This is a blood relations problem. We need to deduce the relationship between B and E by constructing a family tree based on the given statements.
Correct Option: A) Brother's wife
As established, B is married to D, and E is D's brother. Thus, B is E's brother's wife.
Evaluate: \(\frac{1}{\left(\frac{7}{8}+\frac{4}{6}\right)}\) ÷ \( \frac{6}{29} \)
3 \( \frac{5}{37} \)
3 \( \frac{10}{37} \)
3 \( \frac{13}{37} \)
3 \( \frac{1}{37} \)
To evaluate the given expression, we need to follow the order of operations (PEMDAS/BODMAS). First, simplify the sum inside the parentheses, then perform the division. Remember that dividing by a fraction is equivalent to multiplying by its reciprocal.
Simplify the sum inside the parentheses:
The expression inside the parentheses is \[\frac{7}{8} + \frac{4}{6}\]. To add these fractions, we need a common denominator. The least common multiple (LCM) of 8 and 6 is 24.
\[ \frac{7}{8} + \frac{4}{6} = \frac{7 \times 3}{8 \times 3} + \frac{4 \times 4}{6 \times 4} \]\[ = \frac{21}{24} + \frac{16}{24} \]\[ = \frac{21 + 16}{24} \]\[ = \frac{37}{24} \]Substitute the simplified sum back into the original expression:
The expression becomes:
\[ \frac{1}{\left(\frac{37}{24}\right)} \div \frac{6}{29} \]Simplify the reciprocal part:
The term \[\frac{1}{\left(\frac{37}{24}\right)}\] is the reciprocal of \( \frac{37}{24} \), which is \( \frac{24}{37} \).
So the expression is now:
\[ \frac{24}{37} \div \frac{6}{29} \]Perform the division:
Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of \( \frac{6}{29} \) is \( \frac{29}{6} \).
\[ \frac{24}{37} \times \frac{29}{6} \]Multiply the fractions and simplify:
We can simplify by canceling common factors before multiplying. Here, 24 and 6 have a common factor of 6.
\[ \frac{24}{37} \times \frac{29}{6} = \frac{4 \times 6}{37} \times \frac{29}{6} \]\[ = \frac{4 \times 29}{37} \]\[ = \frac{116}{37} \]Convert the improper fraction to a mixed number:
Divide 116 by 37:
\( 116 \div 37 = 3 \) with a remainder of \( 116 - (3 \times 37) = 116 - 111 = 5 \).
So, \[\frac{116}{37} = 3 \frac{5}{37}\].
A) \(3 \frac{5}{37}\) is the correct result obtained after evaluating the expression.
The question asks about India's recent imposition of anti-dumping duties on specific products from a particular country in March 2025. This is a current affairs question related to international trade and economic policy.
Correct Option: D) China — In March 2025, India imposed definitive anti-dumping duties on five products imported from China. This measure was taken to safeguard domestic industries against unfair trade practices, specifically the dumping of goods at prices lower than their normal value.
The question asks about India's literacy rate around the time of its independence in 1947. This requires knowledge of the socio-economic conditions and educational policies during British colonial rule, which generally neglected mass education.
C) Below 20% — Historical data indicates that India's literacy rate around 1947 was indeed very low, typically cited in the range of 12-18%. Therefore, "Below 20%" is the most accurate description.
This problem can be solved by setting up a system of two linear equations with two variables, representing the cost of a chair and a table. We will then solve these equations simultaneously to find the individual costs and finally calculate the cost of 16 chairs and 2 tables.
Let the cost of one chair be \(C\) and the cost of one table be \(T\).
From the first statement, "4 chairs and 4 tables cost ₹7,000", we can form the equation:
\[ 4C + 4T = 7000 \quad \text{(Equation 1)} \]From the second statement, "6 chairs and 5 tables cost ₹8,800", we can form the equation:
\[ 6C + 5T = 8800 \quad \text{(Equation 2)} \]To solve these equations, we can use the elimination method. Multiply Equation 1 by 5 and Equation 2 by 4 to make the coefficients of \(T\) equal:
Multiply Equation 1 by 5:
\[ 5 \times (4C + 4T) = 5 \times 7000 \]\[ 20C + 20T = 35000 \quad \text{(Equation 3)} \]Multiply Equation 2 by 4:
\[ 4 \times (6C + 5T) = 4 \times 8800 \]\[ 24C + 20T = 35200 \quad \text{(Equation 4)} \]Subtract Equation 3 from Equation 4:
\[ (24C + 20T) - (20C + 20T) = 35200 - 35000 \]\[ 4C = 200 \]\[ C = \frac{200}{4} \]\[ C = 50 \]Substitute the value of \(C = 50\) into Equation 1:
\[ 4(50) + 4T = 7000 \]\[ 200 + 4T = 7000 \]\[ 4T = 7000 - 200 \]\[ 4T = 6800 \]\[ T = \frac{6800}{4} \]\[ T = 1700 \]So, the cost of one chair is ₹50 and the cost of one table is ₹1700.
Now, we need to find the cost of 16 chairs and 2 tables:
\[ \text{Cost} = 16C + 2T \]\[ \text{Cost} = 16(50) + 2(1700) \]\[ \text{Cost} = 800 + 3400 \]\[ \text{Cost} = 4200 \]B) ₹4,200 is the correct cost for 16 chairs and 2 tables.
We will solve this problem by setting up equations based on the given conditions. Let the original cost price be \(C_P\). We will express the selling price in two scenarios and then equate them to find the value of \(C_P\).
First Scenario: Original Sale
Let the original cost price of the article be \(C_P\).
Akash sold the article at a loss of 17%. The selling price (\(S_P1\)) in this case is:
\[ S_P1 = C_P - 0.17 C_P = 0.83 C_P \]Second Scenario: Modified Conditions
If he had bought the item at 5% less, the new cost price (\(C_P'\)) would be:
\[ C_P' = C_P - 0.05 C_P = 0.95 C_P \]He sold it for ₹310 more than the original selling price. So, the new selling price (\(S_P2\)) is:
\[ S_P2 = S_P1 + 310 \]In this second scenario, he would have gained 20%. This means the new selling price \(S_P2\) is 20% more than the new cost price \(C_P'\):
\[ S_P2 = C_P' + 0.20 C_P' = 1.20 C_P' \]Formulating the Equation
Substitute the expressions for \(S_P1\) and \(C_P'\) into the equation for \(S_P2\):
From step 3:
\[ S_P2 = 0.83 C_P + 310 \]From step 4:
\[ S_P2 = 1.20 (0.95 C_P) \]Now, equate the two expressions for \(S_P2\):
\[ 0.83 C_P + 310 = 1.20 (0.95 C_P) \]Solving for \(C_P\)
Simplify the equation:
\[ 0.83 C_P + 310 = 1.14 C_P \]Subtract \(0.83 C_P\) from both sides:
\[ 310 = 1.14 C_P - 0.83 C_P \]\[ 310 = 0.31 C_P \]Divide by 0.31 to find \(C_P\):
\[ C_P = \frac{310}{0.31} \]\[ C_P = \frac{310}{\frac{31}{100}} \]\[ C_P = 310 \times \frac{100}{31} \]\[ C_P = 10 \times 100 \]\[ C_P = 1000 \]C) The cost price of the article is ₹1,000.
The question asks for the fundamental difference between input and output devices. To answer this, we need to understand the primary function of each type of device in the context of a computer system's interaction with a user.
D) Input devices enable users to send data into the system, while output devices communicate processed information back to the user. This option perfectly describes the core function of both input and output devices. Input devices facilitate the flow of information from the user to the computer for processing, and output devices facilitate the flow of processed information from the computer back to the user.
The question asks to identify the type of legislative bill that requires the prior approval of the President of India before being introduced in the Lok Sabha. This is a fundamental concept in the Indian parliamentary system, specifically related to financial legislation.
Correct Option: D) Money Bills must be introduced in the Lok Sabha only after obtaining the prior approval of the President of India, as per Article 110 of the Constitution.
Who was elected as the Chair of the Asian Productivity Organization (APO) for the 2025–26 term during the 67th GBM?
Agung Nur Rohmad
Jone Maritino Nemani
Dr. Indra Pradana Singawinata
Amardeep Singh Bhatia
The question asks to identify the individual elected as the Chair of the Asian Productivity Organization (APO) for the 2025–26 term during its 67th Governing Body Meeting (GBM). This requires knowledge of recent appointments within international organizations, specifically the APO.
D) Amardeep Singh Bhatia was elected as the Chair of the Asian Productivity Organization (APO) for the 2025–26 term during the 67th Governing Body Meeting. He is the Secretary, Department of Commerce, Ministry of Commerce & Industry, Government of India.
Rati travels from City A to City B. If Rati drives her car at \( \frac{1}{3} \) of her normal speed, then she reaches City B 50 minutes late. Find the time (in minutes) that Rati would take to travel from City A to City B if she drove at her normal speed.
This problem involves the relationship between speed, distance, and time. When the distance is constant, speed is inversely proportional to time. We can set up equations based on the given information about normal speed and reduced speed, and the corresponding times, to find the normal time taken.
When Rati drives at \( \frac{1}{3} \) of her normal speed, her new speed is \( S' = \frac{1}{3}S \).
She reaches City B 50 minutes late, so her new time is \( T' = T + 50 \) minutes.
Since the distance \(D\) is the same, we can write the equation for the second scenario:
\[ D = S' \times T' \]Now we have two expressions for \(D\):
Equation 1: \(D = S \times T\)
Equation 2: \(D = \frac{1}{3}S (T + 50)\)
Equate the two expressions for \(D\):
\[ S \times T = \frac{1}{3}S (T + 50) \]Since \(S\) cannot be zero (as Rati is traveling), we can divide both sides by \(S\):
\[ T = \frac{1}{3}(T + 50) \]Multiply both sides by 3 to eliminate the fraction:
\[ 3T = T + 50 \]Subtract \(T\) from both sides:
\[ 3T - T = 50 \]Simplify to find \(T\):
\[ 2T = 50 \]Divide by 2:
\[ T = \frac{50}{2} \]Therefore, \( T = 25 \) minutes.