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Test: Pie Chart- 2 (June 27) - CAT MCQ


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10 Questions MCQ Test - Test: Pie Chart- 2 (June 27)

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Test: Pie Chart- 2 (June 27) - Question 1

Directions for Questions: Answer the questions on the basis of the information given below.
Mr. Alfonso has six cars such that each car is of a different brand. The cars with Mr. Alfonso are of the six brands Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai. In the months of January and February in the year 2008, Mr. Alfonso drove exactly one car each day. The following table gives details about the days in January and February 2008 on which he did not drive a car of each of the given six brands. Given that January 1, 2008 was a Tuesday.

The number of days in January and February 2008 on which he drove a car of brand Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai is denoted by CH, FE, HO, ME, BM and HY respectively.
It is also known that HO > ME > CH > BM > HY > FE.
The following pie – chart gives details about the number of days in January and February 2008 on which he drove a car of brand Honda, BMW and Hyundai.

(2014)

Q. Given that the number of days in January 2008 on which Mr. Alfonso drove the car of brand Honda is the maximum. What is the number of days in February 2008 on which he drove the car of brand Honda?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 1

The total number of different days of the week in January and February 2008 are as follows:
Mondays: 8 (4 each in January and February)
Tuesdays: 9 (5 in January and 4 in February)
Wednesdays: 9 (5 in January and 4 in February)
Thursdays: 9 (5 in January and 4 in February)
Fridays: 9 (4 in January and 5 in February)
Saturdays: 8 (4 each in January and February)
Sundays: 8 (4 each in January and February)
Total number of days in January and February 2008 = 60 days
Total number of days on which he drove the car of brand
Honda = 25% of 60 = 60 x 25 / 100 = 15 days
Total number of days on which he drove the car of brand
BMW = 15% of 60 = 60 x 15 / 100 = 9 days
Total number of days on which he drove the car of brand
Hyundai = 10% of 60 = 60 x 10 / 100 = 6 days.
Here, HO > ME > CH > BM > HY > FE (given)
So,we get that 15 > ME > CH > 9 > 6 > FE.
Then, ME + CH + FE= 60 – (15 + 9 + 6) = 30 days
Here that the number of days on which Mr. Alfonso drove the car of brand in January 2008 is the maximum.
Now, the car of brand Honda is only driven on either of the three days of any week, i.e. Monday, Tuesdays and Saturdays.
Total number of Mondays, Tuesdays and Saturdays in January 2008
= 4 + 5 + 4 = 13 days
Total number of days in February 2008 on which he drove the car of brand Honda is = 15 – 13 = 2 days

Test: Pie Chart- 2 (June 27) - Question 2

Directions for Questions: Answer the questions on the basis of the information given below.
Mr. Alfonso has six cars such that each car is of a different brand. The cars with Mr. Alfonso are of the six brands Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai. In the months of January and February in the year 2008, Mr. Alfonso drove exactly one car each day. The following table gives details about the days in January and February 2008 on which he did not drive a car of each of the given six brands. Given that January 1, 2008 was a Tuesday.

The number of days in January and February 2008 on which he drove a car of brand Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai is denoted by CH, FE, HO, ME, BM and HY respectively.
It is also known that HO > ME > CH > BM > HY > FE.
The following pie – chart gives details about the number of days in January and February 2008 on which he drove a car of brand Honda, BMW and Hyundai.

(2014)



Q. If Mr. Alfonso drove the car of brand Chevrolet on all possible Saturdays in January 2008 and all possible Thursdays in January 2008 and February 2008, then what is the number of days on which he drove Ferrari in the given two months?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 2

The total number of different days of the week in January and February 2008 are as follows:
Mondays: 8 (4 each in January and February)
Tuesdays: 9 (5 in January and 4 in February)
Wednesdays: 9 (5 in January and 4 in February)
Thursdays: 9 (5 in January and 4 in February)
Fridays: 9 (4 in January and 5 in February)
Saturdays: 8 (4 each in January and February)
Sundays: 8 (4 each in January and February)
Total number of days in January and February 2008 = 60 days
Total number of days on which he drove the car of brand Honda = 25% of 60 = 60 x 25 / 100 = 15 days
Total number of days on which he drove the car of brand BMW = 15% of 60 = 60 x 15 / 100 = 9 days
Total number of days on which he drove the car of brand
Hyundai = 10% of 60 = 60 x 10 / 100 = 6 days.
Here, HO > ME > CH > BM > HY > FE (given) So,we get that 15 > ME > CH > 9 > 6 > FE.
Then, ME + CH + FE= 60 – (15 + 9 + 6) = 30 days
Mr. Alfonso drove Chevrolet on all possible Saturdays in January 2008 i.e. 4 days and all possible Thursdays in January 2008 and February 2008 i.e. 9 days.
⇒ He drove Chevrolet on 13 days and Mercedes on 14 days as 15 > ME > CH
⇒ Number of days on which he drove Ferrari = 30 – (13 + 14) = 3 days.

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Test: Pie Chart- 2 (June 27) - Question 3

Directions for Questions: Answer the questions on the basis of the information given below.
Mr. Alfonso has six cars such that each car is of a different brand. The cars with Mr. Alfonso are of the six brands Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai. In the months of January and February in the year 2008, Mr. Alfonso drove exactly one car each day. The following table gives details about the days in January and February 2008 on which he did not drive a car of each of the given six brands. Given that January 1, 2008 was a Tuesday.

The number of days in January and February 2008 on which he drove a car of brand Chevrolet, Ferrari, Honda, Mercedes, BMW and Hyundai is denoted by CH, FE, HO, ME, BM and HY respectively.
It is also known that HO > ME > CH > BM > HY > FE.
The following pie – chart gives details about the number of days in January and February 2008 on which he drove a car of brand Honda, BMW and Hyundai.

(2014)



Q. What is the difference between the maximum and minimum possible number of days on which Mr. Alfonso drove the car of brand Mercedes in the given two months?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 3

The total number of different days of the week in January and February 2008 are as follows:
Mondays: 8 (4 each in January and February)
Tuesdays: 9 (5 in January and 4 in February)
Wednesdays: 9 (5 in January and 4 in February)
Thursdays: 9 (5 in January and 4 in February)
Fridays: 9 (4 in January and 5 in February)
Saturdays: 8 (4 each in January and February)
Sundays: 8 (4 each in January and February)
Total number of days in January and February 2008 = 60 days
Total number of days on which he drove the car of brand Honda = 25% of 60 = 60 x 25 / 100 = 15 days
Total number of days on which he drove the car of brand BMW = 15% of 60 = 60 x 15 / 100 = 9 days
Total number of days on which he drove the car of brand
Hyundai = 10% of 60 = 60 x 10 / 100 = 6 days.
Here, HO > ME > CH > BM > HY > FE (given) So,we get that 15 > ME > CH > 9 > 6 > FE.
Then, ME + CH + FE= 60 – (15 + 9 + 6) = 30 days
Maximum number of days on which Mr. Alfonso can drove the Mercedes in two months = 14 days
To minimize the number of number of days when he drove Mercedes in the given two months, we will maximize the number of days on which he drove Ferrari and Chevrolet.
He could drove Ferrari for a maximum of 5 days as FE < 6
⇒ ME + CH = 25 and ME > CH
⇒The minimum value of ME =13 days Difference between maximum and minimum value of ME = 14 – 13 = 1 day

Test: Pie Chart- 2 (June 27) - Question 4

Directions for Questions: Answer the questions on the basis of the information given below.
The subject wise breakup of the marks obtained by 4 students in 5 subjects during their board examination is given below. Assume that all subjects carry equal maximum marks unless specified.

(2014)


Q. If the minimum percentage required to pass in any subject is 33.33 and it is also known that Arya passed in all the subjects, then what can be the minimum percentage that she can score in English?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 4

∵ Minimum percentage required to pass in a subject = 33.33%
Thus, 40° out of 360° represents 33.33%
⇒ 60° represents 50%.
∴ So, minimum percentage scored by Arya = 50%.

Test: Pie Chart- 2 (June 27) - Question 5

Directions for Questions: Answer the questions on the basis of the information given below.
The subject wise breakup of the marks obtained by 4 students in 5 subjects during their board examination is given below. Assume that all subjects carry equal maximum marks unless specified.

(2014)


Q. If the pattern of the examination is changed in such a way that the maximum marks for Maths is double of the maximum marks for any other subject (the breakup of marks remaining unchanged), then what is the ratio of maximum marks, in all the subjects put together, which Geoffrey can score to the maximum marks, in all the subjects put together, which Tommen can score?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 5

Let maximum marks for each subject other than maths. = 100
∴ Maximum marks in maths = 200
For Geoffrey, 144° = 200 marks
⇒ 360° = 500 marks.
For Tommen, if 120° = 200 marks
then 72° > 100 marks,
So, this is not possible, then
∴ For Tommen, 72° = 100 marks
⇒ 360° = 500 marks
∴ Ratio of maximum marks, in all the subjects put together, by Geoffrey and Tommen = 1 : 1.

Test: Pie Chart- 2 (June 27) - Question 6

Directions for Questions: Answer the questions on the basis of the information given below.
The subject wise breakup of the marks obtained by 4 students in 5 subjects during their board examination is given below. Assume that all subjects carry equal maximum marks unless specified.

(2014)


Q. If the pattern of the examination is changed in such a way that the maximum marks for Maths is double of the maximum marks for any other subject (the breakup of marks remaining unchanged), then what is the ratio of maximum marks, in all the subjects put together, which Sansa can score after the change in pattern and before the change in pattern?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 6

Before the change in pattern of examination
For maths:
∵ 144° = 100 marks
⇒ ∴ 360° = 250 marks.
After the change in pattern of examinaiton,
80° = 100 marks
⇒ 360° = 450 marks 
∴ Required ratio = 450 : 250 = 9 : 5.

Test: Pie Chart- 2 (June 27) - Question 7

Directions for Questions: Answer the questions on the basis of the information given below.
The subject wise breakup of the marks obtained by 4 students in 5 subjects during their board examination is given below. Assume that all subjects carry equal maximum marks unless specified.

(2014)


Q. If the marks scored by Geoffrey in Maths is maximum possible then what is the average of percentage marks scored by him in all the subjects?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 7

According to question,
144° = 100 marks
⇒ 360° = 250 marks
∴ Average percentage = 250 / 500 x 100
Marks scored by Geoffrey = 50%.

Test: Pie Chart- 2 (June 27) - Question 8

Directions for Questions: Answer the questions on the basis of the information given below.
The pie charts given below show the distribution of the number of athletes sent by three countries to take part in eight different games in the recently held Olympics. The numbers of athletes sent by USA, China and Russia were in the ratio 8 : 11 : 5. It is also known that the total number of athletes sent by the three countries put together was 2400.

(2013)

Q. The female to male ratio for Cycling athletes from Russia and USA was 1 : 19 and 4 : 1 respectively. If the number of female Cycling athletes was equal to the number of male Cycling athletes for the three countries put together, then what was the number of female Cycling athletes from China?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 8

The total number of athelets sent by:
USA = 2400 × 8 / 24 = 800 athletes
China = 2400 × 11 / 24 = 1100 athletes
Russia = 2400 × 5 /24 = 500 athletes
Total number of Cycling athelets sent by the three countries together = 0.15 × 800 + 0.12 × 1100 + 0.16 × 500 = 332
∴ Total number of female cycling athletes sent by the three countries togeter = 166
Number of female cycling athletes sent by Russia
= 1 / 20 × (0.16 × 500) = 4
Number of female cycling atheletes sent by USA
= 4 / 5 × (0.15 × 800) = 96
Hence, the number of female cycling athletes sent by China =166 – (4 + 96) = 66.

Test: Pie Chart- 2 (June 27) - Question 9

Directions for Questions: Answer the questions on the basis of the information given below.
The pie charts given below show the distribution of the number of athletes sent by three countries to take part in eight different games in the recently held Olympics. The numbers of athletes sent by USA, China and Russia were in the ratio 8 : 11 : 5. It is also known that the total number of athletes sent by the three countries put together was 2400.

(2013)

Q. What was the absolute difference between the maximum number of athletes sent by China for a game and the minimum number of athletes sent by Russia for a game?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 9

The total number of athelets sent by:
USA = 2400 × 8 / 24 = 800 athletes
China = 2400 × 11 / 24 = 1100 athletes
Russia = 2400 × 5 /24 = 500 athletes
Maximum number of athletes sent by China for a game = 0.25 × 1100 = 275
Minimum number of athletes sent by Russia for a game = 0.03 × 500 = 15
∴ The required difference = 275 – 15 = 260

Test: Pie Chart- 2 (June 27) - Question 10

Directions for Questions: Answer the questions on the basis of the information given below.
The pie charts given below show the distribution of the number of athletes sent by three countries to take part in eight different games in the recently held Olympics. The numbers of athletes sent by USA, China and Russia were in the ratio 8 : 11 : 5. It is also known that the total number of athletes sent by the three countries put together was 2400.

(2013)

Q. The number of Water Polo athletes constituted what percent of the total number of athletes sent by the three countries put together?

Detailed Solution for Test: Pie Chart- 2 (June 27) - Question 10

The total number of athelets sent by:
USA = 2400 × 8 / 24 = 800 athletes
China = 2400 × 11 / 24 = 1100 athletes
Russia = 2400 × 5 /24 = 500 athletes
The total number of Water Polo athletes sent by the three countries put together = 0.15 × 800 + 0.25 × 1100 + 0.10 × 500
= 120 + 275 + 50 = 445
∴ Total number of athletes sent by three countries = 2400
So, the required percentage
= 445 / 2400 x 100 = 18.54

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