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QUESTION: 1

**Directions for Questions:** In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.

The height (in metres) of individual platforms are as shown below:

(2017)

Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:

(i) A and B are in the same row or column

(ii) A is at a lower height than B

(iii) If there is/are any individual(s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

**Q. How many individuals in this layout can be reached by just one individual?**

Solution:

The height of the platforms given is as below

The number of person who can be reached by just one individual is circled

A total of 7 persons can be reached by just one individual.

QUESTION: 2

**Directions for Questions:** In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.

The height (in metres) of individual platforms are as shown below:

(2017)

Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:

(i) A and B are in the same row or column

(ii) A is at a lower height than B

(iii) If there is/are any individual(s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

**Q. Which of the following is true for any individual at a platform of height 1 m in this layout?**

Solution:

For individual at a platform of height 1, they cannot be reached by anyone as candition (ii) will be violated.

QUESTION: 3

**Directions for Questions:** In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.

The height (in metres) of individual platforms are as shown below:

(2017)

Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:

(i) A and B are in the same row or column

(ii) A is at a lower height than B

(iii) If there is/are any individual(s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

**Q. We can find two individuals who cannot be reached by anyone in**

Solution:

The height of the platforms given as below.

Only in the fourth column can we find two individuals who can not be reached by anyone. In the fourth column the individual at height 2 and the individual at height 1 can not be reached by any one.

QUESTION: 4

**Directions for Questions:** In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.

The height (in metres) of individual platforms are as shown below:

(2017)

Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:

(i) A and B are in the same row or column

(ii) A is at a lower height than B

(iii) If there is/are any individual(s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

**Q. Which of the following statements is true about this layout?**

Solution:

According to platform height table-Statement 1 is wrong as no individual in row 1 can be reached by 5 or more Same as statement 2 and 1 wrong Only statement 3 correct.

QUESTION: 5

**Directions for Questions:**

There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows.

(a) The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange. one RE is shifted from the former team to the latter team.

(b) After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.

(2017)

**Q. The number of times in which the composition of team T2 and the number of times in which composition of team T4 remained unchanged in two successive months are.**

Solution:

Given that there are 10 SE and 11 RE In the first month, since T1 has one more SE than T2, who in turn has one more SE than T3,... till T5, the number of SEs in T1, T2, T3, T4 and T5 must be 4, 3, 2, 1 and 0.

Also, the team that is assigned the challenging project has one more employees than the rest. Hence, the team that is assigned the challenging project will have 5 employees, while the other teams will have 4 employees.

Since T1 is assigned the challenging project in the first month, T1 will have 5 employees, and the other teams will have 4 employees each.

The following table provides the composition of the teams in the first month:

In the second month, T2 will be allotted the challenging project.

From a, two SEs will be transferred from Tl to T2. One RE is transferred from T2 to T1.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1. Similar transfers will happen between T2 and T4.

The following table provides the number of employees in each team in the second month:

In the third month, T3 will be allotted the challenging project.

From a, two SEs will be transferred from T2 to T3. One RE is transferred from T3 to T2.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the third month:

In the fourth month, T4 will be allotted the challenging project.

From a, two SEs will be transferred from T3 to T4. One RE is transferred from T4 to T3.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen. Also, one SE must be transferred from T2 to T4 and one RE must be transferred from T4 to T2.

However, there are no REs in T4.

Hence, this transfer will not happen.

The following table provides the number of employees in each team in the fourth month: In the fifth month, T5 will be allotted the challenging project.

From a, two SEs will be transferred from T4 to T5. One RE is transferred trom T5 to T4.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the fifth month.

The composition of T2 did not change once between the third and the fourth months.

The composition of T4 changed between any two successive months.

Hence, the answer is (1, 0).

QUESTION: 6

**Directions for Questions:**

There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows.

(a) The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange. one RE is shifted from the former team to the latter team.

(b) After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.

(2017)

**Q. The number of SE in T1 and T5 for the projects in the third month are, respectively.**

Solution:

Given that there are 10 SE and 11 RE In the first month, since T1 has one more SE than T2, who in turn has one more SE than T3,... till T5, the number of SEs in T1, T2, T3, T4 and T5 must be 4, 3, 2, 1 and 0.

Also, the team that is assigned the challenging project has one more employees than the rest. Hence, the team that is assigned the challenging project will have 5 employees, while the other teams will have 4 employees.

Since T1 is assigned the challenging project in the first month, T1 will have 5 employees, and the other teams will have 4 employees each.

The following table provides the composition of the teams in the first month:

In the second month, T2 will be allotted the challenging project.

From a, two SEs will be transferred from Tl to T2. One RE is transferred from T2 to T1.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1. Similar transfers will happen between T2 and T4.

The following table provides the number of employees in each team in the second month:

In the third month, T3 will be allotted the challenging project.

From a, two SEs will be transferred from T2 to T3. One RE is transferred from T3 to T2.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the third month:

In the fourth month, T4 will be allotted the challenging project.

From a, two SEs will be transferred from T3 to T4. One RE is transferred from T4 to T3.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen. Also, one SE must be transferred from T2 to T4 and one RE must be transferred from T4 to T2.

However, there are no REs in T4.

Hence, this transfer will not happen.

The following table provides the number of employees in each team in the fourth month: In the fifth month, T5 will be allotted the challenging project.

From a, two SEs will be transferred from T4 to T5. One RE is transferred trom T5 to T4.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the fifth month.

Number of SE in T1 in third month = 0 Number of SE in T5 in third month = 2.

Hence, the answer is (0, 2)

QUESTION: 7

**Directions for Questions:**

There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows.

(a) The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange. one RE is shifted from the former team to the latter team.

(b) After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.

(2017)

**Q. Which of the following CANNOT be the total credit points earned by any employee from the projects?**

Solution:

Given that there are 10 SE and 11 RE In the first month, since T1 has one more SE than T2, who in turn has one more SE than T3,... till T5, the number of SEs in T1, T2, T3, T4 and T5 must be 4, 3, 2, 1 and 0.

Also, the team that is assigned the challenging project has one more employees than the rest. Hence, the team that is assigned the challenging project will have 5 employees, while the other teams will have 4 employees.

Since T1 is assigned the challenging project in the first month, T1 will have 5 employees, and the other teams will have 4 employees each.

The following table provides the composition of the teams in the first month:

In the second month, T2 will be allotted the challenging project.

From a, two SEs will be transferred from Tl to T2. One RE is transferred from T2 to T1.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1. Similar transfers will happen between T2 and T4.

The following table provides the number of employees in each team in the second month:

In the third month, T3 will be allotted the challenging project.

From a, two SEs will be transferred from T2 to T3. One RE is transferred from T3 to T2.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the third month:

In the fourth month, T4 will be allotted the challenging project.

From a, two SEs will be transferred from T3 to T4. One RE is transferred from T4 to T3.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen. Also, one SE must be transferred from T2 to T4 and one RE must be transferred from T4 to T2.

However, there are no REs in T4.

Hence, this transfer will not happen.

The following table provides the number of employees in each team in the fourth month: In the fifth month, T5 will be allotted the challenging project.

From a, two SEs will be transferred from T4 to T5. One RE is transferred trom T5 to T4.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the fifth month.

Given that challenging projects has 200 credits and standard projects have 100 credits.

In each type of project the credits are equally shared by the employees in the team.

Hence, for a challenging project an employee earns 200/5 = 40 credits.

For a standard project, an employee earns 100/4 = 25 credits.

For the five months, on employee can work in five challenging projects OR four challenging projects and one standard project OR three challenging projects and two standard projects OR two challengmg projects and three challenging projects OR one challenging project and four standard projects OR five standard projects.

In each case, an employee Will earn 200 or 185 or 170 or 155 or 140 or 125 credits.

Hence. it is not possible for an employee to earn 150 credits.

QUESTION: 8

**Directions for Questions:**

There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regularskilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows.

(a) The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month. In exchange. one RE is shifted from the former team to the latter team.

(b) After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team.

(2017)

**Q. One of the employees named Aneek scored 185 points. Which of the following CANNOT be true?**

Solution:

Given that there are 10 SE and 11 RE In the first month, since T1 has one more SE than T2, who in turn has one more SE than T3,... till T5, the number of SEs in T1, T2, T3, T4 and T5 must be 4, 3, 2, 1 and 0.

Also, the team that is assigned the challenging project has one more employees than the rest. Hence, the team that is assigned the challenging project will have 5 employees, while the other teams will have 4 employees.

Since T1 is assigned the challenging project in the first month, T1 will have 5 employees, and the other teams will have 4 employees each.

The following table provides the composition of the teams in the first month:

In the second month, T2 will be allotted the challenging project.

From a, two SEs will be transferred from Tl to T2. One RE is transferred from T2 to T1.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1. Similar transfers will happen between T2 and T4.

The following table provides the number of employees in each team in the second month:

In the third month, T3 will be allotted the challenging project.

From a, two SEs will be transferred from T2 to T3. One RE is transferred from T3 to T2.

From b, one SE will be transferred from T1 to T5, one RE will be transferred from T5 to T1.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the third month:

In the fourth month, T4 will be allotted the challenging project.

From a, two SEs will be transferred from T3 to T4. One RE is transferred from T4 to T3.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen. Also, one SE must be transferred from T2 to T4 and one RE must be transferred from T4 to T2.

However, there are no REs in T4.

Hence, this transfer will not happen.

The following table provides the number of employees in each team in the fourth month: In the fifth month, T5 will be allotted the challenging project.

From a, two SEs will be transferred from T4 to T5. One RE is transferred trom T5 to T4.

From b, one SE must be transferred from T1 to T5.

However, since there are no SEs in T1, this will not happen.

Also, one SE will be transferred from T2 to T4 and one RE will be transferred from T4 to T2.

The following table provides the number of employees in each team in the fifth month.

Since Aneek secured 185 credits he worked in four challenging projects and one standard project. Option A: Aneek could have worked in Tl in first month an challenging project, T2 in second month (in challenging project), T3 in third month (in challenging project), T4 in fourth month (in challenging project) and fifth month (in standard project). Hence, this is possible.

Option B: Aneek could have worked in T1 in first month (in challenging project), T2 in second month (in challenging project), T4 in third month (in standard project), T4 in fourth month (in challenging project) and T5 in fifth month (in challenging project). Hence, this is possible.

Option C: Aneek could have worked in T2 in first month (in standard project), T2 in second month (in challenging project), T3 in third month (in challenging project), T4 in fourth month (in challenging project) and T5 in fifth month (in challenging project).

Hence, this is possible.

Option D: Aneek could hove worked in T1 in first month (in challenging project). He can work in Tl or T5 in the second month. In either case, he cannot work in T3 without working in T2 first. In if we assume, he worked in T3 in the first month, he could not have worked in four teams in the five months. Similarly, we can rule out the other possibilities for this option. Hence, this is the answer.

QUESTION: 9

**Directions for Questions: **Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections. Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 condidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET.

(2017)

(a) No one is below the 80th percentile in all 3 sections.

(b) 150 are at or above the 80th percentile in exactly two sections.

(c) The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.

(d) Number of candidates below 80th percentile in P. Number of candidates below 80th percentile in C. Number of candidates below 80th percentile in M = 4:2:1.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

**Q. What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?**

Solution:

The number of candidates sitting for separate test far BIE who were at or above 90th percentile in CET (a) is either 3 or 10.

From 1, h = 0.

From 2, d + e + f =150

From 3, a = b = c

Since there are a total of 200 candidates,

3a + g = 200 –150 = 50

From 4, (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1 Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7. Since d + e + f =150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7.

Hence, a = 3, 10,17, . ..

Further, since 3a + g = 50, a must be less than 17. Therefore, only two cases are possible for the value of a, i.e.. 3 or 10.

We can calculate the values of the other variables for the two cases.

a = 3 or 10

d = 18 or 10

e = 42 or 40

f = 90 or 100

g = 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.

BIE will consider the candidates who are appearing (for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80. BlE will conduct a separate test for the other students whoare at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

QUESTION: 10

**Directions for Questions: **Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections. Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 condidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET.

(2017)

(a) No one is below the 80th percentile in all 3 sections.

(b) 150 are at or above the 80th percentile in exactly two sections.

(c) The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.

(d) Number of candidates below 80th percentile in P. Number of candidates below 80th percentile in C. Number of candidates below 80th percentile in M = 4:2:1.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

**Q. If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET?**

Solution:

From the given condition. 9 is a multiple of 5, Hence. g = 20.

The number of candidates at or above 90th percentile overall and at or above 80th percentile in both P and M .= e + g = 60.

QUESTION: 11

**Directions for Questions: **Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections. Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 condidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET.

(2017)

(a) No one is below the 80th percentile in all 3 sections.

(b) 150 are at or above the 80th percentile in exactly two sections.

(c) The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.

(d) Number of candidates below 80th percentile in P. Number of candidates below 80th percentile in C. Number of candidates below 80th percentile in M = 4:2:1.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

**Q. If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?**

Solution:

In this case, g = 20. Number of candidates shortlisted for AET = d + e + f + g =10 + 40 + 100 + 20 = 170

QUESTION: 12

**Directions for Questions: **Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections. Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 condidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET.

(2017)

(b) 150 are at or above the 80th percentile in exactly two sections.

(c) The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.

(d) Number of candidates below 80th percentile in P. Number of candidates below 80th percentile in C. Number of candidates below 80th percentile in M = 4:2:1.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

**Q. If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more that 100, how many candidates had to sit for the separate test for BIE?**

Solution:

From the given condition, the number of candidates at or above 90th percentile overall and at or above 80th percentile in P in CET =104.

The number of candidates who have to sit for separate test = 296 + 3 = 299

QUESTION: 13

**Directions for Questions: **

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South(S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O). The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

(2017)

It is also known that:

(I) In S, 60% of the surveyed kids were in G. Moreover, in S, all surveyed kids whose mothers had completed primary education were in school.

(II) In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.

(III) The number of kids in G in NE was the same as the number of kids in G in W.

**Q. What percentage of kids from S were studying in P?**

Solution:

With the table given for kids in different schools whose mothers had dropped out of school we will be adding another value for each value already present and the new value will represent the number of kids in different types of schools for kids whose mothers completed primary education.

QUESTION: 14

**Directions for Questions:**

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South(S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O). The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

(2017)

It is also known that:

(I) In S, 60% of the surveyed kids were in G. Moreover, in S, all surveyed kids whose mothers had completed primary education were in school.

(II) In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.

(III) The number of kids in G in NE was the same as the number of kids in G in W.

**Q. Among the kids in W whose mothers had completed primary education, how many were not in school?**

Solution:

In W, 300 kids whose mothers had completed primary education were not in school.

QUESTION: 15

**Directions for Questions:**

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South(S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O). The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

(2017)

It is also known that:

(I) In S, 60% of the surveyed kids were in G. Moreover, in S, all surveyed kids whose mothers had completed primary education were in school.

(II) In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.

(III) The number of kids in G in NE was the same as the number of kids in G in W.

**Q. In a follow up survey of the same kids two years later, it was found that all the kids were now in school. Of the kids who were not in school earlier, in one region, 25% were in G now, whereas the rest were enrolled in P, in the second region, all such kids were in G now, while in the third region, 50% of such kids has now joined G while the rest had joined P. As a result, in all three regions put together, 50% of the kids who were earlier out of school had joined G. It was also seen that no surveyed kid had changed schools.
What number of the surveyed kids now were in G in W?**

Solution:

As there were initially 2400 students who were not in school and now 1200 of them are in G, with the mentioned percentages the only possibility is 50% of students in W, 25% of students in NE and 100% of students in S who were not going to school shifted to G.

∴ 50% of W = 50% of 1500 = 750

25% of NE = 25% of 600 = 150

100% of S = 100% of 300 = 300

⇒ Total = 750 + 150 + 300

⇒ 1200

∴ now 4200 + 1050 + 750 = 6000 students were in G is W.

QUESTION: 16

**Directions for Questions:**

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South(S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O). The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

(2017)

It is also known that:

(I) In S, 60% of the surveyed kids were in G. Moreover, in S, all surveyed kids whose mothers had completed primary education were in school.

(II) In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.

(III) The number of kids in G in NE was the same as the number of kids in G in W.

**Q. In a follow up survey of the same kids two years later, it was found that all the kids were now in school. Of the kids who were not in school earlier, in one region, 25% were in G now, whereas the rest were enrolled in P, in the second region, all such kids were in G now, while in the third region, 50% of such kids had now joined G while the rest had joined P. As a result, in all three regions put together, 50% of the kids who were earlier out of school had joined G. It was also seen that no surveyed kid had changed schools.
What percentage of the surveyed kids in S, whose mothers had dropped out before completing primary education, were in G now?**

Solution:

As explained in the previous question, all 300 in S who were not going to school, now shifted to G.

Now of the 5700 students whose mothers had dropped out in S regions, 5400 are in G.

i.e = 5400 / 5700 x 100 = 94.70%

QUESTION: 17

**Directions for Questions: **

Healthy Bites is a fast food joint serving three items; burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare up to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However fries cannot be prepared in anticipation of future orders.

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however, the order is considered to be completely served only when all the items of that order are served.

The table below gives the orders of three clients and the times at which they placed their orders:

(2017)

**Q. Assume that only one client’s order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.
At what time is the order placed by Client 1 completely served?**

Solution:

As the item which take the maximum time is burger, client I will be completely served by 10.00 + 10 min = 10 : 10

QUESTION: 18

**Directions for Questions: **

Healthy Bites is a fast food joint serving three items; burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare up to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However fries cannot be prepared in anticipation of future orders.

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however, the order is considered to be completely served only when all the items of that order are served.

The table below gives the orders of three clients and the times at which they placed their orders:

(2017)

**Q. Assume that only one client’s order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.
At what time is the order placed by Client 3 completely served?**

Solution:

Client–1 10 : 00 – 10 : 10 (burger)

Client–2 10 : 10 – 10 : 15 (fries)

Client–3 10 : 15 – 10 : 25 (burger

QUESTION: 19

**Directions for Questions: **

Healthy Bites is a fast food joint serving three items; burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare up to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However fries cannot be prepared in anticipation of future orders.

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however, the order is considered to be completely served only when all the items of that order are served.

The table below gives the orders of three clients and the times at which they placed their orders:

(2017)

**Q. Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.
At what time is the order placed by Client 2 completely served?**

Solution:

Client 1 – 10 : 00 – 10 : 10 (Anish)

Client 2 – 10 : 05 – 10 : 10 (Bani)

The second client can be served by 10 : 10

QUESTION: 20

**Directions for Questions: **

Healthy Bites is a fast food joint serving three items; burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare up to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However fries cannot be prepared in anticipation of future orders.

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however, the order is considered to be completely served only when all the items of that order are served.

The table below gives the orders of three clients and the times at which they placed their orders:

(2017)

**Q. Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.
Also assume that the fourth client came in only at 10 : 35. Between 10: 00 and 10 : 30, for how many minutes is exactly one of the employees idle?**

Solution:

10 : 02 – 10 : 05 (Bani) – 3 minutes

10 : 10 – 10 : 17 (Anish/Bani) – 7 minutes

Total time = 3 + 7 = 10 minutes.

### Data Handling-I (Classification and Tabulation of Data)

Doc | 12 Pages

### Understanding Caselets in Data Interpretation

Video | 12:28 min

### Caselets Set- 1

Video | 14:55 min

- Test: Data Tabulation & Caselets - 1
Test | 20 questions | 35 min

- Test: Data Tabulation & Caselets - 2
Test | 15 questions | 25 min

- Test: Grouping And Tabulation Of Data
Test | 10 questions | 10 min

- Test: Caselets
Test | 10 questions | 10 min

- Test: Tabulation
Test | 18 questions | 20 min