CAT — Arithmetic DI
35 questions, free to view. Click any question to see the answer and explanation.
Answer the questions on the basis of the information given below.
Over the top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii) Elder, and iii) Others.
Some of the subscribers used one app and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the 'Kid' and 'Elder' categories.
The following additional facts are known about the numbers of subscribers.
1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 1/2 of the subscribers from the 'Kid' category and 2/3 of the subscribers from the 'Elder' category subscribers use one app.
3. In 2023, the number of subscribers from the 'Kid' category who used multiple apps was the same as the number of subscribers from the 'Elder' category who used one app.
4. 10,000 subscribers from the 'Kid' category used one app and 15,000 subscribers from the 'Elder' category used multiple apps in 2023.
How many subscribers belonged to the 'Others' category in 2024?
45000
65000
55000
Cannot be determined
55000
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
In 2024 the number of people in others category = 55000.
What percentage of subscribers in the 'Kid' category used multiple apps in 2023?
33.33%
50.00%
5.00%
25.50%
33.33%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The percentage of kids using multiple apps in 2023 is:
= (5000 / 15000) × 100
= 33.33%
What was the percentage increase in the number of subscribers in the 'Elder' category from 2023 to 2024?
60%
50%
65%
40%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The percentage increase in the number of elders from 2023 to 2024 is:
= ((33,000 − 20,000) / 20,000) × 100
= 65%
What could be the minimum percentage of subscribers who used multiple apps in 2024?
20.0%
10.0%
16.5%
22.00%
20.0%
Step 1:
From condition (4), the number of kids using one app in 2023 is 10,000, while the number of elders using multiple apps is 15,000.
According to condition (3), the number of elders using one app is equal to the number of kids using multiple apps. Let this common value be X.
From the graph, in 2023, kids account for 15% of the users and elders account for 20%. Therefore, the remaining 65% correspond to others.
Using this information,
(10000 + X) / (15000 + X) = 15 / 20
⇒ X = 5000
Hence,
- Kids (2023) = 10,000 + 5,000 = 15,000
- Elders (2023) = 5,000 + 15,000 = 20,000
- Others (2023) = 65,000
Thus, the total number of users in 2023 is:
15,000 + 20,000 + 65,000 = 100,000
From condition (1), the total number of users in 2024 is 10% higher than in 2023.
Therefore,
Total users in 2024 = 100,000 × 1.10 = 110,000
Using the bar graph, the category-wise totals for 2024 are:
- Kids = 22,000
- Elders = 33,000
- Others = 55,000
The complete values are:
2023
- Kids: One app = 10,000, Multiple apps = 5,000, Total = 15,000
- Elders: One app = 5,000, Multiple apps = 15,000, Total = 20,000
- Others: Total = 65,000
- Overall total = 100,000
2024
- Kids: One app = 11,000, Multiple apps = 11,000, Total = 22,000
- Elders: One app = 22,000, Multiple apps = 11,000, Total = 33,000
- Others: Total = 55,000
- Overall total = 110,000
The minimum number of people using multiple apps in 2024 is:
= 11,000 + 11,000 + 0
= 22,000
Therefore, the required percentage is:
= (22,000 / 110,000) × 100
= 20%
The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100 grams of nutrients in seven foodgrains. The first column shows the foodgrain category and the second column its codename. The table has some missing values.
The table has some missing values.
The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the foodgrains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the foodgrains are non-zero multiples of 4.
5. P1 contained double the amount of protein that M3 contains.
How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
From the completed table, M1 contains 62 g of carbohydrates per 100 g.
The food grains with a carbohydrate content greater than 62 g per 100 g are:
- C1
- C2
- M2
- P1
- P2
Thus, 5 food grains have a higher carbohydrate content than M1.
Hence, the correct answer is 5.
How many grams of protein were there in 100 grams of nutrients in M2?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
In 100 g of M2, there were 12 g of proteins.
How many grams of other nutrients were there in 100 grams of nutrients in M3?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
In 100 g of M3 there were 24 g of other nutrients.
What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?
Step 1:
From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.
According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.
From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.
If 2x = 24, then the fat content in P1 becomes:
100 − 66 − 24 − 10 = 0,
which is not a valid non-zero multiple of 4.
Similarly, if 2x = 16, then the fat content in P1 becomes:
100 − 66 − 16 − 10 = 8,
forcing x = 8 and leaving the remaining nutrients in M3 as:
100 − 56 − 8 − 12 = 24,
which is feasible.
Hence, x = 8.
Step 2:
From condition (3), all unknown carbohydrate values are non-zero multiples of 5.
Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.
(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)
For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.
However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.
From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.
Since P2 has 70 g of carbohydrates, comparing C1 and C2:
- C1 must have 80 g of carbohydrates, giving 8 g of protein.
- C2 must have 75 g of carbohydrates, giving 12 g of protein.
The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.
The possible pairs are:
- (4, 24)
- (8, 20)
- (12, 16)
in any order.
Hence, the final values are:
- C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
- C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
- M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
- M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
- M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
- P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
- P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8
The median is the value that lies in the middle when all observations are arranged in ascending or descending order.
The protein content (in grams per 100 g) of the seven food grains is:
8, 8, 10, 12, 12, 14, 16
The middle (4th) observation is 12.
Hence, the median protein content is 12 g.
An online e-commerce firm receives daily integer product ratings from 1 through 5 given by buyers. The daily average is the average of the ratings given on that day. The cumulative average is the average of all ratings given on or before that day.
The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively. The distribution of ratings on Day 2 is given in the figure below.
The following information is known about ratings on Day 3.
1. 100 buyers gave product ratings on Day 3.
2. The modes of the product ratings were 4 and 5.
3. The numbers of buyers giving each product rating are non-zero multiples of 10.
4. The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.
How many buyers gave ratings on Day 1?
What is the daily average rating of Day 3?
3.2
3.5
3.0
3.6
3.6
What is the median of all ratings given on Day 3?
Which of the following is true about the cumulative average ratings of Day 2 and Day 3?
The cumulative average of Day 3 decreased from Day 2.
The cumulative average of Day 3 increased by more than 8% from Day 2.
The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
The cumulative average of Day 3 increased by less than 5% from Day 2.
The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
The following information is also known.
1. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
2. The total score on Day 3 is the same as the total score on Day 4.
3. Bimal’s scores are the same on Day 1 and Day 3.
What is Akhil's score on Day 1?
6
5
7
8
7
Step 1:
Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.
From the given information,
- D1 + D2 = 30
- D2 + D3 = 31
- D3 + D4 = 32
- D4 + D5 = 34
Using condition (2),
D3 = D4 = 16
Therefore,
- D1 = 15
- D2 = 15
- D3 = 16
- D4 = 16
- D5 = 18
From conditions (1) and (3), let the unknown scores be represented as a, b, and c.
Since the total score on Day 3 is 16,
2a + c = 16
Also, c > a.
The possible pairs are (4, 8) and (5, 6).
Since all of Chatur's scores are multiples of 3, the only valid choice is:
a = 5, c = 6
This also implies that Chatur's unique highest score occurs on Day 2, so
b = 3.
Step 2:
The known scores are:
- Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
- Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
- Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6
Since Chatur's score of 9 is the unique highest score,
9 > x > y
and
q > 6 > p.
Using the daily totals,
x + y = 6
and
p + q = 12.
The possible values are:
- (x, y) = (5, 1) or (4, 2)
- (p, q) = (5, 7) or (4, 8)
Who attains the maximum total score?
Akhil
Chatur
Bimal
Cannot be determined
Chatur
Step 1:
Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.
From the given information,
- D1 + D2 = 30
- D2 + D3 = 31
- D3 + D4 = 32
- D4 + D5 = 34
Using condition (2),
D3 = D4 = 16
Therefore,
- D1 = 15
- D2 = 15
- D3 = 16
- D4 = 16
- D5 = 18
From conditions (1) and (3), let the unknown scores be represented as a, b, and c.
Since the total score on Day 3 is 16,
2a + c = 16
Also, c > a.
The possible pairs are (4, 8) and (5, 6).
Since all of Chatur's scores are multiples of 3, the only valid choice is:
a = 5, c = 6
This also implies that Chatur's unique highest score occurs on Day 2, so
b = 3.
Step 2:
The known scores are:
- Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
- Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
- Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6
Since Chatur's score of 9 is the unique highest score,
9 > x > y
and
q > 6 > p.
Using the daily totals,
x + y = 6
and
p + q = 12.
The possible values are:
- (x, y) = (5, 1) or (4, 2)
- (p, q) = (5, 7) or (4, 8)
What is the minimum possible total score of Bimal?
Step 1:
Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.
From the given information,
- D1 + D2 = 30
- D2 + D3 = 31
- D3 + D4 = 32
- D4 + D5 = 34
Using condition (2),
D3 = D4 = 16
Therefore,
- D1 = 15
- D2 = 15
- D3 = 16
- D4 = 16
- D5 = 18
From conditions (1) and (3), let the unknown scores be represented as a, b, and c.
Since the total score on Day 3 is 16,
2a + c = 16
Also, c > a.
The possible pairs are (4, 8) and (5, 6).
Since all of Chatur's scores are multiples of 3, the only valid choice is:
a = 5, c = 6
This also implies that Chatur's unique highest score occurs on Day 2, so
b = 3.
Step 2:
The known scores are:
- Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
- Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
- Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6
Since Chatur's score of 9 is the unique highest score,
9 > x > y
and
q > 6 > p.
Using the daily totals,
x + y = 6
and
p + q = 12.
The possible values are:
- (x, y) = (5, 1) or (4, 2)
- (p, q) = (5, 7) or (4, 8)
25 is the minimum possible total score of Bimal when y = 1 and q = 7.
If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?
5
4
6
Cannot be determined
4
Step 1:
Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.
From the given information,
- D1 + D2 = 30
- D2 + D3 = 31
- D3 + D4 = 32
- D4 + D5 = 34
Using condition (2),
D3 = D4 = 16
Therefore,
- D1 = 15
- D2 = 15
- D3 = 16
- D4 = 16
- D5 = 18
From conditions (1) and (3), let the unknown scores be represented as a, b, and c.
Since the total score on Day 3 is 16,
2a + c = 16
Also, c > a.
The possible pairs are (4, 8) and (5, 6).
Since all of Chatur's scores are multiples of 3, the only valid choice is:
a = 5, c = 6
This also implies that Chatur's unique highest score occurs on Day 2, so
b = 3.
Step 2:
The known scores are:
- Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
- Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
- Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6
Since Chatur's score of 9 is the unique highest score,
9 > x > y
and
q > 6 > p.
Using the daily totals,
x + y = 6
and
p + q = 12.
The possible values are:
- (x, y) = (5, 1) or (4, 2)
- (p, q) = (5, 7) or (4, 8)
Total score of Bimal is a multiple of 3 when y = 2 and q = 8.
Then, score of Akhil on Day 2 will be 4.
If Akhil attains a total score of 24, then what is the total score of Bimal?
Step 1:
Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.
From the given information,
- D1 + D2 = 30
- D2 + D3 = 31
- D3 + D4 = 32
- D4 + D5 = 34
Using condition (2),
D3 = D4 = 16
Therefore,
- D1 = 15
- D2 = 15
- D3 = 16
- D4 = 16
- D5 = 18
From conditions (1) and (3), let the unknown scores be represented as a, b, and c.
Since the total score on Day 3 is 16,
2a + c = 16
Also, c > a.
The possible pairs are (4, 8) and (5, 6).
Since all of Chatur's scores are multiples of 3, the only valid choice is:
a = 5, c = 6
This also implies that Chatur's unique highest score occurs on Day 2, so
b = 3.
Step 2:
The known scores are:
- Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
- Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
- Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6
Since Chatur's score of 9 is the unique highest score,
9 > x > y
and
q > 6 > p.
Using the daily totals,
x + y = 6
and
p + q = 12.
The possible values are:
- (x, y) = (5, 1) or (4, 2)
- (p, q) = (5, 7) or (4, 8)
The two plots below give the following information about six firms A, B, C, D, E, and F for 2019 and 2023.
PAT: The firm’s profits after taxes in Rs. crores,
ES: The firm’s employee strength, that is the number of employees in the firm, and
PRD: The percentage of the firm’s PAT that they spend on Research and Development (R&D).
In the plots, the horizontal and vertical coordinates of point representing each firm gives their ES and PAT values respectively. The PRD values of each firm are proportional to the areas around the points representing each firm. The areas are comparable between the two plots, i.e., equal areas in the two plots represent the same PRD values for the two years.
Assume that the annual rate of growth in PAT over the previous year (ARG) remained constant over the years for each of the six firms. Which among the firms A, B, C, and E had the highest ARG?
Firm B
Firm C
Firm A
Firm E
Firm E
Since the annual growth rate is assumed to remain constant, it is sufficient to compare the increase in each firm's PAT between 2019 and 2023.
- Firm A: 3000 → 3900, an increase of 900
- Firm B: 2800 → 3800, an increase of 1000
- Firm C: 2400 → 3000, an increase of 600
- Firm E: 2400 → 3500, an increase of 1100
Among these, Firm E records the highest increase in PAT.
Hence, Firm E has the highest annual growth rate.
The ratio of the amount of money spent by Firm C on R&D in 2019 to that in 2023 is closest to
9 : 4
5 : 9
9 : 5
5 : 6
9 : 5
This question requires a bit of approximation because the bubbles are not completely enclosed within the axes. In such cases, the best approach is to use the most visible bubbles for comparison.
For example, the bubble corresponding to Firm C has an approximate vertical diameter of 3 units in 2019, whereas in 2023 its diameter is about 2 units.
Therefore, the ratio of R&D expenditure in 2019 to that in 2024 is:
= (2400 × (1.5)²) / (3000 × (1)²)
= (8/10) × (9/4)
= 9 : 5
Which among the firms A, C, E, and F had the maximum PAT per employee in 2023?
Firm A
Firm F
Firm E
Firm C
Firm C
The PAT per employee for each firm in 2023 is calculated as follows:
- Firm A: 3900 / 1300 = 3
- Firm C: 3000 / 800 = 3.75
- Firm E: 3500 / 1400 = 2.5
- Firm F: 3200 / 1000 = 3.2
Among these, the highest PAT per employee is 3.75, achieved by Firm C.
Hence, the required answer is Firm C.
Which among the firms C, D, E, and F had the least amount of R&D spending per employee in 2023?
Firm D
Firm C
Firm E
Firm F
Firm D
The PAT per employee for Firms F, C, and E was calculated in the previous question.
Now, for Firm D:
PAT per employee = 2400 / 800 = 3
Among Firms F, C, and D, the bubble sizes are identical. Therefore, the only quantity that needs to be compared is the PAT per employee.
- Firm F = 3.2
- Firm C = 3.75
- Firm D = 3
Thus, Firm D has the lowest PAT per employee among these three.
Now compare Firm D with Firm E by accounting for the proportionality between bubble area and R&D expenditure.
For Firm E:
- PAT per employee = 2.5
- Bubble radius = 1.5 units
R&D expenditure per employee:
= (5/2) × π × (3/2)²
= π × 25/8
For Firm D:
- PAT per employee = 3
- Bubble radius = 1 unit
R&D expenditure per employee:
= 3 × π × (1)²
= 3π
Since 25π/8 > 3π, Firm E has the higher R&D expenditure per employee.
Hence, Firm D had the lowest R&D spending per employee in 2023.
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
For which firm(s) can the amounts raised by them be concluded with certainty in each year?
Only Bzygoo and Czechy and Drjbna
Only Drjbna
Only Czechy and Drjbna
Only Czechy
Only Czechy and Drjbna
Step 1:
Let us determine the annual amount of money raised by each firm.
Brzygoo:
There are two possible distributions of the funds raised (in crores):
Case 1
- 2012: 1
- 2013: 2
- 2014: 3
- 2015: 1
Case 2
- 2012: 1
- 2013: 3
- 2014: 2
- 2015: 1
Czechy:
The amounts raised are:
- 2013: 1
- 2014: a
- 2015: b
- 2016: c
- 2017: 1
Since the total amount raised is 9 crores,
a + b + c = 9 − 2 = 7
The only possible values satisfying this condition are:
(a, b, c) = (2, 3, 2)
(No other arrangement is possible, as changing the last year of existence would not satisfy the total amount of 7.)
Drjbna:
The yearly amounts raised (in crores) are:
- 2011: 1
- 2012: 2
- 2013: 4
- 2014: 2
- 2015: 1
Hence, the exact yearly amounts can be uniquely determined only for Czechy and Drjbna.
What best can be concluded about the total amount of money raised in 2015?
It is either Rs. 7 crores or Rs. 8 crores.
It is either Rs. 8 crores or Rs. 9 crores.
It is exactly Rs. 8 crores.
It is either Rs. 7 crores or Rs. 8 crores or Rs. 9 crores.
It is either Rs. 7 crores or Rs. 8 crores.
Step 1:
Let us determine the possible annual amounts raised by Alfloo and Elavalaki.
Alfloo:
Assume the yearly amounts raised (in crores) are:
- 2009: 1
- 2010: 3
- 2011: 4
- 2012: 5
- 2013: 4
- 2014: 3
- 2015: 2
- 2016: 1
This gives the minimum possible amount in 2010, assuming 2015 = 2 crores.
However, the total comes to 23 crores, which is not feasible.
Therefore, the amount raised in 2010 must increase by 1 crore.
The possible distributions are:
Case 1
- 2009: 1
- 2010: 2
- 2011: 3
- 2012: 4
- 2013: 5
- 2014: 3
- 2015: 2
- 2016: 1
Case 2
- 2009: 1
- 2010: 2
- 2011: 3
- 2012: 5
- 2013: 4
- 2014: 3
- 2015: 2
- 2016: 1
Note: If the amount raised in 2011 were 4 crores, the minimum possible total would be 22 crores (sequence 1-2-4-5-4-3-2-1), which is also not feasible.
Elavalaki:
The possible yearly amounts are:
Case 1
Sub-case 1
- 2010: 1
- 2011: 2
- 2012: 3
- 2013: 4
- 2014: 2
- 2015: 1
Sub-case 2
- 2010: 1
- 2011: 2
- 2012: 4
- 2013: 3
- 2014: 2
- 2015: 1
Case 2
- 2010: 1
- 2011: 3
- 2012: 5
- 2013: 3
- 2014: 1
Hence, the amount of money raised in 2015 could be:
= 2 + 1 + 3 + 1 + 1 or 0
= 7 crores or 8 crores.
What is the largest possible total amount of money (in Rs. crores) that could have been raised in 2013?
The largest possible amount of money raised in 2013 (in Rs. crores) by each firm is:
- Alfloo: 5
- Brzygoo: 3
- Czechy: 1
- Drjbna: 4
- Elavalaki: 4
Therefore, the maximum total amount raised in 2013 is:
= 5 + 3 + 1 + 4 + 4
= 17 crores
Hence, the required answer is 17 crores.
If Elavalaki raised Rs. 3 crores in 2013, then what is the smallest possible total amount of money (in Rs. crores) that could have been raised by all the companies in 2012?
9
11
12
10
11
If Elavalaki raised Rs. 3 crore in 2013, the following two cases are possible:
Case 1
- 2010: 1
- 2011: 3
- 2012: 5
- 2013: 3
- 2014: 1
- 2015: 0
Case 2
- 2010: 1
- 2011: 2
- 2012: 4
- 2013: 3
- 2014: 2
- 2015: 1
Accordingly, the minimum possible amount of money raised by all five companies in 2012 is:
= 4 + 1 + 0 + 2 + 4
= 11 crores.
If the total amount of money raised in 2014 is Rs. 12 crores, then which of the following is not possible?
Alfloo raised the same amount of money as Bzygoo in 2014.
Alfloo raised the same amount of money as Drjbna in 2013.
Bzygoo raised the same amount of money as Elavalaki in 2013.
Bzygoo raised more money than Elavalaki in 2014.
Bzygoo raised the same amount of money as Elavalaki in 2013.
If the total amount of money raised in 2014 is Rs. 12 crores, the possible amounts raised by the firms in 2014 are:
- Alfloo: 3
- Brzygoo: 2 or 3
- Czechy: 2
- Drjbna: 2
- Elavalaki: 2 or 1
These values add up to the required total of 12 crores.
Under this arrangement, the statement "Brzygoo raised the same amount of money as Elavalaki in 2013" cannot be satisfied.
Hence, the given statement is not possible.
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars = 0.5, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order. When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
What is the sum of Travel Costs for all travelers in Zentars?
How many Zentars did Lian spend in the two countries he visited?
What was Jano's total spend in the two countries he visited, in Aurels?
One Brin is equivalent to how many Crowns?
0.5
0.125
4
8
4
Which of the following statements is NOT true about money spent in the local currency?
Jano spent 2000 in Aurevia
Lian spent 2000 in Cyrenia
Jano spent 2000 in Cyrenia
Kira spent 1000 in Aurevia
Jano spent 2000 in Aurevia
Out of 10 countries -- Country 1 through Country 10 -- Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
• Column 1 gives the country's identity.
• Column 2 gives the country's GDP as a fraction of the GDP of Country 9.
• Column 3 gives the country's GDP per capita as a fraction of the GDP per capita of Country 10.
• Column 4 gives the country's annual GDP growth rate.
• Column 5 gives the country's annual population growth rate.
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
Which one among the countries 1 through 8, has the smallest population in 2024?
Country 5
Country 8
Country 3
Country 7
Country 8
Let the GDP of Country 9 be x and its per capita GDP be y.
Then, the population of each country can be expressed as a multiple of x/y:
- Country 5: 10/36
- Country 8: 7/41
- Country 3: 13/20
- Country 7: 8/30
Comparing these fractions, 7/41 is the smallest.
Hence, Country 8 had the smallest population in 2024.
The ratio of Country 4's GDP to Country 5's GDP in 2026 will be closest to
1.314
1.195
0.963
1.032
1.195
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
The ratio of the GDPs of Country 4 and Country 5 in 2026 is:
= (0.12 × 1.005²) / (0.10 × 1.007²)
≈ 1.195
Hence, the required ratio is 1.195.
Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?
Country 4
Country 1
Country 7
Country 5
Country 1
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
The population of each of the given countries can be expressed as a multiple of x/y:
- Country 4: (12/38) × 1.005³
- Country 1: (15/41) × 0.999³
- Country 7: (8/30) × 0.999³
- Country 5: (10/36) × 1.003¹³
Comparing these values, Country 1 has the largest population.
Hence, the correct answer is Country 1.
For how many countries among Country 1 through Country 8 will the GDP per capita in 2027 be lower than that in 2024?
Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.
For each country listed, the GDP growth rate exceeds the population growth rate.
Since the growth in GDP is higher than the growth in population, the per capita GDP of every listed country will be higher in 2027 than it was in 2024.
Hence, none of the given countries will have a lower per capita GDP in 2027 compared to 2024.
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