This problem involves the relationship between efficiency and the time taken to complete a certain amount of work. Efficiency is inversely proportional to the time taken to complete a fixed amount of work. If someone is more efficient, they take less time, and vice-versa.
Understand the relationship: The efficiency of work is inversely proportional to the number of days (or time) taken to finish the work, assuming the total work is constant. This means if the ratio of efficiencies is \(a:b\), then the ratio of days taken will be \(b:a\).
\[ \text{Efficiency} \propto \frac{1}{\text{Time taken}} \]Given information: The ratio of the efficiency of P and Q is given as \(5:7\).
So, \(\text{Efficiency}_P : \text{Efficiency}_Q = 5:7\).
Apply the inverse relationship: Since the number of days taken is inversely proportional to efficiency, the ratio of days taken will be the inverse of the ratio of efficiencies.
\[ \text{Days}_P : \text{Days}_Q = \frac{1}{\text{Efficiency}_P} : \frac{1}{\text{Efficiency}_Q} \] \[ \text{Days}_P : \text{Days}_Q = \frac{1}{5} : \frac{1}{7} \]Simplify the ratio: To remove the fractions, multiply both sides of the ratio by the least common multiple of the denominators (which is \(5 \times 7 = 35\)).
\[ \text{Days}_P : \text{Days}_Q = \left(\frac{1}{5} \times 35\right) : \left(\frac{1}{7} \times 35\right) \] \[ \text{Days}_P : \text{Days}_Q = 7 : 5 \]Correct Option: A) 7:5
Answer the next 3 questions:
Comprehension:
Consider the following information and answer questions based on it.
Among 60 students, 12 like only algebra, 13 like only geometry, 10 like only trigonometry, 5 like both algebra and trigonometry, 8 like only physics, 5 like both physics and geometry and the remaining like both algebra and physics.
Question 1:
The number of students who like physics but not geometry is
This problem can be solved by systematically listing the number of students in each category based on the given information. We need to find the number of students who like physics but do not like geometry. This involves identifying all groups that like physics and then excluding any group that also likes geometry.
The sum of students accounted for so far is:
\[ 12 + 13 + 10 + 5 + 8 + 5 = 53 \]The remaining students are:
\[ 60 - 53 = 7 \]The problem states, "the remaining like both algebra and physics." So, students who like Algebra and Physics (A & P) = 7.
Students who like Physics are in the following categories:
Total students who like Physics = \(8 + 5 + 7 = 20\).
From the groups that like Physics, we need to exclude those who also like Geometry. The groups that like Physics are:
So, the number of students who like physics but not geometry is the sum of "Only Physics" and "Algebra and Physics" groups:
\[ 8 + 7 = 15 \]Correct Option: C) 15
Question 2:
How many students like more than one subject?
The problem asks us to find the number of students who like more than one subject. This can be determined by analyzing the Venn diagram provided (which is implicitly referred to by the question). We need to identify the regions in the Venn diagram that represent the intersection of two or more subjects.
Identify regions representing students liking exactly two subjects:
Identify regions representing students liking all three subjects:
Calculate the total number of students who like more than one subject:
This is the sum of students liking exactly two subjects and students liking all three subjects.
\[ \text{Students liking more than one subject} = \text{(Maths and Physics only)} + \text{(Physics and Chemistry only)} + \text{(Maths and Chemistry only)} + \text{(Maths, Physics, and Chemistry)} \]\[ \text{Students liking more than one subject} = 5 + 7 + 3 + 2 \]\[ \text{Students liking more than one subject} = 17 \]D) 17 — The total number of students who like more than one subject is the sum of students liking exactly two subjects (5 + 7 + 3) and students liking all three subjects (2), which is \(5 + 7 + 3 + 2 = 17\).
Question 3:
The ratio of students who like algebra to those who like geometry is
The problem asks for the ratio of students who like algebra to those who like geometry. To find this, we need to identify the number of students who like each subject from the given data (which is assumed to be available from a preceding context or diagram not provided here, but implied by the question format). Once these numbers are known, we can express their ratio in its simplest form.
Correct Option: B) 4:3
This is a blood relation question. We need to trace the relationships step-by-step from Asha to Aniket, and then determine Sagar's relationship to Aniket's wife. We will break down the complex relationship into simpler parts.
Correct Option: B) Son. Sagar is the son of Aniket's wife.
The problem is a coding-decoding type where letters are assigned numerical values. We need to identify the pattern used to assign values to words and then apply the same pattern to find the value of "TIME". The standard alphabetical position of letters is usually the basis for such problems.
Thus, the value for TIME is 47.
Correct Option: B) 47
This is a coding-decoding problem where letters are represented by digits. We need to find the digit corresponding to the letter 'a' by comparing the given coded words and their numerical representations.
Both words contain the letters 'r', 'i', 'd', 't'.
Common letters: 'r', 'i', 'd'. Common digits: 3, 2, 1. This means 't' corresponds to 4 in 'dirt' and 'e' corresponds to 8 in 'ride'.
Common letters: 'i', 'e'. Common digits: 2, 8.
From step 2, we know 'e' = 8. So, 'i' = 2.
Common letters: 'l', 'k'. Common digits: 5, 6.
From 'talk' = 7564, we know 'l' and 'k' correspond to 5 and 6 (in some order). The remaining letters are 't' and 'a', and the remaining digits are 7 and 4.
From step 2, we found that 't' corresponds to 4.
Therefore, the remaining letter 'a' must correspond to the remaining digit 7.
D) 7 is the correct digit representing 'a'.
The question asks to identify the statement that is NOT true about the Global Positioning System (GPS). We need to evaluate each option based on the known functionalities and limitations of GPS.
Based on this analysis, the statement that GPS can predict cyclones is incorrect.
B) It can predict cyclones. This statement is not true. GPS is a positioning system, not a weather prediction system. While GPS data can contribute to meteorological studies, GPS itself does not predict cyclones.
This question tests knowledge of the Indian constitutional provisions regarding the presidential line of succession. When both the President and Vice-President's offices are vacant, a specific constitutional arrangement comes into play to ensure continuity in the highest office.
Correct Option: C) Chief Justice of India
The average of first three out of four numbers is 18. The average of last three numbers is 14. The sum of first and last number is 16. The last number is:
We are given information about the averages of four numbers. We will represent the four numbers with variables and set up a system of linear equations based on the given average and sum conditions. Solving this system will allow us to find the value of each number, specifically the last number.
The average of the first three numbers is 18. This means:
\[ \frac{a + b + c}{3} = 18 \]\[ a + b + c = 18 \times 3 \]\[ a + b + c = 54 \quad \text{(Equation 1)} \]The average of the last three numbers is 14. This means:
\[ \frac{b + c + d}{3} = 14 \]\[ b + c + d = 14 \times 3 \]\[ b + c + d = 42 \quad \text{(Equation 2)} \]The sum of the first and last number is 16. This means:
\[ a + d = 16 \quad \text{(Equation 3)} \]Now we have a system of three linear equations. Subtract Equation 2 from Equation 1:
\[ (a + b + c) - (b + c + d) = 54 - 42 \]\[ a - d = 12 \quad \text{(Equation 4)} \]We now have two equations involving \(a\) and \(d\):
\[ a + d = 16 \quad \text{(Equation 3)} \]\[ a - d = 12 \quad \text{(Equation 4)} \]Add Equation 3 and Equation 4:
\[ (a + d) + (a - d) = 16 + 12 \]\[ 2a = 28 \]\[ a = \frac{28}{2} \]\[ a = 14 \]Substitute the value of \(a\) into Equation 3 to find \(d\):
\[ 14 + d = 16 \]\[ d = 16 - 14 \]\[ d = 2 \]Correct Option: C) 2