The problem requires us to modify a given number based on specific rules for its odd and even digits. After forming the new number, we need to identify the second digit from the left and the second digit from the right and then calculate their sum.
B) The sum of the digits that are second from the left (6) and second from the right (0) in the new number is \(6 + 0 = 6\).
To identify which species is getting oxidized, we need to understand the definitions of oxidation and reduction in terms of oxygen transfer or change in oxidation states. Oxidation is the gain of oxygen or loss of electrons (increase in oxidation state), while reduction is the loss of oxygen or gain of electrons (decrease in oxidation state).
Analyze the given reaction:
\[ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \]
Examine the change in CuO:
Examine the change in H₂:
Alternatively, consider oxidation states:
D) H₂ is getting oxidized because it gains oxygen to form H₂O, or its oxidation state increases from \(0\) to \(+1\).
The question asks to identify the type of government that is closest to the common people. This requires understanding the different tiers of governance and their proximity and accessibility to citizens in a democratic setup, particularly in the Indian context.
Correct Option: C) Local Government is the closest to the common people. It deals with local issues, provides basic services, and allows for direct participation of citizens in decision-making processes at the village or municipal level.
The problem presents a number series, and we need to find the next number. We will analyze the differences between consecutive terms to identify a pattern. If the first-level differences don't reveal a clear pattern, we will look at the differences of the differences (second-level differences).
A) 124 — The pattern in the differences between consecutive terms is an arithmetic progression of 7, 14, 21, 28. The next difference should be 35. Adding 35 to the last term (89) gives 124.
The problem involves simplifying an equation using the laws of exponents. Specifically, we will use the rule that states: if the bases are the same, then when multiplying, the powers are added (\(a^m \times a^n = a^{m+n}\)). Once the left side of the equation is simplified, we can equate the exponents since the bases are identical on both sides.
\( (-7)^{2k+1} \times (-7)^{5} = (-7)^{12} \)
When multiplying terms with the same base, we add their exponents. Here, the base is \(-7\).
\[ (-7)^{(2k+1) + 5} = (-7)^{12} \]Add the exponents in the power of \(-7\).
\[ (-7)^{2k+6} = (-7)^{12} \]Since the bases are equal on both sides of the equation, their exponents must also be equal.
\[ 2k+6 = 12 \]Subtract 6 from both sides:
\[ 2k = 12 - 6 \]\[ 2k = 6 \]Divide by 2:
\[ k = \frac{6}{2} \]\[ k = 3 \]Correct Option: C) 3
The value of \(k\) that satisfies the given equation is 3.
This problem involves the concepts of relative speed in boats and streams. We need to set up equations based on the given information about downstream and upstream travel times and distances, and then solve for the speed of the boat in still water.
Let the speed of the boat in still water be \(v_b\) km/h.
The speed of the stream is given as \(v_s = 4\) km/h.
When the boat travels downstream, its speed is the sum of its speed in still water and the speed of the stream:
\[ \text{Speed downstream} = v_b + v_s = (v_b + 4) \text{ km/h} \]When the boat travels upstream, its speed is the difference between its speed in still water and the speed of the stream:
\[ \text{Speed upstream} = v_b - v_s = (v_b - 4) \text{ km/h} \]The time taken to travel a certain distance is given by the formula: \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
The boat travels 40 km downstream. The time taken for this journey is:
\[ t_{\text{downstream}} = \frac{40}{v_b + 4} \text{ hours} \]The boat travels 12 km upstream. The time taken for this journey is:
\[ t_{\text{upstream}} = \frac{12}{v_b - 4} \text{ hours} \]The total time taken for both journeys is 4 hours. So, we can write the equation:
\[ t_{\text{downstream}} + t_{\text{upstream}} = 4 \]\[ \frac{40}{v_b + 4} + \frac{12}{v_b - 4} = 4 \]To solve this equation, find a common denominator:
\[ \frac{40(v_b - 4) + 12(v_b + 4)}{(v_b + 4)(v_b - 4)} = 4 \]\[ \frac{40v_b - 160 + 12v_b + 48}{v_b^2 - 16} = 4 \]\[ \frac{52v_b - 112}{v_b^2 - 16} = 4 \]Multiply both sides by \(v_b^2 - 16\):
\[ 52v_b - 112 = 4(v_b^2 - 16) \]\[ 52v_b - 112 = 4v_b^2 - 64 \]Rearrange the terms to form a quadratic equation:
\[ 4v_b^2 - 52v_b - 64 + 112 = 0 \]\[ 4v_b^2 - 52v_b + 48 = 0 \]Divide the entire equation by 4 to simplify:
\[ v_b^2 - 13v_b + 12 = 0 \]Factor the quadratic equation. We need two numbers that multiply to 12 and add up to -13. These numbers are -1 and -12.
\[ (v_b - 1)(v_b - 12) = 0 \]This gives two possible values for \(v_b\):
\[ v_b = 1 \text{ or } v_b = 12 \]Since the speed of the boat in still water must be greater than the speed of the stream (otherwise, it cannot travel upstream), \(v_b > 4\). Therefore, \(v_b = 1\) km/h is not a valid solution.
Thus, the speed of the boat in still water is 12 km/h.
Correct Option: B) 12
The value of \(\frac{1\frac{2}{3}+2\frac{1}{2}\times\frac{2}{3}}{5-4\frac{1}{2}\div\frac{3}{2}}\) is:
3\( \frac{1}{3} \)
1\( \frac{2}{3} \)
2\( \frac{1}{3} \)
4\( \frac{2}{3} \)
To solve this problem, we need to evaluate the given expression by following the order of operations (BODMAS/PEMDAS). We will convert mixed fractions to improper fractions, perform multiplication and division before addition and subtraction, and handle the numerator and denominator separately.
Numerator:
\[ 1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} \]\[ 2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2} \]Denominator:
\[ 4\frac{1}{2} = \frac{4 \times 2 + 1}{2} = \frac{9}{2} \]The numerator is \[1\frac{2}{3}+2\frac{1}{2}\times\frac{2}{3}\]. Substitute the improper fractions:
\[ \frac{5}{3} + \frac{5}{2} \times \frac{2}{3} \]First, perform the multiplication:
\[ \frac{5}{2} \times \frac{2}{3} = \frac{5 \times 2}{2 \times 3} = \frac{10}{6} = \frac{5}{3} \]Now, perform the addition:
\[ \frac{5}{3} + \frac{5}{3} = \frac{5+5}{3} = \frac{10}{3} \]The denominator is \[5-4\frac{1}{2}\div\frac{3}{2}\]. Substitute the improper fraction:
\[ 5 - \frac{9}{2} \div \frac{3}{2} \]First, perform the division. Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[ \frac{9}{2} \div \frac{3}{2} = \frac{9}{2} \times \frac{2}{3} = \frac{9 \times 2}{2 \times 3} = \frac{18}{6} = 3 \]Now, perform the subtraction:
\[ 5 - 3 = 2 \]The expression becomes:
\[ \frac{\frac{10}{3}}{2} = \frac{10}{3} \times \frac{1}{2} = \frac{10}{6} = \frac{5}{3} \]Convert the improper fraction back to a mixed fraction:
\[ \frac{5}{3} = 1\frac{2}{3} \]Correct Option: B) \(1\frac{2}{3}\)
The question asks for the name of a specific helpline launched by the National Commission for Protection of Child Rights (NCPCR) in September 2020 to provide psycho-social support to children affected by the COVID-19 pandemic. This requires knowledge of current affairs and government initiatives related to child welfare during the pandemic.
D) Samvedna is the correct answer. The "Samvedna" helpline was indeed launched by the NCPCR in September 2020 to provide psycho-social first aid and counselling to children during the COVID-19 pandemic.
To find out who is giving the highest total discount, we need to calculate the equivalent single discount for each person's successive discounts. The formula for equivalent single discount for two successive discounts of \(d_1\%\) and \(d_2\%\) is given by: \((d_1 + d_2 - \frac{d_1 \times d_2}{100})\%\).
Ahmed's Discount:
Successive discounts are 6% and 5%.
\[ \text{Equivalent Discount} = \left(6 + 5 - \frac{6 \times 5}{100}\right)\% \] \[ = \left(11 - \frac{30}{100}\right)\% \] \[ = (11 - 0.3)\% \] \[ = 10.7\% \]Suman's Discount:
Successive discounts are 2% and 5%.
\[ \text{Equivalent Discount} = \left(2 + 5 - \frac{2 \times 5}{100}\right)\% \] \[ = \left(7 - \frac{10}{100}\right)\% \] \[ = (7 - 0.1)\% \] \[ = 6.9\% \]Raj's Discount:
Successive discounts are 7% and 2%.
\[ \text{Equivalent Discount} = \left(7 + 2 - \frac{7 \times 2}{100}\right)\% \] \[ = \left(9 - \frac{14}{100}\right)\% \] \[ = (9 - 0.14)\% \] \[ = 8.86\% \]Madhu's Discount:
Successive discounts are 5% and 4%.
\[ \text{Equivalent Discount} = \left(5 + 4 - \frac{5 \times 4}{100}\right)\% \] \[ = \left(9 - \frac{20}{100}\right)\% \] \[ = (9 - 0.2)\% \] \[ = 8.8\% \]Comparison of Discounts:
Comparing these values, Ahmed offers the highest total discount of 10.7%.
C) Ahmed is giving the highest total discount of 10.7%.
The question asks about the nature of the dissolution of the upper house of the Indian Parliament. This refers to the Rajya Sabha. Understanding the fundamental characteristics of the Rajya Sabha as outlined in the Indian Constitution is key to answering this question.
Correct Option: A) It is never dissolved