The question asks to identify the metro city among the given options that has retained its name since India's independence in 1947. This requires knowledge of the historical name changes of major Indian cities.
Correct Option: B) Delhi
The question asks to identify which of the given passes is NOT one of the three business check posts between India and China. This requires knowledge of the major mountain passes that facilitate trade and connectivity between the two countries.
The three operational border passes for trade between India and China are:
The Khyber Pass, on the other hand, is a historical pass connecting Afghanistan and Pakistan, and has no direct trade link between India and China.
D) Khyber Pass is not one of the three business check posts between India and China. It is a historical pass located between Afghanistan and Pakistan, primarily connecting Central Asia with the Indian subcontinent.
The question asks about the ruler who formally introduced the Iqta system in the Delhi Sultanate. We need to identify the specific Sultan responsible for institutionalizing this administrative and revenue assignment system.
C) Iltutmish is the correct answer. He is credited with formally introducing and institutionalizing the Iqta system in the Delhi Sultanate, making it a cornerstone of his administration.
The question asks to identify the specific chemical mixture used for purifying gold. This requires knowledge of the chemical properties of gold and common reagents used in metallurgy and analytical chemistry.
Correct Option: C) Aqua regia is a mixture of concentrated nitric acid and concentrated hydrochloric acid, specifically designed to dissolve noble metals like gold and platinum, thus used in their purification.
Simplify: \( 1.\overline{45} + 0.3\overline{12} - 1.11\overline{2} \)
\( \frac{589}{900} \)
\( \frac{13}{20} \)
\( \frac{374}{495} \)
\( \frac{163}{300} \)
To simplify the given expression, we need to convert each repeating decimal into its equivalent fractional form. Then, we perform the addition and subtraction of these fractions by finding a common denominator.
Let \(x = 1.\overline{45} = 1.454545...\)
Multiply by 100:
\[ 100x = 145.454545... \]\[ 100x - x = 145.454545... - 1.454545... \]\[ 99x = 144 \]\[ x = \frac{144}{99} = \frac{16 \times 9}{11 \times 9} = \frac{16}{11} \]Let \(y = 0.3\overline{12} = 0.3121212...\)
Multiply by 10:
\[ 10y = 3.121212... \]Multiply by 1000:
\[ 1000y = 312.121212... \]\[ 1000y - 10y = 312.121212... - 3.121212... \]\[ 990y = 309 \]\[ y = \frac{309}{990} = \frac{103 \times 3}{330 \times 3} = \frac{103}{330} \]Let \(z = 1.11\overline{2} = 1.112222...\)
Multiply by 100:
\[ 100z = 111.2222... \]Multiply by 1000:
\[ 1000z = 1112.2222... \]\[ 1000z - 100z = 1112.2222... - 111.2222... \]\[ 900z = 1001 \]\[ z = \frac{1001}{900} \]The expression is \( x + y - z \).
\[ \frac{16}{11} + \frac{103}{330} - \frac{1001}{900} \]LCM \(= 2^2 \times 3^2 \times 5^2 \times 11 = 4 \times 9 \times 25 \times 11 = 36 \times 25 \times 11 = 900 \times 11 = 9900\)
Let's recheck the conversion of \(1.11\overline{2}\) to a fraction. A common formula for mixed repeating decimals is:
For \(0.abc\overline{de}\), the fraction is \[\frac{abcde - abc}{\text{number of 9s for repeating part followed by number of 0s for non-repeating part}}\].
For \(1.\overline{45}\): \[1 + 0.\overline{45} = 1 + \frac{45}{99} = 1 + \frac{5}{11} = \frac{11+5}{11} = \frac{16}{11}\]. This is correct.
For \(0.3\overline{12}\): \[\frac{312 - 3}{990} = \frac{309}{990} = \frac{103}{330}\]. This is correct.
For \(1.11\overline{2}\): \(1 + 0.11\overline{2}\). Let \(X = 0.11\overline{2}\). Then \(100X = 11.\overline{2}\) and \(1000X = 112.\overline{2}\). So \(900X = 112 - 11 = 101\). Thus \(X = \frac{101}{900}\).
So \[1.11\overline{2} = 1 + \frac{101}{900} = \frac{900+101}{900} = \frac{1001}{900}\]. This is correct.
Now, let's re-evaluate the sum:
\[ \frac{16}{11} + \frac{103}{330} - \frac{1001}{900} \]LCM(11, 330, 900) = LCM(11, 33*10, 9*100) = LCM(11, 3*11*10, 9*100)
LCM(11, 330, 900) = \(2^2 \times 3^2 \times 5^2 \times 11 = 4 \times 9 \times 25 \times 11 = 900 \times 11 = 9900\).
This LCM is correct.
\[ \frac{16 \times 900}{11 \times 900} + \frac{103 \times 30}{330 \times 30} - \frac{1001 \times 11}{900 \times 11} \]\[ = \frac{14400}{9900} + \frac{3090}{9900} - \frac{11011}{9900} \]\[ = \frac{14400 + 3090 - 11011}{9900} \]\[ = \frac{17490 - 11011}{9900} \]\[ = \frac{6479}{9900} \]Let's re-examine the question and options. Sometimes, the question might imply a simpler common denominator or a slightly different interpretation.
Let's try to simplify the fractions to their lowest terms first and then find LCM.
\(1.\overline{45} = \frac{16}{11}\)
\(0.3\overline{12} = \frac{103}{330}\)
\(1.11\overline{2} = \frac{1001}{900}\)
The expression is \[\frac{16}{11} + \frac{103}{330} - \frac{1001}{900}\].
Let's try to find a common denominator for 11, 330, and 900 again. The LCM is indeed 9900.
Let's re-calculate the numerator:
\(14400 + 3090 - 11011 = 17490 - 11011 = 6479\).
So the result is \(\frac{6479}{9900}\).
Let's check if any option simplifies to this or if there's a typo in the question or options.
Consider the possibility that the question meant \(1.4\overline{5}\) instead of \(1.\overline{45}\) or similar small changes.
If we assume the given options are correct and there's a calculation error on my part, let's re-evaluate the conversions carefully.
Alternative approach for repeating decimals:
\[1.\overline{45} = \frac{145-1}{99} = \frac{144}{99} = \frac{16}{11}\]. Correct.
\[0.3\overline{12} = \frac{312-3}{990} = \frac{309}{990} = \frac{103}{330}\]. Correct.
\[1.11\overline{2} = \frac{1112-111}{900} = \frac{1001}{900}\]. Correct.
The expression is \[\frac{16}{11} + \frac{103}{330} - \frac{1001}{900}\].
We need to find a common denominator for 11, 330, 900.
\(11 = 11\)
\(330 = 2 \times 3 \times 5 \times 11\)
\(900 = 2^2 \times 3^2 \times 5^2\)
LCM \(= 2^2 \times 3^2 \times 5^2 \times 11 = 4 \times 9 \times 25 \times 11 = 900 \times 11 = 9900\).
The calculation is:
\[ \frac{16 \times (9900/11)}{9900} + \frac{103 \times (9900/330)}{9900} - \frac{1001 \times (9900/900)}{9900} \]\[ = \frac{16 \times 900}{9900} + \frac{103 \times 30}{9900} - \frac{1001 \times 11}{9900} \]\[ = \frac{14400 + 3090 - 11011}{9900} \]\[ = \frac{17490 - 11011}{9900} \]\[ = \frac{6479}{9900} \]Let's assume there was a typo in the question and it was \(1.1\overline{2}\) instead of \(1.11\overline{2}\).
If \[1.1\overline{2} = \frac{112-11}{90} = \frac{101}{90}\].
Then the expression would be \[\frac{16}{11} + \frac{103}{330} - \frac{101}{90}\].
LCM(11, 330, 90) = LCM(11, 33*10, 9*10) = \(2 \times 3^2 \times 5 \times 11 = 990\).
\[ \frac{16 \times 90}{990} + \frac{103 \times 3}{990} - \frac{101 \times 11}{990} \]\[ = \frac{1440 + 309 - 1111}{990} \]\[ = \frac{1749 - 1111}{990} \]\[ = \frac{638}{990} = \frac{319}{495} \]This is not among the options. Let's re-examine the options and the result \(\frac{6479}{9900}\). Is there any common factor for 6479 and 9900? \(9900 = 2^2 \times 3^2 \times 5^2 \times 11\). 6479 is not divisible by 2, 3, 5. \(6+4+7+9 = 26\), not divisible by 3. For 11: \(9-7+4-6 = 0\), so 6479 is divisible by 11. \(6479 \div 11 = 589\). So, \[\frac{6479}{9900} = \frac{589 \times 11}{900 \times 11} = \frac{589}{900}\]. My initial calculation of the LCM was correct, and the final fraction \(\frac{6479}{9900}\) was also correct. The error was in not simplifying the final fraction by dividing by 11. So, the simplified result is \(\frac{589}{900}\). This matches option A.A) \( \frac{589}{900} \)
The step-by-step conversion of repeating decimals to fractions and their subsequent arithmetic operation leads to the result \(\frac{6479}{9900}\). Dividing both the numerator and denominator by 11 simplifies this fraction to \(\frac{589}{900}\).
The question asks for the full form of NSDL in economic terminology. This is a direct recall question related to financial market infrastructure in India. Knowing the full forms of key institutions is essential for competitive exams.
A) National Securities Depository Limited is the correct full form of NSDL. NSDL plays a crucial role in the Indian capital market by providing services for holding and transferring securities in electronic form, thereby eliminating the risks associated with physical certificates.
If every 3rd letter from left to right of English alphabet is deleted, then what would be the 12th letter from right in the new series obtained?
The problem asks us to identify the 12th letter from the right in a new series formed by deleting every 3rd letter from the standard English alphabet. We need to first list the original alphabet, then identify and remove the specified letters, and finally count from the right in the resulting series.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
The 3rd letters are C, F, I, L, O, R, U, X.
Deleting these, the new series becomes:
A B D E G H J K M N P Q S T V W Y Z
Original alphabet has 26 letters. We deleted 8 letters (26 / 3 = 8 with a remainder of 2). So, the new series has \(26 - 8 = 18\) letters.
Since there are 18 letters in the new series, the 12th letter from the right is the same as the \((18 - 12 + 1)\)th letter from the left.
Position from left \( = 18 - 12 + 1 = 7\).
Let's count 7 letters from the left in the new series:
1st: A
2nd: B
3rd: D
4th: E
5th: G
6th: H
7th: J
Therefore, the 12th letter from the right is J.
A) J is the 12th letter from the right in the new series.
The question asks to identify the purpose of the Vidyanjali project. This requires knowledge of government educational initiatives in India.
D) Volunteers in government schools to help scholastic work
The Vidyanjali project is an initiative by the Ministry of Education, Government of India, to strengthen government and government-aided schools by connecting them with volunteers from the Indian diaspora, retired teachers, government officials, professionals, and other members of the community. These volunteers contribute to various activities, including academic support (scholastic work), co-scholastic activities, and providing infrastructure support.
This problem involves substituting given mathematical operations for letters and then evaluating the resulting expression using the order of operations (BODMAS/PEMDAS).
The given expression is 48 B 24 C 12 D 3 A 4.
Substituting the operations, we get:
\[ 48 + 24 \div 12 \times 3 - 4 \]BODMAS stands for Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
According to this rule, division and multiplication should be performed before addition and subtraction, from left to right.
The expression becomes:
\[ 48 + 2 \times 3 - 4 \]The expression becomes:
\[ 48 + 6 - 4 \]The expression becomes:
\[ 54 - 4 \]C) 50 is the correct value of the expression after substituting the operations and applying the order of operations.
What is the value of 3 \( \frac{21}{23} \)÷ 3 \( \frac{15}{31} \)?
To solve this problem, we need to convert the given mixed fractions into improper fractions first. Then, we can perform the division operation by multiplying the first fraction by the reciprocal of the second fraction.
The first mixed fraction is \(3 \frac{21}{23}\). To convert it to an improper fraction, we use the formula: \( \text{Whole number} \times \text{Denominator} + \text{Numerator} \) for the new numerator, keeping the original denominator.
\[ 3 \frac{21}{23} = \frac{(3 \times 23) + 21}{23} = \frac{69 + 21}{23} = \frac{90}{23} \]The second mixed fraction is \(3 \frac{15}{31}\).
\[ 3 \frac{15}{31} = \frac{(3 \times 31) + 15}{31} = \frac{93 + 15}{31} = \frac{108}{31} \]Now we need to divide the first improper fraction by the second improper fraction:
\[ \frac{90}{23} \div \frac{108}{31} \]To divide fractions, we multiply the first fraction by the reciprocal of the second fraction:
\[ \frac{90}{23} \times \frac{31}{108} \]We can simplify by finding common factors between the numerators and denominators. Both 90 and 108 are divisible by 18.
\(90 \div 18 = 5\)
\(108 \div 18 = 6\)
So, the expression becomes:
\[ \frac{5}{23} \times \frac{31}{6} \]Now, multiply the numerators and the denominators:
\[ \frac{5 \times 31}{23 \times 6} = \frac{155}{138} \]Correct Option: C) \( \frac{155}{138} \)