The problem asks us to identify symbols in a given series that are immediately preceded by a number and immediately followed by a number. We need to scan the series from left to right and apply this condition to each symbol.
A) 1 is the correct answer because only one symbol (£) in the series fits the criteria of being immediately preceded by a number and immediately followed by a number (6 £ 1).
If the mode of the following data is 165, then what is the value of x?
| Class | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |
| Frequency | 23 | 25 | 37 | x | 28 |
The problem provides a frequency distribution table and states that the mode of the data is 165. We need to find the value of an unknown frequency, 'x'. The mode of grouped data is found using a specific formula. Since the mode is given as 165, the modal class must be the class interval that contains 165. We will identify the modal class and then apply the formula for the mode of grouped data.
Identify the Modal Class: The mode is given as 165. Looking at the class intervals, the value 165 falls into the class 160-165 (upper limit exclusive) or 165-170 (lower limit inclusive). In most grouped data problems, the modal class is the one with the highest frequency. However, if the mode is explicitly given, the modal class is the one that contains the mode. Here, the mode is 165. The class 160-165 has an upper limit of 165, and the class 165-170 has a lower limit of 165. Given that the mode is exactly 165, the modal class is 160-165. This also implies that the frequency of the modal class (37) must be the highest, or at least one of the highest, to be the modal class. If the mode is 165, it means the class 160-165 has the highest frequency. Let's re-evaluate. The mode is 165. This means the modal class is 160-165. The frequency of this class is 37. This implies that 37 must be the highest frequency. However, if 'x' were greater than 37, then 165-170 would be the modal class. This is a common point of confusion. Let's assume the question implies that the given mode 165 is the calculated mode using the formula, and thus the modal class is the one containing 165 and having the highest frequency. If the mode is 165, the modal class is 160-165. Its frequency is 37. This means that 37 must be the highest frequency, so 'x' must be less than or equal to 37. Let's proceed with the formula.
The formula for the mode of grouped data is:
\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]Determine the Modal Class and its parameters:
Given that the mode is 165, the modal class is 160-165.
Substitute the values into the mode formula:
\[ 165 = 160 + \left( \frac{37 - 25}{2(37) - 25 - x} \right) \times 5 \]Solve for x:
\[ 165 - 160 = \left( \frac{12}{74 - 25 - x} \right) \times 5 \]\[ 5 = \left( \frac{12}{49 - x} \right) \times 5 \]Divide both sides by 5:
\[ 1 = \frac{12}{49 - x} \]Multiply both sides by \((49 - x)\):
\[ 49 - x = 12 \]Rearrange to solve for \(x\):
\[ x = 49 - 12 \]\[ x = 37 \]Let's verify this. If \(x=37\), then the frequencies are 23, 25, 37, 37, 28. In this case, the class 160-165 (frequency 37) and 165-170 (frequency 37) both have the highest frequency. However, the mode formula is designed to give a single value. If the mode is 165, it implies that the peak of the distribution is at 165. The formula calculates the mode within the modal class. If \(x=37\), the mode is indeed 165. If the modal class was 165-170 (meaning \(x\) was the highest frequency), then \(L=165\), \(f_1=x\), \(f_0=37\), \(f_2=28\). The mode would be \(165 + \left( \frac{x-37}{2x-37-28} \right) \times 5\). If this mode is 165, then \(\left( \frac{x-37}{2x-65} \right) \times 5\) must be 0, which means \(x-37=0\), so \(x=37\). This confirms that \(x=37\) is the correct value for the mode to be 165. The class 160-165 has frequency 37, and the class 165-170 also has frequency 37. The mode formula correctly identifies 165 as the mode in this scenario.
A) 37. The calculation shows that for the mode to be 165, the value of x must be 37.
The question asks to identify the cave complex where the famous painting 'The Flying Apsara' is located. This requires knowledge of prominent Indian rock-cut architecture and its associated artworks.
Correct Option: D) The Ajanta Caves are a UNESCO World Heritage Site located in Maharashtra, India. They are famous for their rock-cut Buddhist cave monuments which date from the 2nd century BCE to about 480 CE. These caves contain some of the finest surviving examples of ancient Indian art, particularly expressive paintings that depict the Jataka tales (stories of the Buddha's previous lives) and various Buddhist deities and celestial beings, including Apsaras. 'The Flying Apsara' is a well-known masterpiece from these caves, admired for its grace and artistic skill.
The question asks for the venue of the World Economic Forum (WEF) Annual Meeting in January 2025. This is a factual question related to international organizations and their regular events. Knowledge of the WEF's traditional meeting location is key.
Correct Option: C) Davos, Switzerland
To solve this problem, we need to first determine the cost price (CP) and selling price (SP) of a single laptop bag. Then, we calculate the profit per bag. Finally, we use the desired total profit to find out how many bags need to be bought and sold.
The vendor buys 42 laptop bags for ₹41,622.
\[ \text{CP per bag} = \frac{\text{Total Cost}}{\text{Number of bags bought}} = \frac{41622}{42} = \text{₹}991 \]The vendor sells 35 laptop bags for ₹37,345.
\[ \text{SP per bag} = \frac{\text{Total Selling Price}}{\text{Number of bags sold}} = \frac{37345}{35} = \text{₹}1067 \]Profit per bag is the difference between the selling price per bag and the cost price per bag.
\[ \text{Profit per bag} = \text{SP per bag} - \text{CP per bag} = 1067 - 991 = \text{₹}76 \]The desired total profit is ₹3,952. To find the number of bags, divide the total desired profit by the profit per bag.
\[ \text{Number of bags} = \frac{\text{Desired Total Profit}}{\text{Profit per bag}} = \frac{3952}{76} = 52 \]A) 52 is the correct answer. The calculation shows that 52 laptop bags need to be bought and sold to earn a profit of ₹3,952.
The question asks to identify the social reformer who coined the slogan 'One God, One Caste'. This requires knowledge of prominent Indian social reformers and their key philosophies or slogans.
Correct Option: A) Narayana Guru is credited with the slogan 'One God, One Caste' (or more completely, 'One Caste, One Religion, One God for Man'). He worked tirelessly for the upliftment of the downtrodden and against the caste system in Kerala.
The problem requires us to evaluate a given mathematical expression after interchanging certain arithmetic operators. We need to replace '+' with '−', '−' with '+', '×' with '÷', and '÷' with '×'. After performing these substitutions, we will evaluate the new expression following the order of operations (BODMAS/PEMDAS).
\(45 \div 12 - 75 \times 5 + 13 = ?\)
Substitute these into the original equation:
\[ 45 \times 12 + 75 \div 5 - 13 = ? \]First, perform multiplication and division from left to right.
Perform multiplication:
\[ 45 \times 12 = 540 \]Perform division:
\[ 75 \div 5 = 15 \]Substitute these values back into the equation:
\[ 540 + 15 - 13 = ? \]Perform addition:
\[ 540 + 15 = 555 \]Perform subtraction:
\[ 555 - 13 = 542 \]Correct Option: B) 542
The problem states that a person spends 50% of their monthly income and saves ₹7,056. We need to find their monthly expenses. Since income is divided into spending and saving, if 50% is spent, then the remaining percentage must be saved. We can then equate the saving percentage to the given saving amount to find the total income, and subsequently, the expenses.
If the person spends 50% of their monthly income, then the percentage of income saved is:
\[ \text{Percentage Saved} = 100\% - \text{Percentage Spent} \]\[ \text{Percentage Saved} = 100\% - 50\% = 50\% \]We are given that the person saves ₹7,056 per month. From step 1, we know that 50% of the monthly income is saved.
So, 50% of Monthly Income = ₹7,056.
Let the total monthly income be \(I\). Then,
\[ 0.50 \times I = 7056 \]\[ I = \frac{7056}{0.50} \]\[ I = 7056 \times 2 \]\[ I = 14112 \]The problem states that the person spends 50% of their monthly income. We have already calculated the monthly income as ₹14,112.
\[ \text{Monthly Expenses} = 50\% \times \text{Monthly Income} \]\[ \text{Monthly Expenses} = 0.50 \times 14112 \]\[ \text{Monthly Expenses} = 7056 \]A) 7,056
As calculated, the monthly expenses are ₹7,056, which matches option A.
Shubham ranked 19th from the top and 46th from the bottom in his class. How many students are there in his class?
To find the total number of students in the class, we can use the formula that relates a person's rank from both the top and the bottom of a list. When a person's rank from both ends is known, the total number of people can be calculated by adding the two ranks and subtracting 1 (because the person themselves is counted twice).
Thus, there are 64 students in the class.
D) 64 — This is the correct answer obtained by adding Shubham's rank from the top (19) and his rank from the bottom (46), and then subtracting 1 to account for Shubham being counted twice.
A) 62 — This would be incorrect. If we simply subtract 19 from 46, we get 27, which is not relevant. If we subtract 19 from 46 and then subtract 1, it's still not 62. This option does not follow the correct logic for calculating total students.
B) 65 — This is the result of simply adding 19 and 46. However, this counts Shubham twice (once in the top 19 and once in the bottom 46). To get the unique count of students, 1 must be subtracted.
C) 63 — This would be the result if we added 19 and 46 and then subtracted 2, or if there was an error in calculation. The correct formula requires subtracting only 1.
The problem presents a series of letter-number clusters. To find the missing cluster, we need to identify the pattern in the letters and the numbers separately.
C) RNK 26 is the correct option because it follows both the alphabetical decrement pattern for the letters and the arithmetic progression pattern for the numbers.