The problem states that a person spends 32% of his monthly income and saves ₹7,752. We need to find his monthly expenses. We can determine the percentage of income saved and then use this information to calculate the total monthly income. Once the total income is known, the monthly expenses can be found by calculating 32% of the total income.
If a person spends 32% of his income, the remaining percentage is saved.
\[ \text{Percentage saved} = 100\% - \text{Percentage spent} \]\[ \text{Percentage saved} = 100\% - 32\% = 68\% \]We know that 68% of the monthly income is ₹7,752. Let the total monthly income be \(I\).
\[ 0.68 \times I = 7752 \]\[ I = \frac{7752}{0.68} \]\[ I = 11400 \]The monthly expenses are 32% of the total monthly income.
\[ \text{Expenses} = 32\% \times I \]\[ \text{Expenses} = 0.32 \times 11400 \]\[ \text{Expenses} = 3648 \]Thus, the person's monthly expenses are ₹3,648.
C) 3,648 — As calculated, the monthly expenses are ₹3,648.
Cardiomyocytes, or heart muscle cells, are excitable cells that generate and propagate electrical signals to coordinate heart contractions. Maintaining their membrane potential and ensuring proper function relies on precise control of ion movement across the cell membrane. This involves both passive and active transport mechanisms.
Correct Option: D) A combination of voltage-gated ion channels and active ion pumps. This option accurately describes the dual mechanism by which cardiomyocytes maintain their membrane potential and ensure proper contraction. Voltage-gated ion channels are responsible for the rapid changes in membrane potential during an action potential, while active ion pumps maintain the necessary ion gradients over time.
To simplify the given expression, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS. This stands for Brackets/Parentheses, Orders/Exponents, Division/Multiplication (from left to right), and Addition/Subtraction (from left to right).
Let's apply these steps:
Given expression:
\[ 525 \div \{19 + 7 \times (3 - 5)\} \]B) 105 — The step-by-step application of the BODMAS/PEMDAS rule correctly leads to the value 105.
If sinA = \( \frac{3}{4} \), then calculate cosA cotA.
\( \frac{12}{7} \)
\( \frac{7}{12} \)
\( \frac{6}{7} \)
\( \frac{7}{6} \)
We are given the value of \(\sin A\) and need to calculate the value of \(\cos A \cot A\). We can use the fundamental trigonometric identity \(\sin^2 A + \cos^2 A = 1\) to find \(\cos A\). Then, we can use the definition of \(\cot A = \frac{\cos A}{\sin A}\) to find \(\cot A\). Finally, we multiply \(\cos A\) and \(\cot A\).
Find \(\cos A\):
Given \(\sin A = \frac{3}{4}\).
Using the identity \(\sin^2 A + \cos^2 A = 1\):
\[ \left(\frac{3}{4}\right)^2 + \cos^2 A = 1 \]\[ \frac{9}{16} + \cos^2 A = 1 \]\[ \cos^2 A = 1 - \frac{9}{16} \]\[ \cos^2 A = \frac{16 - 9}{16} \]\[ \cos^2 A = \frac{7}{16} \]\[ \cos A = \sqrt{\frac{7}{16}} \]\[ \cos A = \frac{\sqrt{7}}{4} \]Find \(\cot A\):
We know that \(\cot A = \frac{\cos A}{\sin A}\).
Substitute the values of \(\sin A\) and \(\cos A\):
\[ \cot A = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} \]\[ \cot A = \frac{\sqrt{7}}{4} \times \frac{4}{3} \]\[ \cot A = \frac{\sqrt{7}}{3} \]Calculate \(\cos A \cot A\):
Now, multiply the values of \(\cos A\) and \(\cot A\):
\[ \cos A \cot A = \left(\frac{\sqrt{7}}{4}\right) \times \left(\frac{\sqrt{7}}{3}\right) \]\[ \cos A \cot A = \frac{\sqrt{7} \times \sqrt{7}}{4 \times 3} \]\[ \cos A \cot A = \frac{7}{12} \]B) The calculated value of \(\cos A \cot A\) is \(\frac{7}{12}\).
To find the height of the cone, we will use the given curved surface area and diameter. First, we'll calculate the radius from the diameter. Then, using the formula for the curved surface area of a cone, we'll find the slant height. Finally, we'll use the Pythagorean theorem relating the radius, height, and slant height to determine the height of the cone.
The diameter of the base is given as \(102 \text{ cm}\).
\[ \text{Radius} (r) = \frac{\text{Diameter}}{2} = \frac{102 \text{ cm}}{2} = 51 \text{ cm} \]The curved surface area (CSA) of a right circular cone is given by the formula:
\[ \text{CSA} = \pi r l \]We are given \(\text{CSA} = 4335\pi \text{ cm}^2\) and we found \(r = 51 \text{ cm}\).
\[ 4335\pi = \pi \times 51 \times l \]Divide both sides by \(\pi\):
\[ 4335 = 51 \times l \]Solve for \(l\):
\[ l = \frac{4335}{51} \]To simplify the division, we can perform it:
\[ l = 85 \text{ cm} \]For a right circular cone, the radius (\(r\)), height (\(h\)), and slant height (\(l\)) form a right-angled triangle, with \(l\) as the hypotenuse. So, we can use the Pythagorean theorem:
\[ l^2 = r^2 + h^2 \]We have \(l = 85 \text{ cm}\) and \(r = 51 \text{ cm}\). Substitute these values:
\[ (85)^2 = (51)^2 + h^2 \]Calculate the squares:
\[ 7225 = 2601 + h^2 \]Subtract \(2601\) from both sides:
\[ h^2 = 7225 - 2601 \]\[ h^2 = 4624 \]Take the square root of both sides to find \(h\):
\[ h = \sqrt{4624} \]To find the square root:
\[ h = 68 \text{ cm} \]A) 68 cm is the calculated height of the cone.
To find the smallest number by which a given number must be multiplied to make it a perfect square, we need to perform prime factorization of the given number. A number is a perfect square if all the exponents in its prime factorization are even. We will identify the prime factors with odd exponents and multiply the number by the missing factors to make their exponents even.
We start by finding the prime factors of 4232.
\[ 4232 \div 2 = 2116 \]\[ 2116 \div 2 = 1058 \]\[ 1058 \div 2 = 529 \]Now we need to find the prime factors of 529. We can test prime numbers. We find that \(23 \times 23 = 529\).
\[ 529 \div 23 = 23 \]\[ 23 \div 23 = 1 \]Combining these steps, the prime factorization of 4232 is:
\[ 4232 = 2 \times 2 \times 2 \times 23 \times 23 \]\[ 4232 = 2^3 \times 23^2 \]For a number to be a perfect square, all the exponents in its prime factorization must be even. In the factorization \(2^3 \times 23^2\), the prime factor 2 has an exponent of 3 (which is odd), and the prime factor 23 has an exponent of 2 (which is even).
To make the exponent of 2 even, we need to multiply \(2^3\) by \(2^1\), which will result in \(2^{3+1} = 2^4\). The prime factor 23 already has an even exponent, so it does not need any additional factor.
Therefore, the smallest number by which 4232 must be multiplied to make it a perfect square is 2.
If we multiply 4232 by 2, we get:
\[ 4232 \times 2 = 8464 \]\[ 8464 = 2^4 \times 23^2 = (2^2 \times 23)^2 = (4 \times 23)^2 = (92)^2 \]Since \(92^2 = 8464\), 8464 is a perfect square.
D) 2 — Multiplying 4232 by 2 makes the exponent of its prime factor 2 even, thus making the entire number a perfect square.
The question asks to identify the landmark Supreme Court case that clarified the position of the Council of Ministers after the dissolution of the Lok Sabha. This requires knowledge of significant constitutional law judgments in India.
Correct Option: C) UNR Rao vs. Indira Gandhi, 1971
The question asks to identify the leader who traveled extensively in 1857 to gather support from native princes against the British. This requires knowledge of the key figures and their roles during the Indian Rebellion of 1857.
B) Nana Saheb — He was a key leader of the 1857 Rebellion, particularly in Kanpur. He actively sought to unite various Indian rulers against the British and traveled to different regions, including Kalpi, Delhi, and Lucknow, from his base in Bithoor, to garner support for the uprising.
To solve this problem, we first need to find the roots of the first quadratic equation. Since the two quadratic equations share a common root, this common root must satisfy both equations. We can substitute the common root into the second equation and solve for the unknown coefficient 'b'.
The first equation is \( 6x^{2} - 7x - 20 = 0 \).
We can solve this by factoring or using the quadratic formula. Let's use factoring by splitting the middle term:
\[ 6x^{2} - 15x + 8x - 20 = 0 \]\[ 3x(2x - 5) + 4(2x - 5) = 0 \]\[ (3x + 4)(2x - 5) = 0 \]\[ 3x + 4 = 0 \quad \text{or} \quad 2x - 5 = 0 \]\[ x = -\frac{4}{3} \quad \text{or} \quad x = \frac{5}{2} \]The common root can be either \( -\frac{4}{3} \) or \( \frac{5}{2} \). We will test both possibilities.
Substitute \( x = -\frac{4}{3} \) into the second equation \( 6x^{2} + bx - 10 = 0 \):
\[ 6\left(-\frac{4}{3}\right)^{2} + b\left(-\frac{4}{3}\right) - 10 = 0 \]\[ 6\left(\frac{16}{9}\right) - \frac{4b}{3} - 10 = 0 \]\[ \frac{2 \times 16}{3} - \frac{4b}{3} - 10 = 0 \]\[ \frac{32}{3} - \frac{4b}{3} - 10 = 0 \]Multiplying by 3 to clear the denominators:\[ 32 - 4b - 30 = 0 \]\[ 2 - 4b = 0 \]\[ 4b = 2 \]\[ b = \frac{2}{4} \]\[ b = \frac{1}{2} \]Substitute \( x = \frac{5}{2} \) into the second equation \( 6x^{2} + bx - 10 = 0 \):
\[ 6\left(\frac{5}{2}\right)^{2} + b\left(\frac{5}{2}\right) - 10 = 0 \]\[ 6\left(\frac{25}{4}\right) + \frac{5b}{2} - 10 = 0 \]\[ \frac{3 \times 25}{2} + \frac{5b}{2} - 10 = 0 \]\[ \frac{75}{2} + \frac{5b}{2} - 10 = 0 \]Multiplying by 2 to clear the denominators:\[ 75 + 5b - 20 = 0 \]\[ 55 + 5b = 0 \]\[ 5b = -55 \]\[ b = -\frac{55}{5} \]\[ b = -11 \]The calculated values for \( b \) are \( \frac{1}{2} \) and \( -11 \). Comparing these with the given options, \( -11 \) is one of the options.
The value \( b = -11 \) is obtained when the common root is \( x = \frac{5}{2} \), which matches option A.
To find the population of the town in 1988, we need to calculate the population after two successive decades, with a 20% increase in each decade. This is a compound growth problem. We will calculate the population after the first decade (1978) and then use that result to calculate the population after the second decade (1988).
Initial Population:
The population of the town in 1968 was \(P_0 = 4,85,000\).
Population after the first decade (1978):
The population increased by 20% in the first decade (1968-1978).
\[ P_1 = P_0 \times \left(1 + \frac{\text{Percentage Increase}}{100}\right) \]\[ P_1 = 4,85,000 \times \left(1 + \frac{20}{100}\right) \]\[ P_1 = 4,85,000 \times (1 + 0.20) \]\[ P_1 = 4,85,000 \times 1.20 \]\[ P_1 = 5,82,000 \]Population after the second decade (1988):
The population increased by 20% again in the second decade (1978-1988). We use \(P_1\) as the new base population.
\[ P_2 = P_1 \times \left(1 + \frac{\text{Percentage Increase}}{100}\right) \]\[ P_2 = 5,82,000 \times \left(1 + \frac{20}{100}\right) \]\[ P_2 = 5,82,000 \times (1 + 0.20) \]\[ P_2 = 5,82,000 \times 1.20 \]\[ P_2 = 6,98,400 \]Alternatively, using compound growth formula:
The general formula for compound growth is \(P_n = P_0 \times (1 + r)^n\), where \(P_n\) is the population after \(n\) periods, \(P_0\) is the initial population, \(r\) is the growth rate per period, and \(n\) is the number of periods.
Here, \(P_0 = 4,85,000\), \(r = 20\% = 0.20\), and \(n = 2\) decades.
\[ P_2 = 4,85,000 \times (1 + 0.20)^2 \]\[ P_2 = 4,85,000 \times (1.20)^2 \]\[ P_2 = 4,85,000 \times 1.44 \]\[ P_2 = 6,98,400 \]The population of the town in 1988 was 6,98,400.
Correct Option: C) 6,98,400