CAT Mock Test - 2 (New Pattern)


75 Questions MCQ Test CAT Mock Test Series | CAT Mock Test - 2 (New Pattern)


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This mock test of CAT Mock Test - 2 (New Pattern) for CAT helps you for every CAT entrance exam. This contains 75 Multiple Choice Questions for CAT CAT Mock Test - 2 (New Pattern) (mcq) to study with solutions a complete question bank. The solved questions answers in this CAT Mock Test - 2 (New Pattern) quiz give you a good mix of easy questions and tough questions. CAT students definitely take this CAT Mock Test - 2 (New Pattern) exercise for a better result in the exam. You can find other CAT Mock Test - 2 (New Pattern) extra questions, long questions & short questions for CAT on EduRev as well by searching above.
QUESTION: 1

Read the following passage and answer the questions that follows:
The majority of successful senior managers do not closely follow the classical rational model of first clarifying goals, assessing the problem, formulating options, estimating likelihoods of success, making a decision, and only then taking action to implement the decision. Rather, in their day-by-day tactical manoeuvres, these senior executives rely on what is vaguely termed “intuition” to manage a network of interrelated problems that require them to deal with ambiguity, inconsistency, novelty, and surprise; and to integrate action into the process to thinking.
Generations of writers on management have recognized that some practicing managers rely heavily on intuition. In general, however, such writers display a poor grasp of what intuition is. Some see it as the opposite of rationality; others view it as an excuse for capriciousness.
Isenberg’s recent research on the cognitive processes of senior managers reveals that managers’ intuition is neither of these. Rather, senior managers use intuition in at least five distinct ways. First, they intuitively sense when a problem exists. Second, managers rely on intuition to perform well-learned behaviour patterns rapidly. This intuition is not arbitrary or irrational, but is based on years of painstaking practice and hands-on experience that build skills. A third function of intuition is to synthesize isolated bits of data and practice into an integrated picture, often in an “Aha!” experience. Fourth, some managers use intuition as a check on the results of more rational analysis. Most senior executives are familiar with the formal decision analysis models and tools, and those who use such systematic methods for reaching decisions are occasionally leery of solutions suggested by these methods which run counter to their sense of the correct course of action. Finally, managers can use intuition to bypass in-depth analysis and move rapidly to engender a plausible solution. Used in this way, intuition is an almost instantaneous cognitive process in which a manager recognizes familiar patterns.
One of the implications of the intuitive style of executive management is that “thinking” is inseparable from acting. Since managers often “know” what is right before they can analyze and explain it, they frequently act first and explain later. Analysis is inextricably tied to action in thinking/acting cycles, in which managers develop thoughts about their companies and organizations not by analyzing a problematic situation and then acting, but by acting and analyzing in close concert.
Given the great uncertainty of many of the management issues that they face, senior managers often instigate a course of action simply to learn more about an issue. They then use the results of the action to develop a more complete understanding of the issue. One implication of thinking/acting cycles is that action is often part of defining the problem, not just of implementing the solution.

Q. According to the passage, senior managers use intuition in all of the following ways EXCEPT to

Solution:

The question requires you to recognize which of the choices is NOT mentioned in the passage as a way in which senior managers use intuition. The passage does not mention stipulating goals.

QUESTION: 2

Read the following passage and answer the questions that follows:
The majority of successful senior managers do not closely follow the classical rational model of first clarifying goals, assessing the problem, formulating options, estimating likelihoods of success, making a decision, and only then taking action to implement the decision. Rather, in their day-by-day tactical manoeuvres, these senior executives rely on what is vaguely termed “intuition” to manage a network of interrelated problems that require them to deal with ambiguity, inconsistency, novelty, and surprise; and to integrate action into the process to thinking.
Generations of writers on management have recognized that some practicing managers rely heavily on intuition. In general, however, such writers display a poor grasp of what intuition is. Some see it as the opposite of rationality; others view it as an excuse for capriciousness.
Isenberg’s recent research on the cognitive processes of senior managers reveals that managers’ intuition is neither of these. Rather, senior managers use intuition in at least five distinct ways. First, they intuitively sense when a problem exists. Second, managers rely on intuition to perform well-learned behaviour patterns rapidly. This intuition is not arbitrary or irrational, but is based on years of painstaking practice and hands-on experience that build skills. A third function of intuition is to synthesize isolated bits of data and practice into an integrated picture, often in an “Aha!” experience. Fourth, some managers use intuition as a check on the results of more rational analysis. Most senior executives are familiar with the formal decision analysis models and tools, and those who use such systematic methods for reaching decisions are occasionally leery of solutions suggested by these methods which run counter to their sense of the correct course of action. Finally, managers can use intuition to bypass in-depth analysis and move rapidly to engender a plausible solution. Used in this way, intuition is an almost instantaneous cognitive process in which a manager recognizes familiar patterns.
One of the implications of the intuitive style of executive management is that “thinking” is inseparable from acting. Since managers often “know” what is right before they can analyze and explain it, they frequently act first and explain later. Analysis is inextricably tied to action in thinking/acting cycles, in which managers develop thoughts about their companies and organizations not by analyzing a problematic situation and then acting, but by acting and analyzing in close concert.
Given the great uncertainty of many of the management issues that they face, senior managers often instigate a course of action simply to learn more about an issue. They then use the results of the action to develop a more complete understanding of the issue. One implication of thinking/acting cycles is that action is often part of defining the problem, not just of implementing the solution.

Q. The passage suggests which of the following about the “writers on management” mentioned in paragraph 2?

Solution:

The author asserts that the writers in question “display a poor grasp of what intuition is”. The next paragraph presents a view that, according to the author of the passage, characterizes intuition more accurately than the writers on management do.

Isenberg’s research is specifically described as showing the ways in which managers use intuition. Therefore, what Isenberg correctly comprehends, and the writers in question misunderstand, is how managers use intuition, as this choice states.

QUESTION: 3

Read the following passage and answer the questions that follows:
The majority of successful senior managers do not closely follow the classical rational model of first clarifying goals, assessing the problem, formulating options, estimating likelihoods of success, making a decision, and only then taking action to implement the decision. Rather, in their day-by-day tactical manoeuvres, these senior executives rely on what is vaguely termed “intuition” to manage a network of interrelated problems that require them to deal with ambiguity, inconsistency, novelty, and surprise; and to integrate action into the process to thinking.
Generations of writers on management have recognized that some practicing managers rely heavily on intuition. In general, however, such writers display a poor grasp of what intuition is. Some see it as the opposite of rationality; others view it as an excuse for capriciousness.
Isenberg’s recent research on the cognitive processes of senior managers reveals that managers’ intuition is neither of these. Rather, senior managers use intuition in at least five distinct ways. First, they intuitively sense when a problem exists. Second, managers rely on intuition to perform well-learned behaviour patterns rapidly. This intuition is not arbitrary or irrational, but is based on years of painstaking practice and hands-on experience that build skills. A third function of intuition is to synthesize isolated bits of data and practice into an integrated picture, often in an “Aha!” experience. Fourth, some managers use intuition as a check on the results of more rational analysis. Most senior executives are familiar with the formal decision analysis models and tools, and those who use such systematic methods for reaching decisions are occasionally leery of solutions suggested by these methods which run counter to their sense of the correct course of action. Finally, managers can use intuition to bypass in-depth analysis and move rapidly to engender a plausible solution. Used in this way, intuition is an almost instantaneous cognitive process in which a manager recognizes familiar patterns.
One of the implications of the intuitive style of executive management is that “thinking” is inseparable from acting. Since managers often “know” what is right before they can analyze and explain it, they frequently act first and explain later. Analysis is inextricably tied to action in thinking/acting cycles, in which managers develop thoughts about their companies and organizations not by analyzing a problematic situation and then acting, but by acting and analyzing in close concert.
Given the great uncertainty of many of the management issues that they face, senior managers often instigate a course of action simply to learn more about an issue. They then use the results of the action to develop a more complete understanding of the issue. One implication of thinking/acting cycles is that action is often part of defining the problem, not just of implementing the solution.

Q. According to the passage, the classical model of decision analysis includes all of the following EXCEPT

Solution:

The question requires you to recognize which of the choices is NOT mentioned in the passage as a component of the classical model of decision analysis. Only this choice, “action undertaken in order to discover more information about a problem,” does not appear in the passage.

QUESTION: 4

Read the following passage and answer the questions that follows:
The majority of successful senior managers do not closely follow the classical rational model of first clarifying goals, assessing the problem, formulating options, estimating likelihoods of success, making a decision, and only then taking action to implement the decision. Rather, in their day-by-day tactical manoeuvres, these senior executives rely on what is vaguely termed “intuition” to manage a network of interrelated problems that require them to deal with ambiguity, inconsistency, novelty, and surprise; and to integrate action into the process to thinking.
Generations of writers on management have recognized that some practicing managers rely heavily on intuition. In general, however, such writers display a poor grasp of what intuition is. Some see it as the opposite of rationality; others view it as an excuse for capriciousness.
Isenberg’s recent research on the cognitive processes of senior managers reveals that managers’ intuition is neither of these. Rather, senior managers use intuition in at least five distinct ways. First, they intuitively sense when a problem exists. Second, managers rely on intuition to perform well-learned behaviour patterns rapidly. This intuition is not arbitrary or irrational, but is based on years of painstaking practice and hands-on experience that build skills. A third function of intuition is to synthesize isolated bits of data and practice into an integrated picture, often in an “Aha!” experience. Fourth, some managers use intuition as a check on the results of more rational analysis. Most senior executives are familiar with the formal decision analysis models and tools, and those who use such systematic methods for reaching decisions are occasionally leery of solutions suggested by these methods which run counter to their sense of the correct course of action. Finally, managers can use intuition to bypass in-depth analysis and move rapidly to engender a plausible solution. Used in this way, intuition is an almost instantaneous cognitive process in which a manager recognizes familiar patterns.
One of the implications of the intuitive style of executive management is that “thinking” is inseparable from acting. Since managers often “know” what is right before they can analyze and explain it, they frequently act first and explain later. Analysis is inextricably tied to action in thinking/acting cycles, in which managers develop thoughts about their companies and organizations not by analyzing a problematic situation and then acting, but by acting and analyzing in close concert.
Given the great uncertainty of many of the management issues that they face, senior managers often instigate a course of action simply to learn more about an issue. They then use the results of the action to develop a more complete understanding of the issue. One implication of thinking/acting cycles is that action is often part of defining the problem, not just of implementing the solution.

Q. It can be inferred from the passage that which of the following would most probably be one major difference in behaviour between Manager X, who uses intuition to reach decisions, and Manager Y, who uses only formal decision analysis?

Solution:

The question requires you to compare behavior based on intuition with behavior based on formal decision analysis. This choice specifies that the manager who uses intuition incorporates action into the decision-making process, but the manager who uses formal analysis does not. This distinction is made in several places in the passage.

The passage emphasizes that decision-making and actiontaking are separate steps in formal decision analysis: “making a decision, and only then taking action.” On the other hand, those who use intuition “integrate action into the process of thinking”. Again, the author mentions that in the intuitive style of management, “ ‘thinking’ is inseparable from acting”, and “action is often part of defining the problem”.

QUESTION: 5

Read the following passage and answer the questions that follows:
The majority of successful senior managers do not closely follow the classical rational model of first clarifying goals, assessing the problem, formulating options, estimating likelihoods of success, making a decision, and only then taking action to implement the decision. Rather, in their day-by-day tactical manoeuvres, these senior executives rely on what is vaguely termed “intuition” to manage a network of interrelated problems that require them to deal with ambiguity, inconsistency, novelty, and surprise; and to integrate action into the process to thinking.
Generations of writers on management have recognized that some practicing managers rely heavily on intuition. In general, however, such writers display a poor grasp of what intuition is. Some see it as the opposite of rationality; others view it as an excuse for capriciousness.
Isenberg’s recent research on the cognitive processes of senior managers reveals that managers’ intuition is neither of these. Rather, senior managers use intuition in at least five distinct ways. First, they intuitively sense when a problem exists. Second, managers rely on intuition to perform well-learned behaviour patterns rapidly. This intuition is not arbitrary or irrational, but is based on years of painstaking practice and hands-on experience that build skills. A third function of intuition is to synthesize isolated bits of data and practice into an integrated picture, often in an “Aha!” experience. Fourth, some managers use intuition as a check on the results of more rational analysis. Most senior executives are familiar with the formal decision analysis models and tools, and those who use such systematic methods for reaching decisions are occasionally leery of solutions suggested by these methods which run counter to their sense of the correct course of action. Finally, managers can use intuition to bypass in-depth analysis and move rapidly to engender a plausible solution. Used in this way, intuition is an almost instantaneous cognitive process in which a manager recognizes familiar patterns.
One of the implications of the intuitive style of executive management is that “thinking” is inseparable from acting. Since managers often “know” what is right before they can analyze and explain it, they frequently act first and explain later. Analysis is inextricably tied to action in thinking/acting cycles, in which managers develop thoughts about their companies and organizations not by analyzing a problematic situation and then acting, but by acting and analyzing in close concert.
Given the great uncertainty of many of the management issues that they face, senior managers often instigate a course of action simply to learn more about an issue. They then use the results of the action to develop a more complete understanding of the issue. One implication of thinking/acting cycles is that action is often part of defining the problem, not just of implementing the solution.

Q. The passage provides support for which of the following statements?

Solution:

The question requires you to identify a statement that can be inferred from information in the passage but is not explicitly stated. The author asserts that intuitive managers can “move rapidly to engender a plausible solution” and that their intuition is based on “experience that builds skill”. This implies that the combination of skill and rapidity enables mangers to employ their practical experience more efficiently, as this choice states.

QUESTION: 6

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. It can be inferred from the passage that the morphogenetic determinants present in the early embryo are

Solution:

The second and third paragraphs of the passage indicate that morphogenetic determinants are substances in the embryo that are activated after the egg has been fertilized and that “tell a cell what to become”.

If, as the author asserts in the first paragraph, biologists have succeeded in dividing an embryo into two parts, each of which survives and develops into a normal embryo, it can be concluded that the quantity of morphogenetic determinants in the early embryo is greater than that required for the development of a single individual.

QUESTION: 7

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. The main topic of the passage is

Solution:

In identifying the main topic of the passage, you must consider the passage as a whole.

In the first paragraph, the author provides a historical context for the debate described.

In the second paragraph, concerning when and how the determination of embryo cells takes place.

The third and forth paragraphs provide a specific example of the “Recent discoveries in molecular biology” that may lead to the resolution of that debate.

QUESTION: 8

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. According to the passage, when biologists believed that the cells in the early embryo were undetermined, they made which of the following mistakes?

Solution:

According to the author, early investigators arrived at the conclusion that the cells of the embryo are undetermined because they “found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos”.

However, later biologists discovered that when an embryo was cut in places different from the one used by the early investigators, it did not form two whole embryos. Because the earlier biologists apparently arrived at their conclusion without attempting to cut an embryo in different planes, it would appear that they assumed, erroneously, that different ways of separating the embryos would not affect the fate of the two embryo parts.

QUESTION: 9

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. It can be inferred from the passage that the initial production of histones after an egg is fertilized takes place

Solution:

In the third paragraph, the author asserts that substances that function as morphogenetic determinants are located in the cytoplasm of the cell and become active after the cell is fertilized.

In the fourth paragraph we learn that these substances are “maternal messenger RNA’s” and that they “direct, in large part, the synthesis of histones,” which, after being synthesized, “move into the cell nucleus”.

Thus, it can be inferred that after the egg is fertilized, the initial production of histones occurs in the cytoplasm.

QUESTION: 10

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. It can be inferred from the passage that which of the following is dependent on the fertilization of an egg?

Solution:

Lines in the passage indicate that substances that function as morphogenetic determinants are inactive in the unfertilized egg and that when the egg is fertilized, they “become active and, presumably, govern the behavior of the genes they interact with.”

In the fourth paragraph, we learn that these substances exert their control over the fate of the cell by directing “the synthesis of histones.” Because these histones cannot be synthesized until the substances that function as morphogenetic determinants become active, and because these substances do not become active until the egg is fertilized, it can be inferred that the synthesis of the histones is dependent on the fertilization of the egg.

QUESTION: 11

Read the following passage and answer the questions that follows:
Nearly a century ago, biologists found that if they separated an invertebrate animal embryo into two parts at an early stage of its life, it would survive and develop as two normal embryos. This led them to believe that the cells in the early embryo are undetermined in the sense that each cell has the potential to develop in a variety of different ways. Later biologists found that the situation was not so simple. It matters in which plane the embryo is cut. If it is cut in a plane different from the one used by the early investigators, it will not form two whole embryos.

A debate arose over what exactly was happening. Which embryo cells are determined, just when do they become irreversibly committed to their fates, and what are the “morphogenetic determinants” that tell a cell what to become? But the debate could not be resolved because no one was able to ask the crucial questions in a form in which they could be pursued productively. Recent discoveries in molecular biology, however, have opened up prospects for a resolution of the debate. Now investigators think they know at least some of the molecules that act as morphogenetic determinants in early development. They have been able to show that, in a sense, cell determination begins even before an egg is fertilized.

Studying sea urchins, biologist Paul Gross found that an unfertilized egg contains substances that function as morphogenetic determinants. They are located in the cytoplasm of the egg cell; i.e., in that part of the cell’s protoplasm that lies outside of the nucleus. In the unfertilized egg, the substances are inactive and are not distributed homogeneously. When the egg is fertilized, the substances become active and, presumably, govern the behavior of the genes they interact with. Since the substances are unevenly distributed in the egg, when the fertilized egg divides, the resulting cells are different from the start and so can be qualitatively different in their own gene activity.

The substances that Gross studied are maternal messenger RNA’s –products of certain of the maternal genes. He and other biologists studying a wide variety of organisms have found that these particular RNA’s direct, in large part, the synthesis of histones, a class of proteins that bind to DNA. Once synthesized, the histones move into the cell nucleus, where sections of DNA wrap around them to form a structure that resembles beads, or knots, on a string. The beads are DNA segments wrapped around the histones; the string is the intervening DNA. And it is the structure of these beaded DNA strings that guides the fate of the cells in which they are located.

Q. According to the passage, the morphogenetic determinants present in the unfertilized egg cell are which of the following?

Solution:

In his study of sea urchins, Gross “found that an unfertilized egg contains substances that function as morphogetic determinants.”

The passage asserts that the “substances that Gross studied are maternal messenger RNA’s,” and we learn that these maternal messenger RNA’s can be found in “ a wide variety of organisms”.

QUESTION: 12

Read the following passage and answer the questions that follows:
Archaeology as a profession faces two major problems. First, it is the poorest of the poor. Only paltry sums are available for excavating and even less is available for publishing the results and preserving the sites once excavated. Yet archaeologists deal with priceless objects every day.
Second, there is the problem of illegal excavation, resulting in museum-quality pieces being sold to the highest bidder.
I would like to make an outrageous suggestion that would at one stroke provide funds for archaeology and reduce the amount of illegal digging. I would propose that scientific archeological expeditions and governmental authorities sell excavated artifacts on the open market. Such sales would provide substantial funds for the excavation and preservation of archaeological sites and the publication of results. At the same time, they would break the illegal excavator’s grip on the market, thereby decreasing the inducement to engage in illegal activities.
You might object that professionals excavate to acquire knowledge, not money. Moreover, ancient artifacts are part of our global cultural heritage, which should be available for all to appreciate, not sold to the highest bidder. I agree. Sell nothing that has unique artistic merit or scientific value. But, you might reply, everything that comes out of the ground has scientific value. Here we part company. Theoretically, you may be correct in claiming that every artifact has potential scientific value. Practically, you are wrong.
I refer to the thousands of pottery vessels and ancient lamps that are essentially duplicates of one another. In one small excavation in Cyprus, archaeologists recently uncovered 2,000 virtually indistinguishable small jugs in a single courtyard, even precious royal seal impressions known as melekh handles have been found in abundance — more than 4,000 examples so far.
The basement of museums is simply not large enough to store the artifacts that are likely to be discovered in the future. There is not enough money even to catalogue the finds; as a result, they cannot be found again and become as inaccessible as if they had never been discovered. Indeed, with the help of a computer, sold artifacts could be more accessible than are the pieces stored in bulging museum basements. Prior to sale, each could be photographed and the list of the purchasers could be maintained on the computer A purchaser could even be required to agree to return the piece if it should become needed for scientific purposes. It would be unrealistic to suggest that illegal digging would stop if artifacts were sold in the open market. But the demand for the clandestine product would be substantially reduced. Who would want an unmarked pot when another was available whose provenance was known, and that was dated stratigraphically by the professional archaeologist who excavated it?

Q. The primary purpose of the passage is to propose

Solution:

The first paragraph identifies two major problems faced by the archaeological profession: inadequate funding and illegal digging.

The passage indicates that the author is going to suggest how to remedy both problems, thereby benefiting the archaeological profession. The author proceeds to propose allowing the sale of excavated artifacts and to explain how this would solve both problems . The author then supports the proposal by countering possible objections to it, and in the last paragraph explains how the proposal would curb illegal digging.

Thus, the way information is organized in the passage indicates that the author’s purpose is to suggest that allowing the sale of excavated artifacts would provide funds for the archaeological profession and curb illegal digging.

QUESTION: 13

Read the following passage and answer the questions that follows:
Archaeology as a profession faces two major problems. First, it is the poorest of the poor. Only paltry sums are available for excavating and even less is available for publishing the results and preserving the sites once excavated. Yet archaeologists deal with priceless objects every day.
Second, there is the problem of illegal excavation, resulting in museum-quality pieces being sold to the highest bidder.
I would like to make an outrageous suggestion that would at one stroke provide funds for archaeology and reduce the amount of illegal digging. I would propose that scientific archeological expeditions and governmental authorities sell excavated artifacts on the open market. Such sales would provide substantial funds for the excavation and preservation of archaeological sites and the publication of results. At the same time, they would break the illegal excavator’s grip on the market, thereby decreasing the inducement to engage in illegal activities.
You might object that professionals excavate to acquire knowledge, not money. Moreover, ancient artifacts are part of our global cultural heritage, which should be available for all to appreciate, not sold to the highest bidder. I agree. Sell nothing that has unique artistic merit or scientific value. But, you might reply, everything that comes out of the ground has scientific value. Here we part company. Theoretically, you may be correct in claiming that every artifact has potential scientific value. Practically, you are wrong.
I refer to the thousands of pottery vessels and ancient lamps that are essentially duplicates of one another. In one small excavation in Cyprus, archaeologists recently uncovered 2,000 virtually indistinguishable small jugs in a single courtyard, even precious royal seal impressions known as melekh handles have been found in abundance — more than 4,000 examples so far.
The basement of museums is simply not large enough to store the artifacts that are likely to be discovered in the future. There is not enough money even to catalogue the finds; as a result, they cannot be found again and become as inaccessible as if they had never been discovered. Indeed, with the help of a computer, sold artifacts could be more accessible than are the pieces stored in bulging museum basements. Prior to sale, each could be photographed and the list of the purchasers could be maintained on the computer A purchaser could even be required to agree to return the piece if it should become needed for scientific purposes. It would be unrealistic to suggest that illegal digging would stop if artifacts were sold in the open market. But the demand for the clandestine product would be substantially reduced. Who would want an unmarked pot when another was available whose provenance was known, and that was dated stratigraphically by the professional archaeologist who excavated it?

Q. The author implies that all of the following statements about duplicate artifacts are true EXCEPT:

Solution:

The question requires you to identify the answer choice that CANNOT be inferred from the passage. Nothing in the passage implies that duplicate artifacts exceed museum objects in quality.

QUESTION: 14

Read the following passage and answer the questions that follows:
Archaeology as a profession faces two major problems. First, it is the poorest of the poor. Only paltry sums are available for excavating and even less is available for publishing the results and preserving the sites once excavated. Yet archaeologists deal with priceless objects every day.
Second, there is the problem of illegal excavation, resulting in museum-quality pieces being sold to the highest bidder.
I would like to make an outrageous suggestion that would at one stroke provide funds for archaeology and reduce the amount of illegal digging. I would propose that scientific archeological expeditions and governmental authorities sell excavated artifacts on the open market. Such sales would provide substantial funds for the excavation and preservation of archaeological sites and the publication of results. At the same time, they would break the illegal excavator’s grip on the market, thereby decreasing the inducement to engage in illegal activities.
You might object that professionals excavate to acquire knowledge, not money. Moreover, ancient artifacts are part of our global cultural heritage, which should be available for all to appreciate, not sold to the highest bidder. I agree. Sell nothing that has unique artistic merit or scientific value. But, you might reply, everything that comes out of the ground has scientific value. Here we part company. Theoretically, you may be correct in claiming that every artifact has potential scientific value. Practically, you are wrong.
I refer to the thousands of pottery vessels and ancient lamps that are essentially duplicates of one another. In one small excavation in Cyprus, archaeologists recently uncovered 2,000 virtually indistinguishable small jugs in a single courtyard, even precious royal seal impressions known as melekh handles have been found in abundance — more than 4,000 examples so far.
The basement of museums is simply not large enough to store the artifacts that are likely to be discovered in the future. There is not enough money even to catalogue the finds; as a result, they cannot be found again and become as inaccessible as if they had never been discovered. Indeed, with the help of a computer, sold artifacts could be more accessible than are the pieces stored in bulging museum basements. Prior to sale, each could be photographed and the list of the purchasers could be maintained on the computer A purchaser could even be required to agree to return the piece if it should become needed for scientific purposes. It would be unrealistic to suggest that illegal digging would stop if artifacts were sold in the open market. But the demand for the clandestine product would be substantially reduced. Who would want an unmarked pot when another was available whose provenance was known, and that was dated stratigraphically by the professional archaeologist who excavated it?

Q. Which of the following is mentioned in the passage as a disadvantage of storing artifacts in museum basements?

Solution:

The disadvantages of storing artifacts in museum basements are discussed in the fifth paragraph. The passage states that “There is not enough money…to catalogue the finds” and declare that, as a result stored objects cannot be located.

QUESTION: 15

Read the following passage and answer the questions that follows:
Archaeology as a profession faces two major problems. First, it is the poorest of the poor. Only paltry sums are available for excavating and even less is available for publishing the results and preserving the sites once excavated. Yet archaeologists deal with priceless objects every day.
Second, there is the problem of illegal excavation, resulting in museum-quality pieces being sold to the highest bidder.
I would like to make an outrageous suggestion that would at one stroke provide funds for archaeology and reduce the amount of illegal digging. I would propose that scientific archeological expeditions and governmental authorities sell excavated artifacts on the open market. Such sales would provide substantial funds for the excavation and preservation of archaeological sites and the publication of results. At the same time, they would break the illegal excavator’s grip on the market, thereby decreasing the inducement to engage in illegal activities.
You might object that professionals excavate to acquire knowledge, not money. Moreover, ancient artifacts are part of our global cultural heritage, which should be available for all to appreciate, not sold to the highest bidder. I agree. Sell nothing that has unique artistic merit or scientific value. But, you might reply, everything that comes out of the ground has scientific value. Here we part company. Theoretically, you may be correct in claiming that every artifact has potential scientific value. Practically, you are wrong.
I refer to the thousands of pottery vessels and ancient lamps that are essentially duplicates of one another. In one small excavation in Cyprus, archaeologists recently uncovered 2,000 virtually indistinguishable small jugs in a single courtyard, even precious royal seal impressions known as melekh handles have been found in abundance — more than 4,000 examples so far.
The basement of museums is simply not large enough to store the artifacts that are likely to be discovered in the future. There is not enough money even to catalogue the finds; as a result, they cannot be found again and become as inaccessible as if they had never been discovered. Indeed, with the help of a computer, sold artifacts could be more accessible than are the pieces stored in bulging museum basements. Prior to sale, each could be photographed and the list of the purchasers could be maintained on the computer A purchaser could even be required to agree to return the piece if it should become needed for scientific purposes. It would be unrealistic to suggest that illegal digging would stop if artifacts were sold in the open market. But the demand for the clandestine product would be substantially reduced. Who would want an unmarked pot when another was available whose provenance was known, and that was dated stratigraphically by the professional archaeologist who excavated it?

Q. The author’s argument concerning the effect of the official sale of duplicate artifacts on illegal excavation is based on which of the following assumptions?

Solution:

The author’s argument concerning the effect of the official sale of duplicate artifacts on illegal excavation appears in the passage where the author predicts that such official sale would reduce the demand for “the clandestine product.”

The rhetorical question that follows indicates that the author finds it unlikely that any purchaser would prefer objects of unknown provenance to objects of known origin, or, to rephrase, the author assumes that most people would prefer to purchase objects of authenticated provenance, as this choice states.

QUESTION: 16

Direction : Five sentences are given below, labeled A, B, C, D and E. They need to be arranged in a logical order to form a coherent paragraph/passage. From the given options, choose the most appropriate one.

A. This made me think again about my own cynicism about Web activism.

B. Dodd said that he believes that some sort of compromise on the content of the film will be reached so that young people can see the film without seeking their parent’s permission.

C. I called Christopher Dodd, the former senator who now runs the motion picture association.

D. Sure hash tags come and go but they probably make the world, the one beyond the keyboard, a better place.

E. I expected him to suggest that all the online petitioners have failed to grasp the nuance and importance of the rating system.

Solution:

The sequence has a CEB link – the author called a person and expected him to present a particular opinion.

The person presented his own opinion and this made the author rethink his own philosophy – there is a logical flow of thought in the sequence CEBA.

And D helps conclude the sequence. It is also the rethought out opinion of the author and logically follows A.

In statement A “this made me think again” refers to Dodd’s comment.

Hence AD should come after CEB.

Option (d) is the answer.

QUESTION: 17

Direction : Five sentences are given below, labeled A, B, C, D and E. They need to be arranged in a logical order to form a coherent paragraph/passage. From the given options, choose the most appropriate one.

A. These are not art galleries or the studio of an art lover, but restaurants that believe that a dash of artwork can do-up their interiors, simultaneously promoting the works of an artist.

B. For over two years now, Tangerine restaurant has been promoting works of artists who are looking for a platform or budding artists who needs to add to their collection.

C. Restaurants in the city are converting their walls into an exhibition space, displaying the works of different budding artists periodically.

D. The once plain walls here have a new aura as frames holding different colours, themes and mediums adorn them.

E. The works are exhibited on a no-cost basis, but the select works do go through “some amount of objective judging”, by Illango’s Artspace.

Solution:

DA is a mandatory pair as “the once plain walls” of D can be linked to “these are not art galleries” of A. This negates options (c) and (d).

CB is another mandatory pair as “restaurants in the city” in C can be linked to the specific example of “Tangerine restaurant” in B. This negates option (b).

QUESTION: 18

Direction : Five sentences are given below, labeled A, B, C, D and E. They need to be arranged in a logical order to form a coherent paragraph/passage. From the given options, choose the most appropriate one.

A. It is demanding that any party it backs should establish a working group on violence against women and children in the assembly.
B. In the run-up to the January 28 polls, for instance, members of Women Action for Development (WAD) are organising camps in all constituencies.
C. Ironically, Manipur has many activist groups led by women.
D. True empowerment will only happen when women enter the assembly in good numbers.
E. Conflict Widows’ Forum is a group made up of women who have lost their husbands to civil violence in the state.

Solution:

The ‘It’ in A refers to the forum in E. EA forms a mandatory pair and this helps rule out option (b).

Option (a) can be ruled out because C does not follow B; however, B can be used to substantiate C – as done in option (d).

Option (c) can be ruled out for the same reason as B is cited as an example and it follows C.

Option (d) is the answer as C .

QUESTION: 19

Direction : In order to form a coherent paragraph, find the odd sentence from the given set of sentences.

A. This is the country where the leader of the ruling party, the speaker of the lower house of parliament, at least three chief ministers, and a number of sports and business icons are women.

B. It is also a country where a generation of newly empowered young women are going out to work in large number than ever before.

C. It's early days yet, but one hopes these are the first stirrings of change.

D. Trust Law, a news service run by Thomson Reuters, has ranked India as the worst G20 country in which to be a woman.

Solution:

D, A and C makes up a complete story about the country being the worst for women in spite of the facts presented in A and C.

Since there is no connection between this and the “stirrings of change” in C, it is the odd sentence.

QUESTION: 20

Direction : In order to form a coherent paragraph, find the odd sentence from the given set of sentences.

A. For no apparent reason you cannot help yourself from humming or singing a tune by Lady Gaga or Coldplay, or horror upon horrors, the latest American Idol reject.

B. Songs that get stuck in your head and go round and round, sometimes for days, sometimes for months.

C. Some people call them earworms.

D. It there was nothing unique about them they would be swamped by all the other memories that sound similar too.

Solution:

Arranging sentences B, C and A (better than CBA) in that order explains what are “earworms” – Their uniqueness is beside the point even contrary to statement A.

Hence D is the odd sentence.

QUESTION: 21

Direction for question : A paragraph is given below from which the last sentence has been deleted. From the given options, choose the one that completes the paragraph in the most appropriate way.

“Day break” marks the completion of tunneling from two opposite faces. Conventionally, the tunneling work is conducted from both ends of the tunnel. But this being a very long tunnel, Railways decided to engage in simultaneous working at more than two faces by constructing a shaft towards the North end and an adit towards the South end.

Solution:

The paragraph discusses the speeding up of the tunneling work.

Option (a) follows the logical flow by discussing the result of speeding up the work.

Option (b) could have introduced the paragraph but it does not come in logically after the penultimate sentence.

Option (c) is incorrect as it specifies the name of the tunnel and discusses a road for emergencies – both are beyond the scope of the information in the paragraph.

Option (d) deviates from the central idea –the discussion is about the speeding up of work and not about the material used in building the tunnel.

QUESTION: 22

Direction for question : A paragraph is given below from which the last sentence has been deleted. From the given options, choose the one that completes the paragraph in the most appropriate way.

Musharraf has said the scientists were given wide latitude to develop the nuclear program and worked in secret even from top officials. That secrecy also has raised fears that nuclear workers may have transferred technology or equipment to terrorists, either for money or ideological sympathy.
Experts say centrifuge technology would not be of much use to terror groups, who probably could not set up the vast facilities required to enrich useful quantities of uranium, with hundreds of technicians needed to run thousands of centrifuges.

Solution:

The paragraph states the fear that nuclear technology could have been transferred to terrorists. However, the paragraph then changes its tone and states that experts feel that even if this were true, the technology would not be useful to terrorist groups who would need more resources. The penultimate line before the blank indicates that it will be probably impossible for terror groups to take advantage of any leaks in information or even receipt of equipment.

Option (c) substantiates this by saying that it is difficult for countries themselves to gather the resources needed.

Option (a) is incorrect as it brings attention to an irrelevant detail.

Option (b) is out of context as the paragraph is not concerned with the amount of uranium that has been produced by Pakistan.

QUESTION: 23

Read the following passage and answer the questions that follows:
Federal efforts to aid minority businesses began in the 1960’s when the Small Business Administration (SBA) began making federally guaranteed loans and government sponsored management and technical assistance available to minority business enterprises. While this program enabled many minority entrepreneurs to form new businesses, the results were disappointing, since managerial inexperience, unfavorable locations, and capital shortages led to high failure rates. Even 15 years after the program was implemented, minority business receipts were not quite two percent of the national economy’s total receipts. Recently federal policymakers have adopted an approach intended to accelerate development of the minority business sector by moving away from directly aiding small minority enterprises and toward supporting larger, growth-oriented minority firms through intermediary companies. In this approach, large corporations participate in the development of successful and stable minority businesses by making use of government-sponsored venture capital. The capital is used by a participating company to establish a Minority Enterprise Small Business Investment Company or MESBIC. The MESBIC then provides capital and guidance to minority businesses that have potential to become future suppliers or customers of the sponsoring company.
MESBIC’s are the result of the belief that providing established firms with easier access to relevant management techniques and more job-specific experience, as well as substantial amounts of capital, gives those firms a greater opportunity to develop sound business foundations than does simply making general management experience and small amounts of capital available. Further, since potential markets for the minority businesses already exist through the sponsoring companies, the minority businesses face considerably less risk in terms of location and market fluctuation. Following early financial and operating problems, sponsoring corporations began to capitalize MESBIC’s far above the legal minimum of $500,000 in order to generate sufficient income and to sustain the quality of management needed. MESBIC’s are now emerging as increasingly important financing sources for minority enterprises.
Ironically, MESBIC staffs, which usually consist of Hispanic and Black professionals, tend to approach investments in minority firms more pragmatically than do many MESBIC directors, who are usually senior managers from sponsoring corporations. The latter often still think mainly in terms of the “social responsibility approach” and thus seem to prefer deals that are riskier and less attractive than normal investment criteria would warrant. Such differences in viewpoint have produced uneasiness among many minority staff members, who feel that minority entrepreneurs and businesses should be judged by established business considerations. These staff members believe their point of view is closer to the original philosophy of MESBIC’s and they are concerned that, unless a more prudent course is followed, MESBIC directors may revert to policies likely to re-create the disappointing results of the original SBA approach.

Q. Which of the following best states the central idea of the passage?

Solution:

The passage begins by indicating that the results of the SBA approach to aiding minority entrepreneurs “were disappointing” . The passage states that “MESBIC’s are now emerging as increasingly important financing sources for minority enterprises.” Much of the passage is devoted to supporting the author’s view that MESBIC have the greater potential for success, and the last sentence in the passage confirms this view.

QUESTION: 24

Read the following passage and answer the questions that follows:
Federal efforts to aid minority businesses began in the 1960’s when the Small Business Administration (SBA) began making federally guaranteed loans and government sponsored management and technical assistance available to minority business enterprises. While this program enabled many minority entrepreneurs to form new businesses, the results were disappointing, since managerial inexperience, unfavorable locations, and capital shortages led to high failure rates. Even 15 years after the program was implemented, minority business receipts were not quite two percent of the national economy’s total receipts. Recently federal policymakers have adopted an approach intended to accelerate development of the minority business sector by moving away from directly aiding small minority enterprises and toward supporting larger, growth-oriented minority firms through intermediary companies. In this approach, large corporations participate in the development of successful and stable minority businesses by making use of government-sponsored venture capital. The capital is used by a participating company to establish a Minority Enterprise Small Business Investment Company or MESBIC. The MESBIC then provides capital and guidance to minority businesses that have potential to become future suppliers or customers of the sponsoring company.
MESBIC’s are the result of the belief that providing established firms with easier access to relevant management techniques and more job-specific experience, as well as substantial amounts of capital, gives those firms a greater opportunity to develop sound business foundations than does simply making general management experience and small amounts of capital available. Further, since potential markets for the minority businesses already exist through the sponsoring companies, the minority businesses face considerably less risk in terms of location and market fluctuation. Following early financial and operating problems, sponsoring corporations began to capitalize MESBIC’s far above the legal minimum of $500,000 in order to generate sufficient income and to sustain the quality of management needed. MESBIC’s are now emerging as increasingly important financing sources for minority enterprises.
Ironically, MESBIC staffs, which usually consist of Hispanic and Black professionals, tend to approach investments in minority firms more pragmatically than do many MESBIC directors, who are usually senior managers from sponsoring corporations. The latter often still think mainly in terms of the “social responsibility approach” and thus seem to prefer deals that are riskier and less attractive than normal investment criteria would warrant. Such differences in viewpoint have produced uneasiness among many minority staff members, who feel that minority entrepreneurs and businesses should be judged by established business considerations. These staff members believe their point of view is closer to the original philosophy of MESBIC’s and they are concerned that, unless a more prudent course is followed, MESBIC directors may revert to policies likely to re-create the disappointing results of the original SBA approach.

Q. According to the passage, the MESBIC approach differs from the SBA approach in that MESBIC’s

Solution:

In the second paragraph, the author describes the MESBIC approach as one in which “large corporations participate in the development of successful and stable minority businesses by making use of government-sponsored venture capital”. There is no indication in the passage that the SBA approach relies on the participation of large corporations.

QUESTION: 25

Read the following passage and answer the questions that follows:
Federal efforts to aid minority businesses began in the 1960’s when the Small Business Administration (SBA) began making federally guaranteed loans and governmentsponsored management and technical assistance available to minority business enterprises. While this program enabled many minority entrepreneurs to form new businesses, the results were disappointing, since managerial inexperience, unfavorable locations, and capital shortages led to high failure rates. Even 15 years after the program was implemented, minority business receipts were not quite two percent of the national economy’s total receipts. Recently federal policymakers have adopted an approach intended to accelerate development of the minority business sector by moving away from directly aiding small minority enterprises and toward supporting larger, growth-oriented minority firms through intermediary companies. In this approach, large corporations participate in the development of successful and stable minority businesses by making use of government-sponsored venture capital. The capital is used by a participating company to establish a Minority Enterprise Small Business Investment Company or MESBIC. The MESBIC then provides capital and guidance to minority businesses that have potential to become future suppliers or customers of the sponsoring company.
MESBIC’s are the result of the belief that providing established firms with easier access to relevant management techniques and more job-specific experience, as well as substantial amounts of capital, gives those firms a greater opportunity to develop sound business foundations than does simply making general management experience and small amounts of capital available. Further, since potential markets for the minority businesses already exist through the sponsoring companies, the minority businesses face considerably less risk in terms of location and market fluctuation. Following early financial and operating problems, sponsoring corporations began to capitalize MESBIC’s far above the legal minimum of $500,000 in order to generate sufficient income and to sustain the quality of management needed. MESBIC’s are now emerging as increasingly important financing sources for minority enterprises.
Ironically, MESBIC staffs, which usually consist of Hispanic and Black professionals, tend to approach investments in minority firms more pragmatically than do many MESBIC directors, who are usually senior managers from sponsoring corporations. The latter often still think mainly in terms of the “social responsibility approach” and thus seem to prefer deals that are riskier and less attractive than normal investment criteria would warrant. Such differences in viewpoint have produced uneasiness among many minority staff members, who feel that minority entrepreneurs and businesses should be judged by established business considerations. These staff members believe their point of view is closer to the original philosophy of MESBIC’s and they are concerned that, unless a more prudent course is followed, MESBIC directors may revert to policies likely to re-create the disappointing results of the original SBA approach.

Q. Which of the following does the author cite to support the conclusion that the results of the SBA program were disappointing?

Solution:

The author concludes that the results of the SBA approach “were disappointing”. Then he supports the conclusion by citing the fact that “Even 15 years after the program was implemented, minority business receipts were not quite two percent of the national economy’s total receipts”.

QUESTION: 26

Refer to the data below and answer the questions that follow.
In a year prior to the Olympics, all the participating athletes were required to undergo dope tests conducted by the Anti-Doping Association.
The tests of the athletes were done five times. The first test was done in February, the second in April, the third in August, the fourth in October, and the fifth in December. Only the athletes who passed the test in February were eligible for the test in April. Only those who passed the test in April were eligible for the test in August and those who passed in August were tested in October, and so on so forth.
During a test, an athlete was tested for the presence of exactly one of the five banned substances - A, B, C, D or E. An athlete already tested for the presence of a particular substance was not tested for the presence of the same substance in subsequent test(s) during the year.
The table given below shows the number of athletes tested in each of the five months for the different banned substances. Finally, only 91 athletes passed the tests, and were cleared to participate in the Olympics. An athlete is said to pass the test conducted for the presence of a banned substance, when that particular banned substance is not detected in the blood sample of the athlete.

Q. How many athletes were tested for the presence of exactly one banned substance?

Solution:

Number of athletes tested for the presence of at least one banned substance = Total number athletes tested in February = 105 + 245 + 345 + 130 + 95 = 920
Number of athletes tested for the presence of at least two banned substances = Total number athletes tested in April = 75 + 100 + 87 + 100 + 68 = 430
Therefore, the number of athletes who were tested for the presence of exactly one banned substance = 920 – 430 = 490.

QUESTION: 27

Refer to the data below and answer the questions that follow.
In a year prior to the Olympics, all the participating athletes were required to undergo dope tests conducted by the Anti-Doping Association.
The tests of the athletes were done five times. The first test was done in February, the second in April, the third in August, the fourth in October, and the fifth in December. Only the athletes who passed the test in February were eligible for the test in April. Only those who passed the test in April were eligible for the test in August and those who passed in August were tested in October, and so on so forth.
During a test, an athlete was tested for the presence of exactly one of the five banned substances - A, B, C, D or E. An athlete already tested for the presence of a particular substance was not tested for the presence of the same substance in subsequent test(s) during the year.
The table given below shows the number of athletes tested in each of the five months for the different banned substances. Finally, only 91 athletes passed the tests, and were cleared to participate in the Olympics. An athlete is said to pass the test conducted for the presence of a banned substance, when that particular banned substance is not detected in the blood sample of the athlete.

Q. Among all the athletes, how many were not tested for the presence of banned substance D?

Solution:

In February, there were 105 + 245 + 345 + 95 = 790 athletes not tested for the presence of banned substance D.
Athletes who were tested for the presence of banned substance D in April, October or December were among these 790 athletes only.
Therefore, the athletes who were not tested for the presence of banned substance D = 790 – 100 – 58 – 35 – 19 = 578.

QUESTION: 28

Refer to the data below and answer the questions that follow.
In a year prior to the Olympics, all the participating athletes were required to undergo dope tests conducted by the Anti-Doping Association.
The tests of the athletes were done five times. The first test was done in February, the second in April, the third in August, the fourth in October, and the fifth in December. Only the athletes who passed the test in February were eligible for the test in April. Only those who passed the test in April were eligible for the test in August and those who passed in August were tested in October, and so on so forth.
During a test, an athlete was tested for the presence of exactly one of the five banned substances - A, B, C, D or E. An athlete already tested for the presence of a particular substance was not tested for the presence of the same substance in subsequent test(s) during the year.
The table given below shows the number of athletes tested in each of the five months for the different banned substances. Finally, only 91 athletes passed the tests, and were cleared to participate in the Olympics. An athlete is said to pass the test conducted for the presence of a banned substance, when that particular banned substance is not detected in the blood sample of the athlete.

Q. At least how many athletes definitely passed the test conducted in December for the presence of banned substance B?

Solution:

103 athletes were tested in December out of which 91 passed. This means that 12 athletes did not pass the tests in December.

Even if these 12 athletes were out of the 15 who were tested for the presence of the banned substance B, then at least 3 athletes passed the test for the presence of banned substance B.

QUESTION: 29

Refer to the data below and answer the questions that follow.
In a year prior to the Olympics, all the participating athletes were required to undergo dope tests conducted by the Anti-Doping Association.
The tests of the athletes were done five times. The first test was done in February, the second in April, the third in August, the fourth in October, and the fifth in December. Only the athletes who passed the test in February were eligible for the test in April. Only those who passed the test in April were eligible for the test in August and those who passed in August were tested in October, and so on so forth.
During a test, an athlete was tested for the presence of exactly one of the five banned substances - A, B, C, D or E. An athlete already tested for the presence of a particular substance was not tested for the presence of the same substance in subsequent test(s) during the year.
The table given below shows the number of athletes tested in each of the five months for the different banned substances. Finally, only 91 athletes passed the tests, and were cleared to participate in the Olympics. An athlete is said to pass the test conducted for the presence of a banned substance, when that particular banned substance is not detected in the blood sample of the athlete.

Q. Suppose
X = The number of Athletes who failed the test in February due to the presence of substance other than A.
Y = The number of Athletes who failed the test in February due to the presence of any of the five substances.
What can be the approximate minimum value of X, when expressed as a percentage of Y?

Solution:

In Feb, total of 920 Athletes were tested, out of which 430 passed the test and underwent next round of tests in April.

Therefore Y = 920 – 430 = 490. In feb 105 athletes went test for substance A.

If all of them failed the test due to a substance other than A is 490 -105 = 385.

Therefore minimum value of X when expressed as a % of Y is 385 / 490 x 100 = 78.57%. 

QUESTION: 30

Refer to the data table below and answer the questions that follow.
An ice cream manufacturing company named Vadilla manufactures ice cream of three different flavours - Chocolate, Strawberry and Vanilla.The company surveyed a sample of 500 people in each of three cities - New Delhi, Mumbai and Kolkata. In the survey, all participants were asked to select exactly one of the four options - A, B, C and D. Out of these four options, one option indicated that the participant did not like any of the three flavours, while each of the remaining three options indicated the one flavour liked by the participant. No participant liked more than one flavour.
The following bar graph shows the number of participants in the three cities who indicated the responses A, B, C and D.

Q. If the number of participants who liked Vanilla in Mumbai was lower than the number of participants who liked Vanilla in either of the other two cities, then which of the following is definitely true?

Solution:


From (A): In the city of Kolkata, 25% of the participants did not like any of the three flavors 25% of 500 = 125, so either option (B) or option (D) represents “did not like any flavor”

From (B): In all the three cities combined, the difference between the number of participants who liked Strawberry and Vanilla was more than 210.

Using these statements, we can get the following cases: Based on the information provided in points (A), (B) and the above table, we can conclude

Now all the questions can be answered.
Vanilla can be either option (B) or (D).

If the number of people who liked Vanilla in Mumbai is lower than in the other two cities, option (B) represents Vanilla or this corresponds to either case (II) or (III).

In both cases, option (D) represents “None”.

Both Options [1] and [2] are definitely true in both cases.

QUESTION: 31

Refer to the data table below and answer the questions that follow.
An ice cream manufacturing company named Vadilla manufactures ice cream of three different flavours - Chocolate, Strawberry and Vanilla.The company surveyed a sample of 500 people in each of three cities - New Delhi, Mumbai and Kolkata. In the survey, all participants were asked to select exactly one of the four options - A, B, C and D. Out of these four options, one option indicated that the participant did not like any of the three flavours, while each of the remaining three options indicated the one flavour liked by the participant. No participant liked more than one flavour.
The following bar graph shows the number of participants in the three cities who indicated the responses A, B, C and D.

Q. In Mumbai, if the number of participants who liked Chocolate was higher than the number of participants who individually liked Strawberry or Vanilla or did not like any flavour, then which of the following statements is definitely false?

Solution:


From (A): In the city of Kolkata, 25% of the participants did not like any of the three flavors 25% of 500 = 125, so either option (B) or option (D) represents “did not like any flavor”

From (B): In all the three cities combined, the difference between the number of participants who liked Strawberry and Vanilla was more than 210.

Using these statements, we can get the following cases: Based on the information provided in points (A), (B) and the above table, we can conclude

Now all the questions can be answered.
The given condition corresponds to cases I and II. Therefore, option 1 is definitely false. Option 2 may or may not be false. Option 3 is definitely true.

QUESTION: 32

Refer to the data table below and answer the questions that follow.
An ice cream manufacturing company named Vadilla manufactures ice cream of three different flavours - Chocolate, Strawberry and Vanilla.The company surveyed a sample of 500 people in each of three cities - New Delhi, Mumbai and Kolkata. In the survey, all participants were asked to select exactly one of the four options - A, B, C and D. Out of these four options, one option indicated that the participant did not like any of the three flavours, while each of the remaining three options indicated the one flavour liked by the participant. No participant liked more than one flavour.
The following bar graph shows the number of participants in the three cities who indicated the responses A, B, C and D.

Q. A maximum of how many of the statements given below can be simultaneously true?
Statement I: In New Delhi, 200 participants liked the Strawberry flavour.
Statement II: In Mumbai, 160 participants liked the Chocolate flavour.
Statement III: In Kolkata, 30% of the participants liked the Chocolate flavour.
Statement IV: In New Delhi, 50 participants liked the Vanilla flavour.

Solution:


From (A): In the city of Kolkata, 25% of the participants did not like any of the three flavors 25% of 500 = 125, so either option (B) or option (D) represents “did not like any flavor”

From (B): In all the three cities combined, the difference between the number of participants who liked Strawberry and Vanilla was more than 210.

Using these statements, we can get the following cases: Based on the information provided in points (A), (B) and the above table, we can conclude

Statements I, II and III are true for Case III but statement IV is not.

QUESTION: 33

Refer to the data table below and answer the questions that follow.
An ice cream manufacturing company named Vadilla manufactures ice cream of three different flavours - Chocolate, Strawberry and Vanilla.The company surveyed a sample of 500 people in each of three cities - New Delhi, Mumbai and Kolkata. In the survey, all participants were asked to select exactly one of the four options - A, B, C and D. Out of these four options, one option indicated that the participant did not like any of the three flavours, while each of the remaining three options indicated the one flavour liked by the participant. No participant liked more than one flavour.
The following bar graph shows the number of participants in the three cities who indicated the responses A, B, C and D.

Q. Out of the three flavours of the ice cream, only those flavours which were selected by at least 30% of the participants in at least two of the three cities were simultaneously selected for production PAN India. Which of the following flavour(s) could have been selected for production PAN India?

Solution:


From (A): In the city of Kolkata, 25% of the participants did not like any of the three flavors 25% of 500 = 125, so either option (B) or option (D) represents “did not like any flavor”

From (B): In all the three cities combined, the difference between the number of participants who liked Strawberry and Vanilla was more than 210.

Using these statements, we can get the following cases: Based on the information provided in points (A), (B) and the above table, we can conclude

We are looking for the number of participants 150 or more in at least two cities.The flavours representing A and C will be selected as they have been selected by at least 150 participants in at least two cities. In all the three cases, one of A represents Chocolate and the other one represents Strawberry.

Therefore both Chocolate and Strawberry will be selected. 

QUESTION: 34

Refer to the data below and answer the questions that follow.
Two metal traders - Rahul and Anil - trade in metals P and Q only. The chart given below displays the performance (price in Rs. per 10 gm) of the two metals on five days of a particular week.
Rahul: He sold metal P on Monday and Tuesday in the ratio 1:2 by weight. On Wednesday and Friday each, he bought back half of the entire quantity of metal P sold earlier.
He bought metal Q on Tuesday and sold it in equal quantity on Thursday.
Anil: He sold metal P on Monday and Tuesday such that the amounts received on Monday and Tuesday are in the ratio 25 : 28. On Wednesday, he bought back the entire quantity of metal P sold on the previous two days. He bought metal Q on the first three days of the week in equal quantities, and on Thursday, sold 50% of the quantity of metal Q bought over the first three days of the week. On Friday, he sold the remaining 50% of the quantity of metal Q purchased over the first three days of the week.
The profit/loss made by a trader during a particular week is the difference between the amount that he obtains by selling metal and the amount that he spends in buying metal. "Margin" for a week is the profit/loss expressed as a percentage of the total amount obtained by selling metal during that week. The net profit/loss made by Rahul on transactions made for metal P and Q are the same. This holds true for Anil as well. The total net loss on the transactions made by Rahul and Anil is the same.

Q. What is the ratio of the quantity of metal Q bought by Rahul and Anil on Tuesday?

Solution:

For Rahul: Let the quantity of metal P sold on Monday and Tuesday be mp and 2mp respectively.

Therefore, he bought 1.5mp gm each of metal P on Wednesday and Friday.

Net loss = 2500mp + (2800 × 2mp) – (3000 × 1.5mp) – (3200 × 1.5mp) = 1200mp

Let the quantity of metal Q bought on Tuesday be mq.

Net loss = 3600mq – 2000mq = 1600mq

From the given condition, 1200mp = 1600mq or 3mp = 4mq

Total net loss = 3200mq

Total amount obtained by selling the shares = 8100mp + 2000mq = 12800mq

Loss margin = 25%

For Anil: Let the quantity of metal P sold on Monday and Tuesday be gp each.

Therefore, he bought 2gp gm of metal P on Wednesday.

Net loss = (3000 × 2gp) – (2500gp + 2800gp) = 700gp

Let the quantity of metal Q bought on the first three days of the week be gq each.

Therefore, he sold 1.5gq each on Thursday and Friday.

Net loss = (4000 + 3600 + 3200)gq – (2000 + 2400)1.5gq = 4200gq

From the given condition, 700gp = 4200gq or gp = 6gq

Total net loss = 700gp + 4200gq = 8400gq

Total amount obtained by selling the shares = 5300gp + 6600gq = 38400gq

Loss margin = (8400gq / 38400gq) = 22% approximate

It is also given that 3200mq = 8400gq or 8mq = 21gq

Required ratio = 21 : 8.

QUESTION: 35

Refer to the data below and answer the questions that follow.
Two metal traders - Rahul and Anil - trade in metals P and Q only. The chart given below displays the performance (price in Rs. per 10 gm) of the two metals on five days of a particular week.
Rahul: He sold metal P on Monday and Tuesday in the ratio 1:2 by weight. On Wednesday and Friday each, he bought back half of the entire quantity of metal P sold earlier.
He bought metal Q on Tuesday and sold it in equal quantity on Thursday.
Anil: He sold metal P on Monday and Tuesday such that the amounts received on Monday and Tuesday are in the ratio 25 : 28. On Wednesday, he bought back the entire quantity of metal P sold on the previous two days. He bought metal Q on the first three days of the week in equal quantities, and on Thursday, sold 50% of the quantity of metal Q bought over the first three days of the week. On Friday, he sold the remaining 50% of the quantity of metal Q purchased over the first three days of the week.
The profit/loss made by a trader during a particular week is the difference between the amount that he obtains by selling metal and the amount that he spends in buying metal. "Margin" for a week is the profit/loss expressed as a percentage of the total amount obtained by selling metal during that week. The net profit/loss made by Rahul on transactions made for metal P and Q are the same. This holds true for Anil as well. The total net loss on the transactions made by Rahul and Anil is the same.

Q. What is the ratio of the quantity of metal P sold by Rahul and Anil on Monday?

Solution:

For Rahul: Let the quantity of metal P sold on Monday and Tuesday be mp and 2mp respectively.

Therefore, he bought 1.5mp gm each of metal P on Wednesday and Friday.

Net loss = 2500mp + (2800 × 2mp) – (3000 × 1.5mp) – (3200 × 1.5mp) = 1200mp

Let the quantity of metal Q bought on Tuesday be mq.

Net loss = 3600mq – 2000mq = 1600mq

From the given condition, 1200mp = 1600mq or 3mp = 4mq

Total net loss = 3200mq

Total amount obtained by selling the shares = 8100mp + 2000mq = 12800mq

Loss margin = 25%

For Anil: Let the quantity of metal P sold on Monday and Tuesday be gp each.

Therefore, he bought 2gp gm of metal P on Wednesday.

Net loss = (3000 × 2gp) – (2500gp + 2800gp) = 700gp

Let the quantity of metal Q bought on the first three days of the week be gq each.

Therefore, he sold 1.5gq each on Thursday and Friday.

Net loss = (4000 + 3600 + 3200)gq – (2000 + 2400)1.5gq = 4200gq

From the given condition, 700gp = 4200gq or gp = 6gq

Total net loss = 700gp + 4200gq = 8400gq

Total amount obtained by selling the shares = 5300gp + 6600gq = 38400gq

Loss margin = (8400gq / 38400gq) = 22% approximate

It is also given that 3200mq = 8400gq or 8mq = 21gq

Required ratio = (4 / 18) × (21 / 8) = 7 : 12.

QUESTION: 36

Refer to the data below and answer the questions that follow.
Two metal traders - Rahul and Anil - trade in metals P and Q only. The chart given below displays the performance (price in Rs. per 10 gm) of the two metals on five days of a particular week.
Rahul: He sold metal P on Monday and Tuesday in the ratio 1:2 by weight. On Wednesday and Friday each, he bought back half of the entire quantity of metal P sold earlier.
He bought metal Q on Tuesday and sold it in equal quantity on Thursday.
Anil: He sold metal P on Monday and Tuesday such that the amounts received on Monday and Tuesday are in the ratio 25 : 28. On Wednesday, he bought back the entire quantity of metal P sold on the previous two days. He bought metal Q on the first three days of the week in equal quantities, and on Thursday, sold 50% of the quantity of metal Q bought over the first three days of the week. On Friday, he sold the remaining 50% of the quantity of metal Q purchased over the first three days of the week.
The profit/loss made by a trader during a particular week is the difference between the amount that he obtains by selling metal and the amount that he spends in buying metal. "Margin" for a week is the profit/loss expressed as a percentage of the total amount obtained by selling metal during that week. The net profit/loss made by Rahul on transactions made for metal P and Q are the same. This holds true for Anil as well. The total net loss on the transactions made by Rahul and Anil is the same.

Q. What is the loss margin for Rahul?

Solution:

For Rahul: Let the quantity of metal P sold on Monday and Tuesday be mp and 2mp respectively.

Therefore, he bought 1.5mp gm each of metal P on Wednesday and Friday.

Net loss = 2500mp + (2800 × 2mp) – (3000 × 1.5mp) – (3200 × 1.5mp) = 1200mp

Let the quantity of metal Q bought on Tuesday be mq.

Net loss = 3600mq – 2000mq = 1600mq

From the given condition, 1200mp = 1600mq or 3mp = 4mq

Total net loss = 3200mq

Total amount obtained by selling the shares = 8100mp + 2000mq = 12800mq

Loss margin = 25%

For Anil: Let the quantity of metal P sold on Monday and Tuesday be gp each.

Therefore, he bought 2gp gm of metal P on Wednesday.

Net loss = (3000 × 2gp) – (2500gp + 2800gp) = 700gp

Let the quantity of metal Q bought on the first three days of the week be gq each.

Therefore, he sold 1.5gq each on Thursday and Friday.

Net loss = (4000 + 3600 + 3200)gq – (2000 + 2400)1.5gq = 4200gq

From the given condition, 700gp = 4200gq or gp = 6gq

Total net loss = 700gp + 4200gq = 8400gq

Total amount obtained by selling the shares = 5300gp + 6600gq = 38400gq

Loss margin = (8400gq / 38400gq) = 22% approximate

It is also given that 3200mq = 8400gq or 8mq = 21gq

QUESTION: 37

Refer to the data below and answer the questions that follow.
Two metal traders - Rahul and Anil - trade in metals P and Q only. The chart given below displays the performance (price in Rs. per 10 gm) of the two metals on five days of a particular week.
Rahul: He sold metal P on Monday and Tuesday in the ratio 1:2 by weight. On Wednesday and Friday each, he bought back half of the entire quantity of metal P sold earlier.
He bought metal Q on Tuesday and sold it in equal quantity on Thursday.
Anil: He sold metal P on Monday and Tuesday such that the amounts received on Monday and Tuesday are in the ratio 25 : 28. On Wednesday, he bought back the entire quantity of metal P sold on the previous two days. He bought metal Q on the first three days of the week in equal quantities, and on Thursday, sold 50% of the quantity of metal Q bought over the first three days of the week. On Friday, he sold the remaining 50% of the quantity of metal Q purchased over the first three days of the week.
The profit/loss made by a trader during a particular week is the difference between the amount that he obtains by selling metal and the amount that he spends in buying metal. "Margin" for a week is the profit/loss expressed as a percentage of the total amount obtained by selling metal during that week. The net profit/loss made by Rahul on transactions made for metal P and Q are the same. This holds true for Anil as well. The total net loss on the transactions made by Rahul and Anil is the same.

Q. What is the approximate loss margin for Anil?

Solution:

For Rahul: Let the quantity of metal P sold on Monday and Tuesday be mp and 2mp respectively.

Therefore, he bought 1.5mp gm each of metal P on Wednesday and Friday.

Net loss = 2500mp + (2800 × 2mp) – (3000 × 1.5mp) – (3200 × 1.5mp) = 1200mp

Let the quantity of metal Q bought on Tuesday be mq.

Net loss = 3600mq – 2000mq = 1600mq

From the given condition, 1200mp = 1600mq or 3mp = 4mq

Total net loss = 3200mq

Total amount obtained by selling the shares = 8100mp + 2000mq = 12800mq

Loss margin = 25%

For Anil: Let the quantity of metal P sold on Monday and Tuesday be gp each.

Therefore, he bought 2gp gm of metal P on Wednesday.

Net loss = (3000 × 2gp) – (2500gp + 2800gp) = 700gp

Let the quantity of metal Q bought on the first three days of the week be gq each.

Therefore, he sold 1.5gq each on Thursday and Friday.

Net loss = (4000 + 3600 + 3200)gq – (2000 + 2400)1.5gq = 4200gq

From the given condition, 700gp = 4200gq or gp = 6gq

Total net loss = 700gp + 4200gq = 8400gq

Total amount obtained by selling the shares = 5300gp + 6600gq = 38400gq

Loss margin = (8400gq / 38400gq) = 22% approximate

It is also given that 3200mq = 8400gq or 8mq = 21gq

Now all the questions can be answered.

QUESTION: 38

Refer to the data below and answer the questions that follow.

National Institute of Management Ranchi (NIMR) began the first year of its flagship PGP course on June 22, 2017. The institute arranged guest lectures on different subjects, namely Finance, Marketing, Operations, Systems and HR throughout the year. The guest lectures are arranged such that the number of days between the two consecutive guest lectures on the same subject is same. No two subjects had equal gap between their respective two consecutive guest lectures.
The guest lectures on all the five subjects were offered for the first time on June 28th. After that, in the year 2017, the guest lectures on different subjects were offered as per the frequency decided earlier.
It is known that three of the guest lectures on Systems were held on July 18, July 28 and September 6, three of the guest lectures on HR were held on August 12, September 11 and September 26, three of the guest lectures on Marketing were held on July 7, July 13 and July 22, three of the guest lectures on Finance were held on July 2, July 8 and July 12 and three of the guest lectures on Operations were held on July 8, July 23 and August 2.

Q. On which of the following dates was the guest lecture on Systems definitely offered?

Solution:

Suppose we denote June 28 as ‘day 0’.
It can be seen that the guest lectures on Systems were held on days 20, 30 and 70. Therefore, the frequency of guest lectures on Systems was either 2, 5 or 10.

It can be seen that the guest lectures on HR were held on days 45, 75 and 90.
Therefore, the frequency of guest lectures on HR was either 5 or 15.
It can be seen that the guest lectures on Marketing were held on days 9, 15 and 24.

Therefore, the frequency of guest lectures on Marketing was 3.
It can be seen that the guest lectures on Finance were held on days 4, 10 and 14.

Therefore, the frequency of the guest lectures on Finance was 2.
It can be seen that the guest lectures on Operations were held on days 10, 25 and 35.

Therefore, the frequency of the guest lectures on Finance was 5.
Since the frequencies of the guest lectures on different subjects were different, following were the frequencies of the guest lectures on different subjects:

Finance: 2, Marketing: 3, Operations: 5, Systems: 10 and HR: 15.

We know that the frequency of the guest lectures on Systems was 10.

November 14 is day 139, November 15 is day 140, November 16 is day 141 and November 17 is day 142.

Therefore, the guest lecture on Systems was offered on day 140 or November 15. 

QUESTION: 39

Refer to the data below and answer the questions that follow.

National Institute of Management Ranchi (NIMR) began the first year of its flagship PGP course on June 22, 2017. The institute arranged guest lectures on different subjects, namely Finance, Marketing, Operations, Systems and HR throughout the year. The guest lectures are arranged such that the number of days between the two consecutive guest lectures on the same subject is same. No two subjects had equal gap between their respective two consecutive guest lectures.
The guest lectures on all the five subjects were offered for the first time on June 28th. After that, in the year 2017, the guest lectures on different subjects were offered as per the frequency decided earlier.
It is known that three of the guest lectures on Systems were held on July 18, July 28 and September 6, three of the guest lectures on HR were held on August 12, September 11 and September 26, three of the guest lectures on Marketing were held on July 7, July 13 and July 22, three of the guest lectures on Finance were held on July 2, July 8 and July 12 and three of the guest lectures on Operations were held on July 8, July 23 and August 2.

Q. Up-to September 26 (including that day), on how many days were no guest lectures offered by NIMR?

Solution:

Suppose we denote June 28 as ‘day 0’.
It can be seen that the guest lectures on Systems were held on days 20, 30 and 70. Therefore, the frequency of guest lectures on Systems was either 2, 5 or 10.

It can be seen that the guest lectures on HR were held on days 45, 75 and 90.
Therefore, the frequency of guest lectures on HR was either 5 or 15.
It can be seen that the guest lectures on Marketing were held on days 9, 15 and 24.

Therefore, the frequency of guest lectures on Marketing was 3.
It can be seen that the guest lectures on Finance were held on days 4, 10 and 14.

Therefore, the frequency of the guest lectures on Finance was 2.
It can be seen that the guest lectures on Operations were held on days 10, 25 and 35.

Therefore, the frequency of the guest lectures on Finance was 5.
Since the frequencies of the guest lectures on different subjects were different, following were the frequencies of the guest lectures on different subjects:

Finance: 2, Marketing: 3, Operations: 5, Systems: 10 and HR: 15.

Now all the questions can be answered.
September 26 is day 90. We have to count the natural numbers up-to 90 that are not divisible by any of the five numbers 2, 3, 5, 10 and 15. Essentially, we have to count the number of natural numbers up-to 90 that are not divisible by any of 2, 3 and 5 because all the numbers that are divisible by 10 or 15 are divisible by 5.

LCM of 2, 3 and 5 is 30. There are 6 numbers up-to 90 that are divisible by 30.

LCM of 2 and 3 is 6. There are 15 numbers up-to 90 that are divisible by 6.

LCM of 2 and 5 is 10. There are 9 numbers up-to 90 that are divisible by 10.

LCM of 3 and 5 is 15. There are 6 numbers up-to 90 that are divisible by 15. 

The number of numbers up-to 90 that are divisible by 2, 3 and 5 are 45, 30 and 18 respectively. Accordingly, we can construct the following Venn diagram for the numbers up-to 90 that are divisible by at least one of 2, 3 and 5.

The number of numbers up-to 90 that are divisible by at least one of 2, 3 and 5: 24 + 12 + 12 + 6 + 3 + 3 + 6 = 66.

Therefore, the number of numbers up-to 90 that are not divisible by any of these three numbers: 90 – 66 = 24.
Therefore, the required answer is 24.

QUESTION: 40

Refer to the data below and answer the questions that follow.

National Institute of Management Ranchi (NIMR) began the first year of its flagship PGP course on June 22, 2017. The institute arranged guest lectures on different subjects, namely Finance, Marketing, Operations, Systems and HR throughout the year. The guest lectures are arranged such that the number of days between the two consecutive guest lectures on the same subject is same. No two subjects had equal gap between their respective two consecutive guest lectures.
The guest lectures on all the five subjects were offered for the first time on June 28th. After that, in the year 2017, the guest lectures on different subjects were offered as per the frequency decided earlier.
It is known that three of the guest lectures on Systems were held on July 18, July 28 and September 6, three of the guest lectures on HR were held on August 12, September 11 and September 26, three of the guest lectures on Marketing were held on July 7, July 13 and July 22, three of the guest lectures on Finance were held on July 2, July 8 and July 12 and three of the guest lectures on Operations were held on July 8, July 23 and August 2.

Q. Up-to September 26 (including that day), on how many days were guest lectures on exactly three subjects offered by NIMR?

Solution:

Suppose we denote June 28 as ‘day 0’.
It can be seen that the guest lectures on Systems were held on days 20, 30 and 70. Therefore, the frequency of guest lectures on Systems was either 2, 5 or 10.

It can be seen that the guest lectures on HR were held on days 45, 75 and 90.
Therefore, the frequency of guest lectures on HR was either 5 or 15.
It can be seen that the guest lectures on Marketing were held on days 9, 15 and 24.

Therefore, the frequency of guest lectures on Marketing was 3.
It can be seen that the guest lectures on Finance were held on days 4, 10 and 14.

Therefore, the frequency of the guest lectures on Finance was 2.
It can be seen that the guest lectures on Operations were held on days 10, 25 and 35.

Therefore, the frequency of the guest lectures on Finance was 5.
Since the frequencies of the guest lectures on different subjects were different, following were the frequencies of the guest lectures on different subjects:

Finance: 2, Marketing: 3, Operations: 5, Systems: 10 and HR: 15.

We have to count the number of numbers up-to 90 that are divisible by exactly 3 of these numbers. We have the following cases:

Therefore, the required number of number is 9

QUESTION: 41

Refer to the data below and answer the questions that follow.

National Institute of Management Ranchi (NIMR) began the first year of its flagship PGP course on June 22, 2017. The institute arranged guest lectures on different subjects, namely Finance, Marketing, Operations, Systems and HR throughout the year. The guest lectures are arranged such that the number of days between the two consecutive guest lectures on the same subject is same. No two subjects had equal gap between their respective two consecutive guest lectures.
The guest lectures on all the five subjects were offered for the first time on June 28th. After that, in the year 2017, the guest lectures on different subjects were offered as per the frequency decided earlier.
It is known that three of the guest lectures on Systems were held on July 18, July 28 and September 6, three of the guest lectures on HR were held on August 12, September 11 and September 26, three of the guest lectures on Marketing were held on July 7, July 13 and July 22, three of the guest lectures on Finance were held on July 2, July 8 and July 12 and three of the guest lectures on Operations were held on July 8, July 23 and August 2.

Q. Up-to September 26 (including that day), on how many days were guest lectures on Marketing and exactly one other subject offered by NIMR?

Solution:

Suppose we denote June 28 as ‘day 0’.
It can be seen that the guest lectures on Systems were held on days 20, 30 and 70. Therefore, the frequency of guest lectures on Systems was either 2, 5 or 10.

It can be seen that the guest lectures on HR were held on days 45, 75 and 90.
Therefore, the frequency of guest lectures on HR was either 5 or 15.
It can be seen that the guest lectures on Marketing were held on days 9, 15 and 24.

Therefore, the frequency of guest lectures on Marketing was 3.
It can be seen that the guest lectures on Finance were held on days 4, 10 and 14.

Therefore, the frequency of the guest lectures on Finance was 2.
It can be seen that the guest lectures on Operations were held on days 10, 25 and 35.

Therefore, the frequency of the guest lectures on Finance was 5.
Since the frequencies of the guest lectures on different subjects were different, following were the frequencies of the guest lectures on different subjects:

Finance: 2, Marketing: 3, Operations: 5, Systems: 10 and HR: 15.
Now all the questions can be answered.
September 26 is day 90. The frequency of the guest lectures on Marketing is 3. The frequency of guest lectures on other subjects is 2, 5, 10 and 15.

Guest lectures only on Marketing and Finance
LCM of 3 and 2 is 6. We have to look for the number of numbers up-to 90 that are divisible by 6 but not by 5, 10 or 15. There are 15 numbers up-to 90 that are divisible by 6. Out of these, 3 (30, 60 and 90) that are divisible by 5. They need to be taken out. Therefore, the number of days on which guest lectures only on Marketing and Finance were offered = 15 – 3 = 12.

Guest lectures only on Marketing and Operations
LCM of 3 and 5 is 15. The guest lectures on Marketing and Operations both will be offered after every 15 days. But the frequency of guest lectures on HR is 15. Therefore, on those days, the guest lecture on HR will also be offered. Therefore, there is no day on which guest lectures only on Marketing and Operations are offered.

Guest lectures only on Marketing and Systems
LCM of 3 and 10 is 30. We have to look for the number of numbers up-to 90 that are divisible by 30 but not by 2, 5 or 15. There are no such numbers. Therefore there is no day on which guest lectures only on Marketing and Systems are offered.

Guest lectures only on Marketing and HR
LCM of 3 and 15 is 15. We have to look for the number of numbers up-to 90 that are divisible by 15 but not by 2, 5 or 10. There are no such numbers. Therefore, there is no day on which guest lectures only on Marketing and HR are offered.
Therefore, the required number of days is 12.

QUESTION: 42

Refer to the data below and answer the questions that follow.
There are five shops - Ramji Telecom, Star Tarco, Anil Electronics, Mobile Store and EZone - that sell mobile phones. Each of these five shops sell two of the following five mobile companies' phones - J2, J7 Prime, Note 5, C9 Pro and A9 2017, with no two shops selling the same pair of mobile phones. On the eve of Diwali, each shop sold three pieces of mobile phones between 11 AM and 12 PM, selling at least one mobile phone of the two given mobile companies. During this period, exactly three pieces of each of the mobile companies' phones were sold from these five shops.
The price of one piece each of J2, J7 Prime, Note 5, C9 Pro and A9 2017 is Rs. 9000, Rs. 10000, Rs. 11000, Rs. 12000 and Rs. 13000 respectively.
Ramji Telecom, Star Tarco, Anil Electronics and Mobile Store sold mobile phones worth Rs. 29000, Rs. 35000, Rs. 32000 and Rs. 38000 respectively during this period.

Q. E-Zone sells mobiles phones worth Rs. ____.

Solution:

Each mobile shop sold one piece of one mobile phone and two pieces of the other mobile phone that they sell. Now, Mobile Store sold mobile phones worth Rs. 38000. The only possible break up is 2 x 13000 + 12000. Therefore, Mobile Store sold 2 pieces of A9 2017 and one piece of C9 Pro.

There should be one shop that sold two pieces of C9 Pro, which would be worth Rs. 24000. The third piece should be worth (29000, 35000, 32000 or 31000) – 24000 = 5000 or 11000 or 8000 or 7000, out of which only 11000 is possible. Therefore, Star Tarco sold 2 pieces of C9 Pro and 1 piece of Note 5.

There should be one shop that sold 2 pieces of Note 5, which would be worth Rs. 22000. The third piece should be worth (29000, 32000 or 31000) – 22000 = 7000 or 10000 or 9000, out of which both 10000 and 9000 are possible.

Therefore, Anil Electronics sold (2 pieces of Note 5 and 1 piece of J2) or (2 pieces of Note 5 and 1 piece of J7).
Ramji Telecom sold (2 pieces of J2 and 1 piece of Note5) or (2 pieces of J7 Prime and 1 piece of J2).

Now because Star Tarco sold 1 piece of Note 5, therefore Ramji Telecom sold 2 pieces of J7 Prime and 1 piece of J2.

Therefore, Anil Electronics sold 2 pieces of Note 5 and 1 piece of J7 Prime. Thus, EZone sold 2 pieces of J2 and 1 piece of A9 2017.

QUESTION: 43

Refer to the data below and answer the questions that follow.
There are five shops - Ramji Telecom, Star Tarco, Anil Electronics, Mobile Store and EZone - that sell mobile phones. Each of these five shops sell two of the following five mobile companies' phones - J2, J7 Prime, Note 5, C9 Pro and A9 2017, with no two shops selling the same pair of mobile phones. On the eve of Diwali, each shop sold three pieces of mobile phones between 11 AM and 12 PM, selling at least one mobile phone of the two given mobile companies. During this period, exactly three pieces of each of the mobile companies' phones were sold from these five shops.
The price of one piece each of J2, J7 Prime, Note 5, C9 Pro and A9 2017 is Rs. 9000, Rs. 10000, Rs. 11000, Rs. 12000 and Rs. 13000 respectively.
Ramji Telecom, Star Tarco, Anil Electronics and Mobile Store sold mobile phones worth Rs. 29000, Rs. 35000, Rs. 32000 and Rs. 38000 respectively during this period.

Q. E-Zone sells one piece of ____.

Solution:

Each mobile shop sold one piece of one mobile phone and two pieces of the other mobile phone that they sell. Now, Mobile Store sold mobile phones worth Rs. 38000. The only possible break up is 2 × 13000 + 12000.
Therefore, Mobile Store sold 2 pieces of A9 2017 and one piece of C9 Pro.

There should be one shop that sold two pieces of C9 Pro, which would be worth Rs. 24000. The third piece should be worth (29000, 35000, 32000 or 31000) – 24000 = 5000 or 11000 or 8000 or 7000, out of which only 11000 is possible. Therefore, Star Tarco sold 2 pieces of C9 Pro and 1 piece of Note 5.

There should be one shop that sold 2 pieces of Note 5, which would be worth Rs. 22000. The third piece should be worth (29000, 32000 or 31000) – 22000 = 7000 or 10000 or 9000, out of which both 10000 and 9000 are possible.
Therefore, Anil Electronics sold (2 pieces of Note 5 and 1 piece of J2) or (2 pieces of Note 5 and 1 piece of J7).

Ramji Telecom sold (2 pieces of J2 and 1 piece of Note5) or (2 pieces of J7 Prime and 1 piece of J2). Now because Star Tarco sold 1 piece of Note 5, therefore Ramji Telecom sold 2 pieces of J7 Prime and 1 piece of J2.

Therefore, Anil Electronics sold 2 pieces of Note 5 and 1 piece of J7 Prime. Thus, E - Zone sold 2 pieces of J2 and 1 piece of A9 2017.
Now, all the questions can be answered.

QUESTION: 44

Refer to the data below and answer the questions that follow.

There are five shops - Ramji Telecom, Star Tarco, Anil Electronics, Mobile Store and EZone - that sell mobile phones. Each of these five shops sell two of the following five mobile companies' phones - J2, J7 Prime, Note 5, C9 Pro and A9 2017, with no two shops selling the same pair of mobile phones. On the eve of Diwali, each shop sold three pieces of mobile phones between 11 AM and 12 PM, selling at least one mobile phone of the two given mobile companies. During this period, exactly three pieces of each of the mobile companies' phones were sold from these five shops.
The price of one piece each of J2, J7 Prime, Note 5, C9 Pro and A9 2017 is Rs. 9000, Rs. 10000, Rs. 11000, Rs. 12000 and Rs. 13000 respectively.
Ramji Telecom, Star Tarco, Anil Electronics and Mobile Store sold mobile phones worth Rs. 29000, Rs. 35000, Rs. 32000 and Rs. 38000 respectively during this period.

Q. Anil Electronics sells one piece of ____.

Solution:

Each mobile shop sold one piece of one mobile phone and two pieces of the other mobile phone that they sell. Now, Mobile Store sold mobile phones worth Rs. 38000. The only possible break up is 2 x 13000 + 12000.

Therefore, Mobile Store sold 2 pieces of A9 2017 and one piece of C9 Pro.
There should be one shop that sold two pieces of C9 Pro, which would be worth Rs. 24000. The third piece should be worth (29000, 35000, 32000 or 31000) – 24000 = 5000 or 11000 or 8000 or 7000, out of which only 11000 is possible.

Therefore, Star Tarco sold 2 pieces of C9 Pro and 1 piece of Note 5.
There should be one shop that sold 2 pieces of Note 5, which would be worth Rs. 22000. The third piece should be worth (29000, 32000 or 31000) – 22000 = 7000 or 10000 or 9000, out of which both 10000 and 9000 are possible.

Therefore, Anil Electronics sold (2 pieces of Note 5 and 1 piece of J2) or (2 pieces of Note 5 and 1 piece of J7).
Ramji Telecom sold (2 pieces of J2 and 1 piece of Note5) or (2 pieces of J7 Prime and 1 piece of J2).

Now because Star Tarco sold 1 piece of Note 5, therefore Ramji Telecom sold 2 pieces of J7 Prime and 1 piece of J2.

Therefore, Anil Electronics sold 2 pieces of Note 5 and 1 piece of J7 Prime. Thus, EZone sold 2 pieces of J2 and 1 piece of A9 2017.

QUESTION: 45

Refer to the data below and answer the questions that follow.

There are five shops - Ramji Telecom, Star Tarco, Anil Electronics, Mobile Store and EZone - that sell mobile phones. Each of these five shops sell two of the following five mobile companies' phones - J2, J7 Prime, Note 5, C9 Pro and A9 2017, with no two shops selling the same pair of mobile phones. On the eve of Diwali, each shop sold three pieces of mobile phones between 11 AM and 12 PM, selling at least one mobile phone of the two given mobile companies. During this period, exactly three pieces of each of the mobile companies' phones were sold from these five shops.
The price of one piece each of J2, J7 Prime, Note 5, C9 Pro and A9 2017 is Rs. 9000, Rs. 10000, Rs. 11000, Rs. 12000 and Rs. 13000 respectively.
Ramji Telecom, Star Tarco, Anil Electronics and Mobile Store sold mobile phones worth Rs. 29000, Rs. 35000, Rs. 32000 and Rs. 38000 respectively during this period.

Q. Ramji Telecom sells two pieces of ____.

Solution:

Each mobile shop sold one piece of one mobile phone and two pieces of the other mobile phone that they sell. Now, Mobile Store sold mobile phones worth Rs. 38000. 

The only possible break up is 2 x 13000 + 12000. Therefore, Mobile Store sold 2 pieces of A9 2017 and one piece of C9 Pro.

There should be one shop that sold two pieces of C9 Pro, which would be worth Rs. 24000. The third piece should be worth (29000, 35000, 32000 or 31000) – 24000 = 5000 or 11000 or 8000 or 7000, out of which only 11000 is possible.
Therefore, Star Tarco sold 2 pieces of C9 Pro and 1 piece of Note 5.

There should be one shop that sold 2 pieces of Note 5, which would be worth Rs. 22000. The third piece should be worth (29000, 32000 or 31000) – 22000 = 7000 or 10000 or 9000, out of which both 10000 and 9000 are possible.

Therefore, Anil Electronics sold (2 pieces of Note 5 and 1 piece of J2) or (2 pieces of Note 5 and 1 piece of J7).
Ramji Telecom sold (2 pieces of J2 and 1 piece of Note5) or (2 pieces of J7 Prime and 1 piece of J2). Now because Star Tarco sold 1 piece of Note 5, therefore Ramji Telecom sold 2 pieces of J7 Prime and 1 piece of J2.

Therefore, Anil Electronics sold 2 pieces of Note 5 and 1 piece of J7 Prime. Thus, EZone sold 2 pieces of J2 and 1 piece of A9 2017.

QUESTION: 46

The birth and death rates of Ranchi city are given below. Study the chart carefully and answer the given questions.

Q. Calculate the average number of births in the city between 2001 to 2005.

Solution:

From the graph, calculate the values of total births between 2001 to 2005. Then find the average of them.
So, average births = (3.5 + 4 + 5 + 3 + 6.5) / 5 = 4.4.
Therefore, the average births between 2001 to 2005 is 4.4 million (since units are in millions).

QUESTION: 47

The birth and death rates of Ranchi city are given below. Study the chart carefully and answer the given questions.

Q. By what percent (approx.) did the number of births increased in 2005 from the previous year?

Solution:

At first, the increase in births has to be calculated.

So, Increase in deaths in 2005 from 2004 = 6.5 – 3 = 3.5 million.

Now, percentage increase = (3.5 / 3) × 100 = 116.66 %.

As the closest option is 117 %

QUESTION: 48

The birth and death rates of Ranchi city are given below. Study the chart carefully and answer the given questions.

Q. Calculate the difference between the number of deaths and the number of births over the years.

Solution:

First, the total deaths and births over the years have to be calculated.

So, total deaths over the years = 27 and,

Total births over the years = 22

Now, the difference will be = 27 – 22 = 5.

QUESTION: 49

The birth and death rates of Ranchi city are given below. Study the chart carefully and answer the given questions.

Q. In which year the percentage increase in birth rate over the previous year was the maximum?

Solution:

For this question, one can either calculate the percentage increase in each of the years or can just approximate to arrive at the right answer.

To calculate percentage increase in every year, one can use the following formula:
For a certain year, percentage increase or decrease = (difference of values from previous year/ value in that year) × 100.
So, for 2002, percentage increase in = (4 – 3.5) × 100 = 50 %.
Similarly, for 2003, percentage increase = 100 %.

For, 2004, percentage increase = 200 % (decrease).

And for 2005, percentage increase is = 350 %.

It can be seen that the percentage increase was highest in 2005 and so, option (d) is the correct answer.

QUESTION: 50

A private firm AK factory runs with its 11 employees. The average age of the employees of the firm is increased by 2 months when two among them were replaced by 2 new employees of the same age. The ages of the employees who were replaced by the new employees are 18 years and 20 years, respectively. The age of each of the new members is

Solution:

Total increase in age = 11 x 2 = 22 months

Therefore, total age of two new members is 18 years + 20 years + 22 months or 39 years 10 months.
Age of each new member = (39 yrs + 10 months) / 2 = 19 years 11 months

QUESTION: 51

Working together, Anil and Rambo painted a fence in 8 hours. Last year, Rambo painted the fence by himself. The year before, Anil painted it by himself, but took 12 hours less than what Rambo took. How long did Anil and Rambo take (in hours) when each was painting alone? 

Solution:

Anil can do it in b hours, so he does 1 / b per hour. Rambo does it in t hours, so he does 1 / t per hour.

Then, working together, they can do 1 / b  + 1 / t = 1 / 8 per hour (since they took 8 hours working together).

We also know that Rambo takes 12 hours longer than Anil, so t = b + 12.

We get: 1 / b  + 1 / (b + 12) = 1 / 8
⇒ 16b + 96 = b2 + 12b
⇒ b2 - 4b - 96 = 0
(b - 12)(b + 8) = 0

Then, b = 12 and b = - 8. Now, b must be positive, so we'll ignore b = - 8.

Hence, Anil alone takes 12 hours and Rambo alone takes 24 hours. 

QUESTION: 52

At a dairy farm, the 2 inlet pipes to the amul milk cooler (AMC) from the 2 milking parlours can fill the cooler tank in 5 hours and 6 hours, respectively. The outlet pipe from the AMC to the pasteurising unit can empty the cooler tank in 12 hours. If all the 3 pipes are operated simultaneously, then the empty tank will be filled in

Solution:

QUESTION: 53

Mr. and Mrs. Shukla travel from City A to City B and break the journey at City C in between. Somewhere between City A and City B, Mrs. Shukla asks, ''How far have we travelled?'' Mr. Shukla replies, ''Half as far as the distance from here to City C.'' Somewhere between City C and City B, exactly 200 km from the point where she asked the first question, Mrs. Shukla asks, ''How far do we have to go?'' Mr. Shukla replies, Half as far as the distance from City C to here.''

Q. Find the distance (in km) between City A and City B.

Solution:


The figure given above is self-explanatory.
      ► AB = x / 2 + 200 +(200 – x) / 2 = 200 + 100 = 300.

QUESTION: 54

In order to buy a Honda car, Raj borrows Rs. 1,80,000 on the condition that he would pay 7.5% interest every year. He also agrees to repay the principal in equal annual instalments over 21 years. After a certain number of years, however, the rate of interest is reduced to 7%. It is also known that at the end of the agreed period, he will have paid Rs. 2,70,900 in interest in total. For how many years does he pay at the reduced interest rate?

Solution:

Let for n years, he pays an interest of 7.5% and for (21 - n) years, the rate of interest paid is 7%.
So,

Solving it, we get n = 7.
So, for 21 - 7 = 14 years, he pays the reduced interest rate. 

QUESTION: 55

The profit for a private firm Anil Enterprises is given by P = x3 - 6x2 + 12x + 38, where `x` is the output quantity. How much should the firm produce to maximise the profit?

Solution:

We need to find the maximum profit, which can be found by substituting the values given in the option in the given equation.
P = x3 - 6x2 + 12x + 38, if P = 0, that would be our answer.
   ► P (2) = 23 - 6(2)2 + 12 (2) + 38
   ► = 8 - 24 + 24 + 38 = 46
   ► P(3) = (3)3 - 6(3)2 + 12(3) + 38
   ► = 27 - 54 + 36 + 38 = 47
   ► P(4) = 43 - 6(4)2 + 12(4) + 38
   ► = 64 - 96 + 48 + 38 = 54
As x increases, the function P(x) increases. The value at which the function is maximum cannot be determined.

QUESTION: 56

In the elections held during August, 60% of the respondents favoured Mr. Anil, while the balance favoured Mr. Binay. It was found in September polls that 10% of Mr. Anil's supporters switched their preference to Mr. Binay, while the same percentage of Mr. Binay's supporters switched their preference to Mr. Anil. What is the number of the electorate should now switch their preference from Mr. Anil to Mr. Binay so that they are on par?

Solution:

Suppose, the total electorate is 100.
August = Mr. Anil = 60 and Mr. Binay = 40
September 60 - 6 + 4 = 58
    ► 40 + 6 - 4 = 42
Hence, if 8 votes switch over from Mr. Anil to Mr. Binay, the two will be on par.

QUESTION: 57

In Symbiosis, Amit performance in a subject in a semester is evaluated by a class test, a presentation and 6 assignments. All the 6 assignments are of equal weightage. If the presentation counts twice as much as the test and thrice as much as 1 class assignment, then what is the contribution of a class assignment in the student's final performance?

Solution:

Let weightage of class test be t.
Let weightage of presentation be p.
Let weightage of assignment be a.
To find a / (t + p + 6a) .
Given p = 2t = 3a

Thus, a / ( 3 / 2a + 3a + 6a) = 2 / 21.

QUESTION: 58

Smera mixes a 20% alcohol solution and a 30% alcohol solution in the respective ratio of c : d. The new solution thus obtained is mixed with an equal quantity of a 60% alcohol solution to obtain a 42% alcohol solution. Which among the following represents the ratio c : d?

Solution:

Let Smera mix c litres of the 20% solution and d litres of the 30% solution.
Thus, the following equation can be set up:
   ► 20c + 30d + 60(c + d) = 42 x 2(c + d) Or, 3d = 2c Or, c : d = 3 : 2

QUESTION: 59

The Tiara’s shop gives a discount of 12% on a statue of laughing Buddha. Let 'p' denote the percentage, above the cost price, by which the price of the statue is marked up, so that a profit of x% can be made. It is to be noted that p is an integer and x < 100. Which of the following is not a possible value of x?

Solution:

Let the cost price of the statue be Rs. c.
So, (c + (p / 100)c)(1 - 12 / 100) = (1 + x / 100)c
   ►​Or, (1 + p / 100)(1 - 12 / 100)
    ► = (1 + x / 100) Or, (100 + p)(100 - 12)
    ► = 100(100 + x) Or, 88p - 100x
    ► = 1200
If we go with the options by plugging in different values of x in the above equation and solve for p, we shall find that the option (3) does not give an integral value of p, whereas all the other 3 options do.
Thus, 96 is not a possible value of x.

QUESTION: 60

Select the right choice with reference to the following graphs.

Solution:

Check by values:
At x = 1, f(1) = 1 and F(1) = 3
At x = -1, f(-1) = 3 and F(-1) = 1
Now, F(1) = f(-1) and F(-1) = f(1)
So, we can say F(x) = f(-x).

QUESTION: 61

In triangle DEF, points A, B and C are on the sides DE, DF and EF, respectively, such that EC = AC and CF = BC. If < D = 40°, then what is the measure of < ACB?

Solution:


∠CEA = ∠CAE
⇒ ∠ECA = 180 - 2∠CEA

∠BCF = ∠CFB
⇒ ∠BCF = 180 - 2∠CFB

∠ECA + ∠ACB + ∠BCF = 180

∠CEA + ∠CFB = 180 - 40 = 140

∠ACB = 180 – (∠ECA + ∠BCF ) = 180 – ( 180 - 2∠CEA + 180 – 2 ∠CFB) = 100.

QUESTION: 62

The sequence 2, 3, 5, 6, 7, 10, 11, ... consists of all positive integers that are not a square or a cube. Find the 1000th term. 

Solution:

Using the inclusion-exclusion principle, the number of positive integers from 11 to NN that are in your sequence is

Since most numbers are not powers, we might start by computing f(1000)=960.

We are short by 40, so next try f(1040)=999.

Pretty close! The next value is f(1041)=1000 So answer is 1041. 

QUESTION: 63

It is desired to extract the maximum power of 3 from 24!, where n! = n.(n -1).(n - 2) ... 3.2.1. What will be the exponent of 3? 

Solution:

There are eight multiples of 3 up to 24, each of which has 3 as a factor.

There are two more multiples of 3, one in 9 and another in 18.

So, total number of factors of 3 in 24! is 8 + 2 = 10.

QUESTION: 64

Let a1, a2, a3 ... be an AP with a common difference which is not a multiple of 3. The maximum number of consecutive terms, so that all are prime, is __________. 

Solution:

In this type of AP, it can easily be shown that exactly one out of any 3 consecutive terms will be a multiple of 3. So, at most 3 consecutive terms can be prime. This is also true in the case when one is 3. [as 3, 5, 7 or 3, 11, 19].

Another possible series is 5, 11, 17, 23 and 29; however here, the common difference is 6, which is a multiple of 3. Hence, it cannot be considered. 

QUESTION: 65

Rani, Shalini and Babita go out for a picnic. Rani takes along 3 cakes, Shalini takes five cakes and Babita, who did not find the time to go shopping for cakes, contributes Rs. 8. If all of them share the cakes equally, then in what respective ratio should Rani and Shalini divide the money contributed by Babita?

Solution:

Rani`s share = 3 cakes

Shalini`s share = 5 cakes

Babita`s share = Rs. 8

Since all three share the cakes equally, so:

Share of each = 8 / 3

Part of cake that Rani gives to Babita = 3 – 8 / 3 = 1 / 3

Part of cake that Shalini gives to Babita = 5 – 8 / 3 = 7 / 3

Ratio of their contributions to Babita = 1 / 3 : 7 / 3 = 1 : 7

Therefore, Rani and Shalini should share Babita`s money in respective ratio of 1 : 7.

QUESTION: 66

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

Solution:

The sum of all the numbers formed by the digits a1, a2, a3,……….an, without repetition of the digits is given by:
(n - 1)! (a1+ a2 + a3+ ………an) (10n -1) / 9

The sum of the given numbers = 4! (1 + 3 + 5 + 7 + 9) × 11111 = 6666600.

QUESTION: 67

Let N = 553 + 173 – 723. N is divisible by:

Solution:

N = 553 + 173 – 723 = (54 + 1)3 + (18 – 1)3 – 723

When N is divided by 3, we get remainders (1)3 + (- 1)3 – 0 = 0

Hence, the number N is divisible by 3.

Again N = (51 + 4)3 + 173 – (68 + 4)3

When N is divided by 17, the remainder is (4)3 + 0 – (4)3 = 0

Hence, the number is divisible by 17.

Hence, the number is divisible by both 3 and 17. 

QUESTION: 68

Bihar State decided to award two women cricket players, who did well in World Cup. It declared to distribute Rs. 2 Cr between both of them. It is known that Sharman did exceptionally well in the World Cup and deserves more than Mishali, but does not know how much more. So, the State decided to break Rs. 2 Cr into two parts and give Sharman the bigger part. What is the chance that Sharman gets at least twice as much as Mishali?

Solution:

Let Sharman's part be S and Mishali's part be M.
Sharman’s bigger part could be any number between Rs. 1 Cr and Rs. 2 Cr.
Now, if the bigger part is to be at least twice as much as the smaller part, it means
S ≥ 2M or S ≥ 2(2Cr – S) Or,
S ≥ 4Cr / 3
Given that S lies between Rs. 1 Cr and Rs. 2 Cr, then the probability that S lies between 4Cr/3 and 2Cr is equal to (2Cr – 4Cr / 3) / (2Cr – 1Cr) = 2 / 3.

QUESTION: 69

There are 5 consecutive natural numbers with L.C.M. 60. The product of the first two numbers is equal to the 5th number. What is the sum of the numbers?

Solution:

Let n be 1st natural number.
So numbers be
n, n + 1, n + 2, n + 3, n + 4, n + 5
According to question
► n(n + 1) = n + 4
► n2 + n = n + 4
► n2 = 4
► n = 2
– 2 cannot be taken as n is natural number.
So n = 2 numbers be 2, 3, 4, 5, 6
Sum of numbers = 2 + 3 + 4 + 5 + 6 = 20.

QUESTION: 70

A drum contained V litres of a mixture of milk and water, with water and milk in the ratio of 2 : 3. The total volume of the mixture was increased by 60% by adding water. Next, 41.6 litres of the solution in the drum was replaced by water. If the final ratio of water and milk in the drum is 7 : 3, then find the value of V (in litres).

Solution:

The volume of mixture in the drum = V litres

Thus, volume of water = (2 / 5)V litres

Volume of milk = (3 / 5)V litres

Volume of water in the mixture when volume of water equal to 60% of volume of mixture is added to the mixture = ((2 / 5)V + 0.6V) litres = V litres

Total volume of mixture in the drum = (V + (3 / 5)V) litres = ((8/5)V) litres

When 41.6 litres of mixture in the drum is replaced by water, volume of water in the drum = (V - 41.6 x V / (8V / 5) + 41.6) litres = (V + 15.6) litres

Volume of milk in the drum now = ((3 / 5)V - 41.6 x (3 / 5)V / ((8 / 5)V)) litres = (0.6V -15.6) litres

As the ratio of water to milk in the drum is 7 : 3, we have (V + 15.6) / (0.6V - 15.6) = 7 / 3
Or, V = 130

Hence, the initial volume of the mixture in the drum = 130 litres.

QUESTION: 71

ABCDE is a regular pentagon. A point P is taken inside this pentagon such that ∆CDP is an equilateral triangle. What is ∠BPE ?

Solution:


∠BCD = 108°
So, ∠BCP = 48°
ΔBPC is isosceles.
So, ∠BCP = 66°
Hence, θ = 360° - 60° - 2 x 66°
   ► θ = 360° - 60° - 132°
   ► θ = 360° - 192°
   ► θ = 168°

QUESTION: 72

A wooden cylinder of radius 21 cm and height 28 cm is cut into four identical pieces by making two cuts perpendicular to its base. If three of the four pieces are again joined back together in their original position, then what is the total surface area ( sq.cm) of this resulting wooden block ?

Solution:


The sum of the surface areas of BFEC and BCDA (which are rectangles)
  ► = 2 x BF x BC = 2 x 21 cm x 28 cm = 1176 cm2 ..........(1)

Total area of the flat surfaces (circular bottom and top) = 2[3 / 4 x (21)2] cm2
  ► = 2 x 3 / 4 x 22 / 7 x 21 x 21 cm2 = 2079 cm2 ..........(2)

Area of the curved surface = 3 / 4 x 2 x 21 x 28 cm2
  ► = 3 / 4 x 2 x 22 / 7 x 21 x 28 cm2 = 2772 cm2 ..........(3)

So, total surface area of the new block after adding (1), (2) and (3)
  ► = (1176 + 2079 + 2772) cm2 = 6027 cm2.

QUESTION: 73

A bag consists of some red marbles, some blue marbles and two yellow marbles. The probability that a marble picked at random is red is 1 / 2. Also, the probability that two marbles picked simultaneously are blue is 1 / 11. How many marbles are present in the bag?

Solution:

Let the number of marbles in the bag = 2n
Number of yellow marbles = 2
As the probability of drawing a red marble = 1 / 2, number of red marbles in the bag 
= n.
Now, number of blue marbles in the bag = 2n - n - 2 = n - 2
Thus, probability of drawing 2 blue marbles
  ► = n - 2C2nC2 = ((n - 2)(n - 3))/(2n(2n - 1)) = 1/11
  ► Or, 11(n - 2)(n - 3) = 2n(2n - 1)
  ► Or, 7n2 - 53n + 66 = 0
  ► Or, 7n2 - 42n - 11n + 66 = 0
  ► Or, 7n(n - 6) - 11(n - 6) = 0
  ► Or, (7n - 11)(n - 6) = 0 Or, n = 6 (as n is integral)
Thus, number of marbles in the bag = 2n = 2 x 6 = 12.

QUESTION: 74

In a snooker contest, there were a few male and female participants. There was a condition that each of  participants has to play exactly one game with each of others. After the contest is over, it was noticed that  in 45 such games both the participants were female and in 190 games both the players were male. Find   the number of games in which one male and one female participated.

Solution:

Let the total number of female participants = f

Total number of male participants = m

Number of games in which both were female participants = 45

fC2 = 45 ⇒ f(f − 1)2 = 45 ⇒ f(f − 1) = 90 ⇒ f = 10

Number of games in which both players were male participants = 190

mC2 = 190 ⇒ m(m − 1)2 = 190 ⇒ m(m − 1) = 380 ⇒ m = 20

Total number of females = 10

Total number of men = 20

Number of games in which one male and one female participated.

= 20C1 × 10C1 = 20 × 10 = 200. 

QUESTION: 75

In a series of positive numbers, (n + 1)th term is given as t(n + 1) = tn + t(n – 1) (n > 1), where tn is the nth term of the series. If t8 = 136 and t11 = 330, then find t12.

Solution:

t10 = t8 + t9
∴ t8 = t10 - t9 ...(i)
∴ t10 - t9 = 136

Also, t11 = t10 + t9 .... (ii)
∴ t10 + t9 = 330

Solving (i) and (ii) simultaneously, we get
2(t10) = 466
∴ t10 = 233

Now, t12 = t11 + t10 = 330 + 233
t12 = 563. 

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