prepared 4 questions, each prepared 4 questions.
Total number of questions in ET and MT together = 4 × 6 = 24. As ET contained more questions than MT, MT had 11 questions and ET had 13 questions.
Therefore, Beti (6th and 7th), Esha (8th and 9th) and Dave (10th and 11th) prepared exactly two questions for MT. Also, first five questions were of 5 marks, then three 10-mark questions and in the last, three 15- mark questions. As ET has 13 questions and 9 among these contribute to 80 marks, we are left with the possibility that there were four 5-mark questions other than the 9 questions.
Annie prepared 3 questions for ET (i.e., the 5th, 6th and 7th). Esha prepared 8th and 9th questions. And Fakir prepared last two questions. Thus, we have
Only Esha and Dave prepared 15-mark questions for MT while Fakir prepared 15-mark questions for ET.
Hence, option (a).
Workspace:
Which of the following questions did Beti prepare in ET?
- A.
Tenth question
- B.
Fourth question
- C.
Seventh question
- D.
Ninth question
Answer: Option A
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Explanation :
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15-mark questions followed the 10-mark questions.
Therefore, Each of MT and ET had at least 4 + 3 + 2 = 9 questions. Also, these 9 questions were worth 80 marks. So, remaining 20 marks can be from (one 5-mark and one 15-mark questions OR two 10-mark questions OR two 5-mark questions and one 10-mark question OR four 5-mark questions).
From (ii), Annie prepared exactly one question (i.e., fifth question) for MT and minimum number of questions in MT = 1 + 5 × 5 = 11
From (iv) and (v), Fakir and Chetan prepared the first and the third questions respectively. Now from (ii) and (iii), Fakir and Chetan prepared the second and the fourth questions also.
Similarly it can be concluded that Dave prepared first two questions of ET and Chetan prepared the third and the fourth questions of ET.
Also, as each faculty member prepared the same number of questions and Chetan prepared 4 questions, each prepared 4 questions.
Total number of questions in ET and MT together = 4 × 6 = 24. As ET contained more questions than MT, MT had 11 questions and ET had 13 questions.
Therefore, Beti (6th and 7th), Esha (8th and 9th) and Dave (10th and 11th) prepared exactly two questions for MT. Also, first five questions were of 5 marks, then three 10-mark questions and in the last, three 15- mark questions. As ET has 13 questions and 9 among these contribute to 80 marks, we are left with the possibility that there were four 5-mark questions other than the 9 questions.
Annie prepared 3 questions for ET (i.e., the 5th, 6th and 7th). Esha prepared 8th and 9th questions. And Fakir prepared last two questions. Thus, we have
Beti prepared 10th and 11th question in ET.
Hence, option (a).
Workspace:
Answer the following questions based on the information given below.
Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row?
Answer: 13
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Explanation :
Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10
Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13
Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.
As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.
Consider column 2:
C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22
C2R2: All the three pouches have one coin each. Sum = 3
C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4
In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)
Now consider R1
First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.
Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.
Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.
Thus, we have
First column of the second row has 13 coins. i.e., Rs. 13
Answer: 13
Workspace:
How many pouches contain exactly one coin?
Answer: 8
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Explanation :
Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10
Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13
Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.
As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.
Consider column 2:
C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22
C2R2: All the three pouches have one coin each. Sum = 3
C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4
In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)
Now consider R1
First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.
Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.
Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.
Thus, we have
Two pouches in Row 3 Column 1 slot have 1 coin each.
Three pouches in Row 2 Column 2 have 1 coin each.
Two pouches in Row 3 Column 3 slot have 1 coin each.
One pouch in Row 1 Column 3 slot has 1 coin.
Total number of pouches = 2 + 3 + 2 + 1 = 8
Answer: 8
Workspace:
What is the number of slots for which the average amount (in rupees) of its three pouches is an integer?
Answer: 2
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Explanation :
Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10
Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13
Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.
As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.
Consider column 2:
C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22
C2R2: All the three pouches have one coin each. Sum = 3
C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4
In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)
Now consider R1
First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.
Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.
Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.
Thus, we have
For average amount to be an integer, we need to consider slots having total number of coins in the pouches divisible by 3.
There are two such slot i.e., (row 1 column 3) and (row 2 column 2).
Answer: 2.
Workspace:
The number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is
Answer: 3
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Explanation :
Consider C1R1: Maximum coins in a pouch = 4 and minimum coins = 2. The third bag could have 2, 3 or 4 coins. Therefore, sum = 8, 9 or 10
Consider C1R2: Maximum coins in a pouch = 5 and minimum coins = 3. The third bag could have 3, 4 or 5 coins. Therefore, sum = 11, 12 or 13
Consider C1 R3: Maximum coins in a pouch = 2 and minimum coins = 1. As the sum = 4, third pouch has 1 coin.
As the sum of coins in the nine pouches in the column are divisible by 9, the coins in C1R1 and C1R2 has to be 10 (i.e., 2, 4, 4) and 13 (i.e., 3, 5, 5) respectively.
Consider column 2:
C2R1: Maximum coins in a pouch = 8 and minimum coins = 6. The third bag could have 6, 7 or 8 coins. Therefore, sum = 20, 21 or 22
C2R2: All the three pouches have one coin each. Sum = 3
C2R3: Maximum coins in a pouch = 2 and minimum coins = 1. The third bag could have 1 or 2 coins. Therefore, sum = 1 or 4
In order to have number of coins in the cells of the column divisible by 9, sum of the coins in C2R1 = 20(i.e., 6, 6, 8) and in C2R3 = 4(i.e., 1,1,2)
Now consider R1
First two cells together have 30 coins. So the third cell has to have 6(i.e., 1, 2, 3) coins.
Consider R2: First two cells together have 16 coins. Coins in the third cell are in the range 6 + 6 + 20 = 32 to 6 + 20 + 20 = 46. Therefore the third cell has to have 38 (i.e., 6, 12, 20) coins.
Consider R3: First two cells together have 8 coins. So the third cell has to have 10 (i.e., 2, 3, 5) coins.
Thus, we have
The slots for which the total amount in its three pouches strictly exceeds Rs. 10 are Row 1 Column 2, Row 2 Column 1, Row 2 Column 3.
Answer: 3.
Workspace:
Answer the following questions based on the information given below.
In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.
These five people form three teams, Team 1, Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
(1) Each team speaks exactly four languages and has the same number of members.
(2) English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team.
(3) None of the teams include both Quentin and Robert.
(4) Paula and Sally are together in exactly two teams.
(5) Robert is in Team 1 and Quentin is in Team 3.
Who among the following four is not a member of Team 2?
- A.
Quentin
- B.
Paula
- C.
Terence
- D.
Sally
Answer: Option A
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Explanation :
Robert is in Team 1 and speaks Arabic and French.
So from (1) and (2); Team 1 speaks English, Chinese, Arabic and French.
So, Sally cannot be in Team 1. As Arabic is spoken by exactly one team, Robert is only on Team 1.
Therefore from (4), Paula and Sally are in Teams 2 and 3.
Quentin is in Team 3 and speaks Dutch and English.
So from (1) and (2); Team 3 speaks English, Chinese, Dutch and Basque. Also, Paula, Sally and Quentin are members of Team 3.
As Dutch is spoken by exactly one team, Quentin is only on Team 3. Team 3 has exactly 3 members.
Therefore from (1), each team has 3 members. So, Terence is the third member of Team 1 and 2.
Therefore, we have
So, Quentin is not a member of Team 2.
Hence, option (a).
Workspace:
Who among the following four people is a part of exactly two teams?
- A.
Sally
- B.
Paula
- C.
Quentin
- D.
Robert
Answer: Option A
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Explanation :
Consider the solution to first question of this set.
Quentin and Robert are members of exactly one team while Paula is a member of all the three teams. Sally is a part of exactly two teams.
Hence, option (a).
Workspace:
Who among the five people is a member of all teams?
- A.
Terence
- B.
Paula
- C.
No one
- D.
Sally
Answer: Option B
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Explanation :
Consider the solution to first question of this set.
Paula is a member of all teams.
Hence, option (b).
Workspace:
Apart from Chinese and English, which languages are spoken by Team 1?
- A.
Basque and French
- B.
Arabic and French
- C.
Arabic and Basque
- D.
Basque and Dutch
Answer: Option B
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Explanation :
Consider the solution to first question of this set.
Apart from Chinese and English, Team 1 speaks Arabic and French.
Hence, option (b).
Workspace:
Answer the following questions based on the information given below.
A large store has only three departments, Clothing, Produce, and Electronics. The following figure shows the percentages of revenue and cost from the three departments for the years 2016, 2017 and 2018. The dotted lines depict percentage levels. So for example, in 2016, 50% of store's revenue came from its Electronics department while 40% of its costs were incurred in the Produce department.
In this setup, Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost × 100%.
It is known that
- The percentage profit for the store in 2016 was 100%.
- The store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.
- There was no profit from the Electronics department in 2017.
- In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department.
What was the ratio of revenue generated from the Produce department in 2017 to that in 2018?
- A.
4 : 3
- B.
8 : 5
- C.
16 : 9
- D.
9 : 16
Answer: Option B
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Explanation :
Let the total cost of the store in 2016 be 100.
From (1): Total revenue in 2016 = 2 × Total cost in 2016 = 2 × 100 = 200.
From (2): Total revenue in 2017 = 2 × Total revenue in 2016 = 2 × 200 = 400.
From (2): Total cost in 2018 = 2 × Total cost in 2016 = 2 × 100 = 200.
So, we can find the department wise revenue in 2016 and 2017 and cost in 2016 and 2018 using the percentages found from the radar graph.
From (3): Cost from Electronics in 2017 = Revenue from Electronics in 2017 = 120.
So total cost in 2017 = 120/40% = 300. And hence the department wise costs can be found for 2017.
From (4): Revenue from Clothing department in 2018 = Cost for the Produce departement in 2018 = 100.
So total revenue in 2018 = 100/40% = 250.
Revenue generated from the Produce department in 2017 to that in 2018 = 160 : 100 = 8 : 5.
Hence option (b).
Workspace:
What was the percentage profit of the store in 2018?
Answer: 25
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Explanation :
Consider the solution to first question of this set.
Required percentage = (Profit/Cost) × 100 = (50/200) × 100 = 25%.
Hence, 25.
Workspace:
What percentage of the total profits for the store in 2016 was from the Electronics department?
Answer: 70
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Explanation :
Consider the solution to first question of this set.
Total profit in 2016 = 100.
Profit from Electronics in 2016 = 70
Required percentage = (70/100) × 100 = 70%.
Hence, 70.
Workspace:
What was the approximate difference in profit percentages of the store in 2017 and 2018?
- A.
25.0
- B.
33.3
- C.
8.3
- D.
15.5
Answer: Option C
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Explanation :
Consider the solution to first question of this set.
Profit percentage of the store;
In 2017 = (100/300) × 100 = 33.33%.
In 2018 = (50/200) × 100 = 25%.
Required difference = 33.33 − 25 = 8.33 ≈ 8.3.
Hence option (c).
Workspace:
Answer the following questions based on the information given below.
Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2,..., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10, 1 and 2 in Round 7 and so on. The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.
The following information is known about Rounds 1 through 6:
- Gordon did not score consecutively in any two rounds.
- Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
- Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
- Joshin scored in Round 7, while Amita scored in Round 10.
- No player scored in all the four rounds.
What were the scores of Chen, David, and Eric respectively after Round 3?
- A.
3, 0, 3
- B.
3, 3, 0
- C.
3, 3, 3
- D.
3, 6, 3
Answer: Option C
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Explanation :
As the top three performances in each round were awarded 7, 3 and 1 respectively, remaining all performances were awarded 0 points.
Amita was participant in round 1 and in rounds 6 through 10.
So, in all she scored 8(= 1 + 7) points in round 1 and round 6. Also she scored 10 points in rounds 7 through 10 together.
Bala was participant in rounds 1 and 2 and also in rounds 7 through 10.
As he scored 2 points in first two rounds, he must have scored 1 point in each of the two rounds. He scored 3(= 3 or 1+1+1) points in rounds 7 through 10.
Chen scored 3 points in first three rounds and as he scored in three consecutive rounds among 7 through 10, he must have scored 1 point each in round 8, 9 and 10.
David scored 6 points in first four rounds together and 0 (i.e., 6 – 6) each in 9th and 10th rounds.
Eric scored 3(= 3 or 1+1+1) points in five rounds together and hence 10 – 3 = 7 points in the 10th round.
Gordon scored 17 points in rounds 2 through 6 together and hence, 17 – 17 = 0 points in the 7th round.
As he did not scored consecutively in any two rounds, he must have scored 0 in 3rd and 5th round and 7, 7, and 3 points in 2nd, 4th and 6th rounds, not necessarily in the given sequence.
Hansa scored 1 point in one of the rounds 3-6. So, she must have scored 4 – 1 = 3 points in either 7th or 8th rounds.
Ikea scored 2 points in 4th, 5th and 6th rounds together. Therefore, remaining 15 points must have been scored in 7th, 8th and 9th round in some order.
As Chen scored 1 point in the 8th and 9th round, Ikea must have scored 1 point in 7th round and 7 points each in the 8th and 9th round.
Joshin scored 14 points in 5th and 6th rounds. So he scored 7 points in each of 5th and 6th rounds.
Now we need to decide about remaining 3 points in the last four rounds.
As Joshin scored in round 7, and Chen has scored 1 point in each of the last three rounds, Joshin must have scored all 3 points in the 7th round and 0 points in the last three rounds.
Therefore, we can tabulate this information as
As Joshin scored 3 points in the 7th round, Hansa must have scored 3 in 8th round and 0 in the 7th round. Now consider 7th round. Only Amita must have scored 7 in the 7th round.
As she has scored in round 10, the remaining 3 points she must have scored in round 10. Thus, we can now conclude that Bala scored 3 points in the 9th round.
As Joshin scored 7 points in the 6th round Amita must have scored 1 point in this round and hence7 points in the first round.
Gordon scored 7, 7 and 3 points in 2nd, 4th and 6th rounds. Joshin scored 7 points in the 6th round so, Gordon scored 7, 7 and 3 points in 2nd, 4th and 6th rounds respectively.
So we can fill remaining values as 0 in the table for the 6th round.
Thus, Ikea scored 1 point each in 4th and 5th rounds. And hence, Hansa scored 1 point in the 3rd round.
So far we have entered values 1 and 7 for all the rounds. Now we have to enter 3 points of each of Bala, Chen, David and Fatima and 7 points for Fatima.
Eric and Fatima both scored in a round. This is possible only when Fatima scored 7 points and Eric scored 3 points. This must be round 3.
No one other than Fatima scored 3 points in round 5. Similarly, David scored 3 points in the 4th round. Chen and David scored 3 points in 1st and 2nd round in some order.
Thus we have
After round 3, the scores of Chen, David and Eric were 3, 3, 3.
Hence, option (c).
Workspace:
Which three players were in the last three positions after Round 4?
- A.
Bala, Ikea, Joshin
- B.
Bala, Hansa, Ikea
- C.
Bala, Chen, Gordon
- D.
Hansa, Ikea, Joshin
Answer: Option D
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Explanation :
Consider the solution to first question of this set.
The three players in the last position after Round 4 were Hansa, Ikea and Joshin.
Hence, option (d).
Workspace:
Which player scored points in maximum number of rounds?
- A.
Ikea
- B.
Chen
- C.
Amita
- D.
Joshin
Answer: Option A
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Explanation :
Consider the solution to first question of this set.
Ikea scored in maximum number of rounds i.e., in 5 rounds.
Hence, option (a).
Workspace:
Which players scored points in the last round?
- A.
Amita, Chen, David
- B.
Amita, Bala, Chen
- C.
Amita, Chen, Eric
- D.
Amita, Eric, Joshin
Answer: Option C
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Explanation :
Consider the solution to first question of this set.
Amita, Chen and Eric scored in the last round.
Hence, option (c).
Workspace:
Answer the following questions based on the information given below.
Students in a college are discussing two proposals –
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.
A student does not necessarily support either of the two proposals.
In an upcoming election for student union president, there are two candidates in fray: Sunita and Ragini. Every student prefers one of the two candidates.
A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.
- 250 students supported proposal A and 250 students supported proposal B.
- Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.
- Among those who preferred Ragini, 30% supported proposal A.
- 20% of those who supported proposal B preferred Sunita.
- 40% of those who did not support proposal B preferred Ragini.
- Every student who preferred Sunita and supported proposal B also supported proposal A.
- Among those who preferred Ragini, 20% did not support any of the proposals.
What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?
- A.
50
- B.
25
- C.
40
- D.
20
Answer: Option A
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Explanation :
From (1), k + m + p + r = 250 and l + m + q + r = 250
From (4), l + m = 0.2 × 250 = 50
From (6), l = 0 ⇒ m = 50
∴ q + r = 200
From (2), k + l + m + n = 200 and hence p + q + r + s = 300
From (3), p + r = 0.3 × 300 = 90 ⇒ k + m = 160
As l = 0 and k + l + m + n = 200, n = 40
∴ k = 110
From (7), n + s = 100 ⇒ s = 60
∴ p + q + r = 240 and p + r = 90 ⇒ q = 150
∴ q + r = 200 ⇒ r = 50 and hence p = 40
Thus, we have
Out of 100 students who supported both proposals A and B, 50 students preferred Sunita as student union president. i.e., 50%
Workspace:
Among the students surveyed who supported proposal A, what percentage preferred Sunita for student union president?
Answer: 64
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Explanation :
Sunita | Ragini | |
Only proposal A | k | p |
Only proposal B | l | q |
Both A and B | m | r |
Neither A nor B | n | s |
From (1), k + m + p + r = 250 and l + m + q + r = 250
From (4), l + m = 0.2 × 250 = 50
From (6), l = 0 ⇒ m = 50
∴ q + r = 200
From (2), k + l + m + n = 200 and hence p + q + r + s = 300
From (3), p + r = 0.3 × 300 = 90 ⇒ k + m = 160
As l = 0 and k + l + m + n = 200, n = 40
∴ k = 110
From (7), n + s = 100 ⇒ s = 60
∴ p + q + r = 240 and p + r = 90 ⇒ q = 150
∴ q + r = 200 ⇒ r = 50 and hence p = 40
Thus, we have
Sunita | Ragini | |
Only proposal A | 110 | 40 |
Only proposal B | 0 | 150 |
Both A and B | 50 | 50 |
Neither A nor B | 40 | 60 |
Among 250 students who supported proposal A, 110 + 50 = 160 students preferred Sunita for student union president which is equivalent to 64%.
Answer: 64.
Workspace:
What percentage of the students surveyed who did not support proposal A preferred Ragini as student union president?
Answer: 84
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Explanation :
Sunita | Ragini | |
Only proposal A | k | p |
Only proposal B | l | q |
Both A and B | m | r |
Neither A nor B | n | s |
From (1), k + m + p + r = 250 and l + m + q + r = 250
From (4), l + m = 0.2 × 250 = 50
From (6), l = 0 ⇒ m = 50
∴ q + r = 200
From (2), k + l + m + n = 200 and hence p + q + r + s = 300
From (3), p + r = 0.3 × 300 = 90 ⇒ k + m = 160
As l = 0 and k + l + m + n = 200, n = 40
∴ k = 110
From (7), n + s = 100 ⇒ s = 60
∴ p + q + r = 240 and p + r = 90 ⇒ q = 150
∴ q + r = 200 ⇒ r = 50 and hence p = 40
Thus, we have
Sunita | Ragini | |
Only proposal A | 110 | 40 |
Only proposal B | 0 | 150 |
Both A and B | 50 | 50 |
Neither A nor B | 40 | 60 |
250 students did not support proposal A. Out of these, 150 + 60 = 210 students preferred Ragini as student union president, which is equivalent to 84%.
Answer: 84.
Workspace:
How many of the students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president?
- A.
200
- B.
40
- C.
150
- D.
210
Answer: Option C
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Explanation :
From (1), k + m + p + r = 250 and l + m + q + r = 250
From (4), l + m = 0.2 × 250 = 50
From (6), l = 0 ⇒ m = 50
∴ q + r = 200
From (2), k + l + m + n = 200 and hence p + q + r + s = 300
From (3), p + r = 0.3 × 300 = 90 ⇒ k + m = 160
As l = 0 and k + l + m + n = 200, n = 40
∴ k = 110
From (7), n + s = 100 ⇒ s = 60
∴ p + q + r = 240 and p + r = 90 ⇒ q = 150
∴ q + r = 200 ⇒ r = 50 and hence p = 40
Thus, we have
Out of 250 students who supported proposal B, 150 did not support proposal A and preferred Ragini as student union president.
Hence, option (c).
Workspace:
Answer the following questions based on the information given below.
Three doctors, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges Rs. 100/-. Dr. Kane sees each patient for 15 minutes and charges Rs. 200/-, while Dr. Wayne sees each patient for 25 minutes and charges Rs. 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.
The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.
What is the maximum number of patients that the clinic can cater to on any single day?
- A.
31
- B.
15
- C.
30
- D.
12
Answer: Option A
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Explanation :
The clinic is open from 9 a.m. to 11:30 a.m. i.e., for 150 minutes.
In these 150 minutes, Dr. Ben can see 15 patients, Dr Kale can see 10 patients and Dr Wayne can see 6 patients.
Thus, the clinic can cater to 15 + 10 + 6 = 31 patients.
Hence, option (a).
Workspace:
The queue is never empty on one particular Saturday. Which of the three doctors would earn the maximum amount in consultation charges on that day?
- A.
Dr. Wayne
- B.
Dr. Kane
- C.
Both Dr. Wayne and Dr. Kane
- D.
Dr. Ben
Answer: Option B
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Explanation :
The clinic is open from 9 a.m. to 11:30 a.m. i.e., for 150 minutes.
As the queue is never empty, in these 150 minutes, Dr. Ben can see 15 patients, Dr Kale can see 10 patients and Dr Wayne can see 6 patients and thereby earn Rs. 1,500, Rs. 2,000 and Rs. 1,800 respectively.
i.e. Dr. Kane earns the maximum amount in consultation charges on that day.
Hence, option (b).
Workspace:
Mr. Singh visited the clinic on Monday, Wednesday, and Friday of a particular week, arriving at 8:50 a.m. on each of the three days. His token number was 13 on all three days. On which day was he at the clinic for the maximum duration?
- A.
Wednesday
- B.
Monday
- C.
Friday
- D.
Same duration on all three days
Answer: Option B
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Explanation :
In first 50 minutes, Dr. Ben sees 5patients; Dr. Kane sees 3 patients and he must be with his four patients while Dr. Wayne sees 2 patients.
These must be first 11 patients. Now about patients with token numbers 12 and 13.
On Monday patient with token number 13 will go to Dr. Wayne, on Wednesday and Friday he will go to Dr. Ben.
Hence, on Monday, he will spend 10 + 50 + 25 = 85 minutes, while on the remaining two days, he will spend 10 + 50 + 10 = 70 minutes.
So, Mr. Singh was at the clinic for the maximum duration on Monday.
Hence, option (b).
Workspace:
On a slow Thursday, only two patients are waiting at 9 a.m. After that two patients keep arriving at exact 15 minute intervals starting at 9:15 a.m. -- i.e. at 9:15 a.m., 9:30 a.m., 9:45 a.m. etc. Then the total duration in minutes when all three doctors are simultaneously free is
- A.
0
- B.
30
- C.
10
- D.
15
Answer: Option A
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Explanation :
Patient with token no. 1 will be seen by Dr. Wayne for 25 minutes and that with token number 2 will be seen by Dr. Ben for 10 minutes.
Now the two patients those entered at 9:15 a.m were given tokens with numbers 3 and 4.
These will be seen by Dr. Ben and Dr. Kane respectively.
As Dr. Kane takes 15 minutes to examine one patient, he will be free at 9:30 am.
Thus, in first half an hour three doctors will not be simultaneously free.
This is true for duration of 30 minutes henceforth.
Thus the total duration in minutes when all three doctors are simultaneously free will be 0.
Hence, option (a).
Workspace:
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