This is a blood relation problem. We need to deduce the relationship between Siri and Sree based on the given statements. We will build a family tree or list the relationships step-by-step.
Correct Option: D) Sister-in-law
If α & β are the zeros of the polynomial 5x² - 7x + 2 then the sum of their reciprocals is:
\( \frac{5}{2} \)
\( \frac{2}{5} \)
\( \frac{7}{5} \)
\( \frac{7}{2} \)
The problem asks for the sum of the reciprocals of the zeros of a given quadratic polynomial. We can use the relationships between the coefficients of a quadratic polynomial and its zeros (sum of zeros and product of zeros) to find the required sum.
Thus, the sum of the reciprocals of the zeros is \( \frac{7}{2} \).
Correct Option: D) \( \frac{7}{2} \)
The question asks to identify the permanent tissue where the cells are dead. We need to recall the characteristics of different types of permanent plant tissues, specifically focusing on whether their cells are living or dead at maturity.
Correct Option: C) Sclerenchyma is a simple permanent tissue consisting of dead cells with thick, lignified walls. These cells provide mechanical support and protection to the plant.
The question asks for the location of the testes outside the abdominal cavity. This is a fundamental concept in human male reproductive anatomy. The key is to identify the structure that houses the testes externally.
Correct Option: D) Scrotum
The testes are located outside the abdominal cavity within a pouch of skin called the scrotum. This external placement is crucial because sperm production (spermatogenesis) requires a temperature that is 2-3 degrees Celsius lower than the normal internal body temperature. The scrotum acts as a thermoregulator, maintaining this optimal temperature for sperm viability.
To find the smallest number by which 72 must be multiplied to obtain a perfect square, we need to perform prime factorization of 72. A number is a perfect square if all the exponents in its prime factorization are even.
First, we find the prime factors of 72.
\[ 72 = 2 \times 36 \]\[ 72 = 2 \times 2 \times 18 \]\[ 72 = 2 \times 2 \times 2 \times 9 \]\[ 72 = 2 \times 2 \times 2 \times 3 \times 3 \]So, the prime factorization of 72 is \(2^3 \times 3^2\).For a number to be a perfect square, all the exponents in its prime factorization must be even. In the prime factorization of 72, which is \(2^3 \times 3^2\):
To make the exponent of 2 even, we need to multiply \(2^3\) by \(2^1\) to get \(2^{3+1} = 2^4\). The exponent of 3 is already even, so we don't need to multiply by any more factors of 3.
Therefore, the smallest number by which 72 must be multiplied to make it a perfect square is 2.
\[ 72 \times 2 = (2^3 \times 3^2) \times 2 \]\[ = 2^4 \times 3^2 \]\[ = (2^2)^2 \times (3)^2 \]\[ = (2^2 \times 3)^2 \]\[ = (4 \times 3)^2 \]\[ = (12)^2 \]\[ = 144 \]Since 144 is \(12^2\), it is a perfect square.Correct Option: D) 2
Multiplying 72 by 2 gives 144, which is \(12^2\), a perfect square. This is the smallest number required to achieve a perfect square.
This problem involves finding the shortest distance between a starting point and an ending point after a series of movements in different directions. This can be solved by breaking down the movements into horizontal (East-West) and vertical (North-South) components and then using the Pythagorean theorem to find the resultant displacement.
The total displacement in the East direction (x-component) from the starting point is:
\[ \text{Net East displacement} = 15 \text{ km (initial East)} + 30 \text{ km (later East)} = 45 \text{ km} \]The total displacement in the South direction (y-component) from the starting point is:
\[ \text{Net South displacement} = 20 \text{ km (initial South)} + 40 \text{ km (later South)} = 60 \text{ km} \]So, the final position relative to the starting point is 45 km East and 60 km South.
The shortest distance between the starting point and the destination is the hypotenuse of a right-angled triangle formed by the net East displacement and the net South displacement. We use the Pythagorean theorem:
\[ \text{Shortest Distance} = \sqrt{(\text{Net East displacement})^2 + (\text{Net South displacement})^2} \]\[ \text{Shortest Distance} = \sqrt{(45)^2 + (60)^2} \]\[ \text{Shortest Distance} = \sqrt{2025 + 3600} \]\[ \text{Shortest Distance} = \sqrt{5625} \]\[ \text{Shortest Distance} = 75 \text{ km} \]B) 75 km is the shortest distance between the starting point and the destination, calculated using the Pythagorean theorem.
This problem involves understanding the relationship between wave speed, wavelength, and time period. The distance between two consecutive wave crests is the wavelength, and the time taken for a wave crest to travel this distance is the time period of the wave. We can use the formula relating speed, distance, and time to find the required time.
Identify the given information:
Understand what needs to be found:
The time required by the crests to reach the boat refers to the time it takes for one complete wave (from one crest to the next) to pass a fixed point, which is the time period (\(T\)) of the wave.
Recall the relationship between wave speed, wavelength, and time period:
The wave speed (\(v\)) is given by the formula:
\[ v = \frac{\lambda}{T} \]where \(\lambda\) is the wavelength and \(T\) is the time period.
Rearrange the formula to solve for the time period (\(T\)):
\[ T = \frac{\lambda}{v} \]Substitute the given values into the formula:
\[ T = \frac{80 \text{ m}}{20 \text{ m/s}} \]Calculate the time period:
\[ T = 4 \text{ s} \]Correct Option: D) 4 s
The time required by the crests to reach the boat, which is the time period of the wave, is 4 seconds.
To solve this problem, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)).
The expression inside the parentheses is \(7 - 3\).
\[ 7 - 3 = 4 \]Now substitute the result back into the original expression:
\[ -70 + 28 \div 4 \]Perform the division:
\[ 28 \div 4 = 7 \]Substitute the result of the division back into the expression:
\[ -70 + 7 \]Perform the addition:
\[ -70 + 7 = -63 \]Thus, the final answer is \(-63\).
D) -63 is the correct answer following the order of operations.
This problem involves understanding percentages. We are given a specific mark obtained and its equivalent percentage. We need to find the total marks the test was out of. We can set up a simple equation where the given marks represent a certain percentage of the total marks.
We know that 63 marks represent 84% of the total marks \(T\). This can be written as:
\[ 84\% \text{ of } T = 63 \]Or, in mathematical terms:
\[ \frac{84}{100} \times T = 63 \]To find \(T\), we can rearrange the equation:
\[ T = 63 \times \frac{100}{84} \]Simplify the fraction:
\[ T = \frac{6300}{84} \]We can divide both numerator and denominator by common factors. For example, both are divisible by 7:
\[ T = \frac{900}{12} \]Now, divide 900 by 12:
\[ T = 75 \]So, the test was out of 75 marks.
Correct Option: C) 75
This problem involves work and time. We are given the time taken by pairs of individuals (A and B, B and C, C and A) to complete a task. We need to find the time each individual (A, B, and C) takes to complete the task alone. We will use the concept of work rate, where work rate is the reciprocal of the time taken to complete the task.
Let the time taken by A, B, and C to complete the task individually be \(T_A\), \(T_B\), and \(T_C\) days respectively.
Their individual work rates per day will be \(\frac{1}{T_A}\), \(\frac{1}{T_B}\), and \(\frac{1}{T_C}\).
From the given information, we can form equations based on their combined work rates:
Add Equation 1, Equation 2, and Equation 3:
\[ \left( \frac{1}{T_A} + \frac{1}{T_B} \right) + \left( \frac{1}{T_B} + \frac{1}{T_C} \right) + \left( \frac{1}{T_C} + \frac{1}{T_A} \right) = \frac{1}{40} + \frac{1}{30} + \frac{1}{24} \]\[ 2 \left( \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C} \right) = \frac{3 + 4 + 5}{120} \]\[ 2 \left( \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C} \right) = \frac{12}{120} \]\[ 2 \left( \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C} \right) = \frac{1}{10} \]\[ \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C} = \frac{1}{20} \quad \text{(Equation 4)} \]
Now, subtract each of the original equations (1, 2, 3) from Equation 4 to find the individual work rates:
Thus, A takes 60 days, B takes 120 days, and C takes 40 days to complete the task individually.
Correct Option: C) 60, 120 and 40