The question asks for a specific statistic from the Global Risks Perception Survey (GRPS) 2025: the percentage of respondents who anticipate an unsettled global outlook over the next two years. This requires knowledge of the findings of this particular report.
C) 52% — The Global Risks Report 2025, based on the GRPS, indicates that 52% of respondents anticipate an unsettled global outlook over the next two years.
The question asks to identify the mountain pass located specifically in the Pir Panjal range. This requires knowledge of the geographical locations of major mountain passes in the Indian subcontinent.
C) Banihal Pass — The Banihal Pass is a significant mountain pass across the Pir Panjal range in the Himalayas. It connects the Kashmir Valley with the outer Himalayas and the plains to the south. The Jawahar Tunnel, and later the Banihal Road Tunnel, are located beneath this pass, facilitating year-round connectivity.
The question asks about the year when the provision for free legal aid to the poor was added to the Directive Principles of State Policy (DPSP) in the Indian Constitution. This requires knowledge of constitutional amendments, specifically those related to DPSPs.
Therefore, the provision for free legal aid to the poor was added to the Directive Principles of State Policy in 1976.
Correct Option: D) 1976
The problem requires us to modify a given number by adding 2 to each of its digits, based on whether the digit is even or odd. After forming the new number, we need to find the sum of its first two digits and its last two digits.
Correct Option: B) 27
The problem asks us to find a pair of letter sets that follows the same pattern as the two given pairs: PNQ : QQO and UQR : VTP. We need to identify the transformation rule applied to the letters in the first set to get the letters in the second set for both given examples, and then apply that rule to the options.
A) OHM : PKK
Which positive number should replace both question marks in the following equation?
\( \frac{?}{56} \) = \( \frac{504}{?} \)
The problem presents an equation with a missing positive number represented by a question mark. To solve for this number, we can set up an algebraic equation by replacing the question mark with a variable and then solve for that variable. The equation involves fractions, so cross-multiplication will be a useful technique.
Let the positive number that replaces both question marks be \(x\). The given equation can then be written as:
\[ \frac{x}{56} = \frac{504}{x} \]To solve for \(x\), we can cross-multiply the terms:
\[ x \times x = 56 \times 504 \]Simplify both sides of the equation:
\[ x^2 = 56 \times 504 \]Calculate the product on the right side:
\[ 56 \times 504 = 28224 \]So, the equation becomes:
\[ x^2 = 28224 \]To find \(x\), we need to take the square root of both sides. Since the problem states that \(x\) is a positive number, we only consider the positive square root:
\[ x = \sqrt{28224} \]Calculate the square root of 28224:
We can estimate or use prime factorization. Let's try to estimate. We know that \(100^2 = 10000\) and \(200^2 = 40000\). So \(x\) should be between 100 and 200. The last digit of 28224 is 4, which means the last digit of its square root must be 2 or 8.
Let's try numbers ending in 2 or 8 in the range of 100-200. For example, \(160^2 = 25600\). Let's try \(168^2\):
\[ 168 \times 168 = 28224 \]So, \(x = 168\).
C) 168 is the correct positive number that satisfies the given equation.
We need to find the length of the longer side of a rectangle. We are given two conditions: the relationship between its sides and the relationship between its area and perimeter. We will set up equations based on these conditions and solve for the dimensions of the rectangle.
Let the length of the longer side of the rectangle be \(L\) cm and the length of the shorter side be \(W\) cm.
According to the first condition, "The shorter side of a rectangle is 21 cm less than the longer side."
\[ W = L - 21 \quad \text{(Equation 1)} \]The area of the rectangle is given by \(A = L \times W\).
The perimeter of the rectangle is given by \(P = 2(L + W)\).
According to the second condition, "The numerical value of its area is equal to 5 times the numerical value of its perimeter."
\[ A = 5P \]\[ L \times W = 5 \times 2(L + W) \]\[ LW = 10(L + W) \quad \text{(Equation 2)} \]Now, substitute Equation 1 into Equation 2:
\[ L(L - 21) = 10(L + (L - 21)) \]\[ L^2 - 21L = 10(2L - 21) \]\[ L^2 - 21L = 20L - 210 \]Rearrange the terms to form a quadratic equation:
\[ L^2 - 21L - 20L + 210 = 0 \]\[ L^2 - 41L + 210 = 0 \]Solve the quadratic equation for \(L\). We can use factorization. We need two numbers that multiply to 210 and add up to -41. These numbers are -35 and -6.
\[ L^2 - 35L - 6L + 210 = 0 \]\[ L(L - 35) - 6(L - 35) = 0 \]\[ (L - 35)(L - 6) = 0 \]This gives two possible values for \(L\):
\[ L = 35 \quad \text{or} \quad L = 6 \]If \(L = 6\) cm, then from Equation 1, \(W = 6 - 21 = -15\) cm. A side length cannot be negative, so \(L = 6\) is not a valid solution.
If \(L = 35\) cm, then from Equation 1, \(W = 35 - 21 = 14\) cm. Both \(L\) and \(W\) are positive, so this is a valid solution.
The length of the longer side is \(L = 35\) cm.
C) 35
The problem asks us to identify a pair of letter groups that follows the same pattern as the two given pairs: MWC : JYA and KUR : HWP. We need to find the positional difference between corresponding letters in each pair.
C) JVP : GXN
The question asks to identify the winner of the Best Actress (Critics) award at the South Indian International Movie Awards (SIIMA) 2024. This requires knowledge of recent awards and their recipients.
Correct Option: C) Aishwarya Rai Bachchan won the Best Actress (Critics) award at SIIMA 2024 for her role in Ponniyin Selvan 2.
An electric pump can fill a tank in 6 hours. Due to a leakage in the tank, it takes 7\( \frac{1}{2} \) hours to fill the tank. How much time will this leak take to empty the full tank if water does not get in or out of the tank through any other point during this period?
This problem involves the concept of work and time, specifically related to filling and emptying a tank. We can determine the rate at which the pump fills the tank and the combined rate at which the pump fills and the leak empties the tank. By finding the difference between these rates, we can calculate the rate at which the leak alone empties the tank. Once we have the leak's emptying rate, we can find the time it takes to empty the full tank.
Calculate the filling rate of the pump:
The electric pump can fill the tank in 6 hours. Therefore, in 1 hour, the pump fills \(\frac{1}{6}\) of the tank.
\[\text{Pump filling rate} = \frac{1}{6} \text{ tank/hour}\]Calculate the combined filling rate (pump - leak):
Due to a leakage, it takes \(7\frac{1}{2}\) hours to fill the tank. Convert this mixed fraction to an improper fraction:
\[7\frac{1}{2} = \frac{(7 \times 2) + 1}{2} = \frac{14 + 1}{2} = \frac{15}{2} \text{ hours}\]So, in 1 hour, the pump (with the leak) fills \[\frac{1}{\frac{15}{2}} = \frac{2}{15}\] of the tank.
\[\text{Combined filling rate} = \frac{2}{15} \text{ tank/hour}\]Calculate the emptying rate of the leak:
The combined rate is the pump's filling rate minus the leak's emptying rate. Let \(R_L\) be the rate at which the leak empties the tank.
\[\text{Pump filling rate} - \text{Leak emptying rate} = \text{Combined filling rate}\] \[\frac{1}{6} - R_L = \frac{2}{15}\]Now, solve for \(R_L\):
\[R_L = \frac{1}{6} - \frac{2}{15}\]To subtract these fractions, find a common denominator, which is 30.
\[R_L = \frac{1 \times 5}{6 \times 5} - \frac{2 \times 2}{15 \times 2}\] \[R_L = \frac{5}{30} - \frac{4}{30}\] \[R_L = \frac{5 - 4}{30}\] \[R_L = \frac{1}{30} \text{ tank/hour}\]This means the leak empties \(\frac{1}{30}\) of the tank in 1 hour.
Calculate the time taken by the leak to empty the full tank:
If the leak empties \(\frac{1}{30}\) of the tank in 1 hour, then it will empty the full tank (1 unit) in the reciprocal of this rate.
\[\text{Time taken by leak} = \frac{1}{\text{Leak emptying rate}} = \frac{1}{\frac{1}{30}} = 30 \text{ hours}\]C) 30 hours