Question-16
Read the information given below carefully and answer the questions that follow:
A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:
1. Are you shopping at the website for the first time? (YES or NO);
2. Specify your gender. (MALE or FEMALE);
3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)
240 shoppers answer the survey, among whom 65 are first time shoppers.
Furthermore:
1. The ratio of the numbers of male to female shoppers is 1:2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3: 4:5
ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1:1:4:4:6:6
iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1:1:1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers.
Determine the number of happy male shoppers:
a)19
b)18
c)10
d)20
e)22
Explanation
SOLUTION-[20] D
From the given data the following table can be created
Hence the value of x=10
Question-17
Read the information given below carefully and answer the questions that follow:
A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:
1. Are you shopping at the website for the first time? (YES or NO);
2. Specify your gender. (MALE or FEMALE);
3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)
240 shoppers answer the survey, among whom 65 are first time shoppers.
Furthermore:
1. The ratio of the numbers of male to female shoppers is 1:2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3: 4:5 ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers,
unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1:1:4:4:6:6
iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1:1:1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers.
Which among the following cannot be determined uniquely:
a)All the numericaldata can be determined uniquely
b)The number of first-time happy male shoppers.
c)The number of returning male shoppers
d)The number of first-time neutral male shoppers
e)The number of returning unhappy female shoppers
Explanation
SOLUTION[A] -All the numericaldata can be determined uniquely
From the given data the following table can be created
Hence the value of x=10
Question-18
Read the information given below carefully and answer the questions that follow:
A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:
1. Are you shopping at the website for the first time? (YES or NO);
2. Specify your gender. (MALE or FEMALE);
3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)
240 shoppers answer the survey, among whom 65 are first time shoppers.
Furthermore:
1. The ratio of the numbers of male to female shoppers is 1:2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3: 4:5
ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1:1:4:4:6:6 iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1:1:1 : 2, while the number of happy
first-time female shoppers is equal to the number of unhappy first-time male shoppers.
Which amongst the following is the lowest in number ?
a)Number of happy returning male shoppers
b)Number of neutral first timemale shoppers
c)Number of unhappy first timefemale shoppers
d)Number of unhappy first time male shoppers
e)Number of neutral first time female shoppers
Explanation
SOLUTION[E]Number of neutral first time female shoppers
From the given data the following table can be created
Hence the value of x=10
Question-19
Read the information provided carefully and answer the questions that follow.
In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science. Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations.
The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks awarded in each of the examinations were on the basis of the scores obtained by them in those examinations.
If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.
If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for
The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations.
The following facts are also known:
1. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations.
II. The student who missed the Mathematics examination did not miss any other examination.
III. One of the students who missed the Hindi examination did not miss any other examination. The other student who missed the Hindi examination also missed the Science examination.
Which of the following students did not appear for the English examination?
a)ALVA AND BAL
b)FOXXY AND ELGAR
c)DEEPALI AND CARDY
d)BAL AND FOXXY
e)CAN NOT BE DETERMINED
Explanation
SOLUTION[B] FOXXY AND ELGAR
CONSIDER-
ALVA=ALVA
BITHI=BAL
CARL=CARDY
DEEP=DEEPALI
ESHA=ELGAR
FONI=FOXXY Based on Condition II, we understand that the student who missed the Mathematics examination did not miss any other examination. This indicates that the Maths score is bound to be
the average of the best 3 out of the 4 exam scores obtained by this candidate. Based on this inference, we can proceed with identifying the math score that can be represented as an average of the rest of the scores. We can straightaway eliminate Deep and Esha as potential candidates, given that their Mathematics score is greater than the rest of the exam scores. After estimating the average scores for the rest of the candidates, we observe that only Carl has missed his Mathematics examination.
For Carl: best 3 out of 4-80(Hindi), 90(Social Science), 100(Science)
Avg. = 270/3 = 90 which matches the given value
… Carl missed his Mathematics examination.
Further, based on Condition III, we can surmise that the student who missed Hindi and Science should have similar average scores in these two subjects. We notice that Alva has the same score of 75 in both Hindi and Science. The same can be said about Deep, who has a score of 90 in both these subjects. Thus, one out of Alva and Deep missed out on Hindi and Science examination, while the second individual missed out only on the Hindi examination.
Since we know that Carl, Alva and Deep are unlikely to have missed out on the English exam, we can divert our attention to determining which individual out of Bithi, Esha and Foni failed to appear for this subject. However, we notice that Bithi’s English score is greater than the rest of her scores, thereby helping us eliminate her as the potential candidate.
For Esha: best 3 out of 4-85(Hindi), 95(Mathematics), 60(Science)
Avg. = 240/3 = 80 which matches the given value
… Esha most likely missed her English examination.
For Foni: best 3 out of 4-78(Mathematics), 83(Social Science), 88(Science)
Avg. 249/3 = 83 which matches the given value
… Foni most likely missed her English examination.
Based on Condition I, we know that exactly two candidates missed the examinations for English, Hindi, Science, and Social Science. For English, we determined these individuals to be Esha and Foni. For Hindi, we determined these
individuals to be Alva and Deep. For Science, we know one of the individuals is either Alva or Deep. Given that Carl, Alva and Deep cannot be a part of the group that missed Science or Social Science exam, we can proceed by carefully scrutinizing the rest of the group that includes Bithi, Esha and Foni.
We notice that Bithi has a similar score in both Science and Social Science examination. Assuming that she did miss these exams, let us proceed to check if this was actually the case.
For Bithi: Best 2 out 3-90(English), 80(Hindi)
Avg = 170/2 = 85 which matches the given value
… Bithi is likely to have missed her Science and Social Science examinations.
We additionally notice that Foni has a similar score in English and Social Science. On considering the
best 2 out of 3 scores, the average value of the score for both the subject holds (equal to 83). Thus, we can conclude that Bithi and Foni missed their Social Science examination.
Thus, the students who missed just one exam were: Carl (Mathematics); Esha (English) and one out of Alva and Deep (Hindi).
Hence of the six students, we can correctly determine the missed subjects for four of them (except Alva and Deep):
Mathematics: Carl; English: Esha &Foni; Hindi: Alva & Deep; Science: Bithi& one out of Alva and Deep; Social Science: Foni&Bithi
Hence, the correct answer to this question is Option : Esha and Foni.
Question-20
Read the information provided carefully and answer the questions that follow.
In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science. Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations.
The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks awarded in each of the examinations were on the basis of the scores obtained by them in those examinations.
If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for. If a student missed two examinations, then the marks awarded in each of these examinations
was the average of the best two among the three scores in the examinations they appeared for
The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations.
The following facts are also known:
1. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations.
II. The student who missed the Mathematics examination did not miss any other examination.
III. One of the students who missed the Hindi examination did not miss any other examination. The other student who missed the Hindi examination also missed the Science examination.
Who among the following did not appear for the Maths examination?
a)FOXXY
b)ELGAR
c)DEEPALI
d)BAL
e)CARDY
Explanation
SOLUTION[E] CARDY
CONSIDER-
ALVA=ALVA
BITHI=BAL
CARL=CARDY
DEEP=DEEPALI
ESHA=ELGAR
FONI=FOXXY
Based on Condition II, we understand that the student who missed the Mathematics examination did not miss any other examination. This indicates that the Maths score is bound to be the average of the best 3 out of the 4 exam scores obtained by this candidate. Based on this inference, we can proceed with identifying the math score that can be represented as an average of the rest of the scores. We can straightaway eliminate Deep and Esha as potential candidates, given that their Mathematics score is greater than the rest of the exam scores.
For Alva: best 3 out of 4-80(English), 75(Hindi), 75(Science)
Avg. = 230/3 = 76.67 ≠ 70
For Carl: best 3 out of 4-80(Hindi), 90(Social Science), 100(Science)
Avg. = 270/3 = 90 which matches the given value
… Carl most likely missed his Mathematics examination.
For Foni: best 3 out of 4-83(English), 83 (Social Science), 88(Science) Avg. = 254/3 = 84.67 ≠ 78
Hence, we observe that only Carl has missed his Mathematics examination.
Question-21
Read the information provided carefully and answer the questions that follow.
In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science. Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations. The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks
awarded in each of the examinations were on the basis of the scores obtained by them in those examinations.
If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.
If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for
The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations.
The following facts are also known:
1. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations.
II. The student who missed the Mathematics examination did not miss any other examination.
III. One of the students who missed the Hindi examination did not miss any other examination. The other student who missed the Hindi examination also missed the Science examination.
Which of the following students might have missed the General Sciences examination?
a)FOXXY
b)CARDY
c)DEEPALI
d)EITHER CARDY OR FOXXY
e)CANNOT BE DETERMINED
Explanation
SOLUTION[C] ALVA AND DEEPALI
Based on Condition II, we understand that the student who missed the Mathematics examination did
not miss any other examination. This indicates that the Maths score is bound to be the average of th
best 3 out of the 4 exam scores obtained by this candidate. Based on this inference, we can proceed
with identifying the math score that can be represented as an average of the rest of the scores. We
can straightaway eliminate Deep and Esha as potential candidates, given that their Mathematics sco
is greater than the rest of the exam scores. After estimating the average scores for the rest of the
candidates, we observe that only Carl has missed his Mathematics examination.
For Carl: best 3 out of 4-80(Hindi), 90(Social Science), 100(Science) Avg. = 270/3 = 90 which matches the given value
… Carl missed his Mathematics examination.
Further, based on Condition III, we can surmise that the student who missed Hindi and Science shoul
have similar average scores in these two subjects. We notice that Alva has the same score of 75 in
both Hindi and Science. The same can be said about Deep, who has a score of 90 in both these
subjects. Thus, one out of Alva and Deep missed out on Hindi and Science examination, while the
second individual missed out only on the Hindi examination.
Since we know that Carl, Alva and Deep are unlikely to have missed out on the English exam, we can divert our attention to determining which individual out of Bithi, Esha and Foni failed to appear for thi subject. However, we notice that Bithi’s English score is greater than the rest of her scores, thereby helping us eliminate her as the potential candidate.
For Esha: best 3 out of 4-85(Hindi), 95(Mathematics), 60(Science)
Avg. = 240/3 = 80 which matches the given value
… Esha most likely missed her English examination.
For Foni: best 3 out of 4-78(Mathematics), 83 (Social Science), 88(Science)
Avg. = 249/3 = 83 which matches the given value
… Foni most likely missed her English examination.
Based on Condition I, we know that exactly two candidates missed the examinations for English, Hindi, Science, and Social Science.
For English, we determined these individuals to be Esha and Foni. For Hindi, we determined these individuals to be Alva and Deep. For Science, we know one of the individuals is either Alva or Deep. Given that Carl, Alva and Deep cannot be a part of the group that missed Science or Social Science exam, we can proceed by carefully scrutinizing the rest of the group that includes Bithi, Esha and Foni. We notice that Bithi has a similar score in both Science and Social Science examination.
Assuming that she did miss these exams, let us proceed to check if this was actually the case.
For Bithi: Best 2 out 3 – 90(English), 80(Hindi)
Avg = 170/2 = 85 which matches the given value
… Bithi is likely to have missed her Science and Social Science examinations.
We additionally notice that Foni has a similar score in English and Social Science. On considering the best 2 out of 3 scores, the average value of the score for both the subject holds (equal to 83). Thus, wecan conclude that Bithi and Foni missed their Social Science examination.
Thus, the students who missed just one exam were: Carl (Mathematics); Esha (English) and one out of Alva and Deep (Hindi).
Hence of the six students, we can correctly determine the missed subjects for four of them (except Alva and Deep):
Mathematics: Carl; English: Esha &Foni; Hindi: Alva & Deep; Science: Bithi& one out of Alva and Deep; Social Science: Foni&Bithi
Hence, the correct answer to this question is Option B: Alva and Deep.
CONSIDER-
ALVA=ALVA
BITHI=BAL
CARL=CARDY
DEEP=DEEPALI
ESHA=ELGAR
FONI=FOXXY
Question-22
Read the information provided carefully and answer the questions that follow. In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science.
Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations.
The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks awarded in each of the examinations were on the basis of the scores obtained by them in those examinations.
If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.
If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for
The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations.
The following facts are also known:
1. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations.
II. The student who missed the Mathematics examination did not miss any other examination.
III. One of the students who missed the Hindi examination did not miss any other examination. The other student who missed the Hindi examination also missed the Science examination.
Which amongst the following students did not appear for Hindi examination?
a)BAL AND FOXXY
b)ALVA AND BAL
c)ALVA AND DEEPALI
d)DEEPALI AND BAL
e)CARDY AND DEEPALI
Explanation
SOLUTION [A] FIVE
From statement 6, it can be concluded that the total number of new cases is equal to 12+12+5+14 = 43.
Also, since the total number of cases in Kitmisto is 14, it can be concluded that the number of cases each day is either 2 or 3, where 3 cases are recorded on 4 days and 2 cases are recorded on 1 day. From statement 4, it can be concluded that the number of new cases for Pesmisto will be 0,1,1,1, and 2, in any order(Since the total number of
cases is 5, and the maximum number of new cases is 2).
In statement 3, it is given that the number of new cases kept increasing during the 5-day period.
Now, as it is already known that the maximum number of cases for Pesmisto is 2, the maximum total number of cases in a day(or on day 5) will be less than 12.
Let us consider the maximum number of cases on Day 5 as 10.
Thus the maximum number of cases possible for the remaining days will be 9, 8, 7, and 6. So, the total number of maximum cases possible for this case will be 40(less than 43).
Thus, the number of cases on Day 5 will be 11(i.e., b/w 10 and 11)
Now, if the number of cases on Day 4 is 9, the maximum number of cases possible for the remaining
days will be 8, 7, and 6.
Thus, the maximum number of cases, in this case, will be 41 (less than 43).
So, the number of cases on day 4 will be 10.
Now, if the number of cases on Day 3 is 8, the number of cases on day 2 will be 7, and the maximum
possible number of cases on Day 1 will be 6.
Thus, the number of cases, in this case, will be 42(less than 43).
Thus, the number of cases on day 3 will be 9, the number of cases on day 2 will be 8, and the number of cases on day 1 will be 5.
| NEIGH | DAY 1 | 2 | 3 | 4 | 5 | TOTAL |
| LEV | 12 | |||||
| THRY | 12 | |||||
| PES | 5 | |||||
| KIT | 14 | |||||
| TOTAL | 5 | 8 | 9 | 10 | 11 | 43 |
Since all the neighbourhood has at least one case on Day 1, the only possible combination will be 1, 1, 1, and 2 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Now, for the other 4 days, the number of cases in Kitmisto will be 3.
For day 5, the number of cases will be 3, 3, 2, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively(since the maximum number of cases in Pesmisto is 2).
And since Pesmisto only has the maximum number of cases on one day, the number of cases on day 4 will be 3, 3, 1, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
On day 2, since Kitmisto is the only neighbourhood to have 3 cases, the number of cases on day 2 will be 2, 2, 1, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Now, on day 3, the number of cases will be 3, 3, 0, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Thus, the final table will be as follows
| NEIGH | DAY 1 | 2 | 3 | 4 | 5 | TOTAL |
| LEV | 1 | 2 | 3 | 3 | 3 | 12 |
| THRY | 1 | 1 | 3 | 1 | 2 | 12 |
| PES | 1 | 3 | 0 | 3 | 3 | 5 |
| KIT | 2 | 3 | 3 | 3 | 3 | 14 |
| TOTAL | 5 | 8 | 9 | 10 | 11 | 43 |
Question-23
Read the contents of the passage carefully and answer the questions that follow:
There are only four neighbourhoods in a city – Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1. 2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
On how many days did Levmisto and Tyhrmisto have the same number of new cases ?
a)FIVE
b)TWO
c)ONE
d)SIX
e)NIL
Explanation
SOLUTION[E] ONLY ON DAY 3
Question-24
Read the contents of the passage carefully and answer the questions that follow: There are only four neighbourhoods in a city – Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of
new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1. 2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
On which day(s) did Pesmisto have no new case at all ?
a)on day 1 and day 2
b)only on day 2
c)only on day 1
d)on day 1 and day 4
e)only on day 3
Explanation
SOLUTION[B]
There have been exactly 3 cases in Levmisto on Day 3
Question-25
Read the contents of the passage carefully and answer the questions that follow:
There are only four neighbourhoods in a city – Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1. 2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
Which of the following can be correctly concluded about the number of new cases in Levmisto on Day 3 ?
a)There have been exactly 1 case in Levmisto on Day 3
b)There have been exactly 3 cases in Levmisto on Day 3
c)There have been exactly 7 cases in Levmisto on Day 3
d)There have been exactly 4 cases in Levmisto on Day 3
e)there have been either 4 or 7 casr in levimisto on day 3
Explanation
SOLUTION[B]-
There have been exactly 8 cases on Day 2
Question-26
Read the contents of the passage carefully and answer the questions that follow:
There are only four neighbourhoods in a city – Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1. 2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
What can be concluded most correctly about the total number of new cases in the city on Day 2?
a)There have been exactly 6 cases on Day 2
b)There have been exactly 8 cases on Day 2
c)There have been exactly 7 cases on Day 2
d)There have been exactly 6or 7 cases on Day 2
e)There have been exactly 13 cases on Day 2
Explanation
SOLUTION[E] 90
Let us compute the total asset base = This is Rs. 70 lakhs + Rs. 50 lakhs + Rs. 3*30 lakhs + n (Where number of gold coins is n). Or, 210 lakhs + n is what the total asset base. This is equally distributed, or n should be a multiple of 3. So, let us rewrite n as 3m. Let us say the total asset base is 210 + 3m. Each child should
get 70 + m. The gold coins were also equally distributed. This tells us that each child got m gold coins each. This makes life simple. Now, the non-gold coin part should have been equally distributed. So, each child should have gotten Rs. 70 lakhs worth stuff. What can we infer from here?
So, each child should have gotten Rs. 70 lakhs worth stuff. 1. Whoever gets the house cannot get a flat. Or, the person who gets the house should get Rs. 20 akhs in cash.
2. No one can get all three flats. So, the flats get distributed as 1 and 2 to the wo remaining children. So, one child gets House + Rs. 20 lakhs, one gets one flat plus Rs. 40 lakhs, and the third received two flats and Rs. 10 lakhs. 3. Beyond all this they receive an equal number of gold coins.
No one can get all three flats. So, the flats get distributed as 1 and 2 to the two
remaining children. So, one child gets House + Rs. 20 lakhs, one gets one flat
plus Rs. 40 lakhs, and the third received two flats and Rs. 10 lakhs.
Neeta receives 2 flats and Rs. 10 lakhs, Geeta receives one flat and Rs. 40
lakhs and Seeta gets house and Rs. 20 lakhs.
The value of the assets were distributed in the ratio of 1:2:3. If we take them as 1x, 2x and 3x, then Total value = x + 2x + 3x = 6x .
If there are ” m ^ prime prime gold coins, then the total 210 + m is a multiple of 6. 210 is a multiple of 6. So, m is a multiple of 6=> m = 6a .
Also, the gold coins were distributed in the ratio of 2/3 / 4 =>No of gold coins is 2y + 3y + 4y = 9y m = 9y .
m is a multiple of 6 and 9. Therefore, m is a multiple of 18. Hence, m can be expressed as 18k for some k. Now, total value of assets = 210 + 18k This is split in the ratio of 1:2:3 among Neeta, Seeta and Geeta.
Neeta gets =1/6^ * (210 + 18K) = 35 + 3k = (35 – k) + 4k .
Seeta gets =2/6^ * (210 + 18K) = 70 + 6k .
Geeta gets =3/6^ * (210 + 18k) = 105 + 9k = (105 + k) + 8k .
We also know that, One child got all three flats and she did not get the house.
One child, other than Geeta, got Rs. 30 lakh in bank deposits.
One of Seeta and Neeta got Rs. 30 lakh in bank deposits.
30 lakhs. Then the remaining amount she will have is 40 + 6k which is not possible through any combination.
Hence, Neeta gets Rs. 30 lakh in bank deposits. Neeta got totally (35 – k) + 4k worth of assets. So, k = 5 Neeta gets Rs. 30 lakh in bank deposits and 4*5 = 20 gold coins.
Seeta gets one house and Rs. 20 lakh in bank deposits and 30 gold coins. Geeta gets 3 flats and Rs. 20 lakh in bank deposits and 40 gold coins. Therefore, total no. of gold coins = 20 + 30 + 40 = 90 .
The question is “How many gold coins did the old woman have?”
Hence, the answer is “90”.
Question-27
Read the information given below carefully and answer the questions that follow:
An old woman had the following assets:
(a) Rs. 70 lakh in Bank deposits
(b) 1 house worth Rs. 50 lakh
(c) 3 flats, each worth Rs. 30 lakh
(d) Certain number of gold coins, each worth Rs. 1 lakh
She wanted to distribute her assets among her three children, Neeta, Seeta and Geeta.
The house, any of the flats or any of the coins were not to be split. That is, the house went entirely to one child; a flat went to one child and similarly, gold coins went to one child.
Among the three, Neeta received the least amount in bank deposits, while Geeta received the highest.
The value of the assets was distributed equally among the children, as were the gold coins.
The value of the assets distributed among Neeta, Seeta and Geeta was in the ratio of 1:2:3, while the gold coins were distributed among them in the ratio of 2:3:4. One child got all three flats and she did not get the house. One child, other than Geeta, got Rs. 30 lakh in bank deposits.
How many gold coins in all did the old woman have?
a)99
b)81
c)72
d)102
e)90
Explanation
SOLUTION[D] 20 L
Let us compute the total asset base = This is Rs. 70 lakhs + Rs. 50 lakhs + Rs. 3 * 30 lakhs + n (Where number of gold coins is n).
Or, 210 lakhs + n is what the total asset base. This is equally distributed, or n should be a multiple of 3. So, let us rewrite n as 3m. Let us say the total asset base is 210 + 3m. Each child should get 70 + m. The gold coins were also equally distributed. This tells us that each child got m gold coins each. This makes life simple. Now, the non-gold coin part should have been equally distributed. So, each child should have gotten Rs. 70 lakhs worth stuff. What can we infer from here?
So, each child should have gotten Rs. 70 lakhs worth stuff. 1. Whoever gets the house cannot get a flat. Or, the person who gets the house should get Rs. 20 akhs in cash.
2. No one can get all three flats. So, the flats get distributed as 1 and 2 to the two remaining children. So, one child gets House + Rs. 20 lakhs, one gets one flat plus Rs. 40 lakhs, and the third received two flats and Rs. 10 lakhs.
3. Beyond all this they receive an equal number of gold coins.
No one can get all three flats. So, the flats get distributed as 1 and 2 to the two remaining children. So, one child gets House + Rs. 20 lakhs, one gets one flat
plus Rs. 40 lakhs, and the third received two flats and Rs. 10 lakhs.
Neeta receives 2 flats and Rs. 10 lakhs, Geeta receives one flat and Rs. 40
lakhs and Seeta gets house and Rs. 20 lakhs.
From the stem, it is clear that Seeta receives a house and Rs. 20 lakhs in bank deposits.
The question is “How much did Seeta receive in bank deposits (in lakhs of rupees)?”
Hence, the answer is “20 lakhs”.
Question-28
Read the information given below carefully and answer the questions that follow:
An old woman had the following assets:
(a) Rs. 70 lakh in Bank deposits
(b) 1 house worth Rs. 50 lakh
(c) 3 flats, each worth Rs. 30 lakh
(d) Certain number of gold coins, each worth Rs. 1 lakh
She wanted to distribute her assets among her three children, Neeta, Seeta and Geeta.
The house, any of the flats or any of the coins were not to be split. That is, the house went entirely to one child; a flat went to one child and similarly, gold coins went to one child.
Among the three, Neeta received the least amount in bank deposits, while Geeta received the highest.
The value of the assets was distributed equally among the children, as were the gold coins.
How much did Seeta receive in bank deposits ?
a)RS 48 L
b)RS 40 L
c)RS 30 L
d)RS 20 L
e)RS 10 L
Explanation
SOLUTION[A] – A
Company A added Rs. 300 crores, the maximum addition in absolute terms
Question-29
Read the information given carefully and answer the question that follows:
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015.
Company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B’s revenues in 2016 was equal to that of Company D in 2015, while Company C’s 2016 revenue was equal to that of Company B’s in 2015, Company A’s 2016 revenue was equal to that of Company E in 2015.
Ananyaa, an accountant observes that one of the companies has twice the growth rate of another. Mushtaq, her colleague corrects her and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D. and growth rate equal to the average growth rate of A and D.
Ravinder, known for his cryptic-speak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
In absolute terms, which Company added the maximum revenue in the year 2016?
Explanation
SOLUTION[A] – A
Company A added Rs. 300 crores, the maximum addition in absolute terms
Question- 30
Read the information given carefully and answer the question that follows:
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. Company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B’s revenues in 2016 was equal to that of Company D in 2015, while Company C’s 2016
revenue was equal to that of Company B’s in 2015, Company A’s 2016 revenue was equal to that of Company E in 2015.
Ananyaa, an accountant observes that one of the companies has twice the growth rate of another. Mushtaq, her colleague corrects her and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D. and growth rate equal to the average growth rate of A and D.
Ravinder, known for his cryptic-speak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
Which Company had the third highest growth rate ?
a)D
b)E
c)A
d)C
e)B
Explanation
SOLUTION [D] COMPANY C
Question-31
Read the information given carefully and answer the question that follows:
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015.
Company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B’s revenues in 2016 was equal to that of Company D in 2015, while Company C’s 2016 revenue was equal to that of Company B’s in 2015, Company A’s 2016 revenue was equal to that of Company E in 2015.
Ananyaa, an accountant observes that one of the companies has twice the growth rate of another. Mushtaq, her colleague corrects her and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D. and growth rate equal to the average growth rate of A and D.
Ravinder, known for his cryptic-speak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
What is the overall average growth rate seen by all 5 companies put together?
a)23.5%
b)28.5%
c)24.2%
d)27%
e)18.5%
Explanation
SOLUTION [C] 24.2%
