To find the fifth term of a geometric progression (GP), we use the formula for the \(n\)-th term of a GP, which is given by \(a_n = a \cdot r^{(n-1)}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Thus, the fifth term of the geometric progression is 162.
Correct Option: D) 162
We are given the ratio of two numbers and how they change after certain operations, leading to a new relationship between them. We will set up algebraic equations based on the given information and solve for the numbers.
Let the two numbers be \(9x\) and \(5x\), since their ratio is \(9:5\). The greater number is \(9x\) and the smaller number is \(5x\).
According to the problem, 9 is added to the greater number, so the new greater number is \(9x + 9\).
5 is subtracted from the smaller number, so the new smaller number is \(5x - 5\).
The problem states that the new greater number becomes thrice the new smaller number. We can write this as an equation:
\[ 9x + 9 = 3(5x - 5) \]Now, we solve this equation for \(x\):
\[ 9x + 9 = 15x - 15 \]Subtract \(9x\) from both sides:
\[ 9 = 15x - 9x - 15 \]\[ 9 = 6x - 15 \]Add 15 to both sides:
\[ 9 + 15 = 6x \]\[ 24 = 6x \]Divide by 6:
\[ x = \frac{24}{6} \]\[ x = 4 \]Now substitute the value of \(x\) back into our original expressions for the numbers:
Greater number \(= 9x = 9 \times 4 = 36\)
Smaller number \(= 5x = 5 \times 4 = 20\)
Let's verify the conditions: If 9 is added to 36, it becomes 45. If 5 is subtracted from 20, it becomes 15. Is 45 thrice 15? Yes, \(45 = 3 \times 15\). So, the numbers are 36 and 20.
Correct Option: D) 36, 20
The problem involves setting up equations based on the given information about the total number of students and the ratio of girls to boys. We will use variables to represent the number of boys and girls and then solve for the number of boys.
Let the total number of students in the class be \(T\).
Given: \(T = 40\)
Let the number of boys be \(B\) and the number of girls be \(G\).
The total number of students is the sum of boys and girls:
\[ B + G = T \]\[ B + G = 40 \quad \text{(Equation 1)} \]The problem states that the number of girls is three-fifths of the number of boys.
\[ G = \frac{3}{5} B \quad \text{(Equation 2)} \]Substitute Equation 2 into Equation 1:
\[ B + \left(\frac{3}{5} B\right) = 40 \]Combine the terms involving \(B\):
\[ B \left(1 + \frac{3}{5}\right) = 40 \]\[ B \left(\frac{5}{5} + \frac{3}{5}\right) = 40 \]\[ B \left(\frac{8}{5}\right) = 40 \]Solve for \(B\):
\[ B = 40 \times \frac{5}{8} \]\[ B = \frac{40 \times 5}{8} \]\[ B = 5 \times 5 \]\[ B = 25 \]Thus, the number of boys in the class is 25.
We can also find the number of girls to verify:
\[ G = \frac{3}{5} B = \frac{3}{5} \times 25 = 3 \times 5 = 15 \]Check the total number of students:
\[ B + G = 25 + 15 = 40 \]The total matches the given information, so our calculation is correct.
Correct Option: B) 25 — The number of boys is 25, which satisfies the given conditions that the total students are 40 and girls are three-fifths of boys.
This problem involves the concept of work and time, specifically the relationship between the number of men, days, and hours worked to complete a certain amount of work. The total work done can be considered constant in both scenarios. We can use the formula \(\text{Work} = \text{Men} \times \text{Days} \times \text{Hours per day}\) or a combined proportionality approach.
Since the work is the same in both scenarios, we can equate the total work done. The total work is proportional to the product of men, days, and hours per day.
\[ M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2 \]First, calculate the product on the left side:
\[ 24 \times 15 \times 8 = 360 \times 8 = 2880 \]\[ 2880 = M_2 \times 60 \]Therefore, 48 men will be required to complete the same work in 10 days working 6 hours per day.
C) 48 — As calculated, 48 men are required to complete the work under the new conditions.
This question asks for the full form of the acronym BARC. This is a factual recall question related to scientific organizations in India.
Correct Option: D) Bhabha Atomic Research Center
The area of the walls of a room is also known as the lateral surface area of the room. For a rectangular room, this can be calculated by multiplying the perimeter of the floor by the height of the room.
The area of the walls (lateral surface area) of a room is given by the formula:
\[ \text{Area of walls} = \text{Perimeter of floor} \times \text{Height of room} \]Correct Option: C) 54 m²
The question asks to identify the vitamin that exists in three active forms: retinal, retinol, and retinoic acid. This requires knowledge of the different forms and functions of various vitamins.
B) Vitamin A — Vitamin A (retinoids) indeed has three primary active forms: retinol, retinal, and retinoic acid. These forms are crucial for various biological processes including vision, immune function, and cell growth.
The _____ Five-Year Plan of India completed its term in March 2017.
The question asks to identify which Five-Year Plan of India concluded in March 2017. To answer this, one needs to know the duration of the various Five-Year Plans in India.
B) 12th — The 12th Five-Year Plan (2012-2017) was the last Five-Year Plan of India and concluded in March 2017.
To find the quadratic equation given its roots, we can use the relationship between the roots and the coefficients of a quadratic equation. If \(\alpha\) and \(\beta\) are the roots of a quadratic equation \(ax^2 + bx + c = 0\), then the equation can be written as \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).
The given roots are \(\alpha = -2\) and \(\beta = 3\).
\[ \text{Sum of roots} = \alpha + \beta = -2 + 3 = 1 \]
\[ \text{Product of roots} = \alpha \beta = (-2) \times (3) = -6 \]
The general form of a quadratic equation with roots \(\alpha\) and \(\beta\) is:
\[ x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \]
Substitute the calculated sum and product:
\[ x^2 - (1)x + (-6) = 0 \]
\[ x^2 - x - 6 = 0 \]
Correct Option: B) \( x^{2} - x - 6 = 0 \)
Solve the following.
1 + \(\frac{1+\cos\theta}{\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}-\frac{\sin\theta}{1-\cos\theta}\)= ?
The problem asks us to simplify a trigonometric expression. We will use fundamental trigonometric identities such as \(\sin^2\theta + \cos^2\theta = 1\) and \(\sin^2\theta = (1-\cos\theta)(1+\cos\theta)\) to simplify the terms and combine them.
Let the given expression be \(E\):
\[ E = 1 + \frac{1+\cos\theta}{\sin\theta} - \frac{\sin^2\theta}{1+\cos\theta} - \frac{\sin\theta}{1-\cos\theta} \]Consider the third term, \(\frac{\sin^2\theta}{1+\cos\theta}\). We know that \(\sin^2\theta = 1 - \cos^2\theta\). We can factor \(1 - \cos^2\theta\) as \((1-\cos\theta)(1+\cos\theta)\).
\[ \frac{\sin^2\theta}{1+\cos\theta} = \frac{1-\cos^2\theta}{1+\cos\theta} = \frac{(1-\cos\theta)(1+\cos\theta)}{1+\cos\theta} \]Assuming \(1+\cos\theta \neq 0\), we can cancel out the \((1+\cos\theta)\) term:
\[ \frac{\sin^2\theta}{1+\cos\theta} = 1-\cos\theta \]Substitute this back into the original expression:
\[ E = 1 + \frac{1+\cos\theta}{\sin\theta} - (1-\cos\theta) - \frac{\sin\theta}{1-\cos\theta} \]Simplify the expression by combining the constant terms and rearranging:
\[ E = 1 + \frac{1+\cos\theta}{\sin\theta} - 1 + \cos\theta - \frac{\sin\theta}{1-\cos\theta} \]\[ E = \frac{1+\cos\theta}{\sin\theta} + \cos\theta - \frac{\sin\theta}{1-\cos\theta} \]Now, let's combine the terms with \(\sin\theta\) in the denominator. Find a common denominator for \(\frac{1+\cos\theta}{\sin\theta}\) and \(\frac{\sin\theta}{1-\cos\theta}\). The common denominator is \(\sin\theta(1-\cos\theta)\).
\[ E = \frac{(1+\cos\theta)(1-\cos\theta)}{\sin\theta(1-\cos\theta)} - \frac{\sin\theta \cdot \sin\theta}{\sin\theta(1-\cos\theta)} + \cos\theta \]\[ E = \frac{1-\cos^2\theta}{\sin\theta(1-\cos\theta)} - \frac{\sin^2\theta}{\sin\theta(1-\cos\theta)} + \cos\theta \]Since \(1-\cos^2\theta = \sin^2\theta\), the first two terms combine:
\[ E = \frac{\sin^2\theta - \sin^2\theta}{\sin\theta(1-\cos\theta)} + \cos\theta \]\[ E = \frac{0}{\sin\theta(1-\cos\theta)} + \cos\theta \]Thus, the expression simplifies to:
\[ E = 0 + \cos\theta \]\[ E = \cos\theta \]A) \(\cos\theta\) is the correct simplification of the given expression.