The question asks for the origin village of the Chipko Andolan. This requires knowledge of the historical details of this environmental movement in India.
Correct Option: C) Reni in Garhwal is the correct answer. The Chipko Andolan originated in the Reni village of the Garhwal region (then Uttar Pradesh, now Uttarakhand) in 1973.
Study the following table carefully and answer the questions based on it. Number of visitors who visit different tourist places in different months in the same year are given in the following table.
| Months | Visitors | Total | ||||
| Tourist Place A | Tourist Place B | Tourist Place C | Tourist Place D | Tourist Place E | ||
| September | 200 | 70 | 30 | 290 | 10 | 600 |
| October | 120 | 130 | 70 | 150 | 290 | 760 |
| November | 45 | 35 | 25 | 125 | 160 | 390 |
| December | 160 | 110 | 40 | 115 | 130 | 555 |
| January | 80 | 90 | 70 | 100 | 140 | 480 |
| February | 130 | 150 | 30 | 40 | 390 | 740 |
What is the difference in the number of visitors to the place A between the months of December and January?
Correct Option: B) 80
If x = \( \frac{-1}{2} \) and y = 2, find the value of
Correct Option: A) 94
The question asks to identify the Indian Premier League (IPL) franchise for which South African cricketer Dewald Brevis made his debut in the fifteenth edition (IPL 2022). This requires knowledge of player signings and debuts in recent IPL seasons.
Correct Option: B) Mumbai Indians. Dewald Brevis was purchased by the Mumbai Indians in the IPL 2022 mega auction for INR 3 crore and made his debut for the franchise in the same season.
The given series is a number series where we need to find the pattern of differences or operations between consecutive terms to determine the missing number. We will analyze the differences between terms to identify a repeating or sequential pattern.
Therefore, the number that replaces the question mark is 204.
Correct Option: D) 204
This problem involves analyzing inequalities to determine the truthfulness of given conclusions. We need to combine the provided statements and derive relationships between the variables to check each conclusion.
Combine the given statements:
Statements are: \(A = B \geq C = D\) and \(E > F \geq G = D\).
From the first statement, we have \(A = B \geq C = D\).
From the second statement, we have \(E > F \geq G = D\).
Since both statements share \(D\), we can combine them:
\(A = B \geq C = D = G \leq F < E\)
This combined inequality gives us a clear relationship between all variables.
Evaluate Conclusion I: \(E > C\)
From the combined inequality, we have \(C = D\) and \(D = G \leq F < E\). Therefore, \(C = D < E\). This means \(C < E\), or \(E > C\).
So, Conclusion I is true.
Evaluate Conclusion II: \(A \geq G\)
From the combined inequality, we have \(A = B \geq C = D = G\). This directly implies \(A \geq G\).
So, Conclusion II is true.
Evaluate Conclusion III: \(E \geq B\)
From the combined inequality, we have \(B \geq C = D = G \leq F < E\).
We have \(B \geq C\) and \(C = D\). We also have \(D = G \leq F < E\).
So, \(B \geq C = D\). And \(D < E\).
This gives us \(B \geq D\) and \(D < E\). We cannot establish a definite relationship between \(B\) and \(E\). For example, if \(B=5\), \(C=4\), \(D=4\), \(G=4\), \(F=6\), \(E=7\), then \(B < E\). If \(B=7\), \(C=6\), \(D=6\), \(G=6\), \(F=8\), \(E=9\), then \(B < E\). If \(B=5\), \(C=5\), \(D=5\), \(G=5\), \(F=6\), \(E=7\), then \(B < E\). If \(B=7\), \(C=6\), \(D=6\), \(G=6\), \(F=7\), \(E=8\), then \(B < E\).
Consider the path from B to E: \(B \geq C = D < E\). This only tells us that \(B\) can be greater than, equal to, or less than \(E\). For example, if \(B=5, C=4, D=4, E=6\), then \(B < E\). If \(B=7, C=6, D=6, E=5\), this is not possible as \(D < E\). If \(B=7, C=6, D=6, E=8\), then \(B < E\). If \(B=5, C=5, D=5, E=6\), then \(B < E\). If \(B=6, C=6, D=6, E=7\), then \(B < E\).
There is no direct relationship \(B \geq E\) or \(E \geq B\) that can be concluded. The relationship is \(B \geq D\) and \(D < E\). This does not guarantee \(E \geq B\).
So, Conclusion III is false.
Correct Option: B) Only conclusions I and II are true
Conclusions I (\(E > C\)) and II (\(A \geq G\)) are true based on the combined statements. Conclusion III (\(E \geq B\)) is not necessarily true.
Express 123.4\(\overline{7}\) in \( \frac{p}{q} \) form, where p and q are integers and q ≠ 0.
\(\frac{12347}{99}\)
\(\frac{11113}{90}\)
\(\frac{12347}{90}\)
\(\frac{11113}{9}\)
To express a repeating decimal in the form \(\frac{p}{q}\), we use an algebraic method. We set the given number equal to a variable, multiply it by powers of 10 to align the repeating part, and then subtract the equations to eliminate the repeating decimal, solving for the variable.
Let the given number be \(x\):
\[ x = 123.4\overline{7} \]Write out the repeating decimal:
\[ x = 123.4777... \quad \text{(Equation 1)} \]Multiply Equation 1 by a power of 10 such that the non-repeating part after the decimal point is moved to the left of the decimal point. Here, we need to move '4' to the left, so we multiply by \(10^1 = 10\):
\[ 10x = 1234.777... \quad \text{(Equation 2)} \]Multiply Equation 1 by a power of 10 such that one full repeating block is moved to the left of the decimal point. The repeating block is '7'. To move one '7' to the left of the decimal point, we need to move the decimal two places to the right (past '4' and past '7'). So, we multiply by \(10^2 = 100\):
\[ 100x = 12347.777... \quad \text{(Equation 3)} \]Subtract Equation 2 from Equation 3 to eliminate the repeating part:
\[ 100x - 10x = 12347.777... - 1234.777... \]\[ 90x = 11113 \]Solve for \(x\):
\[ x = \frac{11113}{90} \]B) \(\frac{11113}{90}\) is the correct representation of \(123.4\overline{7}\) in \(\frac{p}{q}\) form.
A) \(\frac{12347}{99}\) — This would be the result if the number was \(123.\overline{47}\) (where both 4 and 7 repeat) or if the calculation was done incorrectly by not accounting for the non-repeating digit '4'.
C) \(\frac{12347}{90}\) — This fraction is obtained if the number was \(1234.7\overline{7}\) and only the '7' repeated, or if the subtraction step was incorrect, specifically if \(10x\) was subtracted from \(100x\) as \(12347 - 1234.7\) incorrectly, or if the original number was treated as \(1234.7\overline{7}\) and the non-repeating part was not handled correctly.
D) \(\frac{11113}{9}\) — This would imply that the denominator was 9, which happens when there is only one repeating digit and no non-repeating digits after the decimal, and the multiplication was by 10 and 1. For example, \(123.\overline{7}\) would lead to a denominator of 9.
If the area of an equilateral triangle is 25\(\sqrt{3}\) cm2, then the length (in cm) of each side of this equilateral triangle is:
To find the length of each side of an equilateral triangle when its area is given, we use the formula for the area of an equilateral triangle. We will set up an equation with the given area and solve for the side length.
The area of an equilateral triangle with side length \(a\) is given by the formula:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]We are given that the area is \(25\sqrt{3}\) cm\(^2\). So,
\[ 25\sqrt{3} = \frac{\sqrt{3}}{4} a^2 \]Divide both sides by \(\sqrt{3}\):
\[ 25 = \frac{1}{4} a^2 \]Multiply both sides by 4:
\[ 25 \times 4 = a^2 \]\[ 100 = a^2 \]Take the square root of both sides:
\[ a = \sqrt{100} \]\[ a = 10 \]The length of each side of the equilateral triangle is 10 cm.
A) 10 — As calculated, the side length \(a\) is 10 cm.
What will be the area (in cm²) of a semicircle whose perimeter is 72 cm? (Take π = \( \frac{22}{7} \))
To find the area of a semicircle, we first need to determine its radius. The perimeter of a semicircle consists of the curved arc length and the diameter. We can set up an equation using the given perimeter and solve for the radius. Once the radius is known, we can calculate the area of the semicircle.
Understand the formula for the perimeter of a semicircle:
The perimeter of a semicircle is the sum of the length of its curved arc and its diameter. If \(r\) is the radius, the length of the arc is half the circumference of a full circle, which is \(\frac{1}{2} (2\pi r) = \pi r\). The diameter is \(2r\).
So, the perimeter \(P\) of a semicircle is given by:
\[ P = \pi r + 2r \]Substitute the given perimeter and solve for the radius:
We are given that the perimeter \(P = 72 \text{ cm}\) and \(\pi = \frac{22}{7}\).
\[ 72 = \pi r + 2r \]\[ 72 = r \left( \pi + 2 \right) \]\[ 72 = r \left( \frac{22}{7} + 2 \right) \]\[ 72 = r \left( \frac{22}{7} + \frac{14}{7} \right) \]\[ 72 = r \left( \frac{22 + 14}{7} \right) \]\[ 72 = r \left( \frac{36}{7} \right) \]\[ r = \frac{72 \times 7}{36} \]\[ r = 2 \times 7 \]\[ r = 14 \text{ cm} \]Calculate the area of the semicircle:
The area of a full circle is \(\pi r^2\). The area of a semicircle is half the area of a full circle.
Area of semicircle \(A = \frac{1}{2} \pi r^2\)
Substitute the value of \(r = 14 \text{ cm}\) and \(\pi = \frac{22}{7}\):
\[ A = \frac{1}{2} \times \frac{22}{7} \times (14)^2 \]\[ A = \frac{1}{2} \times \frac{22}{7} \times 196 \]\[ A = \frac{1}{2} \times 22 \times \frac{196}{7} \]\[ A = \frac{1}{2} \times 22 \times 28 \]\[ A = 11 \times 28 \]\[ A = 308 \text{ cm}^2 \]A) 308 cm² is the calculated area of the semicircle.
This problem involves calculating the time it takes for a faulty clock, which loses a certain amount of time every hour, to fall behind by a specific total amount. We need to determine the rate of loss per day and then use this rate to find the number of days required to accumulate the total desired loss.
The clock loses 45 minutes every hour.
There are 24 hours in a day. So, the total time lost per day will be:
\[ \text{Time lost per day} = \text{Loss per hour} \times \text{Number of hours in a day} \]\[ \text{Time lost per day} = 45 \text{ minutes/hour} \times 24 \text{ hours/day} \]\[ \text{Time lost per day} = 1080 \text{ minutes} \]Since there are 60 minutes in an hour:
\[ \text{Time lost per day (in hours)} = \frac{1080 \text{ minutes}}{60 \text{ minutes/hour}} \]\[ \text{Time lost per day (in hours)} = 18 \text{ hours} \]The question asks in how many days the clock will be 18 hours slow. We just calculated that the clock loses 18 hours in 1 day.
Therefore, the clock will be 18 hours slow in 1 day.
Correct Option: C) 1