The problem asks us to find a relationship between two sets of letters based on their positions in the English alphabet and then apply the same logic to a new set of letters. We need to identify the pattern of letter shifts (forward or backward) for each letter in the given pairs.
A) PMRO is the correct option as derived from applying the +4 shift to each letter of LINK.
The problem asks us to find the values of X and Y such that the mathematical operation relating the first number to the second number in the pair X : 243 is the same as the operation relating 17 to Y. We need to identify this common pattern.
If \(X = 25\), and we know \(X-k=3\), then \(25-k=3 \implies k=22\).
Now, apply this \(k=22\) to the second pair: \(Y = (17-k)^5 = (17-22)^5 = (-5)^5 = -3125\). This does not match \(Y=171\).
Let's look at the numbers again. X : 243. We know \(243 = 3^5\).
17 : Y.
Consider the possibility that the first number is related to the base and the second number is related to the power.
If \(X=25\), then \(25-22 = 3\). And if \(25\) is related to the power 5, maybe \(25/5 = 5\). This is not consistent.
If the pattern is \(N \rightarrow (N-8)^3\)?
For X : 243, if \(X=17\), then \((17-8)^3 = 9^3 = 729\). This is not 243.
For X : 243, if \(X=25\).
We know \(243 = 3^5\). How can 25 relate to 3 and 5?
What if the operation is \((\sqrt{N})^P\)?
Let's consider the pattern: \(N \rightarrow (N-k)\) and then some power.
Let's try to find a pattern that works for X=25 and 243.
If \(X=25\), then \(25-2 = 23\). Not useful.
What if the pattern is \((N-k)^m\)?
For X : 243, if \(X=25\), then \(25-k\) raised to some power gives 243.
We know \(3^5 = 243\). So, \(25-k = 3 \implies k = 22\). And the power is 5.
So the pattern is \((N-22)^5\).
Let's check this pattern for 17 : Y.
\(Y = (17-22)^5 = (-5)^5 = -3125\). This does not match \(Y=171\).
The question asks for the venue of the FIH Men's Hockey World Cup 2023. This is a factual question related to current affairs and major sporting events. To answer correctly, one needs to recall the host city/state for this specific event.
The FIH Men's Hockey World Cup 2023 was jointly hosted by two cities in Odisha, India: Bhubaneswar and Rourkela.
Correct Option: C) Rourkela, Odisha
Rourkela, along with Bhubaneswar, was one of the host cities for the FIH Men's Hockey World Cup 2023. The newly built Birsa Munda International Hockey Stadium in Rourkela was one of the primary venues for the tournament.
This problem involves substituting given symbols with mathematical operators and then evaluating the resulting expression using the order of operations (BODMAS/PEMDAS).
The original equation is: \(15 \text{ P } 6 \text{ R } 11 \text{ P } 4 \text{ S } 38 \text{ R } 41 \text{ S } 2 \text{ P } 5 = ?\)
Substituting the operators, we get:
\[ 15 \times 6 - 11 \times 4 + 38 - 41 + 2 \times 5 = ? \]First, perform all multiplications from left to right.
The equation now becomes:
\[ 90 - 44 + 38 - 41 + 10 = ? \]The value of the expression is \(53\).
D) 53 is the correct answer, as derived from the step-by-step calculation.
The question asks to identify which of the given cells exhibits changing shapes. We need to recall the characteristics and typical morphology of each cell type listed in the options.
C) Amoeba — Amoeba is a protozoan that moves and feeds by extending and retracting pseudopods, which results in continuous changes in its body shape. This characteristic is unique among the given options.
The problem asks us to identify the letter-cluster pair that does not follow the same pattern as the other three, based on English alphabetical order. We need to find the relationship between the letters within each pair and between the two clusters in the pair.
Correct Option: C) CH – MR
The question asks to identify the recipient of the Vigyan Ratna Award in the Rashtriya Vigyan Puraskar 2024 from the given options. This requires knowledge of the awardees announced for the Rashtriya Vigyan Puraskar.
D) Prof. Govindarajan Padmanabhan — Prof. Govindarajan Padmanabhan is a distinguished biochemist and molecular biologist known for his significant contributions to the field of life sciences, particularly in malaria research. He was indeed one of the recipients of the Vigyan Ratna Award in the Rashtriya Vigyan Puraskar 2024 for his lifetime achievements.
A shopkeeper, on selling a pan for ₹10, loses \( \frac{1}{11} \) of what it cost him. The cost price of a pan is:
The problem states that a shopkeeper loses a fraction of the cost price when selling an item. We are given the selling price and the fraction of loss relative to the cost price. We need to find the cost price. We can set up an equation relating the selling price, cost price, and the loss percentage.
B) ₹11 is the correct answer. This is derived by setting up the equation where the selling price plus the loss (which is a fraction of the cost price) equals the cost price.
The question asks about the products formed when an acid reacts with a base. This is a fundamental concept in chemistry known as a neutralization reaction. We need to recall the general equation for such reactions.
Correct Option: D) Both salt and water
To find the cost of the canvas needed to cover the cuboidal box, we first need to calculate the total surface area of the box. Since the box is closed, we need to consider all six faces. We will convert all dimensions to a consistent unit (meters) and then use the formula for the total surface area of a cuboid. Finally, we will multiply the total surface area by the cost per square meter to find the total cost.
The length and breadth are already in meters. Convert the height from centimeters to meters.
\[ H = 90 \text{ cm} = \frac{90}{100} \text{ m} = 0.9 \text{ m} \]The formula for the total surface area of a cuboid is:
\[ \text{TSA} = 2(LB + BH + HL) \]\[ \text{TSA} = 2((2.5 \text{ m} \times 2 \text{ m}) + (2 \text{ m} \times 0.9 \text{ m}) + (0.9 \text{ m} \times 2.5 \text{ m})) \]\[ \text{TSA} = 2(5 \text{ m}^2 + 1.8 \text{ m}^2 + 2.25 \text{ m}^2) \]\[ \text{TSA} = 2(9.05 \text{ m}^2) \]\[ \text{TSA} = 18.1 \text{ m}^2 \]The cost of the canvas is ₹70 per \(\text{m}^2\).
\[ \text{Total Cost} = \text{TSA} \times \text{Cost per m}^2 \]\[ \text{Total Cost} = 18.1 \text{ m}^2 \times \text{₹}70/\text{m}^2 \]\[ \text{Total Cost} = \text{₹}1267 \]B) ₹1,267 is the correct answer, as calculated by finding the total surface area of the cuboid and multiplying it by the given cost per square meter.