The value of the expression (1 + \( \frac{1}{3} \))(1 + \( \frac{1}{4} \))(1 + \( \frac{1}{5} \))…(1 + \(\frac{1}{n-1}\)) is:
\( \frac{n}{3} \)
\( \frac{1}{3} \)
1 + \( \frac{1}{n} \)
\(\frac{n}{n-1}\)
The problem asks us to find the value of a product of terms. Each term is of the form \((1 + \frac{1}{k})\). We will simplify each term and then observe the pattern in the product to find a general formula.
Each term in the expression is of the form \((1 + \frac{1}{k})\). Let's simplify this general form:
\[ 1 + \frac{1}{k} = \frac{k}{k} + \frac{1}{k} = \frac{k+1}{k} \]Now, substitute this simplified form back into the given expression:
\[ \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) \left(1 + \frac{1}{5}\right) \dots \left(1 + \frac{1}{n-1}\right) \]This becomes:
\[ \left(\frac{3+1}{3}\right) \left(\frac{4+1}{4}\right) \left(\frac{5+1}{5}\right) \dots \left(\frac{(n-1)+1}{n-1}\right) \]Which simplifies to:
\[ \left(\frac{4}{3}\right) \left(\frac{5}{4}\right) \left(\frac{6}{5}\right) \dots \left(\frac{n}{n-1}\right) \]Observe the pattern in the product. This is a telescoping product, where the numerator of one term cancels out the denominator of the next term:
\[ \frac{4}{\cancel{3}} \times \frac{5}{\cancel{4}} \times \frac{6}{\cancel{5}} \times \dots \times \frac{\cancel{n-1}}{\cancel{n-2}} \times \frac{n}{\cancel{n-1}} \]More precisely, the numerator of the first term (4) cancels with the denominator of the second term (4). The numerator of the second term (5) cancels with the denominator of the third term (5), and so on.
After all the cancellations, only the denominator of the first term and the numerator of the last term will remain.
The denominator of the first term is \(3\).
The numerator of the last term is \(n\).
So, the product simplifies to:
\[ \frac{n}{3} \]Correct Option: A) \( \frac{n}{3} \)
The question asks to identify the organization with which prominent revolutionaries like Ram Prasad Bismil, Chandrashekhar Azad, and Bhagat Singh were associated. This requires knowledge of the history of revolutionary movements in India.
Correct Option: A) Hindustan Socialist Republican Association
Ram Prasad Bismil, Chandrashekhar Azad, and Bhagat Singh were all prominent members of the Hindustan Socialist Republican Association (HSRA). The HSRA was formed in 1928 by reorganizing the Hindustan Republican Association (HRA) with a new socialist objective. Chandrashekhar Azad was a key leader, and Bhagat Singh was a significant ideologue and activist within the organization. Ram Prasad Bismil, though executed before the HSRA was formally constituted, was a leading figure of its predecessor, the HRA, and his revolutionary spirit aligned with the HSRA's goals.
The question asks us to identify implicit assumptions based on a given statement. An assumption is something taken for granted or accepted as true without proof. It must be a logical consequence or a necessary condition for the statement to be made. We need to analyze each assumption to see if it is directly implied by the statement.
A) Neither I nor II is implicit. As analyzed above, neither of the assumptions can be logically derived or taken for granted from the given statement. The statement is a policy announcement, not a justification or an observation about the economic state.
This problem involves calculating the cost price of items when the selling price and profit percentage are known. We need to find out how many clips were bought for a rupee, given that selling 16 clips for a rupee yields a 25% profit.
Substituting the given values:
\[ \frac{1}{16} = \text{CP} \times \left(1 + \frac{25}{100}\right) \]\[ \frac{1}{16} = \text{CP} \times \left(1 + \frac{1}{4}\right) \]\[ \frac{1}{16} = \text{CP} \times \left(\frac{5}{4}\right) \]Now, solve for CP:
\[ \text{CP} = \frac{1}{16} \times \frac{4}{5} \]\[ \text{CP} = \frac{1}{4 \times 5} \]\[ \text{CP} = \frac{1}{20} \text{ Rupee} \]So, the cost price of 1 clip is \( \frac{1}{20} \) Rupee.
A) 20 clips. This is derived from calculating the cost price per clip and then determining how many clips could be bought for one rupee at that cost price.
The sum of 3.\(\overline{26}\) - 2.\(\overline{14}\)+ 1.\(\overline{33}\) is:
2.\(\overline{45}\)
2.\(\overline{25}\)
2.\(\overline{15}\)
2.\(\overline{55}\)
To solve this problem, we need to convert the given repeating decimals into fractions, perform the addition and subtraction, and then convert the resulting fraction back into a repeating decimal or a mixed number with a repeating decimal part.
The question asks to identify one of the defence attorneys in the famous 'Sehgal–Dhillon–Shah Nawaz' trials, also known as the Indian National Army (INA) trials. This requires knowledge of prominent figures involved in the Indian independence movement and their specific roles in historical events.
C) Jawaharlal Nehru was indeed one of the prominent defence attorneys who represented the INA officers in the trials. He, along with other legal luminaries, formed the defence committee.
If (x + \( \frac{1}{x} \)) = 7, then (x - \( \frac{1}{x} \)) is equal to:
2\(\sqrt{5}\)
\(\sqrt{5}\)
3\(\sqrt{5}\)
5\(\sqrt{5}\)
This problem can be solved using algebraic identities. We are given the value of \((x + \frac{1}{x})\) and need to find the value of \((x - \frac{1}{x})\). A useful identity relating these two expressions is \((a+b)^2 - (a-b)^2 = 4ab\), or more specifically, \((a-b)^2 = (a+b)^2 - 4ab\).
Let the given expression be \(a = x\) and \(b = \frac{1}{x}\). We are given \((x + \frac{1}{x}) = 7\).
We need to find \((x - \frac{1}{x})\).
Recall the algebraic identity:
\[ (a - b)^2 = (a + b)^2 - 4ab \]Substitute \(a = x\) and \(b = \frac{1}{x}\) into the identity:
\[ \left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \cdot x \cdot \frac{1}{x} \]Simplify the right side of the equation:
\[ \left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \]Substitute the given value \((x + \frac{1}{x}) = 7\) into the equation:
\[ \left(x - \frac{1}{x}\right)^2 = (7)^2 - 4 \]Calculate the square and subtract:
\[ \left(x - \frac{1}{x}\right)^2 = 49 - 4 \]\[ \left(x - \frac{1}{x}\right)^2 = 45 \]To find \((x - \frac{1}{x})\), take the square root of both sides:
\[ x - \frac{1}{x} = \pm\sqrt{45} \]Simplify the square root of 45. We know that \(45 = 9 \times 5\):
\[ \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5} \]Therefore,
\[ x - \frac{1}{x} = \pm 3\sqrt{5} \]Since the options provided are positive, we take the positive value.
Correct Option: C) \(3\sqrt{5}\)
To determine which of the given numbers is NOT prime, we need to test each number for divisibility by prime numbers up to its square root. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. If a number can be divided evenly by any other number (besides 1 and itself), it is a composite number, not a prime number.
First, find the approximate square root of 811:
\[ \sqrt{811} \approx 28.47 \]We need to check for divisibility by prime numbers up to 28: 2, 3, 5, 7, 11, 13, 17, 19, 23.
Since 811 is not divisible by any prime number up to its square root, 811 is a prime number.
First, find the approximate square root of 317:
\[ \sqrt{317} \approx 17.8 \]We need to check for divisibility by prime numbers up to 17: 2, 3, 5, 7, 11, 13, 17.
Since 317 is not divisible by any prime number up to its square root, 317 is a prime number.
First, find the approximate square root of 817:
\[ \sqrt{817} \approx 28.58 \]We need to check for divisibility by prime numbers up to 28: 2, 3, 5, 7, 11, 13, 17, 19, 23.
Since 817 is divisible by 19 (and \(19 \times 43 = 817\)), 817 is a composite number, not a prime number.
First, find the approximate square root of 313:
\[ \sqrt{313} \approx 17.69 \]We need to check for divisibility by prime numbers up to 17: 2, 3, 5, 7, 11, 13, 17.
Since 313 is not divisible by any prime number up to its square root, 313 is a prime number.
Correct Option: C) 817 is not a prime number because it can be factored as \(19 \times 43\).
The question asks to identify the first Indian woman elected as a Fellow of the Royal Society (FRS) in April 2019. This is a factual recall question related to significant achievements in science.
B) Dr. Gagandeep Kang became the first Indian woman to be elected to the Fellow of the Royal Society (FRS) in April 2019. She is a clinical scientist known for her research on enteric infections and vaccines.
Solve the given equation.
\(\frac{3+4\div2\times3}{4+3\times2\div3}\) =
\( \frac{3}{5} \)
\( \frac{5}{4} \)
\( \frac{3}{4} \)
\( \frac{3}{2} \)
To solve the given expression, we need to follow the order of operations, commonly known as BODMAS/PEMDAS. This acronym stands for Brackets/Parentheses, Orders/Exponents, Division, Multiplication, Addition, and Subtraction. Operations at the same level (like division and multiplication, or addition and subtraction) are performed from left to right.
The given expression is:
\[ \frac{3+4\div2\times3}{4+3\times2\div3} \]
Thus, the solution to the given equation is \( \frac{3}{2} \).
Correct Option: D) \( \frac{3}{2} \)