To find the ratio of distances travelled in one hour, we first need to calculate the speed of the bus and the train. Speed is defined as distance divided by time. Once we have the speeds, which represent the distance travelled in one hour, we can form the ratio.
The bus travels 160 km in 4 hours.
\[ \text{Speed of bus} = \frac{\text{Distance}}{\text{Time}} = \frac{160\ \text{km}}{4\ \text{hours}} = 40\ \text{km/hour} \]The train travels 320 km in 5 hours.
\[ \text{Speed of train} = \frac{\text{Distance}}{\text{Time}} = \frac{320\ \text{km}}{5\ \text{hours}} = 64\ \text{km/hour} \]The speed itself represents the distance travelled in one hour.
Ratio = (Distance by bus in one hour) : (Distance by train in one hour)
\[ \text{Ratio} = 40 : 64 \]Divide both numbers by their greatest common divisor, which is 8.
\[ \text{Ratio} = \frac{40}{8} : \frac{64}{8} = 5 : 8 \]Correct Option: B) 5:08
This problem can be solved using the principle of inclusion-exclusion for sets or by using percentages. We need to find the number of students who play neither game. This can be found by subtracting the percentage of students who play at least one game from the total percentage (100%).
Identify the given percentages:
Calculate the percentage of students who play at least one game:
We use the formula for the union of two sets:
\[ P(C \cup F) = P(C) + P(F) - P(C \cap F) \]\[ P(\text{at least one game}) = 60\% + 30\% - 10\% \]\[ P(\text{at least one game}) = 90\% - 10\% \]\[ P(\text{at least one game}) = 80\% \]Calculate the percentage of students who play neither game:
The percentage of students who play neither game is the total percentage minus the percentage of students who play at least one game.
\[ P(\text{neither game}) = 100\% - P(\text{at least one game}) \]\[ P(\text{neither game}) = 100\% - 80\% \]\[ P(\text{neither game}) = 20\% \]Calculate the number of students who play neither game:
Since there are 100 students in total, 20% of 100 students will be:
\[ \text{Number of students (neither game)} = 20\% \times 100 \]\[ \text{Number of students (neither game)} = \frac{20}{100} \times 100 \]\[ \text{Number of students (neither game)} = 20 \]B) 20
To determine how long the water in the tank will last, we need to calculate the total water requirement of the village per day and the total volume of water the tank can hold. Then, we divide the total tank volume by the daily water requirement.
Correct Option: B) 3 days
Directions for the following 4 (four) items:(Covers Questions 44 to 47)
Read the following passage and answer the items that follow. Your answers to these items should be based on the passage only.
Passage
A species that exerts an influence out of proportion to its abundance in an ecosystem is called a keystone species. The keystone species may influence both the species richness of communities and the flow of energy and materials through ecosystems. The sea star Pisaster ochraceous, which lives in rocky intertidal ecosystems on the Pacific coast of North America is also an example of a keystone species. Its preferred prey is the mussel Mytilus californianus. In the absence of sea stars, these mussels crowd out other competitors in a broad belt of the intertidal zone. By consuming mussels, sea star creates bare spaces that are taken over by a variety of other species.
A study at the University of Washington demonstrated the influence of Pisaster on species richness by removing sea stars from selected parts of the intertidal zone repeatedly over a period of five years. Two major changes occurred in the areas from which sea stars were removed. First, the lower edge of the mussel bed extended father down into the intertidal zone, showing that sea stars are able to eliminate mussels completely where they are covered with water most of the time. Second, and more dramatically, 28 species of animals and algae disappeared from the sea star removal zone Eventually only Mytilus, the dominant competitor, occupied the entire substratum. Through its effect on competitive relationships, predation by Pisaster largely determines which species live in these rocky intertidal ecosystems.
What is the crux of the passage?
Sea star has a preferred prey.
A preferred prey determines the survival of a keystone species.
Keystone species ensures species diversity.
Sea star is the only keystone species on the Pacific coast of North America.
The question asks for the crux or main point of the passage. To identify the crux, one must understand the central theme or the most significant conclusion drawn from the information provided in the passage.
Correct Option: C) Keystone species ensures species diversity.
The passage clearly illustrates that the sea star, a keystone species, prevents the dominance of mussels, thereby creating space for a variety of other species. The removal of the sea star led to a significant loss of species, confirming its role in maintaining species diversity. This is the central theme and the most important conclusion presented in the passage.
With reference to the passage, consider the following statements:
1. Musset are generally the dominant species in intertidal ecosystems.
2. The survival of sea stars is generally determined by the abundance of mussels.
Which of the statements given above is/are correct?
The question asks us to evaluate two statements regarding mussels and sea stars in intertidal ecosystems. Since no passage is provided, we must rely on general ecological knowledge about these species and their interactions.
Correct Option: D) Neither 1 nor 2
Which of the following is/are implied by the passage?
1. Mussels are always hard competitors for sea stars.
2. Sea stars of the Pacific coasts have reached the climax of their evolution.
3. Sea stars constitute an important component in the energy flow in intertidal ecosystems.
Which of the statements given above is/are correct?
The question asks us to identify which statements are implied by a given passage. Since the passage is not provided, we must assume that the question is designed to test general knowledge about sea stars and intertidal ecosystems, or that the passage is implicitly understood to convey common ecological principles related to these organisms. We will evaluate each statement based on general ecological understanding.
D) 3 only
Consider the following assumptions:
1. The food chains/food web in an ecosystem are influenced by keystone species.
2. The presence of keystone species is a specific characteristic of aquatic ecosystems.
3. If the keystone species is completely removed from an ecosystem, it will lead to the collapse of the ecosystem.
With reference to the passage, which of the above assumptions is/are valid?
The question asks to identify the valid assumptions regarding keystone species in an ecosystem. We need to evaluate each given assumption based on ecological principles.
Correct Option: C) 1 and 3 only
Consider the following argument.
"In order to be a teacher one must graduate from college. All poets are poor. Some Mathematicians are poets. No college graduate is poor."
Which one of the following is not a valid conclusion regarding the above argument?
Some Mathematicians are not teachers.
Some teachers are not Mathematicians.
Teachers are not poor.
Poets are not teachers.
This problem involves deductive reasoning based on a set of given statements. We need to identify which conclusion cannot be logically derived from the premises. We can use Venn diagrams or logical syllogisms to represent the relationships between the categories (Teacher, College Graduate, Poet, Poor, Mathematician) and then check each option.
Correct Option: B) Some teachers are not Mathematicians.
This conclusion cannot be definitively drawn from the given statements. We know that Teachers are College Graduates and Mathematicians can be Poets (who are Poor). Since College Graduates are not Poor, any Mathematician who is a Poet cannot be a Teacher. However, this doesn't tell us anything about Teachers who are not Poets, or Mathematicians who are not Poets. It is possible, based on the given information, that all teachers are mathematicians, or that no teachers are mathematicians, or that some teachers are mathematicians. We cannot conclude that some teachers are definitely not mathematicians.
This problem involves calculating the average score. We are given the average score for the first three tests and information about the score on the fourth test. We need to find the new average score for all four tests.
So, for the first 3 tests:
\[ N = \frac{\text{Sum of scores for 3 tests}}{3} \]Correct Option: D) N + 5
We need to find the fraction of females who are single. We can assume a total number of persons (e.g., 100) to make calculations easier. Then, we will calculate the number of males, females, married persons, married males, and married females. Finally, we can determine the number of single females and express it as a fraction of the total number of females.
Let the total number of persons in the group be 100.
Correct Option: D) 2/3