The value of \( \frac{1}{4} \) + \( \frac{2}{5} \) ÷ [{2 \( \frac{1}{5} \) − 2} × 5] − \( \frac{2}{3} \) × \( \frac{3}{5} \) is:
\( \frac{3}{5} \)
\( \frac{2}{3} \)
\( \frac{1}{4} \)
\( \frac{5}{6} \)
To solve this problem, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS (Brackets/Parentheses, Orders/Exponents, Division and Multiplication (from left to right), Addition and Subtraction (from left to right)).
Let's break down the calculation:
The given expression is:
\[ \frac{1}{4} + \frac{2}{5} \div \left[\left(2 \frac{1}{5} - 2\right) \times 5\right] - \frac{2}{3} \times \frac{3}{5} \]Step 1: Simplify the mixed fraction inside the brackets.
\[ 2 \frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5} \]Substitute this back into the expression inside the brackets:
\[ \left[\left(\frac{11}{5} - 2\right) \times 5\right] \]Step 2: Perform the subtraction inside the parentheses.
\[ \frac{11}{5} - 2 = \frac{11}{5} - \frac{2 \times 5}{5} = \frac{11}{5} - \frac{10}{5} = \frac{1}{5} \]Now the expression inside the brackets becomes:
\[ \left[\frac{1}{5} \times 5\right] \]Step 3: Perform the multiplication inside the brackets.
\[ \frac{1}{5} \times 5 = 1 \]So, the original expression simplifies to:
\[ \frac{1}{4} + \frac{2}{5} \div 1 - \frac{2}{3} \times \frac{3}{5} \]Step 4: Perform the division.
\[ \frac{2}{5} \div 1 = \frac{2}{5} \]Step 5: Perform the multiplication.
\[ \frac{2}{3} \times \frac{3}{5} = \frac{2 \times 3}{3 \times 5} = \frac{6}{15} = \frac{2}{5} \]Now substitute these simplified terms back into the main expression:
\[ \frac{1}{4} + \frac{2}{5} - \frac{2}{5} \]Step 6: Perform the addition and subtraction from left to right.
\[ \frac{1}{4} + \frac{2}{5} - \frac{2}{5} = \frac{1}{4} + 0 = \frac{1}{4} \]Thus, the value of the given expression is \( \frac{1}{4} \).
C) \( \frac{1}{4} \) is the correct answer, as derived from the step-by-step calculation following the order of operations.
Find the sum of the roots of the quadratic equation 7x² + 28x + 1 = 0.
To find the sum of the roots of a quadratic equation, we can use Vieta's formulas. For a standard quadratic equation of the form \(ax^2 + bx + c = 0\), the sum of the roots is given by the formula \( -b/a \).
Correct Option: D) −4
The problem asks us to identify a set of numbers that follows the same logical relationship as the given sets: (4, 50, 6) and (2, 18, 4). We need to find a pattern involving the first, second, and third numbers in each set.
Let the numbers be \(A\), \(B\), and \(C\).
We need to find a relationship between \(A=4\), \(B=50\), and \(C=6\).
Let's try operations involving \(A\) and \(C\) to get \(B\).
Consider the sum of \(A\) and \(C\): \(4 + 6 = 10\). If we square this sum and multiply by a factor, or multiply by 5, we get \(10 \times 5 = 50\). This seems plausible.
Alternatively, consider the squares of \(A\) and \(C\):
Sum of squares: \(16 + 36 = 52\). This is close to 50, but not exact.
Let's try another approach: \((A+C) \times k = B\). Here, \((4+6) \times k = 50 \implies 10 \times k = 50 \implies k = 5\).
So, the pattern could be: \((A+C) \times 5 = B\).
Here, \(A=2\), \(B=18\), \(C=4\).
Using the proposed pattern: \((A+C) \times 5 = B\)
\((2+4) \times 5 = 6 \times 5 = 30\).
However, \(B\) is 18, not 30. So, the pattern \((A+C) \times 5 = B\) is incorrect.
Let's look for a different relationship. What if we consider the squares of \(A\) and \(C\), and their sum or difference, or some multiple?
For (4, 50, 6):
Try \((A^2 + C^2) / 2\)? \((16 + 36) / 2 = 52 / 2 = 26\). Not 50.
What if the middle number is related to the sum of the squares of the first and third numbers, multiplied by some factor or added to some constant?
Let's consider the sum of the numbers: \(4+50+6 = 60\).
Let's consider the product of the first and third numbers: \(4 \times 6 = 24\).
For (4, 50, 6):
Consider the relationship: \(B = (A^2 + C^2)/2 + X\)
\(50 = (4^2 + 6^2)/2 + X = (16+36)/2 + X = 52/2 + X = 26 + X \implies X = 24\).
So, the pattern could be \(B = (A^2 + C^2)/2 + 24\).
Let's check this with the second set (2, 18, 4):
\(A=2\), \(B=18\), \(C=4\).
\((2^2 + 4^2)/2 + 24 = (4+16)/2 + 24 = 20/2 + 24 = 10 + 24 = 34\).
This is not 18. So, this pattern is also incorrect.
Let's look at the numbers again: (4, 50, 6) and (2, 18, 4).
For (4, 50, 6):
If we take the square of the first number and multiply it by some factor, or the square of the third number.
Let's try: \(A^2 + C^2/2 = B\)? \(4^2 + 6^2/2 = 16 + 36/2 = 16 + 18 = 34\). Not 50.
What about: \(A \times C + X = B\)? \(4 \times 6 = 24\). \(24 + 26 = 50\).
So, \(B = A \times C + 26\).
Let's check this with the second set (2, 18, 4):
\(A \times C + 26 = 2 \times 4 + 26 = 8 + 26 = 34\). Not 18.
For (4, 50, 6):
Consider \(A^2 \times k_1 + C^2 \times k_2 = B\).
What if the middle number is related to the square of the first number and the third number?
Pattern: \(A^2 + C^2 + X = B\)? \(4^2 + 6^2 = 16 + 36 = 52\). \(52 - 2 = 50\). So \(B = A^2 + C^2 - 2\).
Let's check this with the second set (2, 18, 4):
\(A=2\), \(B=18\), \(C=4\).
\(A^2 + C^2 - 2 = 2^2 + 4^2 - 2 = 4 + 16 - 2 = 20 - 2 = 18\).
This pattern works for both sets!
The rule is: The middle number is equal to the sum of the squares of the first and third numbers, minus 2.
Mathematically: \(B = A^2 + C^2 - 2\).
A) (15, 162, 3)
Let's recheck the pattern. There might be a simpler one.
Let's look at the numbers again: (4, 50, 6) and (2, 18, 4).
For (4, 50, 6): \(4^2 = 16\), \(6^2 = 36\). \(16+36 = 52\). \(50\) is close to \(52\).
For (2, 18, 4): \(2^2 = 4\), \(4^2 = 16\). \(4+16 = 20\). \(18\) is close to \(20\).
It seems the pattern is \(B = A^2 + C^2 - 2\).
Let's re-evaluate option
The question asks about monetary measures to correct excess demand. Excess demand occurs when aggregate demand exceeds aggregate supply at full employment, leading to inflation. To correct this, the central bank (RBI in India) needs to reduce the money supply and credit availability in the economy. This is a contractionary monetary policy.
B) Neither A nor B
The question asks us to evaluate two statements about washing soda: its chemical formula and one of its uses. We need to determine if each statement is true or false and then select the option that correctly reflects this.
Correct Option: C) Both the statements are true.
The problem asks us to find the value of \(a^3 + b^3\) given the sum \(a+b\) and the product \(ab\). We can use the algebraic identity for the sum of cubes, which relates \(a^3 + b^3\) to \((a+b)\) and \(ab\).
The identity for the sum of cubes is:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]We know that \((a+b)^2 = a^2 + 2ab + b^2\). Rearranging this, we get:
\[ a^2 + b^2 = (a+b)^2 - 2ab \]Substitute the expression for \(a^2 + b^2\) into the identity for \(a^3 + b^3\):
\[ a^3 + b^3 = (a+b)((a+b)^2 - 2ab - ab) \]\[ a^3 + b^3 = (a+b)((a+b)^2 - 3ab) \]We are given \(a+b = 48\) and \(ab = 56\). Substitute these values into the derived formula:
\[ a^3 + b^3 = (48)((48)^2 - 3 \times 56) \]A) 1,02,528 is the correct value calculated.
To determine the number of \(\sigma\)-bonds, \(\pi\)-bonds, and total bonds in a benzene molecule, we need to recall its chemical structure. Benzene (\(\text{C}_6\text{H}_6\)) is a cyclic hydrocarbon with alternating single and double bonds, forming a hexagonal ring.
So, the number of \(\sigma\)-bonds is 12, \(\pi\)-bonds is 3, and the total bonds is 15.
Correct Option: A) 12, 3, 15
The question asks to identify the recipient of the Padma Shri award in 2022 in the field of Science and Engineering. This requires knowledge of the Padma Awards given in that specific year and category.
Correct Option: A) Subbanna Ayyappan — Dr. Subbanna Ayyappan, a distinguished agricultural scientist and former Director General of the Indian Council of Agricultural Research (ICAR), was awarded the Padma Shri in 2022 for his significant contributions to Science and Engineering, particularly in fisheries and aquaculture.
The question asks to identify a double displacement reaction from the given options. A double displacement reaction is a type of chemical reaction where two compounds react, and the positive ions (cations) and negative ions (anions) of the two reactants switch places, forming two new compounds. This often results in the formation of a precipitate, a gas, or water.
D) \( \text{Na}_2\text{SO}_4\text{(aq) + BaCl}_2\text{(aq)} \rightarrow \text{BaSO}_4\text{(s) + 2NaCl(aq)} \). This is a double displacement reaction because the cations (\(\text{Na}^+\) and \(\text{Ba}^{2+}\)) and anions (\(\text{SO}_4^{2-}\) and \(\text{Cl}^-\)) of the two reactant compounds exchange places to form two new compounds.
The question asks to identify the movement also known as the 'Salt Satyagraha'. This requires knowledge of major Indian independence movements and their alternative names or key events.
Correct Option: C) Civil Disobedience Movement
The Salt Satyagraha (Dandi March), initiated by Mahatma Gandhi in March 1930, was the most significant event that launched the Civil Disobedience Movement. It was a direct act of civil disobedience against the British salt monopoly and tax, symbolizing the broader struggle for Indian independence.