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# CAT LRDI MCQ - 3

## 32 Questions MCQ Test CAT Mock Test Series | CAT LRDI MCQ - 3

Description
This mock test of CAT LRDI MCQ - 3 for CAT helps you for every CAT entrance exam. This contains 32 Multiple Choice Questions for CAT CAT LRDI MCQ - 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this CAT LRDI MCQ - 3 quiz give you a good mix of easy questions and tough questions. CAT students definitely take this CAT LRDI MCQ - 3 exercise for a better result in the exam. You can find other CAT LRDI MCQ - 3 extra questions, long questions & short questions for CAT on EduRev as well by searching above.
*Answer can only contain numeric values
QUESTION: 1

### Group Question Answer the following question based on the information given below. Bandhan Furniture received orders from six new schools that are going to start their operations shortly. The bars in the below chart represent the number of pairs (chair + table) required per class and the line graph represents the number of classrooms in the particular school. There are two types of furniture available: type A and type B. Type A costs Rs 450 per pair while type B costs Rs 375 per pair. Schools C and E opted for type A furniture while the rest ordered type B furniture. What is the total revenue generated from the six schools (in Rs.)?

Solution:

As calculated in the table, total revenue = Rs. 29,23,500

*Answer can only contain numeric values
QUESTION: 2

### Bandhan Furniture received orders from six new schools that are going to start their operations shortly. The bars in the below chart represent the number of pairs (chair + table) required per class and the line graph represents the number of classrooms in the particular school. There are two types of furniture available: type A and type B. Type A costs Rs 450 per pair while type B costs Rs 375 per pair. Schools C and E opted for type A furniture while the rest ordered type B furniture. What percent of the furniture ordered constituted type A furniture? (Round off your answer 3 upto two decimal places.)

Solution:

Considering solution to the first question,

Total furniture ordered = 42 x 40 + 36 x 35 + 28 x 35 + 28 x 50 + 30 x 30 + 24 x 50 = 7420

Type A furniture = 28 x 35 + 30 x 30 = 1880

Required % = 1880/7420 x 100 = 25.34%

*Answer can only contain numeric values
QUESTION: 3

### Bandhan Furniture received orders from six new schools that are going to start their operations shortly. The bars in the below chart represent the number of pairs (chair + table) required per class and the line graph represents the number of classrooms in the particular school. There are two types of furniture available: type A and type B. Type A costs Rs 450 per pair while type B costs Rs 375 per pair. Schools C and E opted for type A furniture while the rest ordered type B furniture. What is the difference in the amount received from school B and school C (in Rs.)?

Solution:

From the solution to the first question, required difference = 472500 - 441000 = Rs. 31500

*Answer can only contain numeric values
QUESTION: 4

Bandhan Furniture received orders from six new schools that are going to start their operations shortly. The bars in the below chart represent the number of pairs (chair + table) required per class and the line graph represents the number of classrooms in the particular school.

There are two types of furniture available: type A and type B. Type A costs Rs 450 per pair while type B costs Rs 375 per pair. Schools C and E opted for type A furniture while the rest ordered type B furnitureSchool A made a change in their order. The school has 14 classes (LKG + UKG + grades 1 to 12), with 3 classrooms for each class. They decided to purchase Type A furniture for grades 9 to 12, while Type B for rest of the classes. By how much percent will their expense increase now? (Round off your answer upto two decimal places.)

Solution:

Previous expenditure = Rs. 630000

New expenditure = 30 x 40 x 375 + 12 x 40 x 450 = 450000 + 216000 = Rs. 666000

Percent increase in expenditure = 36000/630000 x 100 = 5.71%

QUESTION: 5

Answer the following question based on the information given below.

The census officer provided the data regarding changes in population of three major towns - A, B and C for three years. Population of town A was 180600 in third year and it increased by 5% and 7.5% in second and third year respectively. Population of town B increased by 25% in second year which was same as 150% of population of town A in first year. After taking population control measures, town B succeeded in controlling population as growth rate in third year was half of that of previous year. The area of town C is 1250 km2 and population density for second year was 250. Growth rate for town C was 11.11% and 10% for second and third year respectively.

Note : Population density = Total population / Total area. Growth rate is change in population over previous year for all towns and for all years.

1. Population of town B in third year exceeds how much by that of town A in second year?
Solution:

Let the Population of Town A in first year be 100.

Thus, population of town A in third year = 105% of 107.50% of 100 = 112.875

But, population of town A in third year = 180600

Population of Town A in first year = (180600 x 100) / 112.875 = 160000

Population of town A in second year = 105% of 160000 = 168000

Hence, Population of town B in second year = 150% of 160000 = 240000

As growth rate of population for town B in the second year was 25%, population of town B in first year = (240000 x 100) /125 = 192000

Also, growth rate becames half in third years of that of in second year.

••• Population of town B in third year = (240000 x 112.5) /100= 270000

For town C, population in second year = Population density x Area = 250 x 1250 = 312500

As growth rate for town C was 11.11% for second year, population of C in first year = (312500 x 100) / 111.11 » 281250

Also, growth rate for town C was 10% for third year, Population of C in third year = 110% of 312500 = 343750

Thus, we can present above data in tabular form as follows:

Required difference = Population of town B in third year - Population of town A in second year

= 270000 - 168000 =102000 Hence, option 3.

QUESTION: 6

The census officer provided the data regarding changes in population of three major towns - A, B and C for three years. Population of town A was 180600 in third year and it increased by 5% and 7.5% in second and third year respectively. Population of town B increased by 25% in second year which was same as 150% of population of town A in first year. After taking population control measures, town B succeeded in controlling population as growth rate in third year was half of that of previous year. The area of town C is 1250 km2 and population density for second year was 250. Growth rate for town C was 11.11% and 10% for second and third year respectively.

Note : Population density = Total population / Total area. Growth rate is change in population over previous year for all towns and for all years.

The average population of town B for three years forms what percentage of average 3 population of town C for three years?

Solution:

Considering solution to the first question,

Average population of town B = (192000 + 240000 + 270000) / 3 = 702000 / 3 = 234000

Average population of town C = (281250 + 312500 + 343750) / 3 = 937500 / 3 = 312500

Required percentages = (234000 x 100) / 312500 = 74.88 Hence, option 2.

QUESTION: 7

The census officer provided the data regarding changes in population of three major towns - A, B and C for three years. Population of town A was 180600 in third year and it increased by 5% and 7.5% in second and third year respectively. Population of town B increased by 25% in second year which was same as 150% of population of town A in first year. After taking population control measures, town B succeeded in controlling population as growth rate in third year was half of that of previous year. The area of town C is 1250 km2 and population density for second year was 250. Growth rate for town C was 11.11% and 10% for second and third year respectively.

Note : Population density = Total population / Total area. Growth rate is change in population over previous year for all towns and for all years.

For town B, male to female ratio for last two year was 7 : 5 and literate male and illiterate male are in the ratio of 4 : 1 for same years. Find the ratio between illiterate male in second year and literate male in third year.

Solution:

Considering solution to the first question,

Number of male in town B for Second year = (7 x 240000) /12 = 140000

Number of male in town B for third year = (7 x 270000) / 12 = 157500

Number of illiterate male in second year = (1 x 140000) 15 = 28000

Number of literate male in third year = (4 x 157500) 15 = 126000

Thus, required ratio = 28000 : 126000 i.e. 2 : 9 Hence, option 4.

QUESTION: 8

The census officer provided the data regarding changes in population of three major towns - A, B and C for three years. Population of town A was 180600 in third year and it increased by 5% and 7.5% in second and third year respectively. Population of town B increased by 25% in second year which was same as 150% of population of town A in first year. After taking population control measures, town B succeeded in controlling population as growth rate in third year was half of that of previous year. The area of town C is 1250 km2 and population density for second year was 250. Growth rate for town C was 11.11% and 10% for second and third year respectively.

Note : Population density = Total population / Total area. Growth rate is change in population over previous year for all towns and for all years.

In the third year, if 3/8th part of population of town A is not above 20 years, 33% of population of town B is of the age 20 years or less and 70% of population of town C is above 20 years, how much population of three towns is above 20 years in the third year?

Solution:

Considering solution to the first question, population above 20 years for

Town A = 180600 - [(180600 x 3)/8] = 112875

Town B = (100 - 33)% of 270000 = 180900

Town C = 70% of 343750 = 240625

Thus, required total = 112875 + 180900 + 240625 = 534400

Hence, option 3.

QUESTION: 9

Group Question

Answer the following question based on the information given below

The graph given above pertains to the recruitment and promotion details of a company called TFC. Every year, the company takes in new recruits, known popularly as “Nanos”, in the company for its off-shore projects. These Nanos remain Nanos for two years during which they undergo a rigorous training. Immediately after completion of two years of training, these Nanos appear for a screening test, following which they get promoted as Vets if they pass the test. For the number of Nanos that are recruited in a particular year, the number of Nanos getting converted into Vets is between 60% to 80% (only integral values of percentage), both values included. On conversion to Vets, these employees leave TFC operations in the country for on-shore projects. The Nanos who fail to get promoted as Vets continue as Nanos for one more year following which they are labeled as Vets without any test and are then transferred to its on-shore operations.

In the above chart, the innermost circle represents the year of intake of the Nanos, while the outer circles represent the numbers of Nanos and Vets for that year. Further for any given year, the number of Nanos always exceeds the number of Vets in the same year.

For example, in the year 2005, the number of Nanos that TFC had was 65 and the number of Vets in the same year was 42.

No employee can bypass the process and get recruited as a Vet directly.

The number of Nanos from the 2006 batch who were promoted as Vets without a test 3 was....

Solution:

The data that we have can be tabulated as follows:

Now, since the number of Nanos that get promoted as Vets in a year is between 60% - 80%, both included, the number of Vets so produced for each year can be represented as follows:

Now if we compare the number of Vets in year 2005, it becomes evident that the number of Nanos getting promoted as Vets cannot be 44 since there were only 42 Vets in 2005.

Hence the number of Nanos from 2003 that were promoted as Vets in 2005 was 33. This is because no other percentage will give an integral number of Nanos.

Thus, the remaining Nanos, i.e. (55 - 33) = 22 were promoted as Vets in next year, i.e. in year 2006.

Now, since in year 2006, there were a total of 70 Vets of which 22 were from the 2003 batch, the number of Nanos from 2004 who got promoted as Vets in 2006 would be (70 - 22) = 48, which is within the required range of (45 to 60) for the year 2004.

Therefore, the remaining 27 Nanos (75 - 48) will get promoted as Vets in the next year i.e. in 2007.

Now, in 2007, the number of Vets is 79, of which 27 have been promoted from the 2004 batch.

Hence, the number of Nanos from the 2005 batch who were promoted as Vets in 2007 would be (79 - 27) = 52.

Using the logic explained earlier, the remaining 65 - 52 = 13 Nanos will be promoted in 2008.

The number of Vets in 2008 is 77, of which the number 13 are Nanos promoted from the 2005 batch.

Therefore, the remaining 77 - 13 = 64 Nanos are promoted from the 2006 batch.

There are 100 Nanos in the 2006 batch.

Therefore, the remaining 100 - 64 = 36 Nanos are promoted as Vets without a test in 2009. Hence, option 3.

QUESTION: 10

The graph given above pertains to the recruitment and promotion details of a company called TFC. Every year, the company takes in new recruits, known popularly as “Nanos”, in the company for its off-shore projects. These Nanos remain Nanos for two years during which they undergo a rigorous training. Immediately after completion of two years of training, these Nanos appear for a screening test, following which they get promoted as Vets if they pass the test. For the number of Nanos that are recruited in a particular year, the number of Nanos getting converted into Vets is between 60% to 80% (only integral values of percentage), both values included. On conversion to Vets, these employees leave TFC operations in the country for on-shore projects. The Nanos who fail to get promoted as Vets continue as Nanos for one more year following which they are labeled as Vets without any test and are then transferred to its on-shore operations.

In the above chart, the innermost circle represents the year of intake of the Nanos, while the outer circles represent the numbers of Nanos and Vets for that year. Further for any given year, the number of Nanos always exceeds the number of Vets in the same year.

For example, in the year 2005, the number of Nanos that TFC had was 65 and the number of Vets in the same year was 42.

No employee can bypass the process and get recruited as a Vet directly.

The percentage of Nanos getting promoted as Vets in the first instance was highest for which of the following batches of Nanos?

Solution:

Consider the solution to the previous question. A Nano gets promoted as a Vet if the Nano is promoted at the end of the two year period.

The percentage of Nanos in each batch that are promoted as Vets can be calculated as shown below:

2003: 33 out of 5 i.e. 60%

2004: 48 out of 70 i.e. 68.55%

2005: 52 out of 65 i.e. 80%

2006: 64 out of 100 i.e. 64%

Hence, option 3.

QUESTION: 11

The graph given above pertains to the recruitment and promotion details of a company called TFC. Every year, the company takes in new recruits, known popularly as “Nanos”, in the company for its off-shore projects. These Nanos remain Nanos for two years during which they undergo a rigorous training. Immediately after completion of two years of training, these Nanos appear for a screening test, following which they get promoted as Vets if they pass the test. For the number of Nanos that are recruited in a particular year, the number of Nanos getting converted into Vets is between 60% to 80% (only integral values of percentage), both values included. On conversion to Vets, these employees leave TFC operations in the country for on-shore projects. The Nanos who fail to get promoted as Vets continue as Nanos for one more year following which they are labeled as Vets without any test and are then transferred to its on-shore operations.

In the above chart, the innermost circle represents the year of intake of the Nanos, while the outer circles represent the numbers of Nanos and Vets for that year. Further for any given year, the number of Nanos always exceeds the number of Vets in the same year.

For example, in the year 2005, the number of Nanos that TFC had was 65 and the number of Vets in the same year was 42.

No employee can bypass the process and get recruited as a Vet directly.

What was the number of Nanos who did not get promoted as Vets at the first instance from the ones present in 2005?

Solution:

Consider the solution to the first question.

From the 65 Nanos in 2005, 13 Nanos are promoted to Vets in 2008.

Thus, these are not promoted at the first instance.

Hence, option 1.

QUESTION: 12

The graph given above pertains to the recruitment and promotion details of a company called TFC. Every year, the company takes in new recruits, known popularly as “Nanos”, in the company for its off-shore projects. These Nanos remain Nanos for two years during which they undergo a rigorous training. Immediately after completion of two years of training, these Nanos appear for a screening test, following which they get promoted as Vets if they pass the test. For the number of Nanos that are recruited in a particular year, the number of Nanos getting converted into Vets is between 60% to 80% (only integral values of percentage), both values included. On conversion to Vets, these employees leave TFC operations in the country for on-shore projects. The Nanos who fail to get promoted as Vets continue as Nanos for one more year following which they are labeled as Vets without any test and are then transferred to its on-shore operations.

In the above chart, the innermost circle represents the year of intake of the Nanos, while the outer circles represent the numbers of Nanos and Vets for that year. Further for any given year, the number of Nanos always exceeds the number of Vets in the same year.

For example, in the year 2005, the number of Nanos that TFC had was 65 and the number of Vets in the same year was 42.

No employee can bypass the process and get recruited as a Vet directly.

If we consider the period from 2007 - 2009, then in which of the following years was the number of Vets the highest?

Solution:

Number of Vets in 2007 : 79 Number of Vets in 2008 : 77 Hence, option 3 can be eliminated.

The number of Nanos from the 2006 batch who got promoted as Vets in 2009 is 36.

Further as there were 80 Nanos who were recruited in 2007, atleast 60% of them, i.e. 48 of them will be promoted as Vets in 2009

Thus, the minimum number of Nanos in 2009 will be 36 + 48 i.e. 84.

Thus, the maximum number of Vets was in 2009.

Hence, option 2.

QUESTION: 13

Group Question

Answer the following question based on the information given below.

Badrinath College of Engineering (BCE) was established on 1st May 1993 on the 60th birth anniversary of Mr. Prakash Badrinath with 4, 5, 3 and 2 professors in Civil, Mechanical, Electrical and Information Technology department respectively. On 1st of June 1993 one more professor joined. On every 1st June of the subsequent years up to 1996 one professor is recruited for one of the departments with the salary of Rs. 15,000 per month. So, at the end of 4 years, 4 new professors were recruited across 4 departments. Professors who joined in May 1993 were placed in different pay scales based on their experience and qualifications, with the minimum of Rs. 15,000 per month and difference between any two pay scales is in multiples of Rs. 5,000. Each department has only one head of the department and he is paid more than the others. After completion of one year of service, each professor is given an annual increment of Rs. 3,000 effective from 1st June every year. During the period from 1st May 1993 to 31st May 1996 only one professor resigned across the four departments.

The table given below shows the annual average monthly salary of the professors of BCE for each department as on 1st June every year.

When did a new professor join the civil department?

Solution:

Every professor gets an increment of Rs. 3000 on 1st June on completion of one year, so the average salary should also increase by 3000 after a year.

There is no change in average salary from May-93 and Jun-93 in Civil, Electrical and Information Technology (IT) departments. So, the new professor with salary Rs. 15,000, must have joined Mechanical department.

Also, the increase in the salary is not 3000 for Civil department in Jun-94. So, a new professor must have joined Civil department on 1st June 1994. On the similar lines, we can conclude that a new professor with Rs. 15000 in departments Electrical and IT on 1st June 1995 and 1st June 1996 respectively. Average salary of Mechanical department was not changed by 3000 in Jun-96. So, the professort who has resigned, must have resigned from Mechanical department.

Hence, option 4.

QUESTION: 14

Badrinath College of Engineering (BCE) was established on 1st May 1993 on the 60th birth anniversary of Mr. Prakash Badrinath with 4, 5, 3 and 2 professors in Civil, Mechanical, Electrical and Information Technology department respectively. On 1st of June 1993 one more professor joined. On every 1st June of the subsequent years up to 1996 one professor is recruited for one of the departments with the salary of Rs. 15,000 per month. So, at the end of 4 years, 4 new professors were recruited across 4 departments. Professors who joined in May 1993 were placed in different pay scales based on their experience and qualifications, with the minimum of Rs. 15,000 per month and difference between any two pay scales is in multiples of Rs. 5,000. Each department has only one head of the department and he is paid more than the others. After completion of one year of service, each professor is given an annual increment of Rs. 3,000 effective from 1st June every year. During the period from 1st May 1993 to 31st May 1996 only one professor resigned across the four departments.

The table given below shows the annual average monthly salary of the professors of BCE for each department as on 1st June every year.

In which period and from which department did a professor resign? (Note: 1st June, 1993 to 31st May, 1994 is considered as period 93-94.)

Solution:

Referring the solution of the first question of the set, a professor resigned from mechanical department in 95-96.

Hence, option 3.

QUESTION: 15

Badrinath College of Engineering (BCE) was established on 1st May 1993 on the 60th birth anniversary of Mr. Prakash Badrinath with 4, 5, 3 and 2 professors in Civil, Mechanical, Electrical and Information Technology department respectively. On 1st of June 1993 one more professor joined. On every 1st June of the subsequent years up to 1996 one professor is recruited for one of the departments with the salary of Rs. 15,000 per month. So, at the end of 4 years, 4 new professors were recruited across 4 departments. Professors who joined in May 1993 were placed in different pay scales based on their experience and qualifications, with the minimum of Rs. 15,000 per month and difference between any two pay scales is in multiples of Rs. 5,000. Each department has only one head of the department and he is paid more than the others. After completion of one year of service, each professor is given an annual increment of Rs. 3,000 effective from 1st June every year. During the period from 1st May 1993 to 31st May 1996 only one professor resigned across the four departments.

The table given below shows the annual average monthly salary of the professors of BCE for each department as on 1st June every year.

What is the salary of the professor when he left the Mechanical department in the given period?

Solution:

Average salary of five professors on 1st June 1996 = 26000 /. Average salary of five professors on 1st June 1995 = 23000

Average salary of six professors on 1st June 1995 = 23500

Average salary of the sixth professor who has resigned before 1st June 1996 = 6 x 23500 - 5 x 23000 = 26000

Hence, option 2.

QUESTION: 16

Badrinath College of Engineering (BCE) was established on 1st May 1993 on the 60th birth anniversary of Mr. Prakash Badrinath with 4, 5, 3 and 2 professors in Civil, Mechanical, Electrical and Information Technology department respectively. On 1st of June 1993 one more professor joined. On every 1st June of the subsequent years up to 1996 one professor is recruited for one of the departments with the salary of Rs. 15,000 per month. So, at the end of 4 years, 4 new professors were recruited across 4 departments. Professors who joined in May 1993 were placed in different pay scales based on their experience and qualifications, with the minimum of Rs. 15,000 per month and difference between any two pay scales is in multiples of Rs. 5,000. Each department has only one head of the department and he is paid more than the others. After completion of one year of service, each professor is given an annual increment of Rs. 3,000 effective from 1st June every year. During the period from 1st May 1993 to 31st May 1996 only one professor resigned across the four departments.

The table given below shows the annual average monthly salary of the professors of BCE for each department as on 1st June every year.

Electrical department has a tradition that none of the professor gets the same salary. Professor Kulkami and Professor Rathwa are in the electrical department of the college from its inception and none from them is the head of the department. What is the salary of the third professor who joined along with Prof. Kulkami and Prof. Rathwa, for the month of June 1996?

Solution:

Total salary of 3 professors of Electrical department in 1993 = 3 x 20000 = 60000 The possible salary of all the three professors is 15000, 20000 and 25000.

The third professor must be the head of the department and must be recruited at the starting salary of Rs. 25,000 in May 1993.

Therefore, salary of the third professor of Electrical department in June 1996 = 25000 + 3000 + 3000 + 3000 = Rs. 34,000.

Hence, option 1.

*Answer can only contain numeric values
QUESTION: 17

Group Question

Answer the following question based on the information given below.

In a T20 match between New Zealand and Australia, the top 4 batsmen were Guptill, Anderson from New Zealand and Maxwell and Warner from Australia. Following information regarding them is given below:

1. All runs scored by them were in Is (Single), 4s or 6s. Each of the four players scored at least 1 six, 1 four and 1 single.
2. The ratio of 6s hit by Guptill and Anderson was 2 : 3 which is the same for Maxwell and Warner.
3. A total of fifteen 6s were hit by these players.
4. Guptil hit three more 4s than Anderson while Maxwell hit three more 4s than Warner.
5. The number of singles scored by each of them was in A.R The descending order of these numbers was Guptill, Anderson, Maxwell and Warner.
6. Warner played the highest number of dot balls (balls in which no runs were scored) which were equal to half the number of runs scored by him and 9 more than those played by Anderson.
7. Half of the balls played by Maxwell were dot balls and the number of dots played by Guptill were equal to the number of singles scored by him.
8. Maxwell scored 50 off 30 balls while Warner scored 46 off 33.
9. The number of singles taken by Guptill was twice the number of 4s hit by him.

What percentage of the balls faced by the four batsmen were dot balls? (Round off your

Solution:

From the given data the table can be filled up as shown below,

We are given that 15 sixes were hit by these players.

Hence 5m + 5n = 15 or m + n = 3 m and n can take values 1 or 2.

Warner played 23 dot balls, hence he scored 46 runs in 10 balls. Thus, there are two possible cases:

Case 1: Warner hits 3 sixes.

The equation turns down to remaining 28 from 7 balls, only possibility being 7 fours. But all batsmen took at least one single. Hence, this case is invalid.

Thus, Warner hits 6 sixes.

••• From the remaining 4 balls, he hits 2 fours and takes 2 singles.

So we get the values of m, n, a andy as 1, 2, 2 and 2 respectively.

Now, Maxwell has 50 runs off 30 balls. Hence, he scored six singles i.e. d = 4.

From (9), we can see that x = 4

Thus the complete table is as follows,

Total balls faced = 37+ 31+30+ 33 = 131 Dot balls = 14 + 14 + 15 + 23 = 66 ••• Percentage of dot balls = 66/131 x 100 = 50.38%

*Answer can only contain numeric values
QUESTION: 18

In a T20 match between New Zealand and Australia, the top 4 batsmen were Guptill, Anderson from New Zealand and Maxwell and Warner from Australia. Following information regarding them is given below:

1. All runs scored by them were in Is (Single), 4s or 6s. Each of the four players scored at least 1 six, 1 four and 1 single.
2. The ratio of 6s hit by Guptill and Anderson was 2 : 3 which is the same for Maxwell and Warner.
3. A total of fifteen 6s were hit by these players.
4. Guptil hit three more 4s than Anderson while Maxwell hit three more 4s than Warner.
5. The number of singles scored by each of them was in A.R The descending order of these numbers was Guptill, Anderson, Maxwell and Warner.
6. Warner played the highest number of dot balls (balls in which no runs were scored) which were equal to half the number of runs scored by him and 9 more than those played by Anderson.
7. Half of the balls played by Maxwell were dot balls and the number of dots played by Guptill were equal to the number of singles scored by him.
8. Maxwell scored 50 off 30 balls while Warner scored 46 off 33.
9. The number of singles taken by Guptill was twice the number of 4s hit by him.

What percentage of the runs scored by these batsmen were scored in boundaries (4s and 6s)? (Round off your answer upto two decimal places.)

Solution:

Considering the solution to the first question of the set:

Total runs = 54 + 44 + 50 + 46 = 194

Runs in boundaries =18x4+15x6 = 162

Required % = 162/194 x 100 = 83.51%

*Answer can only contain numeric values
QUESTION: 19

In a T20 match between New Zealand and Australia, the top 4 batsmen were Guptill, Anderson from New Zealand and Maxwell and Warner from Australia. Following information regarding them is given below:

1. All runs scored by them were in Is (Single), 4s or 6s. Each of the four players scored at least 1 six, 1 four and 1 single.
2. The ratio of 6s hit by Guptill and Anderson was 2 : 3 which is the same for Maxwell and Warner.
3. A total of fifteen 6s were hit by these players.
4. Guptil hit three more 4s than Anderson while Maxwell hit three more 4s than Warner.
5. The number of singles scored by each of them was in A.R The descending order of these numbers was Guptill, Anderson, Maxwell and Warner.
6. Warner played the highest number of dot balls (balls in which no runs were scored) which were equal to half the number of runs scored by him and 9 more than those played by Anderson.
7. Half of the balls played by Maxwell were dot balls and the number of dots played by Guptill were equal to the number of singles scored by him.
8. Maxwell scored 50 off 30 balls while Warner scored 46 off 33.
9. The number of singles taken by Guptill was twice the number of 4s hit by him

Strike rate is defined as the runs scored per 100 balls. What was the highest strike rate by any batsman? (Round off your answer to the nearest integer.)

Solution:

Considering the solution to the first question of the set, the highest strike rate was 167 of Marks Maxwell.

*Answer can only contain numeric values
QUESTION: 20

In a T20 match between New Zealand and Australia, the top 4 batsmen were Guptill, Anderson from New Zealand and Maxwell and Warner from Australia. Following information regarding them is given below:

1. All runs scored by them were in Is (Single), 4s or 6s. Each of the four players scored at least 1 six, 1 four and 1 single.
2. The ratio of 6s hit by Guptill and Anderson was 2 : 3 which is the same for Maxwell and Warner.
3. A total of fifteen 6s were hit by these players.
4. Guptil hit three more 4s than Anderson while Maxwell hit three more 4s than Warner.
5. The number of singles scored by each of them was in A.R The descending order of these numbers was Guptill, Anderson, Maxwell and Warner.
6. Warner played the highest number of dot balls (balls in which no runs were scored) which were equal to half the number of runs scored by him and 9 more than those played by Anderson.
7. Half of the balls played by Maxwell were dot balls and the number of dots played by Guptill were equal to the number of singles scored by him.
8. Maxwell scored 50 off 30 balls while Warner scored 46 off 33.
9. The number of singles taken by Guptill was twice the number of 4s hit by him.

What is the number of singles taken by the given four batsmen?

Solution:

Marks Considering the solution to the first question of the set,

Number of singles = 14 + 10 + 6 + 2 = 32

QUESTION: 21

Group Question

Answer the following question based on the information given below.

A cricket coaching academy has twelve rooms distributed equally on three floors -1,2 and 3. Four of the twelve rooms are unoccupied and the remaining rooms are occupied by eight cricketers falling into three different categories - batsman, bowler or allrounder. Only one cricketer stays per room and no cricketer can fall into more than one category. There are at least two and at most three cricketers per floor as well as per category. Following instruction regarding them is given below:

1. Vikas and Mahesh are on different floors, but both not on the second, and are in the same category.
2. Deepak and Rakesh are on the same floor and in the same category, but both are not allrounders.
3. Shyam and Praveen are on different floors, but not on the first and belong to different categories, but are not allrounders.
4. Sanket is on the second floor and Shyam is not on the same floor as Sanket.
5. Rahul is an allrounder and does not stay on the first floor.
6. Both players on the second floor are bowlers.
7. Among the three allrounders, two live on the third floor.
8. Rahul and Mahesh stay on the same floor and there are three batsmen.

Who among the following is neither a bowler nor a batsman?

Solution:

Ignore the four unoccupied flats on the three floors.

Since it is known that there are three allrounders and three batsmen, there can be only two bowlers.

Similarly, since both players on the second floor are bowlers, there are only two players on the second floor and three players each on the first and third floors.

Hence, there cannot be a batsman or allrounder on the second floor. Also, among the three allrounders, two live on the third floor.

Thus, the floor and category wise break-up is:

Floor 3-3 players : 2 allrounders + 1 batsman Floor 2-2 players : 2 bowlers

Floor 1-3 players : 1 allrounder + 2 batsmen

From 3 and 4, Sanket = bowler = second floor and Shyam = batsman = third floor and Praveen = bowler = second floor

Since Rahul is an allrounder and does not stay on the first floor, Rahul is on the third floor. Hence, Rahul = allrounder = third floor

Deepak and Rakesh are on the same floor and in the same category, but are not allrounders.

Based on the floor and category wise break-up obtained above, they can only be batsmen on the first floor.

Hence, Deepak = batsman = first floor and Rakesh = batsman = first floor

Since the three batsmen have been identified, the remaining two players i.e. Vikas and Mahesh have to be allrounders.

Since Mahesh is on the same floor as Rahul, Mahesh = allrounder = third floor and Vikas = allrounder = first floor

The entire player-wise arrangement is as shown below:

Among the players mentioned, only Vikas is neither a batsmen nor a bowler.

Hence, option 4.

QUESTION: 22

A cricket coaching academy has twelve rooms distributed equally on three floors -1,2 and 3. Four of the twelve rooms are unoccupied and the remaining rooms are occupied by eight cricketers falling into three different categories - batsman, bowler or allrounder. Only one cricketer stays per room and no cricketer can fall into more than one category. There are at least two and at most three cricketers per floor as well as per category. Following instruction regarding them is given below:

1. Vikas and Mahesh are on different floors, but both not on the second, and are in the same category.
2. Deepak and Rakesh are on the same floor and in the same category, but both are not allrounders.
3. Shyam and Praveen are on different floors, but not on the first and belong to different categories, but are not allrounders.
4. Sanket is on the second floor and Shyam is not on the same floor as Sanket.
5. Rahul is an allrounder and does not stay on the first floor.
6. Both players on the second floor are bowlers.
7. Among the three allrounders, two live on the third floor.
8. Rahul and Mahesh stay on the same floor and there are three batsmen.

Who among these are the allrounders?

Solution:

Considering the solution to the first question of the set, Vikas, Mahesh and Rahul are the three allrounders.

Hence, option 3.

QUESTION: 23

A cricket coaching academy has twelve rooms distributed equally on three floors -1,2 and 3. Four of the twelve rooms are unoccupied and the remaining rooms are occupied by eight cricketers falling into three different categories - batsman, bowler or allrounder. Only one cricketer stays per room and no cricketer can fall into more than one category. There are at least two and at most three cricketers per floor as well as per category. Following instruction regarding them is given below:

1. Vikas and Mahesh are on different floors, but both not on the second, and are in the same category.
2. Deepak and Rakesh are on the same floor and in the same category, but both are not allrounders.
3. Shyam and Praveen are on different floors, but not on the first and belong to different categories, but are not allrounders.
4. Sanket is on the second floor and Shyam is not on the same floor as Sanket.
5. Rahul is an allrounder and does not stay on the first floor.
6. Both players on the second floor are bowlers.
7. Among the three allrounders, two live on the third floor.
8. Rahul and Mahesh stay on the same floor and there are three batsmen.

Which of the following is true about the bowlers?

Solution:

Consider the solution to the first question of the set.

Sanket and Praveen are the only bowlers, and they stay on the second floor.

Hence, only the statement in option 4 is true.

Hence, option 4.

QUESTION: 24

A cricket coaching academy has twelve rooms distributed equally on three floors -1,2 and 3. Four of the twelve rooms are unoccupied and the remaining rooms are occupied by eight cricketers falling into three different categories - batsman, bowler or allrounder. Only one cricketer stays per room and no cricketer can fall into more than one category. There are at least two and at most three cricketers per floor as well as per category. Following instruction regarding them is given below:

1. Vikas and Mahesh are on different floors, but both not on the second, and are in the same category.
2. Deepak and Rakesh are on the same floor and in the same category, but both are not allrounders.
3. Shyam and Praveen are on different floors, but not on the first and belong to different categories, but are not allrounders.
4. Sanket is on the second floor and Shyam is not on the same floor as Sanket.
5. Rahul is an allrounder and does not stay on the first floor.
6. Both players on the second floor are bowlers.
7. Among the three allrounders, two live on the third floor.
8. Rahul and Mahesh stay on the same floor and there are three batsmen.

To which category and floor does Shyam belong?

Solution:

Consider the solution to the first question of the set.

Shyam is a batsman staying on the third floor.

Hence, option 2.

QUESTION: 25

Group Question

Answer the following question based on the information given below.

For his exams, Charlie has to study for 4 papers - Statistics, Mathematics, Economics and Foundation Course. He is left with 5 days from Tuesday to Saturday to prepare for his exams. He goes to the library and studies from 10 a.m. to 4 p.m. He has to make a time table such that:

1. He gets to study Statistics on all 5 days from 10 a.m. to 12 p.m. everyday or 2 p.m. to 4 p.m.everyday.
2. He studies Mathematics on Tuesday, Thursday and Saturday from 10.30 a.m. to 12.30 a.m. or on Wednesday and Friday from 1 p.m. to 4 p.m.
3. He can study Economics on Tuesday and Thursday from 12.30 p.m. to 2.30 p.m. or on Wednesday and Friday from 10 a.m. to 12 p.m.
4. He studies Foundation Course from 12 p.m. to 2 p.m. for 2 hour sessions on two days or on only 1 day for 4 hours.
5. He can study one subject at a time and he has to maximize his studying hours.
6. He can choose only one of the above mentioned studying hours for all subject throughout the week.

If Charlie studies Mathematics on Wednesday, on which day does he study Foundation Course?

Solution:

From 1,

Case 1: Charlie studies statistics from 10 a.m. to 12 p.m. everyday.

From 2, he cannot study Mathematics from 10:30 a.m. to 12:30 p.m. and so has to study it on Wednesday and Friday from 1 p.m. to 4 p.m.

Similarly, From 3, he has to study Economics on Tuesday and Thursday from 12:30 p.m. to 2:30 p.m. And from 4, Foundation Course on Saturday from 12 p.m. to 4 p.m.

Final arrangement is given below,

Case 2: Charlie studies statistics from 2 p.m. to 4 p.m. everyday.

Then, he has to study Mathematics on Tuesday, Thursday and Saturday from 10:30 a.m. to 12:30 p.m.

And, Economics on Wednesday and Friday from 10 a.m. to 12 p.m.

Hence, Foundation Course on Wednesday and Friday from 12 p.m. to 2:00 p.m.

Final arrangement is given below,

Considering the above tables,

If Charlie studies Mathematics on Wednesday then he studies Foundation Course on Saturday. Hence, option 4.

QUESTION: 26

For his exams, Charlie has to study for 4 papers - Statistics, Mathematics, Economics and Foundation Course. He is left with 5 days from Tuesday to Saturday to prepare for his exams. He goes to the library and studies from 10 a.m. to 4 p.m. He has to make a time table such that:

1. He gets to study Statistics on all 5 days from 10 a.m. to 12 p.m. everyday or 2 p.m. to 4 p.m.everyday.
2. He studies Mathematics on Tuesday, Thursday and Saturday from 10.30 a.m. to 12.30 a.m. or on Wednesday and Friday from 1 p.m. to 4 p.m.
3. He can study Economics on Tuesday and Thursday from 12.30 p.m. to 2.30 p.m. or on Wednesday and Friday from 10 a.m. to 12 p.m.
4. He studies Foundation Course from 12 p.m. to 2 p.m. for 2 hour sessions on two days or on only 1 day for 4 hours.
5. He can study one subject at a time and he has to maximize his studying hours.
6. He can choose only one of the above mentioned studying hours for all subject throughout the week.

Q. If Charlie studies Foundation Course for 4 hours on the same day, which subject does he 3 study on Thursday other than Statistics?

Solution:

Considering solution to the first question of the set, he studies Economics on Thursday.

Hence, option 2.

QUESTION: 27

For his exams, Charlie has to study for 4 papers - Statistics, Mathematics, Economics and Foundation Course. He is left with 5 days from Tuesday to Saturday to prepare for his exams. He goes to the library and studies from 10 a.m. to 4 p.m. He has to make a time table such that:

1. He gets to study Statistics on all 5 days from 10 a.m. to 12 p.m. everyday or 2 p.m. to 4 p.m.everyday.
2. He studies Mathematics on Tuesday, Thursday and Saturday from 10.30 a.m. to 12.30 a.m. or on Wednesday and Friday from 1 p.m. to 4 p.m.
3. He can study Economics on Tuesday and Thursday from 12.30 p.m. to 2.30 p.m. or on Wednesday and Friday from 10 a.m. to 12 p.m.
4. He studies Foundation Course from 12 p.m. to 2 p.m. for 2 hour sessions on two days or on only 1 day for 4 hours.
5. He can study one subject at a time and he has to maximize his studying hours.
6. He can choose only one of the above mentioned studying hours for all subject throughout the week.

If Charlie studies Statistics from 2 p.m. to 4 p.m., on which of the following days does he study Foundation Course?

1. Wednesday
2. Friday
3. Saturday
Solution:

Considering solution to the first question of the set, Charlie studies Foundation Course on Wednesday and Friday.

Hence, option 4.

QUESTION: 28

For his exams, Charlie has to study for 4 papers - Statistics, Mathematics, Economics and Foundation Course. He is left with 5 days from Tuesday to Saturday to prepare for his exams. He goes to the library and studies from 10 a.m. to 4 p.m. He has to make a time table such that:

1. He gets to study Statistics on all 5 days from 10 a.m. to 12 p.m. everyday or 2 p.m. to 4 p.m.everyday.
2. He studies Mathematics on Tuesday, Thursday and Saturday from 10.30 a.m. to 12.30 a.m. or on Wednesday and Friday from 1 p.m. to 4 p.m.
3. He can study Economics on Tuesday and Thursday from 12.30 p.m. to 2.30 p.m. or on Wednesday and Friday from 10 a.m. to 12 p.m.
4. He studies Foundation Course from 12 p.m. to 2 p.m. for 2 hour sessions on two days or on only 1 day for 4 hours.
5. He can study one subject at a time and he has to maximize his studying hours.
6. He can choose only one of the above mentioned studying hours for all subject throughout the week.

If Charlie studies Foundation Course twice in the week, then which subject does he study on Tuesday, other than Statistics?

Solution:

Considering solution to the first question of the set, Charlie studies mathematics on Tuesday other than Statistics.

Hence, option 1.

QUESTION: 29

Group Question

Answer the following question based on the information given below.
Six friends A, B, C, D, E and F go shopping to a mall. Four of them bought at least two items except E and F who bought only 1 item each. In all, 3 bags, 3 pairs of shoes, 4 pens and 2 pairs of trousers were bought. All those who bought shoes also bought pens, however, vice versa is not necessarily true. Two people bought trousers. Except B and E, all others bought pens. Out of the three people who bought bags, only one of them also bought a pair of trousers. C and D bought bags.

Q. Who else besides B bought a pair of trousers?

Solution:

We know that all others except B and E bought pens. All those who bought shoes also bought pens. 3 pairs of shoes were purchased and F bought only one item.

••• A, C and D purchased shoes.

All of them except E and F bought atleast 2 items. ••• B bought a bag and a pair of trousers.

Only one person bought a bag and a trouser.

3 people bought bags.

••• E must have bought only trousers.

••• We have:

Only B and E bought trousers.

Hence, option 3.

QUESTION: 30

Six friends A, B, C, D, E and F go shopping to a mall. Four of them bought at least two items except E and F who bought only 1 item each. In all, 3 bags, 3 pairs of shoes, 4 pens and 2 pairs of trousers were bought. All those who bought shoes also bought pens,  however, vice versa is not necessarily true. Two people bought trousers. Except B and E, all others bought pens. Out of the three people who bought bags, only one of them also bought a pair of trousers. C and D bought bags.

Who all bought shoes?

Solution:

From the table in the answer to the first question of this set, we can determine that A, C and D bought shoes.

Hence, option 4.

QUESTION: 31

Six friends A, B, C, D, E and F go shopping to a mall. Four of them bought at least two items except E and F who bought only 1 item each. In all, 3 bags, 3 pairs of shoes, 4 pens and 2 pairs of trousers were bought. All those who bought shoes also bought pens,  however, vice versa is not necessarily true. Two people bought trousers. Except B and E, all others bought pens. Out of the three people who bought bags, only one of them also bought a pair of trousers. C and D bought bags.

Which two people buy exactly the same things?

Solution:

Both C and D buy shoes, pens and bags.

Hence, option 2.

QUESTION: 32

Six friends A, B, C, D, E and F go shopping to a mall. Four of them bought at least two items except E and F who bought only 1 item each. In all, 3 bags, 3 pairs of shoes, 4 pens and 2 pairs of trousers were bought. All those who bought shoes also bought pens,  however, vice versa is not necessarily true. Two people bought trousers. Except B and E, all others bought pens. Out of the three people who bought bags, only one of them also bought a pair of trousers. C and D bought bags.

Apart from C and D, who bought a bag?

Solution:

From the table in the answer to the first question of the set, we can determine that B, C and

D bought bags. Hence, option 2.