The question asks to identify the Indian dance form featured in Michael Jackson's 1991 music video 'Black or White'. This requires specific knowledge of the cultural elements incorporated into the video.
D) Odissi — The music video for Michael Jackson's 1991 hit single 'Black or White' famously featured a segment showcasing the Indian classical dance form Odissi. A young girl is seen performing Odissi in a traditional costume, representing Indian culture in the video's global montage.
This problem involves substituting new meanings for mathematical operators and then evaluating the given expressions to find the correct one. We need to apply the BODMAS/PEMDAS rule after substitution.
First, let's list the given substitutions:
Now, we will evaluate each option by substituting the new operators and then performing the calculations.
Substitute the operators:
\[ 12 \times 4 + 7 \div 2 - 3 \]\[ 48 + 3.5 - 3 \]\[ 51.5 - 3 \]\[ 48.5 \]Since \(48.5 \neq 4\), Option A is incorrect.Substitute the operators:
\[ 8 \times 4 - 2 + 6 \div 2 \]\[ 32 - 2 + 3 \]\[ 30 + 3 \]\[ 33 \]Since \(33 \neq 4\), Option B is incorrect.This is the same expression as Option B. As calculated above, the result is 33.
\[ 8 \times 4 - 2 + 6 \div 2 \]\[ 32 - 2 + 3 \]\[ 30 + 3 \]\[ 33 \]Since \(33 \neq 5\), Option C is incorrect.Substitute the operators:
\[ 12 \div 4 + 7 \times 2 - 3 \]\[ 3 + 14 - 3 \]\[ 17 - 3 \]\[ 14 \]Since \(14 = 14\), Option D is correct.D) 12 × 4 − 7 ÷ 2 + 3 = 14 is the correct equation after substituting the given operator meanings and evaluating the expression.
To find the degree measure of an angle subtended at the center of a circle by an arc, we use the formula relating arc length, radius, and the angle in radians. Once the angle is found in radians, it can be converted to degrees using the conversion factor \(\pi \text{ radians} = 180^\circ\).
Identify the given values:
Recall the formula for arc length:
The length of an arc (\(l\)) is given by the formula:
where \(r\) is the radius of the circle and \(\theta\) is the angle subtended at the center in radians.
Calculate the angle in radians:
Rearrange the formula to solve for \(\theta\):
Substitute the given values:
\[ \theta = \frac{22 \text{ cm}}{28 \text{ cm}} = \frac{11}{14} \text{ radians} \]Convert the angle from radians to degrees:
To convert an angle from radians to degrees, multiply by \(\frac{180^\circ}{\pi}\). We know that \(\pi \approx \frac{22}{7}\).
Substitute the value of \(\theta\) and \(\pi\):
\[ \text{Angle in degrees} = \frac{11}{14} \times \frac{180^\circ}{\frac{22}{7}} \]Simplify the expression:
\[ \text{Angle in degrees} = \frac{11}{14} \times \frac{180^\circ}{1} \times \frac{7}{22} \]Cancel out common factors:
\[ \text{Angle in degrees} = \frac{11}{14} \times \frac{180^\circ}{1} \times \frac{7}{2 \times 11} \]Cancel 11 from numerator and denominator:
\[ \text{Angle in degrees} = \frac{1}{14} \times 180^\circ \times \frac{7}{2} \]Cancel 7 from 14:
\[ \text{Angle in degrees} = \frac{1}{2} \times 180^\circ \times \frac{1}{2} \]Perform the multiplication:
\[ \text{Angle in degrees} = \frac{180^\circ}{4} = 45^\circ \]Correct Option: B) 45°
The question asks for the location where India's first solar-powered train was launched in July 2017. This is a factual question related to current affairs and technological advancements in Indian railways.
Correct Option: C) Delhi
In July 2017, India's first solar-powered train was flagged off from Safdarjung Railway Station in Delhi. This was a significant step towards green energy initiatives in Indian Railways.
If \( a^{2} + b^{2} + c^{2} + d^{2} = 1 \), what will be the maximum value of the product abcd?
This problem can be solved using the AM-GM (Arithmetic Mean - Geometric Mean) inequality. The AM-GM inequality states that for a set of non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. Equality holds when all the numbers are equal.
We are given the condition \( a^{2} + b^{2} + c^{2} + d^{2} = 1 \). We need to find the maximum value of the product \( abcd \).
Consider the four non-negative terms \( a^2, b^2, c^2, d^2 \). Applying the AM-GM inequality to these terms:
\[ \frac{a^2 + b^2 + c^2 + d^2}{4} \ge \sqrt[4]{a^2 b^2 c^2 d^2} \]Substitute the given condition \( a^{2} + b^{2} + c^{2} + d^{2} = 1 \) into the inequality:
\[ \frac{1}{4} \ge \sqrt[4]{(abcd)^2} \]Simplify the right side:
\[ \frac{1}{4} \ge (abcd)^{2/4} \]\[ \frac{1}{4} \ge (abcd)^{1/2} \]\[ \frac{1}{4} \ge \sqrt{abcd} \]To find the maximum value of \( abcd \), we need to square both sides of the inequality:
\[ \left(\frac{1}{4}\right)^2 \ge (\sqrt{abcd})^2 \]\[ \frac{1}{16} \ge abcd \]This inequality shows that \( abcd \) is always less than or equal to \( \frac{1}{16} \). Therefore, the maximum value of \( abcd \) is \( \frac{1}{16} \).
The equality in AM-GM holds when \( a^2 = b^2 = c^2 = d^2 \). Given \( a^2 + b^2 + c^2 + d^2 = 1 \), this means \( 4a^2 = 1 \), so \( a^2 = \frac{1}{4} \). Thus, \( a = b = c = d = \pm \frac{1}{2} \). For the product \( abcd \) to be positive and maximized, we can choose \( a=b=c=d=\frac{1}{2} \). In this case, \[abcd = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\].
A) \( \frac{1}{16} \) is the maximum value of the product \( abcd \), as derived using the AM-GM inequality.
This is a blood relation question. To solve such problems, we break down the statement into smaller, more manageable parts, starting from the end of the sentence and working backward to establish the relationship with the main person (Roma in this case).
Therefore, the boy is the son of Roma's daughter.
D) Daughter's son — As derived in the step-by-step analysis, the boy is the son of Roma's daughter.
The question asks to identify the endemic species that is NOT found in the Western Ghats. This requires knowledge of the distribution of the listed species, particularly their endemic regions within India.
Correct Option: B) Hispid Hare is not found in the Western Ghats. It is primarily found in the grasslands of the Himalayan foothills.
We are given the initial ratio of two quantities and the new ratio after both quantities are decreased by a certain value. We can represent the quantities using a variable and set up an equation based on the given ratios to find the value of the variable, and subsequently, the smaller quantity.
Let the two quantities be \(3x\) and \(5x\), since their initial ratio is \(3:5\).
When each quantity is decreased by 9, the new quantities become \((3x - 9)\) and \((5x - 9)\).
The new ratio is given as \(5:9\). So, we can set up the equation:
\[ \frac{3x - 9}{5x - 9} = \frac{5}{9} \]Now, we cross-multiply to solve for \(x\):
\[ 9(3x - 9) = 5(5x - 9) \]\[ 27x - 81 = 25x - 45 \]Rearrange the terms to isolate \(x\):
\[ 27x - 25x = 81 - 45 \]\[ 2x = 36 \]\[ x = \frac{36}{2} \]\[ x = 18 \]The smaller quantity was initially \(3x\). Substitute the value of \(x\):
\[ \text{Smaller quantity} = 3 \times 18 = 54 \]The larger quantity was initially \(5x\):
\[ \text{Larger quantity} = 5 \times 18 = 90 \]Let's verify the new ratio: \((54 - 9) : (90 - 9) = 45 : 81\). Dividing both by 9, we get \(5:9\), which matches the given information.
Correct Option: C) 54
To solve this series problem, we need to identify the pattern or rule governing the sequence of numbers. We will analyze the relationship between consecutive terms or look for a mathematical formula that generates these numbers.
Alternative Approach (Recognizing a pattern of cubes and numbers):
Observe that the numbers can be expressed in the form \(n^3 - n\):
Following this pattern, the next term for \(n=7\) would be:
Both methods yield the same result.
A) 336 is the correct number that replaces the question mark, as derived from the identified pattern.
The question asks to identify the National Highway that connected Delhi and Mumbai according to the old numbering system. This requires knowledge of the historical National Highway network in India.
National Highway 8 (NH-8) was a major highway in India, connecting the national capital Delhi to the financial capital Mumbai. It passed through important cities like Jaipur, Ajmer, Udaipur, Ahmedabad, Vadodara, and Surat. This highway was part of the Golden Quadrilateral project. With the renumbering of National Highways in 2010, the old NH-8 has largely been renumbered as NH-48.
Correct Option: D) NH8 — Old NH-8 was the primary National Highway connecting Delhi and Mumbai. It has since been renumbered as NH-48.