Answer the following question based on the information given below.
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 order 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:
1.
CAT 2017 LRDI Slot 1
| LR - Selection & Distribution
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 pervious order is being prepared. At what time is the order placed by Client 1 completely served?
Now since only one client's order can be processed at any given point of time, the burger preparation for the first client which takes 10 minutes, will get completed by 10 : 10 by one of the 2 employees.
In the meantime, the 1 order of ice-cream will be completed by the 2nd employee in 2 minutes and will be completed by 10 : 02.
One of the 2 employees can put the 3 portions of french fries in the fryer at 10 and will be completed by 10 : 05.
[He can put the fries in fryer and immediately start with either burger or ice-cream]
As can be seen, out of all 3 items, the burger will be the last item to be prepared for the first client and the preparation of that item will be completed by 10 : 10 am.
So the order of the first client will be completed by 10 : 10.
2.
CAT 2017 LRDI Slot 1
| LR - Selection & Distribution
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 pervious order is being prepared. At what time is the order placed by Client 3 completely served?
As per the answer to the previous question order of client 1 will be completed by 10.10.
For 2nd client:
Since order of only one client can be processed at a time, the 2 portions of fries can be prepared by 10:15 and 1 order of ice-cream by 10:12.
So the order of client 2 will be complete by 10:15.
For 3rd client:
Now, the burger preparation of client 3 can be completed by 10:25 and the preparation of fries can be completed by 10:20.
3.
CAT 2017 LRDI Slot 1
| LR - Selection & Distribution
Suppose the employees are allowed to process multiple order 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?
Suppose Anish will complete preparation of the burger of client 1 by 10:10. In the meantime, Bani will complete the order of ice-cream of client 1 by 10:02. 3 portions of fries would have been prepared by 10:05.
At 10:05 when 2nd client's order comes, Bani would be free and the fryer would also be free.
Bani can put the 2 portions of fries in the fryer (completes at 10:10) and start the order of 1 ice-cream of client 2 which will be completed by by 10.07.
So the entire order of client 2 will complete by 10:10.
4.
CAT 2017 LRDI Slot 1
| LR - Selection & Distribution
Suppose the employees are allowed to process multiple order 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?
Suppose Anish will complete preparation of the burger of client 1 by 10:10. In the meantime, Bani will complete the order of ice-cream of client 1 by 10:02. 3 portions of fries would have been prepared by 10:05.
Bani would be free from 10:02 till 10:05 i.e., for 3 minutes.
At 10:05 when 2nd client's order comes, Bani and fryper would be free but Anish would be occupied.
Bani can put the 2 portions of fries in the fryer (completes at 10:10) and start the order of 1 ice-cream of client 2 which will be completed by by 10.07. So the entire order of client 2 will complete by 10:10.
At 10:07 when 3rd client's order comes, Bani would be free but fryer and Anish would be occupied.
Bani would start the order of burger and complete it by 10:17.
Anish and fryer both would be free at 10:10 and hence Anish can start the order of 1 portion of fries which would complete by 10:15.
Bani would be free from 10:17 till 10:30. Anish would be free from 10:10 till 10:30.
Between 10:10 and 10:30 exactly one employee would be free only between 10:10 and 10:17 i.e., for 7 minutes.
Therefore exactly one employee if idle between 10:02 - 10:05 and 10:10 - 10:17 i.e., total 10 minutes.
Answer the following question based on the information given below.
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:
It is also known that:
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.
In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.
The number of kids in G in NE was the same as the number of kids in G in W.
5.
CAT 2017 LRDI Slot 1
| DI - Tables & Graphs
What percentage of kids from S were studying in P?
As per the information given in the question, since 50 kids were sampled from each of the village, number of kids in the villages in each of the regions is as below:
NE Region ⇒ 150 × 50 = 7500
W Region ⇒ 250 × 50 = 12500
E Region ⇒ 200 × 50 = 10000
Further, 55% of 30000 or 16500 kids studied in G, 37% of 30000 or 11100 studied in P and 8% of 30000 or 2400 did not go to school. This further implies that in G, mothers of 3000 kids completed their primary education and in P and O, mothers of 8400 and 600 kids completed their education.
Now we need to bifurcate mothers of all these children into those who had dropped out and those who completed their primary education for each of the 3 regions i.e., G, P and O.
Let us construct the table as shown below
Following condition (1) we know that in S, 60% of the surveyed kids are in G. So this would mean that in S in G we have 60% of 10000 or 6000 kids.
Now out of these 6000 kids, 5100 kids mothers dropped out of school before completing primary education.
This means that mothers of 900 kids in S in G completed their primary education.
Now following condition (3), as the number of kids in G in NE was the same as the number of kids in G in W, the number of kids in each of these 2 categories will be (16500 – 6000)/2 or 5250 kids.
This means that the number of kids in G in NE and W whose mothers had completed primary education is 5250-4200 or 1050.
Now in S, out of 10000 kids, mothers of 5700 kids had dropped out of school before completing primary education. So mothers of 10000 – 5700 or 4300 kids completed their primary education. Out of these 4300, 900 were in G in S.
Now as per condition (1), all kids whose mothers had completed primary education, were in school.
This means that all kids in S whose mothers had completed their primary education were either in G or P.
Out of 4300, 900 were in G. So the remaining 4300 – 900 or 3400 will be in P. So a total of 300 + 3400 or 3700 kids.
This further implies that S in as the mothers of all those kids in S who dropped out of school, none of them completed their education. Now as per condition (2), as in NE among the O kids, 50% had mothers who dropped out, it would imply that the remaining 50% have completed their education. So if 300 mothers had dropped, it would imply that 300 mothers have completed their education in NE in O.
As a total of 600 mothers in O have completed their education and 300 out of these are from NE and none from S, it would mean that 300 mothers of kids from W in O completed their education. So in all in O, the number of mothers who completed their education from NE, W and S are 600, 1500 and 300 respectively.
Now the total number of kids in W that are in P is 12500 – 1500 – 5250 = 5750 and in P in NE is 7500 – 600 – 5250 = 1650.
Out of these 5750 kids in W that are in P, 1900 kids mothers have dropped out of primary education, which means 5750 – 1900 = 3850 kids mothers have not completed their primary education.
Also out of 1650 kids who in NE who are in P, 1650 – 500 or 1150 kids’ mothers have not completed their primary education. Our table is now complete. Let us draw up the final table
Looking at the table, we can see that out of 10000 kids in S, 3700 were studying in P.
As per the information given in the question, since 50 kids were sampled from each of the village, number of kids in the villages in each of the regions is as below:
NE Region ⇒ 150 × 50 = 7500
W Region ⇒ 250 × 50 = 12500
E Region ⇒ 200 × 50 = 10000
Further, 55% of 30000 or 16500 kids studied in G, 37% of 30000 or 11100 studied in P and 8% of 30000 or 2400 did not go to school. This further implies that in G, mothers of 3000 kids completed their primary education and in P and O, mothers of 8400 and 600 kids completed their education.
Now we need to bifurcate mothers of all these children into those who had dropped out and those who completed their primary education for each of the 3 regions i.e., G, P and O.
Let us construct the table as shown below
Following condition (1) we know that in S, 60% of the surveyed kids are in G. So this would mean that in S in G we have 60% of 10000 or 6000 kids.
Now out of these 6000 kids, 5100 kids mothers dropped out of school before completing primary education.
This means that mothers of 900 kids in S in G completed their primary education.
Now following condition (3), as the number of kids in G in NE was the same as the number of kids in G in W, the number of kids in each of these 2 categories will be (16500 – 6000)/2 or 5250 kids.
This means that the number of kids in G in NE and W whose mothers had completed primary education is 5250-4200 or 1050.
Now in S, out of 10000 kids, mothers of 5700 kids had dropped out of school before completing primary education. So mothers of 10000 – 5700 or 4300 kids completed their primary education. Out of these 4300, 900 were in G in S.
Now as per condition (1), all kids whose mothers had completed primary education, were in school.
This means that all kids in S whose mothers had completed their primary education were either in G or P.
Out of 4300, 900 were in G. So the remaining 4300 – 900 or 3400 will be in P. So a total of 300 + 3400 or 3700 kids.
This further implies that S in as the mothers of all those kids in S who dropped out of school, none of them completed their education. Now as per condition (2), as in NE among the O kids, 50% had mothers who dropped out, it would imply that the remaining 50% have completed their education. So if 300 mothers had dropped, it would imply that 300 mothers have completed their education in NE in O.
As a total of 600 mothers in O have completed their education and 300 out of these are from NE and none from S, it would mean that 300 mothers of kids from W in O completed their education. So in all in O, the number of mothers who completed their education from NE, W and S are 600, 1500 and 300 respectively.
Now the total number of kids in W that are in P is 12500 – 1500 – 5250 = 5750 and in P in NE is 7500 – 600 – 5250 = 1650.
Out of these 5750 kids in W that are in P, 1900 kids mothers have dropped out of primary education, which means 5750 – 1900 = 3850 kids mothers have not completed their primary education.
Also out of 1650 kids who in NE who are in P, 1650 – 500 or 1150 kids’ mothers have not completed their primary education. Our table is now complete. Let us draw up the final table
As can be seen in the ‘O’, Category of kids there are 300 mothers who had completed their primary education.
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 number of the surveyed kids now were in G in W?
Consider the final table obtained in the solution to the first question.
Now since 50% of 2400 i.e. 1200 kids were in G now and from one of the regions all had joined G, obviously it cannot ne from W, since in O in W region, there are 1500 kids. Also, if 25% of the kids from ‘O’ in the W region
it would mean that the 50% kids who joined G would have to be from NE or S region, which would mean that the total number of kids who joined G from O would never total upto 1200. So 50% of the kids from O in the W regions join G which means
kids from W region ‘O’join G. Now 100% of the kids from the NE, region in O (i.e., 600 kids) cannot join G as then total of the kids from the NE region in O (i.e., 600 kids) cannot join G as then total of the kids from ‘O’ region joining G region would exceed 1200.
This implies that 25% of the kids from the NE region in O
join G which means that 100% of the kids from S region in
'O' join G.
So, the number of kids from ‘NE’, ‘W’ and ‘S’ region in O joining G is 150, 750 and 300 respectively and this totals upto 1200 kids, which is the required number of students joining G from O.
Number of kids now in G in W region is 5250 + 750 = 6000.
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?
Consider the final table obtained in the solution to the first question.
Now since 50% of 2400 i.e. 1200 kids were in G now and from one of the regions all had joined G, obviously it cannot ne from W, since in O in W region, there are 1500 kids. Also, if 25% of the kids from ‘O’ in the W region
it would mean that the 50% kids who joined G would have to be from NE or S region, which would mean that the total number of kids who joined G from O would never total upto 1200. So 50% of the kids from O in the W regions join G which means
kids from W region ‘O’join G. Now 100% of the kids from the NE, region in O (i.e., 600 kids) cannot join G as then total of the kids from the NE region in O (i.e., 600 kids) cannot join G as then total of the kids from ‘O’ region joining G region would exceed 1200.
This implies that 25% of the kids from the NE region in O
join G which means that 100% of the kids from S region in
'O' join G.
Number of mothers in S who dropped out = 5100 + 300 + 300 = 5700
Mothers who are in the ‘S’ region in G = 5100 + 300 = 5400
Percentage of mothers who are in the 'S' region in G = × 100 ≅ 94.7%
Answer the following question based on the information given below.
Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Enginneering (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 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
No one is below the 80th percentile in all 3 sections.
150 are at or above the 80th percentile in exactly two sections.
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.
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 candidates is appearing in the AET by AIE, BIE consider their AET score for final selection provided the candidates 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 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.
9.
CAT 2017 LRDI Slot 1
| LR - Venn Diagram
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?
Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.
The number of candidates scoring 80 percentile and above in only Physics, only Chemistry and only Math is the same. Let this be ‘d’
Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.
Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.
Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.
Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.
a + b + c = 150
Also a + b + c + 3d + e = 200
⇒ 3d + e = 50
Given that (2d + c) : (2d + a) : (2d + b) = 4 : 2 : 1
This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c = 150. So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.
The number of students who scored 90 percentile and above and scored at least 80 percentile in Physics (but not in Chemistry and Math) will be eligible for the BIE entrance test. This is equal to d which is either 3 or 10.
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?
Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.
The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’
Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.
Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.
Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.
Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.
a + b + c = 150
Also a + b + c + 3d + e = 200
⇒3d + e=50
Given that (2d + c) : (2d + a) : (2d + b) = 4: 2: 1
This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.
So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.
Now 3d + e = 50
Also, d = 3 or 10
But it is given that e is a multiple of 5, so
e = 20
Now and
Solving the above expression we get the following equations:
c – 2a = 20 … (I)
c – 4b = 60 … (II)
a – 2b = 20 … (III)
Adding (I), (II) and (III) we get
–5b + 3c = 250 … (VI)
Solving (II) and (VI) we get
b = 10 and c = 100
∴ a =150 – 10 – 100 = 40
Now the number of candidates who scored 90 percentile overall and above and who score 80 percentile and above in P and M is a + e = 40 + 20 = 60
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?
Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.
The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’
Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.
Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.
Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.
Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.
a +b + c = 150
Also a + b + c + 3d + e = 200
⇒3d + e=50
Given that (2d + c) : (2d + a) : (2d + b)
= 4: 2: 1
This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.
So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.
Since the number of candidates who are at 90 percentile and above and also at or 80th percentile in all 3 sections is a multiple of 5 (which in the Venn Diagram is ‘e’), the Value of e(as explained in the previous questions) will have to be 20. Now the number of candidates who get 80 percentile in atleast 2 out of P, C and M which in the Venn Diagram will be the sum of a, b, c and e.
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 than 100, how many candidates had to sit for the separate test for BIE?
Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.
The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’
Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.
Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.
Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.
Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.
a +b + c = 150
Also a + b + c + 3d + e = 200
⇒3d+e=50
Given that (2d + c) : (2d + a) : (2d + b)
= 4: 2: 1
This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.
So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.
The number of candidates who are at or above 90th percentile overall and also at or above 80th percentile in P (as indicated in the Venn diagram) = a + b + d + e. As indicated in the answers to the previous question a = 3 or 10. Now we have already seen that if d = 10, a = 40, e As indicated in the answers to the previous question a = 3 or 10. Now we have So then a + b + d + e
= 42 + 18 + 3 + 41 = 104, which satisfies the condition given in the question.
Now the number of candidates who appear separately for the BIE test will be those who got 90 percentile and above overall and got 80 percentile and above only in P plus there who got 80 percentile only in P but got less than 90th percentile overall. Candidates who got 80th percentile only in P and got less than 90 in percentile overall = 400 – 104 =296
Number of candidates who get 80 percentile and above only in P and got 90 in percentile and above overall = d = 3
So, the number of candidates who sit for the separate test for BIE = 296 + 3 = 299
Answer the following question based on the information given below.
Simple Happiness index (SHI) of a country is computed on the basis of three parameters: social support (S), freedom to life choices (F) and corruption perception (C). Each of these three parameters is measured on a scale of 0 to 8 (integers only). A country is then categorized based on the total score obtained by summing the scores of all the three parameters, as shown in the following table:
Following diagram depicts the frequency distribution of the scores in S, F and C of 10 countries – Amda, Benga, Calla, Delma, Eppa, Varsa, Wanna, Xanda, Yanga and Zooma:
Further, the following are known:
Amda and Calla jointly have the lowest total score, 7, with identical scores in all the three parameters.
Zooma has a total score of 17.
All the 3 countries, which are categorized as happy, have the highest score in exactly one parameter.
Let us tabulate the information given in the bar graph depicting the number of countries against the happiness index score for each attribute.
There are 2 possibilities for Amda and Calla’s scores. Since each of them has identical scores for each of the 3 attributes, we need a pair of identical scores for each of the 3 attributes.
Since the total of the scores for each of them is 7, this is possible if they get scores of 1, 2 and 4 or 1, 3 and 3. If they get scores of 1, 2 and 4, then the only way this is possible is if each of them gets 1 in F, 2 in C and 4 in S as it is only for these scores in the respective attributes that there are 2 or more people. If they get scores of 1, 3 and 3 then each of them gets 1 in F and 3 in S and C each. In either case, Amda and Calla score 1 in F.
Let us tabulate the information given in the bar graph depicting the number of countries against the happiness index score for each attribute.
There are 2 possibilities for Amda and Calla’s scores. Since each of them has identical scores for each of the 3 attributes, we need a pair of identical scores for each of the 3 attributes.
Zooma has a total score of 17 and is hence in the category of Happy. Now each of the 3 countries in the category of Happy has the highest score for exactly one parameter.
There are 2 ways for Zooma to get a score of 17. Either Zooma gets scores of 6, 6 and 5 in attiributes C, S and F respectively or gets scores of 7 in F, 6 in S and 4 in C.
Let us tabulate the information given in the bar graph depicting the number of countries against the happiness index score for each attribute.
There are 2 possibilities for Amda and Calla’s scores. Since each of them has identical scores for each of the 3 attributes, we need a pair of identical scores for each of the 3 attributes.
If Zooma had scores of 6, 6 and 5 in C, S and F respectively, then Benga and Delma would have their highest scores in S and F (not nesessarily in that order).
Further, the country which scored the highest is S could score 5 in F and the country which scored the highest in F could score 5 in S.
Now since both Benga and Delma have equal scores and the equal score has to be in the category of C for both of them, the only possibility for this is when each of them scores 3 in C (as for no score higher than 3 do we have a pair of identical scores for C). So the maximum possible score when Benga and Delma have equal scores is 7 + 5 + 3 = 15
If Zooma gets a score of 7 in F, 6 in S and 4 in C, then one amongst Benga and Delma will get a score of 7 in S and the other a score of 6 in C.
The country with a score of 7 in S could score 5 in F. However since Zooma has scored 4 in C, the country with a score of 7 in S and 5 in F can only have a score of 3 in C.
This gives a total of 15. So the country with a score of 6 in C can have a score of 5 and 4 in F and S (though not essentially in that order). So in this case too the maximum possible score is 15.
Let us tabulate the information given in the bar graph depicting the number of countries against the happiness index score for each attribute.
There are 2 possibilities for Amda and Calla’s scores. Since each of them has identical scores for each of the 3 attributes, we need a pair of identical scores for each of the 3 attributes.
If we take scores of Zooma as 7 in F, 6 in S and 4 in C, then Benga can get a score of 16 only by scoring 6 in C and 5 each in S and F.
Benga getting a score of by 16 by scoring 7 in S, 5 in F and 4 in C is not possible as Zooma scores 4 in C. Then Delma can score 15 by scoring 7 in S, 5 in F and 3 in C.
If we assume Zooma scored 6 in C, 6 in S and 5 in F, then Benga can get a score of 16 by scoring 7 and 5 in S and F (not essentially in that order) and 4 in C.
Delma then can also get a score of 15 by scoring 7 and 5 in S and F (again not essentially in that order) and 3 in C. The 2 cases can be as listed below:
So we will have one score each of 5 in F and S each. From other scores of 4 and 3 we will have either three 4’s in S, one 4 in F, one 3 in S and two 3’s in F and C or one 4 in S, F and C each, two 3’s in F and four 3’s in C.
These 2 cases are also dependent on the individual score S of Amda and Calla. In either case, as there is only one 5 in S and F and only one 4 in C, there is a possibility of only are score of 13 by having S-5, F-5, C-3 in the 1st case or a score of S-5, F-4, C-4 or F-5, S-4, C-4 in the 2nd case.
Answer the following question based on the information given below.
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 fromT5 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.
17.
CAT 2017 LRDI Slot 1
| LR - Selection & Distribution
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:
1. Let number of SE’s (special skilled employees) in T5 in the first month be ‘x’. Then number of SE’s in T4, T3, T2 and T1 will be x + 1, x + 2, x + 3 and x + 4 respectively. As per given data, number of SE’s is 10.
∴ x + x + 1 + x + 2 + x + 3 + x + 4 = 10
∴ x = 0
So number of SE’s in T1, T2, T3, T4 and T5 will be 4, 3, 2, 1 and 0 respectively. Now T1, T2, T3, T4 and T5 get the challenging projects in months.
1, 2, 3, 4 and 5 respectively.
Also as total number of employees in the team having the challenging projects in a particular month is 1 more than the other teams the standard projects in the first month, T1 will have a total of 5 employees in the 1st month and hence 1 RE (Regular Skilled Employee). Since the other teams have a total of 4 employees each in the 1st month, the number of RE’s in T2, T3, T4 and T5 will be 1, 2, 3 and 4 respectively in the first month. So in the 1st month break up of employees in each of the teams will be as follows
In the second month, the number of team members following condition (a) will be as follows
Following condition (b) number of employees in month 2 will be
For month 3 following condition (a)
Following condition (b)
For month 4, following condition (a)
his is the final number of SE’s and RE’s in month4 as after following condition (a) as T1 has no SE and T4 has no RE, no further interchange can happen
For month 5, following condition (a)
Following condition (b)
Using this information let us answer the questions.
The composition of T2 remains unchanged only in months 3 and 4 and that of T4 changes every month. So the correct option to choose is (1, 0).
1. Let number of SE’s (special skilled employees) in T5 in the first month be ‘x’. Then number of SE’s in T4, T3, T2 and T1 will be x + 1, x + 2, x + 3 and x + 4 respectively. As per given data, number of SE’s is 10.
∴ x + x + 1 + x + 2 + x + 3 + x + 4 = 10
∴ x = 0
So number of SE’s in T1, T2, T3, T4 and T5 will be 4, 3, 2, 1 and 0 respectively. Now T1, T2, T3, T4 and T5 get the challenging projects in months.
1, 2, 3, 4 and 5 respectively.
Also as total number of employees in the team having the challenging projects in a particular month is 1 more than the other teams the standard projects in the first month, T1 will have a total of 5 employees in the 1st month and hence 1 RE (Regular Skilled Employee). Since the other teams have a total of 4 employees each in the 1st month, the number of RE’s in T2, T3, T4 and T5 will be 1, 2, 3 and 4 respectively in the first month. So in the 1st month break up of employees in each of the teams will be as follows
In the second month, the number of team members following condition (a) will be as follows
Following condition (b) number of employees in month 2 will be
For month 3 following condition (a)
Following condition (b)
For month 4, following condition (a)
his is the final number of SE’s and RE’s in month4 as after following condition (a) as T1 has no SE and T4 has no RE, no further interchange can happen
For month 5, following condition (a)
Following condition (b)
Using this information let us answer the questions.
The number of SE in T1 and T5 for month 3 is 0 and 2 respectively.
1. Let number of SE’s (special skilled employees) in T5 in the first month be ‘x’. Then number of SE’s in T4, T3, T2 and T1 will be x + 1, x + 2, x + 3 and x + 4 respectively. As per given data, number of SE’s is 10.
∴ x + x + 1 + x + 2 + x + 3 + x + 4 = 10
∴ x = 0
So number of SE’s in T1, T2, T3, T4 and T5 will be 4, 3, 2, 1 and 0 respectively. Now T1, T2, T3, T4 and T5 get the challenging projects in months.
1, 2, 3, 4 and 5 respectively.
Also as total number of employees in the team having the challenging projects in a particular month is 1 more than the other teams the standard projects in the first month, T1 will have a total of 5 employees in the 1st month and hence 1 RE (Regular Skilled Employee). Since the other teams have a total of 4 employees each in the 1st month, the number of RE’s in T2, T3, T4 and T5 will be 1, 2, 3 and 4 respectively in the first month. So in the 1st month break up of employees in each of the teams will be as follows
In the second month, the number of team members following condition (a) will be as follows
Following condition (b) number of employees in month 2 will be
For month 3 following condition (a)
Following condition (b)
For month 4, following condition (a)
his is the final number of SE’s and RE’s in month4 as after following condition (a) as T1 has no SE and T4 has no RE, no further interchange can happen
For month 5, following condition (a)
Following condition (b)
Using this information let us answer the questions.
As 200 points are divided equally between 5 team members working on a challenging project and 100 points are divided equally between 4 team members working on a standard Project, each team
member working on a challenging project gets or 40 points and each team member working on a standard project gets or 25 points. Now each member works on exactly 5 projects.
Let us assume that ‘x’ out of these 5 projects are standard projects and the remaining ‘5 – x’ are challenging projects, so total points in terms of x will be 40(50 – x) + 25x
⇒ 200 – 40 + 25x
⇒ 200 – 15x
Putting values of x as 0, 1, 2, 3, 4 and 5, we get total points as 200, 185, 170, 155, 140 and 125. Looking at the options we can see options (1), (3) and (4) are possible.
1. Let number of SE’s (special skilled employees) in T5 in the first month be ‘x’. Then number of SE’s in T4, T3, T2 and T1 will be x + 1, x + 2, x + 3 and x + 4 respectively. As per given data, number of SE’s is 10.
∴ x + x + 1 + x + 2 + x + 3 + x + 4 = 10
∴ x = 0
So number of SE’s in T1, T2, T3, T4 and T5 will be 4, 3, 2, 1 and 0 respectively. Now T1, T2, T3, T4 and T5 get the challenging projects in months.
1, 2, 3, 4 and 5 respectively.
Also as total number of employees in the team having the challenging projects in a particular month is 1 more than the other teams the standard projects in the first month, T1 will have a total of 5 employees in the 1st month and hence 1 RE (Regular Skilled Employee). Since the other teams have a total of 4 employees each in the 1st month, the number of RE’s in T2, T3, T4 and T5 will be 1, 2, 3 and 4 respectively in the first month. So in the 1st month break up of employees in each of the teams will be as follows
In the second month, the number of team members following condition (a) will be as follows
Following condition (b) number of employees in month 2 will be
For month 3 following condition (a)
Following condition (b)
For month 4, following condition (a)
his is the final number of SE’s and RE’s in month4 as after following condition (a) as T1 has no SE and T4 has no RE, no further interchange can happen
For month 5, following condition (a)
Following condition (b)
Using this information let us answer the questions.
To score 185 points Aneek must have worked in 4 challenging projects and 1 standard projects
1] One way Aneek (as an SE) could have worked is by working on challenging projects in T1, T2, T3 and T4 in months 1, 2, 3 and 4 respectively and worked on a standard project in T4 in month 5. This is possible if Aneek was an SE who was moved from T1 to T2 in month 2, T2 to T3 in month 3 and then T3 to T4 in month 4. Subsequently in month 5, he was retained in T4 for the standard project. So option [1] is possible.
2] Another possibility is if Aneek as an SE was moved from T1 to T2 in month 2 to work on a challenging project, then moved from T2 to T4 in month 3 to work on a standard Project, retained in T4 in month 4 to work on a challenging project and then moved from T4 to T5 in month 5 to work on a challenging project. So option [2] is also possible.
3] Aneek could be an SE working on a standard project in T2 in month 1 and subsequently retained in T2 in month 2, and moved to T3, T4 and T5 in months 3, 4 and 5 to work on challenging projects. So this option too is possible.
4] This option is not possible because Aneek could not have worked in T3 without working in T2 prior to that Hence option [4] is ruled out.
Answer the following question based on the information given below.
In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.
The heights (in metres) of individual platforms are as shown below:
6 1 2 4 3
9 5 3 2 8
7 8 4 6 5
3 9 5 1 2
1 7 6 3 9
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.
21.
CAT 2017 LRDI Slot 1
| LR - Arrangements
How many individuals in this layout can be reached by just one individual?
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 5 people i.e., 2, 3 and 5 in the same row, 3 in the column above above and 5 in the column below.
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 5 people i.e., 2, 3 and 5 in the same row, 3 in the column above above and 5 in the column below.
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
A person having a height of 1 m cannot be reached by anyone.
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 5 people i.e., 2, 3 and 5 in the same row, 3 in the column above above and 5 in the column below.
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
In the last row there is only one individual with height 3 m who cannot be reached.
So option [1] is ruled out. Similarly, in the 4th row there is only one person with height 1 m who cannot be reached by anyone. So option [2] is also incorrect.
In the 4th column, there are 2 people, one with 2 m height and one with 1 m height who cannot be reached by anyone.
So option [3] is correct. In the middle column there is no one who cannot be reached by anyone. So option [4] is also incorrect.
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 5 people i.e., 2, 3 and 5 in the same row, 3 in the column above above and 5 in the column below.
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
In row 1, the maximum number of people who can reach a particular individual is 3, which is for the person with height 6 m and for the person with height 4 m. So statement (1) is false.
Row 3 does not have any individual who cannot be reached by anyone. Hence statement [2] is false.
In row 1, person with height 6 m and person with height 4 m can each be reached by 3 people. In row 2 people with height 5 m and 3 m can each be reached by 2 people.
In Row 3, people with height 7 m and 3 m can be reached by only 1 person. In Row 4, people with height 3 m and 2 m can be reached by only 1 person. In Row 5, people with height 7 m and 6 m can be reached by 2 people each. So in each row we have 2 individuals who can be reached by an equal number of individuals.
So statement 3 is true. Individual with height 9 m in row 2 can be reached by 4 people.
Answer the following question based on the information given below.
A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.
The underlying principle that they are working on is the following:
Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day.
25.
CAT 2017 LRDI Slot 1
| DI - Routes & Networks
If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is:
Since there are a total of 10 cities we can have a combination of 2 cities in 10C2 or 45 ways. Now, if for example we have 2 cities City 1 and City 2, then the cities can be connected in the following 4 ways:
Morning flight from city 1 to city 2
Morning flight from city 2 to city 1
Evening flight from city 1 to city 2
Evening flight from city 2 to city 1
So the minimum number of direct flights to connect all cities is 45 × 4 or 180 ways.
Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and / or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:
Let us suppose that City 1, City 2 and City 3 are the hubs and City 4, City 5... upto City 10 are 7 of the other 10 cities.
Now City 1, City 2 and City 3 connect with each other in 4 possible ways (as mentioned in the answer to the previous question).
Now 2 out of 3 cities can be a chosen in 3C2 or 3 ways. So total of no ways City 1, City 2 and City 3 connect with each other is 3 × 4 or 12 ways.
Now City 1 will connect with each of City 4, City 5, City 6 ..... City 10 in 4 possible ways (as explained in the previous questions answer).
So, total number of flights between City 1 and the cities 4 to 10 is 28.
Similarly there will be 28 flights each for City 2 and City 3 that will connect it with the 7 cities. So total minimum number of flights between 2 cities will be 12 + 28 + 28 + 28 = 96.
Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further suppose that direct flights are allowed only between two cities satisfying one of the following:
Both cities are in G1
Between A and any city in G2
Between B and any city in G3
Between C and any city in G4
Then the minimum number of direct fights that satisfies the underlying principle of the airline is:
Now A, B and C are in G1. As there is no restriction of flights that can connect with each other in G1, the total number of flights connecting 2 cities in G1 is 3C2 × 4
= 12
A can connect with any of the 3 cities in G1 in 3C1 × 4 = 12 ways
B can connect with any of the 2 cities in G2 in 2C1 × 4 = 8 ways
C can connect with any of the 3 cities in G3 in 2C1 × 4 = 8 ways
Total minimum number of direct flights that satisfy the underlying principle is 12 + 12 + 8 + 8 = 40
Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:
Both cities are in G1
Between A and any city in G2
Between B and any city in G3
Between C and any city in G4
However, due to operational difficulties at A, it was later decided that the only flights that would operate at A would be those to and from B. Cities in G2 would have to be assigned to G3 or to G4.
What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose?
Given the above conditions, we know that in G1, A would connect to B and B would connect with C. B would connect with any of the 2 cities in G3 and C would connect with any of the 2 cities in G4.
Now the 2 cities from G2 that were connected to A earlier can now connects to one of the cities in G3 or G4. There is no flight connection possible now between A and C in G1.
So the total number of flight connections as compared to the number of flight connections in the previous question will be less by number of flights connecting A and C i.e., 4.
Answer the following question based on the information given below.
Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N). The roads from A to M, and from N to B, are both short and narrow. In each case, one car takes 6 minutes to cover the distance, and each additional car increases the travel time per car by 3 minutes because of congestion. (For example, if only two cars drive from A to M, each car takes 9 minutes.) On the road from A to N, one car takes 20 minutes, and each additional car increases the travel time per car by 1 minute. On the road from M to b, one car takes 20 minutes, each additional car increases the travel time per car by 0.9 minute.
The police department orders each car to take a particular route in such a manner that it is not possible for any car to reduce its travel time by not following the order, while the other cars are following the order.
29.
CAT 2017 LRDI Slot 1
| DI - Routes & Networks
How many cars would be asked to take the route A-N-B, that is Akala-Nanur-Bakala route, by the police department?
The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.
Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes. So 2 cars would take the A-N-B route.
If all the cars follow the police order, what is the difference in travel time (in minutes) between a car which takes the route A-N-B and a car that takes the route A-M-B?
The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.
Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.
Minimum travel time would be when 2 cars are assigned to each of the 2 routes by the police department.
The difference in travel time when the cars follow the police order is 30-29.9 or 0.1 minutes.
A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M-N-B. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.
How many cars would the police department order to take the A-M-N-B route so that it is not possible for any car to reduce its travel time by not following the order while the other cars follow the order? (Assume that the police department would never order all the cars to take the same route:)
The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.
Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.
The police department would order two cars to take the A-M-N-B route
A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M-N-B. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.If all the cars follow the police order, what is the minimum travel time (in minutes) from A to B? (Assume that the police department would never order all the cars to take the same route.)
The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.
Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.
The minimum travel time from A to B is 32 minutes.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.
Solution
Discuss
Report
