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CAT 2022 Question Paper Slot 3 | All Questions

Previous Year Questions

    01.

    The passage below is accompanied by a set of questions. Choose the best answer to each question.



    Nature has all along yielded her flesh to humans. First, we took nature's materials as food, fibers, and shelter. Then we learned to extract raw materials from her biosphere to create our own new synthetic materials. Now Bios is yielding us her mind-we are taking her logic.

     

    Clockwork logic-the logic of the machines-will only build simple contraptions. Truly complex systems such as a cell, a meadow, an economy, or a brain (natural or artificial) require a rigorous nontechnological logic. We now see that no logic except bio-logic can assemble a thinking device, or even a workable system of any magnitude.

     

    It is an astounding discovery that one can extract the logic of Bios out of biology and have something useful. Although many philosophers in the past have suspected one could abstract the laws of life and apply them elsewhere, it wasn't until the complexity of computers and human-made systems became as complicated as living things, that it was possible to prove this. It's eerie how much of life can be transferred. So far, some of the traits of the living that have successfully been transported to mechanical systems are: self-replication, self-governance, limited self-repair, mild evolution, and partial learning.

     

    We have reason to believe yet more can be synthesized and made into something new. Yet at the same time that the logic of Bios is being imported into machines, the logic of Technos is being imported into life. The root of bioengineering is the desire to control the organic long enough to improve it. Domesticated plants and animals are examples of technos-logic applied to life. The wild aromatic root of the Queen Anne's lace weed has been fine-tuned over generations by selective herb gatherers until it has evolved into a sweet carrot of the garden; the udders of wild bovines have been selectively enlarged in a "unnatural" way to satisfy humans rather than calves. Milk cows and carrots, therefore, are human inventions as much as steam engines and gunpowder are. But milk cows and carrots are more indicative of the kind of inventions humans will make in the future: products that are grown rather than manufactured.

     

    Genetic engineering is precisely what cattle breeders do when they select better strains of Holsteins, only bioengineers employ more precise and powerful control. While carrot and milk cow breeders had to rely on diffuse organic evolution, modern genetic engineers can use directed artificial evolution-purposeful design-which greatly accelerates improvements.

     

    The overlap of the mechanical and the lifelike increases year by year. Part of this bionic convergence is a matter of words. The meanings of "mechanical" and "life" are both stretching until all complicated things can be perceived as machines, and all self-sustaining machines can be perceived as alive. Yet beyond semantics, two concrete trends are happening: (1) Human-made things are behaving more lifelike, and (2) Life is becoming more engineered. The apparent veil between the organic and the manufactured has crumpled to reveal that the two really are, and have always been, of one being.

     

    01.

    The author claims that, "Part of this bionic convergence is a matter of words". Which one of the following statements best expresses the point being made by the author?

    Option D is the correct answer.

    Video Explanation

    Explanatory Answer

    “The overlap of the mechanical and the lifelike increases year by year. Part of this bionic convergence is a matter of words . The meanings of “mechanical” and “life” are both stretching until all complicated things can be perceived as machines, and all self-sustaining machines can be perceived as alive.”

     

    From the above line, the author tries to show the increasing similarities between ‘mechanical’ and ‘lifelike’ with the passage of time. He states that this increase in similarities will continue till the meanings and the perception of the words become synonymous.

     

    Option A: This option states the opposite of what the author tried to convey and hence is not the correct option. Option B: This option is distorted and can be rejected on the same grounds as option A.

     

    Option C: This is a distorted inference, and the author did not use the above statement to show the meeting grounds of ‘genetic engineering’ and ‘mechanical engineering’. Thus, this is not the correct option.

     

    Option D: This option aptly expresses the point made by the author in the last paragraph, and hence is the correct option. Thus, the correct option is D. 

    02.

    Which one of the following sets of words/phrases best serves as keywords to the passage?

    Option D is the correct answer.

    Video Explanation

    Explanatory Answer

    The starting two paragraphs discuss the complexity of the biosphere and how it is impossible to build a thinking device without bio-logic in the next paragraphs, the author describes how with the increasing complexity of human-made systems(not until it was comparable to living things), it has become possible to transfer these traits into mechanical systems. Examples of these are bioengineering and genetic engineering. Then in the concluding paragraph, the author discusses about the convergence of these two logics(Biologic and Techno logic).

     

    Options B and C do not talk about the conclusion of the passage(convergence of the logics), and hence can be eliminated. Out of options A and D, we should select the option with bio-logic and techno-logic instead of carrots and cows, because the broader idea is about bio and techno, not carrots and cows.

     

    Thus, the correct option is D. 

    03.

    None of the following statements is implied by the arguments of the passage, EXCEPT:

    Option C is the correct answer.

    Video Explanation

    Explanatory Answer

    "Although many philosophers in the past have suspected one could abstract the laws of life and apply them elsewhere, it wasn't until the complexity of computers and human-made systems became as complicated as living things that it was possible to prove this."

     

    Option A can be easily rejected from the above excerpt from the passage. Also, it can be inferred that now(not before), since the complexity of computers and human-made systems are comparable, the logic of Bios can be applied to machines. Although option C seems to convey the same meaning, it generalises the complexity and is a distorted inference.

     

    The author has nowhere mentioned or implied in the passage that purposeful design represents the pinnacle of scientific expertise in the service of human betterment and civilisational progress.

     

    Thus, option D can also be rejected.

     

    "Genetic engineering is precisely what cattle breeders do when they select better strains of Holsteins, only bioengineers employ more precise and powerful control. While carrot and milk cow breeders had to rely on diffuse organic evolution, modern genetic engineers can use directed artificial evolution—purposeful design—which greatly accelerates improvements."

     

    From the above excerpt from the penultimate paragraph, it can be inferred that although genetic engineering has less control over the products than bioengineering, they both try to evolve the product artificially. Thus, option B can be inferred from the passage.

     

    Thus, the correct option is B. 

    04.

    The author claims that, "The apparent veil between the organic and the manufactured has crumpled to reveal that the two really are, and have always been, of one being." Which one of the following statements best expresses the point being made by the author here?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

    "Yet beyond semantics, two concrete trends are happening: (1)Human-made things are behaving more lifelike, and (2) Life is becoming more engineered. The apparent veil between the organic and the manufactured has crumpled to reveal that the two really are, and have always been, of one being .."

     

    The main argument made by the author in the last paragraph is regarding the increasing similarities between manufactured and organic(lifelike) reality. According to the author, the growing similarities(because of the scientific advances) have distorted the understanding of the realities and have made us think that perhaps these two are and have always been the same.

     

    Option A: This is a distorted inference. It is not that the Organic reality has crumpled under the veil of manufacturing; instead, their meanings are converging mutually. Thus, this is not the correct option.

     

    Option B: This is again a distorted inference. It is not the organic veil that has crumpled; instead, it is the apparent veil. Similarly, in the second half of the option, the organic reality is replaced with the apparent reality.

     

    Option C: This option aptly expresses the main point of the author and is the correct option.

     

    Option D: The author nowhere stated or implied this, and hence this option can be easilyeliminated. Thus, the correct option is D. 

    02.

    Read the following information carefully, analyze it, and answer the question based on it. 

     

    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.

    01.

    What BEST can be concluded about the total number of new cases in the city on Day 2?

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

    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.

     

    Screenshot_74.png     

     

    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: 

     

    Screenshot_75.png     

     

    From the data, it can be concluded that the total number of cases on Day 2 is equal to 8. Thus, the correct option is D.

    02.

    What BEST can be concluded about the number of new cases in Levmisto on Day 3?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

    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.

     

    Screenshot_74.png     

     

    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: 

     

    Screenshot_75.png     

    From the final table, it can be concluded that the total number of cases in Levmisto is 3 on day 3.
    Thus, the correct option is C.

    03.

    On which day(s) did Pesmisto not have any new case?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

    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.

     

    Screenshot_74.png     

     

    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: 

     

    Screenshot_75.png     

     

    From the final table, it can be concluded that on Day 3, the number of cases will be zero for Pesmisto. Thus, the correct option is A. 

    04.

    Which of the two statements below is/are necessarily false?
    Statement A: There were 2 new cases in Tyhrmisto on Day 3.
    Statement B: There were no new cases in Pesmisto on Day 2.

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

    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.

     

    Screenshot_74.png     

     

    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: 

     

    Screenshot_75.png     

    From the final table, it can be concluded that both statements are false. Thus the correct option is D. 

     

    05.

    On how many days did Levmisto and Tyhrmisto have the same number of new cases?

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

    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.

     

    Screenshot_74.png     

     

    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: 

     

    Screenshot_75.png     

     

    It can be concluded from the final table that the number of cases will be the same for all the days.
    Thus, the correct option is D

    03.

    Read the following information carefully, analyze it, and answer the question based on it. 

    In the following, a year corresponds to 1st of January of that year.

    A study to determine the mortality rate for a disease began in 1980. The study chose 1000 males and 1000 females and followed them for forty years or until they died, whichever came first. The 1000 males chosen in 1980 consisted of 250 each of ages 10 to less than 20, 20 to less than 30, 30 to less than 40, and 40 to less than 50. The 1000 females chosen in 1980 also consisted of 250 each of ages 10 to less than 20, 20 to less than 30, 30 to less than 40, and 40 to less than 50.

    The four figures below depict the age profile of those among the 2000 individuals who were still alive in 1990, 2000, 2010, and 2020. The blue bars in each figure represent the number of males in each age group at that point in time, while the pink bars represent the number of females in each age group at that point in time. The numbers next to the bars give the exact numbers being represented by the bars. For example, we know that 230 males among those tracked and who were alive in 1990 were aged between 20 and 30.

    01.

    How many individuals who were being tracked and who were less than 30 years of age in 1980 survived until 2020?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

    The total number of male and female test cases in 1980 = 1000 

     

    Screenshot_73.png     

     

    The total number of males alive in 2000 = 180 + 205 + 160 + 100 = 645 Thus, the number of dead males in 2000 = 1000 - 645 = 355

    Similarly, the total number of dead females in 2000 = 1000 - (210 + 175 + 150 + 120) = 1000 - 655 = 345

    Thus, the required ratio = 355 : 345 = 71 : 69. Thus, the correct option is A. 

    02.

    How many of the males who were being tracked and who were between 20 and 30 years of age in 1980 died in the period 2000 to 2010?

    Answer : 40

    Video Explanation

    Explanatory Answer

    The total number of male and female test cases in 1980 = 1000 

     

    Screenshot_73.png     

     

    The total number of males between 20 and 30 years of age in 1980 who died in 2000 = 205 The total number of males between 20 and 30 years of age in 1980 who died in 2010 = 165

    Thus, the total number of males between 20 and 30 years of age in 1980 who died in the period 2000 to 2010 = 205 - 165 = 40

    Hence, 40 

    03.

    How many of the females who were being tracked and who were between 20 and 30 years of age in 1980 died between the ages of 50 and 60?

    Answer : 30

    Video Explanation

    Explanatory Answer

    The total number of male and female test cases in 1980 = 1000 

     

    Screenshot_73.png     

     

    We are given that there are 250 females from age 20-30 in 1980 and in 2000 these females age are from 40-50 but only 175 are alive in 2000.

     

    In 2000 there were 175 females from age 40-50. If we assume that out of these, 30 females were of age 48 years in 2000 and they died in 2005, then there are 30 females who died at the age of 53.

     

    If we assume that out of the 175 females, 30 females were of age 42 years in 2000, and they died in 2005, then 30 females died at the age of 47. Now, if we assume that there are 15 females of age 42 and 15 females of age 48 in the year 2000, and they all died in 2005, then we have 15 females who died at the age of 47 and 15 females who died at the age of 53.

     

    So we can see that there are many cases possible. We are given that there were 250 females aged 20-30 in 1980, and in 2010, these females ages are from 50-60, but only 145 are alive in 2010.

     

    In 2010 there were 145 females from age 50-60. If we assume that out of these, 40 females were of age 58 years in 2010 and they died in 2015, then there are 40 females who died at the age of 63.

     

    If we assume that out of the 145 females, 40 females are of age 52 years age in 2010, and they died in 2015, then 40 females died at the age of 57. Now, if we assume that there are 15 females of age 52 and 25 females of age 58 in the year 2010, and they all died in 2015, then we have 15 females who died at the age of 57 and 25 females who died at the age of 63.

     

    So we can see that again, there are many cases possible. In the first case, the range of values possible is from 0 to 30. In the second case, the range of values possible is from 0 to 40. So in total, we get a range of possible values from 0 to 70.

     

    Thus, only one possible value of this question is not possible.

     

    Hence, 30 

    04.

    Read the following information carefully, analyze it, and answer the question based on it. 

     

    Pulak, Qasim, Ritesh, and Suresh participated in a tournament comprising of eight rounds. In each round, they formed two pairs, with each of them being in exactly one pair. The only restriction in the pairing was that the pairs would change in successive rounds. For example, if Pulak formed a pair with Qasim in the first round, then he would have to form a pair with Ritesh or Suresh in the second round. He would be free to pair with Qasim again in the third round. In each round, each pair decided whether to play the game in that round or not. If they decided not to play, then no money was exchanged between them. If they decided to play, they had to bet either ₹1 or ₹2 in that round. For example, if they chose to bet ₹2, then the player winning the game got ₹2 from the one losing the game.

     

    At the beginning of the tournament, the players had ₹10 each. The following table shows partial information about the amounts that the players had at the end of each of the eight rounds. It shows every time a player had ₹10 at the end of a round, as well as every time, at the end of a round, a player had either the minimum or the maximum amount that he would have had across the eight rounds. For example, Suresh had ₹10 at the end of Rounds 1, 3, and 8 and not after any of the other rounds. The maximum amount that he had at the end of any round was ₹13 (at the end of Round 5), and the minimum amount he had at the end of any round was ₹8 (at the end of Round 2). At the end of all other rounds, he must have had either ₹9, ₹11, or ₹12.

     

    It was also known that Pulak and Qasim had the same amount of money with them at the end of Round 4.

     

    Screenshot_52.png

    01.

    What BEST can be said about the amount of money that Ritesh had with him at the end of Round 8?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

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    02.

    What BEST can be said about the amount of money that Pulak had with him at the end of Round 6?

    Option A is the correct answer.

    Video Explanation

    Explanatory Answer

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    03.

    How much money (in ₹) did Ritesh have at the end of Round 4?

    Answer : 6

    Video Explanation

    Explanatory Answer

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    04.

    How many games were played with a bet of ₹2?

    Answer : 6

    Video Explanation

    Explanatory Answer

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    05.

    Which of the following pairings was made in Round 5?

    Option C is the correct answer.

    Video Explanation

    Explanatory Answer

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    06.

    In 2000, what was the ratio of the number of dead males to dead females among those being tracked?

    Option C is the correct answer.

    Video Explanation

    Explanatory Answer

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    07.

    How many people who were being tracked and who were between 30 and 40 years of age in 1980 survived until 2010?

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

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    05.

    Read the following information carefully, analyze it, and answer the question based on it. 

     

    All the first-year students in the computer science (CS) department in a university take both the courses (i) AI and (ii) ML. Students from other departments (non-CS students) can also take one of these two courses, but not both. Students who fail in a course get an F grade; others pass and are awarded A or B or C grades depending on their performance. The following are some additional facts about the number of students who took these two courses this year and the grades they obtained.

    1. The numbers of non-CS students who took AI and ML were in the ratio 2 : 5.
    2. The number of non-CS students who took either AI or ML was equal to the number of CS students.
    3. The numbers of non-CS students who failed in the two courses were the same and their total is equal to the number of CS students who got a C grade in ML.
    4. In both the courses, 50% of the students who passed got a B grade. But, while the numbers of students who got A and C grades were the same for AI, they were in the ratio 3 : 2 for ML.
    5. No CS student failed in AI, while no non-CS student got an A grade in AI.
    6. The numbers of CS students who got A, B and C grades respectively in AI were in the ratio 3 : 5 : 2, while in ML the ratio was 4 : 5 : 2.
    7. The ratio of the total number of non-CS students failing in one of the two courses to the number of CS students failing in one of the two courses was 3 : 1.
    8. 30 students failed in ML.

    01.

    How many students took AI?

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

    The numbers of CS students who got A, B and C grades respectively in AI were in t
    he ratio 3a, 5a and 2a, while in ML the ratio was 4b, 5b and 2b.

    From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
    n each category. Now from statement 8, we can say that number of CS students who failed in MI i
    s equal to 30-b. 

     

    Screenshot_69.png

     

    2 b 3

    From statement 7, we can say that, 30− b = 1

    or, 2b = 90-3b or, b= 18

    CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

    From statement 2, we can say that 10a = 7x or, x = 30 

    Substituting the values of a, b and x in the above table. 

     

    Screenshot_71.png

     

    A total of 270 students took AI out of which 252 students passed and a total of 360 students took
    ML out of which 330 students passed.

    From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
    de B and 63 got Grade A and 63 got Grade C.

    Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
    99 got Grade A and 66 got Grade C. Therefore, the final table which we get is

     

    Screenshot_70.png 

    02.

    How many CS students failed in ML?

    Answer : 12

    Video Explanation

    Explanatory Answer

    The numbers of CS students who got A, B and C grades respectively in AI were in t
    he ratio 3a, 5a and 2a, while in ML the ratio was 4b, 5b and 2b.

    From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
    n each category. Now from statement 8, we can say that number of CS students who failed in MI i
    s equal to 30-b. 

     

    Screenshot_69.png

     

    2 b 3

    From statement 7, we can say that, 30− b = 1

    or, 2b = 90-3b or, b= 18

    CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

    From statement 2, we can say that 10a = 7x or, x = 30 

    Substituting the values of a, b and x in the above table. 

     

    Screenshot_71.png

     

    A total of 270 students took AI out of which 252 students passed and a total of 360 students took
    ML out of which 330 students passed.

    From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
    de B and 63 got Grade A and 63 got Grade C.

    Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
    99 got Grade A and 66 got Grade C. Therefore, the final table which we get is

     

    Screenshot_70.png 

    03.

    How many non-CS students got A grade in ML?

    Answer : 27

    Video Explanation

    Explanatory Answer

    The numbers of CS students who got A, B and C grades respectively in AI were in t
    he ratio 3a, 5a and 2a, while in ML the ratio was 4b, 5b and 2b.

    From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
    n each category. Now from statement 8, we can say that number of CS students who failed in MI i
    s equal to 30-b. 

     

    Screenshot_69.png

     

    2 b 3

    From statement 7, we can say that, 30− b = 1

    or, 2b = 90-3b or, b= 18

    CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

    From statement 2, we can say that 10a = 7x or, x = 30 

    Substituting the values of a, b and x in the above table. 

     

    Screenshot_71.png

     

    A total of 270 students took AI out of which 252 students passed and a total of 360 students took
    ML out of which 330 students passed.

    From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
    de B and 63 got Grade A and 63 got Grade C.

    Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
    99 got Grade A and 66 got Grade C. Therefore, the final table which we get is

     

    Screenshot_70.png 

    04.

    How many students got A grade in AI?

    Option D is the correct answer.

    Video Explanation

    Explanatory Answer

    The numbers of CS students who got A, B and C grades respectively in AI were in t
    he ratio 3a, 5a and 2a, while in ML the ratio was 4b, 5b and 2b.

    From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
    n each category. Now from statement 8, we can say that number of CS students who failed in MI i
    s equal to 30-b. 

     

    Screenshot_69.png

     

    2 b 3

    From statement 7, we can say that, 30− b = 1

    or, 2b = 90-3b or, b= 18

    CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

    From statement 2, we can say that 10a = 7x or, x = 30 

    Substituting the values of a, b and x in the above table. 

     

    Screenshot_71.png

     

    A total of 270 students took AI out of which 252 students passed and a total of 360 students took
    ML out of which 330 students passed.

    From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
    de B and 63 got Grade A and 63 got Grade C.

    Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
    99 got Grade A and 66 got Grade C. Therefore, the final table which we get is

     

    Screenshot_70.png 

    05.

    How many non-CS students got B grade in ML?

    Option D is the correct answer.

    Video Explanation

    Explanatory Answer

    The numbers of CS students who got A, B and C grades respectively in AI were in t
    he ratio 3a, 5a and 2a, while in ML the ratio was 4b, 5b and 2b.

    From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
    n each category. Now from statement 8, we can say that number of CS students who failed in MI i
    s equal to 30-b. 

     

    Screenshot_69.png

     

    2 b 3

    From statement 7, we can say that, 30− b = 1

    or, 2b = 90-3b or, b= 18

    CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

    From statement 2, we can say that 10a = 7x or, x = 30 

    Substituting the values of a, b and x in the above table. 

     

    Screenshot_71.png

     

    A total of 270 students took AI out of which 252 students passed and a total of 360 students took
    ML out of which 330 students passed.

    From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
    de B and 63 got Grade A and 63 got Grade C.

    Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
    99 got Grade A and 66 got Grade C. Therefore, the final table which we get is

     

    Screenshot_70.png 

    06.

    The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

     

    Option C is the correct answer.

    Video Explanation

    Explanatory Answer

    In any polygon, the largest side length should be less than the sum of the lengths of the other sides.
    Let the fourth side be ‘x’.
    Case i) x is the largest side.
    x < 1 + 2 + 6
    x < 7
    x can take the following values, 4, 5, 6
    Case ii)
    4 is the largest side and x is less than 4.
    4 < x + 1 +2
    x > 1
    x can take the values 2 or 3
    So, x can take the values 2, 3, 4, 5, 6.
    x can take 5 values.

    08.

    A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is

     
    Answer : The answer is '12'

    Video Explanation

    Explanatory Answer

    Since the total donation amount is 15,250 rupees, there should be at least one 250 rupee note. The remaining 15000 can exist as thirty 500 rupee notes.

    Since the total donation amount is 15,250 rupees, there should be at least one 250 rupee note. The remaining 15000 can exist as thirty 500 rupee notes.

    Rs. 250 Rs. 500 Rs. 100 Number of notes
    1 30 0 31

    But the total number of notes will only be 31. We have 100 notes. So to keep the number of 500 notes as high as possible let’s convert the 500 rupee notes to 100 rupee notes. Every time we do this conversion we add 4 new notes.

    Rs. 250 Rs. 500 Rs. 100 Number of notes
    1 30 0 31
    1 29 5 35
    We add 4 new notes in each conversion, Our target is to reach 100 notes from 35 notes, the closest we can got to 100 by adding only 4’s to 35 is 99
    99 = 35 + 4 (16)
    So this happens after 16 steps
    1 13 85 99

    From here, we can convert one 500 rupee note to two 250 rupee notes.

    3 12 85 100

    So, the maximum number of 500 rupee notes is 12

    09.

    A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

     

    Answer : The answer is '150'

    Video Explanation

    Explanatory Answer

    Let the number of students in the school be N.
    N < 5000
    N leaves a remainder of 4 when divided by 9, 10, 12, or 25.
    Since 4 is less than 9, 10, 12 and 25
    N leaves a remainder of 4 when divided by LCM(9, 10, 12, 25).
    N leaves a remainder of 4 when divided by 900.
    N = 900(x) + 4
    Since N < 5000
    x can range from 0 to 5
    But 900(x) + 4 is a multiple of 11 only when x = 2
    N = 900(2) + 4 = 1804
    When we divide these 1804 students into groups of 12, we get, 150 groups.
    Because, 1804 = 12(150) + 4

    10.

    In a triangle ABC, AB = AC = 8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is

     

    Option B is the correct answer.

    Video Explanation

    Explanatory Answer

    Overlapping circles