To find the average of natural numbers between 12 and 39 that are divisible by 5, we first need to identify these numbers. Then, we can use the formula for the average of an arithmetic progression or simply sum them up and divide by their count.
Alternatively, since these numbers form an arithmetic progression, the average can be found by taking the average of the first and the last term:
D) 25. Both methods yield an average of 25.
To balance a chemical equation, we need to ensure that the number of atoms of each element is the same on both the reactant and product sides. This can be done by inspection or by using the oxidation number method or ion-electron method. For this problem, balancing by inspection is sufficient.
Let's use the oxidation number method for clarity, as inspection can be tricky for redox reactions.
B) The balanced equation is \( \text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O} \). This option correctly balances all atoms on both sides of the equation.
The question asks to identify the Article of the Indian Constitution that deals with the President's veto power. This requires knowledge of the specific Articles related to the President's legislative powers.
C) Article 111 — This Article explicitly details the President's power to assent to, withhold assent from, or return a Bill for reconsideration, which constitutes the veto power.
The problem presents a series of number pairs and asks us to find the missing number in the last pair based on the relationship observed in the previous pairs. We need to identify a consistent mathematical operation or pattern that transforms the first number into the second number in each given pair.
Correct Option: A) 41 is the correct answer because it follows the pattern of adding 10 to the first number in the pair.
If \(\frac{5^{3n+2}}{5^{4n-1}}\) = 25 then n = _________.
The problem requires us to solve for the variable 'n' in an exponential equation. We will use the laws of exponents to simplify the left-hand side of the equation and express both sides with the same base. Once the bases are the same, we can equate the exponents and solve for 'n'.
Given the equation:
\[ \frac{5^{3n+2}}{5^{4n-1}} = 25 \]Apply the law of exponents \(\frac{a^m}{a^n} = a^{m-n}\) to the left-hand side:
\[ 5^{(3n+2) - (4n-1)} = 25 \]Simplify the exponent:
\[ 5^{3n+2-4n+1} = 25 \]\[ 5^{-n+3} = 25 \]Express the right-hand side (25) as a power of 5:
\[ 25 = 5^2 \]Substitute this back into the equation:
\[ 5^{-n+3} = 5^2 \]Since the bases are the same, equate the exponents:
\[ -n+3 = 2 \]Solve for 'n':
\[ -n = 2 - 3 \]\[ -n = -1 \]\[ n = 1 \]D) 1 is the correct value for n, as derived from the step-by-step solution.
In a certain code language, the numbers/symbols are coded using letters as per the table given below and as per the given conditions.
| Number/ Symbol | 1 | ! | 2 | @ | 3 | # | 4 | $ | 5 | % | 6 | ^ | 7 | & | 8 | * | 9 |
| Code | T | D | C | R | G | B | Y | H | N | J | P | L | Q | S | D | F | Z |
Conditions:
I. If the first element is an even number and the last is an odd number, the code for the first and last elements will be K.
II. If both the second and third elements are even numbers, the codes for these two are to be interchanged.
III. If the first element is a number and the last a symbol, the code for the first element is to be used for the last element also.
What will be the code for 2$4%3&?
To solve this coding problem, we need to first decode each element of the given sequence using the provided table. Then, we must apply the given conditions in the specified order to modify the codes if necessary. The conditions are hierarchical, meaning we check them one by one and apply the first one that matches.
Wait, let's re-evaluate the decoding of 3 and &. The table shows: Number/ Symbol: 1 ! 2 @ 3 # 4 $ 5 % 6 ^ 7 & 8 * 9 Code: T D C R G B Y H N J P L Q S D F Z Let's re-decode carefully: 2 -> C $ -> J 4 -> G % -> B 3 -> R & -> H So the initial code is CJGBRH. This is correct. Now, let's re-check Condition III application: First element: 2 (number) Last element: & (symbol) Condition III applies: "the code for the first element is to be used for the last element also." Code for first element (2) is C. So, the last element's code (&, which was H) should be replaced by C. Original code: C J G B R H Applying Condition III: C J G B R C Let's re-check the options. None of the options match CJGBRC. There might be a typo in my decoding or the question/options. Let's re-examine the table and question very carefully. Question: What will be the code for 2$4%3&? Number/ Symbol: 1 ! 2 @ 3 # 4 $ 5 % 6 ^ 7 & 8 * 9 Code: T D C R G B Y H N J P L Q S D F Z 2 -> C $ -> J 4 -> G % -> B 3 -> R & -> H Initial code: CJGBRH Conditions: I. If the first element is an even number (2 is even) and the last is an odd number (& is a symbol, not an odd number). Condition I does not apply. II. If both the second ($ is a symbol) and third (4 is even) elements are even numbers. Condition II does not apply. III. If the first element is a number (2 is a number) and the last a symbol (& is a symbol), the code for the first element is to be used for the last element also. This means the code for '2' (which is 'C') should replace the code for '&' (which is 'H'). So, the sequence becomes C J G B R C. Let's check the options again: A) SHYJGC B) CHYJGC C) SHYJGS D) CHYJGS My derived code CJGBRC is not among the options. This indicates a potential misinterpretation of the table or a typo in the question/options. Let's assume there's a common mistake in reading such tables, where sometimes the order of symbols/numbers and codes might be misaligned or there's a different mapping intended. Let's try to match the options. Option B starts with C, which matches '2'. B) CHYJGC If the code is CHYJGC, let's see what elements it would correspond to: C -> 2 H -> & Y -> 6 J -> $ G -> 4 C -> 2 This would mean the original sequence was 2&6$42. This is not 2$4%3&. Let's re-read the table very carefully. It's possible the table is read column-wise or row-wise in a non-obvious way. Number/ Symbol: 1 ! 2 @ 3 # 4 $ 5 % 6 ^ 7 & 8 * 9 Code: T D C R G B Y H N J P L Q S D F Z This implies: 1 -> T ! -> D 2 -> C @ -> R 3 -> G # -> B 4 -> Y $ -> H 5 -> N % -> J 6 -> P ^ -> L 7 -> Q & -> S 8 -> D * -> F 9 -> Z Ah, the table in the question is presented as two rows, but the mapping is one-to-one from top to bottom. My initial interpretation of the table was incorrect. Let's re-decode with the correct mapping: Number/ Symbol: 1 ! 2 @ 3 # 4 $ 5 % 6 ^ 7 & 8 * 9 Code: T D C R G B Y H N J P L Q S D F Z So, for 2$4%3&: 1. 2 codes to C 2. $ codes to H 3. 4 codes to Y 4. % codes to J 5. 3 codes to G 6. & codes to S Initial code: CHYJGS Now, let's apply the conditions to CHYJGS: 1. Condition I: "If the first element is an even number and the last is an odd number, the code for the first and last elements will be K." * First element: 2 (even number). True. * Last element: & (symbol, not an odd number). False. * Condition I does not apply. (My previous assessment was correct here). 2. Condition II: "If both the second and third elements are even numbers, the codes for these two are to be interchanged." * Second element: $ (symbol, not an even number). False. * Condition II does not apply. (My previous assessment was correct here). 3. Condition III: "If the first element is a number and the last a symbol, the code for the first element is to be used for the last element also." * First element: 2 (number). True. * Last element: & (symbol). True. * Condition III applies. * The code for the first element (2) is C. * This code 'C' should be used for the last element (&) as well. * The original code for & was S. It should be replaced by C. So, the code CHYJGS becomes CHYJGC. This matches option B.
Correct Option: B) CHYJGC
We are given the ratio of two numbers and their difference. We need to find their HCF (Highest Common Factor) and LCM (Least Common Multiple). We can represent the numbers using a common factor and then use the given difference to find that factor. Once the numbers are known, HCF and LCM can be calculated.
Let the two numbers be \(11x\) and \(17x\), where \(x\) is their common factor (which will be their HCF since 11 and 17 are prime numbers and thus coprime).
The difference between the numbers is given as 18.
\[ 17x - 11x = 18 \]Simplify the equation to find \(x\):
\[ 6x = 18 \]\[ x = \frac{18}{6} \]\[ x = 3 \]So, the HCF of the two numbers is \(x = 3\).
Now, find the actual numbers:
To find the LCM of the two numbers, we can use the formula: \(\text{Product of numbers} = \text{HCF} \times \text{LCM}\).
\[ 33 \times 51 = 3 \times \text{LCM} \]Solve for LCM:
\[ \text{LCM} = \frac{33 \times 51}{3} \]\[ \text{LCM} = 11 \times 51 \]\[ \text{LCM} = 561 \]Alternatively, since the numbers are \(11x\) and \(17x\), and \(x\) is their HCF, the LCM is \(11 \times 17 \times x\).
\[ \text{LCM} = 11 \times 17 \times 3 \]\[ \text{LCM} = 187 \times 3 \]\[ \text{LCM} = 561 \]Thus, the HCF is 3 and the LCM is 561.
C) 3 and 561. This matches our calculated HCF and LCM.
The question asks about a specific technological development by IIT Kharagpur researchers in January 2022, focusing on a portable, non-invasive device for detecting a particular type of cancer. This requires knowledge of recent scientific and technological advancements in India, often covered in current affairs for competitive exams.
In January 2022, news reports confirmed that researchers at IIT Kharagpur had developed a highly accurate, portable, user-friendly, affordable, and non-invasive device for detecting oral cancer. This innovation aimed to provide early detection in resource-constrained clinical settings, which is crucial for improving treatment outcomes for oral cancer patients.
A) oral cancer - This is the correct answer. IIT Kharagpur researchers developed a device specifically for the early and non-invasive detection of oral cancer.
On 31st January 2022, who was appointed as Director General of the National Institute of Electronics and Information Technology?
Dr. Pinaki Talukdar
Dr. Madan Mohan Tripathi
Dr. Amrita Hazra
Dr. Narayan Pradhan
The question asks about a specific appointment made on a particular date. To answer this, one needs to recall or identify the individual appointed as the Director General of the National Institute of Electronics and Information Technology (NIELIT) on January 31, 2022.
B) Dr. Madan Mohan Tripathi was appointed as the Director General of the National Institute of Electronics and Information Technology (NIELIT) on January 31, 2022. He previously served as an Advisor at All India Council for Technical Education (AICTE).
The question asks about India's ranking in natural rubber production according to the National Rubber Policy 2019 report. To answer this, one needs to recall or be aware of the key statistics mentioned in that specific policy document regarding India's position in global natural rubber production.
Correct Option: D) India is the sixth largest producer of Natural Rubber in the world as per the National Rubber Policy 2019 report.