The question asks about the deity to whom the Dilwara Temples in Mount Abu are dedicated. This requires knowledge of Indian temple architecture and religious affiliations.
Correct Option: B) Adinath
The Dilwara Temples are a group of five Jain temples located near Mount Abu, Rajasthan. The most famous among them is the Vimal Vasahi Temple, which is dedicated to Adinath (Rishabhanatha), the first Jain Tirthankara. These temples are known for their exquisite marble carvings and intricate architecture.
The problem requires us to find the correct combination of mathematical operators that, when placed in the blanks of the given equation, will make the equation true. We will test each option by substituting the operators into the equation and performing the calculations according to the order of operations (BODMAS/PEMDAS).
The given equation is: \(9 \_ 3 \_ 0 \_ 8 \_ 2 = 23\)
Substitute the operators into the equation:
\[ 9 \div 3 + 0 - 8 \times 2 \]\[ 3 + 0 - 16 \]\[ 3 - 16 \]\[ -13 \]Since \(-13 \neq 23\), Option A is incorrect.Substitute the operators into the equation:
\[ 9 \times 3 + 0 - 8 \div 2 \]\[ 27 + 0 - 4 \]\[ 27 - 4 \]\[ 23 \]Since \(23 = 23\), Option B is correct.Substitute the operators into the equation:
\[ 9 - 3 + 0 \div 8 \times 2 \]\[ 9 - 3 + 0 \times 2 \]\[ 9 - 3 + 0 \]\[ 6 \]Since \(6 \neq 23\), Option C is incorrect.Substitute the operators into the equation:
\[ 9 + 3 - 0 \times 8 \div 2 \]\[ 9 + 3 - 0 \div 2 \]\[ 9 + 3 - 0 \]\[ 12 \]Since \(12 \neq 23\), Option D is incorrect.B) When the symbols \(\times, +, -, \div\) are sequentially placed in the blanks, the equation becomes \(9 \times 3 + 0 - 8 \div 2\). Following the order of operations:
The result \(23\) matches the right-hand side of the equation, so this option is correct.
What smallest positive fraction should be added to 3 \( \frac{2}{3} \) + 6 \( \frac{7}{12} \) + 4 \( \frac{9}{36} \) + 5 + 7 \( \frac{1}{12} \) to make the sum a whole number?
\( \frac{5}{12} \)
\( \frac{11}{12} \)
\( \frac{7}{12} \)
\( \frac{13}{12} \)
To find the smallest positive fraction that needs to be added to the given sum to make it a whole number, we first need to calculate the sum of the given mixed fractions and whole numbers. After finding the total sum, we will identify its fractional part. The smallest positive fraction to be added will be the difference between 1 and this fractional part (if the fractional part is not zero).
Convert mixed fractions to improper fractions or separate whole and fractional parts:
The given expression is \[3 \frac{2}{3} + 6 \frac{7}{12} + 4 \frac{9}{36} + 5 + 7 \frac{1}{12}\].
We can separate the whole number parts and the fractional parts:
Simplify the fractional parts:
First, simplify the fraction \( \frac{9}{36} \):
Find a common denominator for the fractional parts:
The denominators are 3, 12, 4, and 12. The least common multiple (LCM) of these denominators is 12.
Add the fractional parts:
\[ \text{Sum of fractional parts} = \frac{8}{12} + \frac{7}{12} + \frac{3}{12} + \frac{1}{12} \]\[ = \frac{8 + 7 + 3 + 1}{12} = \frac{19}{12} \]Convert the improper fraction to a mixed number:
\[ \frac{19}{12} = 1 \frac{7}{12} \]Add the sum of whole parts and the sum of fractional parts:
\[ \text{Total Sum} = 25 + 1 \frac{7}{12} = 26 \frac{7}{12} \]Determine the smallest positive fraction to make the sum a whole number:
The fractional part of the total sum is \( \frac{7}{12} \). To make this a whole number, we need to add a fraction that completes it to the next whole number (1).
The required fraction is \( 1 - \frac{7}{12} \).
A) \( \frac{5}{12} \)
As calculated, adding \( \frac{5}{12} \) to \( 26 \frac{7}{12} \) results in \[26 + \frac{7}{12} + \frac{5}{12} = 26 + \frac{12}{12} = 26 + 1 = 27\], which is a whole number.
The question asks to identify which of the given cities is located on the Hudson River. This requires knowledge of world geography, specifically the major rivers and the cities they flow through.
Correct Option: C) New York — The Hudson River flows through New York State and empties into New York Bay, adjacent to New York City. The city's western boundary of Manhattan is formed by the Hudson River.
In 2019, ______ became the 24th Chief of the Naval Staff of India.
Admiral Karambir Singh
Admiral Bimal Verma
S Jaishankar
Admiral Sunil Lamba
The question asks to identify the 24th Chief of the Naval Staff of India who took office in 2019. This is a factual question related to current affairs and appointments in the Indian armed forces.
A) Admiral Karambir Singh became the 24th Chief of the Naval Staff of India on May 31, 2019. He served in this position until November 30, 2021.
This is an analogy question where we need to find the relationship between the first pair of words and apply the same relationship to the second pair to find the missing term.
Correct Option: B) Catalogue — A catalogue (or catalog) is a list of all the books, periodicals, and other materials in a library, organized in a systematic way to help users find what they are looking for. It serves as a guide or reference, just as a manual does in a laboratory.
To determine the time taken for sunlight to reach Earth, we need to know the average distance between the Sun and Earth and the speed of light. We can then use the fundamental formula relating distance, speed, and time.
The relationship between distance, speed, and time is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]Substitute the known values into the formula:
\[ \text{Time} = \frac{1.496 \times 10^{11}\ \text{m}}{3 \times 10^8\ \text{m/s}} \]Calculate the result:
\[ \text{Time} \approx 498.67\ \text{seconds} \]To convert seconds to minutes, divide by 60:
\[ \text{Minutes} = \frac{498.67}{60} \approx 8.31\ \text{minutes} \]This means 8 full minutes and a fraction of a minute. To find the remaining seconds:
\[ \text{Remaining seconds} = (0.31 \times 60)\ \text{seconds} \approx 18.6\ \text{seconds} \]Rounding this, we get approximately 8 minutes and 19-20 seconds.
The calculated time is approximately 8 minutes and 20 seconds, which matches option D.
D) 8 min 20 sec is the correct answer. This is the widely accepted average time for sunlight to travel from the Sun to Earth, based on the average distance and the speed of light.
This problem involves solving an arithmetic addition puzzle where letters represent digits. We need to find the maximum possible value for the digit Q by analyzing the addition column by column, starting from the units place.
Analyze the units column:
We have \(9 + 7 + 8\).
\(9 + 7 + 8 = 24\).
This means the units digit of the sum is \(4\), and there is a carry-over of \(2\) to the tens column.
Analyze the tens column: In the tens column, we have \(P + R + Q\) plus the carry-over \(2\) from the units column. The sum of these must result in a number whose units digit is \(1\) (from \(1114\)). So, \(P + R + Q + 2 = \text{_1}\) (where \(\text{_1}\) means a number ending in 1, like 1, 11, 21, etc.). This simplifies to \(P + R + Q = \text{_1} - 2\), which means \(P + R + Q = \text{_9}\) (a number ending in 9, like 9, 19, 29, etc.).
Analyze the hundreds column: In the hundreds column, we have \(5 + 3 + 2\) plus any carry-over from the tens column. The sum is \(11\). \(5 + 3 + 2 = 10\). Since the sum in the hundreds place is \(11\), this implies that there must have been a carry-over of \(1\) from the tens column to the hundreds column. This means that \(P + R + Q + 2\) must have been \(11\) or \(21\) or \(31\), etc., such that a \(1\) is carried over.
Combine information from tens and hundreds columns: From step 2, we know \(P + R + Q\) must end in \(9\). From step 3, we know there was a carry-over of \(1\) from the tens column. This means \(P + R + Q + 2\) must have been \(11\) or \(21\). If \(P + R + Q + 2 = 11\), then \(P + R + Q = 9\). If \(P + R + Q + 2 = 21\), then \(P + R + Q = 19\). If \(P + R + Q + 2 = 31\), then \(P + R + Q = 29\).
Determine the maximum value of Q:
P, R, and Q are single digits (0-9). To maximize Q, we should minimize P and R.
The minimum value for P and R (if they are distinct from Q and each other, though not specified, usually they are distinct digits or can be repeated) is \(0\) and \(1\), or \(0\) and \(0\) if repetition is allowed.
Let's assume P and R can be any digit from 0 to 9. To maximize Q, P and R should be as small as possible, i.e., \(P=0\) and \(R=0\).
Case 1: \(P + R + Q = 9\)
If \(P=0\) and \(R=0\), then \(Q = 9\). This is a valid digit.
Case 2: \(P + R + Q = 19\)
If \(P=0\) and \(R=0\), then \(Q = 19\), which is not a single digit. So this case is not possible with \(P=0, R=0\).
The maximum sum for \(P+R+Q\) is \(9+9+9 = 27\). So \(P+R+Q=19\) is possible (e.g., \(P=1, R=9, Q=9\)).
Case 3: \(P + R + Q = 29\)
The maximum sum for \(P+R+Q\) is \(9+9+9 = 27\). So \(P+R+Q=29\) is not possible.
So, the possible sums for \(P+R+Q\) are \(9\) or \(19\).
If \(P+R+Q = 9\), the maximum Q (when \(P=0, R=0\)) is \(9\).
If \(P+R+Q = 19\), the maximum Q (when \(P=0, R=0\)) would be \(19\), which is not a single digit. To make \(P+R+Q=19\), if we want to maximize Q, we must minimize P and R. The smallest distinct digits for P and R would be \(0\) and \(1\). So \(0+1+Q=19 \Rightarrow Q=18\), which is not a single digit. If P and R can be the same, \(P=0, R=0 \Rightarrow Q=19\), not a single digit. The smallest possible sum for P+R is \(0+0=0\). So, \(Q\) can be at most \(19-0=19\), which is not a single digit. Thus, \(P+R+Q=19\) cannot yield a single digit Q if P and R are also single digits.
Let's re-evaluate. \(P, R, Q\) are digits from \(0\) to \(9\).
We have \(P+R+Q = 9\) or \(P+R+Q = 19\).
If \(P+R+Q = 9\):
To maximize Q, we set P and R to their minimum values, which are \(0\).
So, \(0 + 0 + Q = 9 \Rightarrow Q = 9\).
This is a valid digit.
If \(P+R+Q = 19\):
To maximize Q, we set P and R to their minimum values, which are \(0\).
So, \(0 + 0 + Q = 19 \Rightarrow Q = 19\).
This is not a single digit, so this case cannot yield a valid Q.
Therefore, the only valid scenario that allows Q to be a single digit is when \(P+R+Q=9\). In this case, the maximum value of Q is \(9\).
(Example: \(P=0, R=0, Q=9\). Check: \(509 + 307 + 298 = 1114\). Units: \(9+7+8=24\) (4, carry 2). Tens: \(0+0+9+2=11\) (1, carry 1). Hundreds: \(5+3+2+1=11\). This works.)
Correct Option: B) 9
The number of colours a computer monitor can display is determined by its colour depth, which is measured in bits per pixel. Each bit can represent two states (0 or 1). Therefore, for a given number of bits, the total number of distinct colours is calculated as \(2^{\text{number of bits}}\).
\[ \text{Total Colours} = 2^{\text{number of bits}} \]
\[ \text{Total Colours} = 2^{16} \]
\[ 2^{16} = 65536 \]
Correct Option: A) 65536
What is the value of the following expression?
50 - [8{9 - (6 - (\(\overline{5+3}\)))}]
To solve this expression, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS (Brackets/Parentheses, Orders/Exponents, Division/Multiplication, Addition/Subtraction). We will start with the innermost brackets and work our way outwards, paying special attention to the vinculum (bar over numbers) which acts as a grouping symbol.
B) -38 is the correct value of the given expression after following the order of operations.