The question asks about the standard domestic power supply specifications in India. This involves identifying the correct voltage and frequency for alternating current (AC) used in Indian households.
B) 220 V, 50 Hz is the correct specification for domestic power supply in India. This voltage and frequency combination is standard for most household appliances and electrical systems across the country.
This is a linear arrangement puzzle. We need to deduce the seating order of the five friends based on the given clues, keeping in mind they are facing North. We will build the arrangement step-by-step.
A) Madavi is sitting at the rightmost end, as determined by the combined seating arrangement P-S-R-T-M.
To find the breadth of the rectangular park, we will set up an algebraic equation based on the given information. We are given the relationship between the length and breadth, and the area of the park. We will form a quadratic equation and solve for the breadth.
Define variables: Let the breadth of the rectangular park be \(b\) meters. According to the problem, the length \(l\) is 22 m more than its breadth. So, \(l = b + 22\) meters.
Formulate the area equation: The area of a rectangle is given by the formula: \(\text{Area} = \text{length} \times \text{breadth}\). We are given that the area is 1400 m². So, we can write the equation:
\[ (b + 22) \times b = 1400 \]Solve the quadratic equation: Expand the equation:
\[ b^2 + 22b = 1400 \]Rearrange it into a standard quadratic form \(ax^2 + bx + c = 0\):
\[ b^2 + 22b - 1400 = 0 \]We need to find two numbers that multiply to -1400 and add up to 22. These numbers are 50 and -28.
\[ b^2 + 50b - 28b - 1400 = 0 \]Factor by grouping:
\[ b(b + 50) - 28(b + 50) = 0 \] \[ (b - 28)(b + 50) = 0 \]This gives two possible values for \(b\):
\[ b - 28 = 0 \implies b = 28 \] \[ b + 50 = 0 \implies b = -50 \]Choose the valid solution: Since breadth cannot be negative, we discard \(b = -50\). Therefore, the breadth of the park is \(b = 28\) meters.
Verify the answer (optional): If breadth \(b = 28\) m, then length \(l = 28 + 22 = 50\) m. Area \(= l \times b = 50 \times 28 = 1400\) m². This matches the given area.
Correct Option: C) 28
To solve this problem, we need to apply the given rules to each letter in the word "FRANTIC". We will identify vowels and consonants, change them according to the rules (vowel to the next letter, consonant to the previous letter), and then check which letter appears twice in the newly formed group.
The new group of letters formed is E, Q, B, M, S, J, B.
In the group {E, Q, B, M, S, J, B}, the letter B appears twice.
B) The letter 'B' appears twice in the new group of letters (E, Q, B, M, S, J, B).
We need to find a 2-digit number that satisfies two conditions: the sum of its digits is 9, and the digit at the tens place is double the digit at the units place. We can solve this by setting up a system of equations or by checking the given options against these conditions.
From the first condition, "The sum of the digits of a 2-digit number is 9", we can write the equation:
\[ t + u = 9 \quad \text{(Equation 1)} \]From the second condition, "The number at the tens place is double of the number at the units place", we can write the equation:
\[ t = 2u \quad \text{(Equation 2)} \]Now, substitute Equation 2 into Equation 1:
\[ (2u) + u = 9 \]\[ 3u = 9 \]Solve for \(u\):
\[ u = \frac{9}{3} \]\[ u = 3 \]Substitute the value of \(u\) back into Equation 2 to find \(t\):
\[ t = 2u \]\[ t = 2(3) \]\[ t = 6 \]The 2-digit number is \(10t + u = 10(6) + 3 = 60 + 3 = 63\).
Alternatively, we can check the given options:
Correct Option: D) 63
We are given three numbers that are co-prime to one another. We are also given the product of the first two numbers and the product of the last two numbers. We need to find the LCM of these three numbers. Since the numbers are co-prime to one another, their HCF is 1. We can find the individual numbers by using the given products and then calculate their LCM.
Let the three co-prime numbers be \(a\), \(b\), and \(c\).
Given that the product of the first two numbers is 432:
\(a \times b = 432 \quad \text{(Equation 1)}\)
Given that the product of the last two numbers is 945:
\(b \times c = 945 \quad \text{(Equation 2)}\)
Since \(a\), \(b\), and \(c\) are co-prime to one another, it means that \(\text{HCF}(a, b) = 1\), \(\text{HCF}(b, c) = 1\), and \(\text{HCF}(a, c) = 1\). This implies that \(b\) is the common factor between the two products, and \(b\) must be the \(\text{HCF}\) of 432 and 945.
Find the HCF of 432 and 945:
Prime factorization of \(432\):
\[ 432 = 2^4 \times 3^3 \]Prime factorization of \(945\):
\[ 945 = 3^3 \times 5 \times 7 \]The common prime factors with the lowest powers give the HCF:
\[ \text{HCF}(432, 945) = 3^3 = 27 \]So, \(b = 27\).
Now, substitute the value of \(b\) into Equation 1 to find \(a\):
\[ a \times 27 = 432 \] \[ a = \frac{432}{27} = 16 \]Substitute the value of \(b\) into Equation 2 to find \(c\):
\[ 27 \times c = 945 \] \[ c = \frac{945}{27} = 35 \]The three numbers are \(a=16\), \(b=27\), and \(c=35\).
Verify that they are co-prime to one another:
The condition is satisfied.
Find the LCM of the three numbers \(16, 27, 35\).
Since the numbers are pairwise co-prime, their LCM is simply their product:
\[ \text{LCM}(16, 27, 35) = 16 \times 27 \times 35 \] \[ 16 \times 27 = 432 \] \[ 432 \times 35 = 15120 \]Correct Option: B) 15120
To find the magnification produced by a concave mirror, we first need to determine the image distance using the mirror formula. Once the image distance is known, we can calculate the magnification using the magnification formula.
Identify the given values and sign conventions:
For a concave mirror:
Apply the mirror formula to find the image distance (\(v\)):
The mirror formula is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]Substitute the values into the mirror formula:
\[ \frac{1}{-18} = \frac{1}{v} + \frac{1}{-12} \]\[ -\frac{1}{18} = \frac{1}{v} - \frac{1}{12} \]Rearrange the formula to solve for \(v\):
\[ \frac{1}{v} = -\frac{1}{18} + \frac{1}{12} \]Find a common denominator, which is 36:
\[ \frac{1}{v} = \frac{-2}{36} + \frac{3}{36} \]\[ \frac{1}{v} = \frac{1}{36} \]\[ v = 36 \text{ cm} \]Calculate the magnification (\(m\)):
The magnification formula is:
\[ m = -\frac{v}{u} \]Substitute the values of \(v\) and \(u\) into the magnification formula:
\[ m = -\frac{36 \text{ cm}}{-12 \text{ cm}} \]\[ m = -(-3) \]\[ m = +3 \]D) The magnification produced by the mirror is \(+3\). The positive sign indicates that the image is virtual and erect, and the value 3 indicates that the image is three times larger than the object.
To find the approximate value, we will first round off each number to its nearest integer. Then, we will perform the operations following the BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction) rule.
Thus, the closest approximate value is 21.
Correct Option: A) 21
The given series is an alternating series where two different patterns are interleaved. We need to identify these patterns to find the missing number.
D) -3 is the correct answer because the even-positioned terms follow a pattern of subtracting 2 from the previous term.
The question asks for the specific population threshold that defines an endangered language under the Scheme for Protection and Preservation of Endangered Languages of India (SPPEL). This is a factual question requiring knowledge of the scheme's criteria.
C) 10,000. The SPPEL defines endangered languages as those spoken by less than 10,000 people. This threshold is crucial for identifying which languages qualify for protection, preservation, and documentation efforts by the Central Institute of Indian Languages (CIIL).