Three friends arranged a party. Tanveer paid \( \frac{2}{3} \) as much as Yusuf paid. Yusuf paid \( \frac{1}{2} \) as much as Sachin paid. The fraction of the total expenditure by Yusuf was:
\( \frac{2}{11} \)
\( \frac{3}{11} \)
\( \frac{7}{11} \)
\( \frac{5}{11} \)
To find the fraction of the total expenditure paid by Yusuf, we need to express the amounts paid by Tanveer and Yusuf in terms of Sachin's payment. Then, we can find the ratio of their payments and calculate the total expenditure. Finally, we determine Yusuf's share of the total.
Let \(T\) be the amount Tanveer paid.
Let \(Y\) be the amount Yusuf paid.
Let \(S\) be the amount Sachin paid.
From the first statement: "Tanveer paid \( \frac{2}{3} \) as much as Yusuf paid."
\[ T = \frac{2}{3} Y \quad \text{(Equation 1)} \]From the second statement: "Yusuf paid \( \frac{1}{2} \) as much as Sachin paid."
\[ Y = \frac{1}{2} S \quad \text{(Equation 2)} \]From Equation 2, we already have \(Y\) in terms of \(S\):
\[ Y = \frac{1}{2} S \]Substitute \(Y = \frac{1}{2} S\) into Equation 1 to find \(T\) in terms of \(S\):
\[ T = \frac{2}{3} \left( \frac{1}{2} S \right) \]\[ T = \frac{2}{6} S \]\[ T = \frac{1}{3} S \]The total expenditure \(E\) is the sum of payments by Tanveer, Yusuf, and Sachin:
\[ E = T + Y + S \]Substitute the expressions in terms of \(S\):
\[ E = \frac{1}{3} S + \frac{1}{2} S + S \]To add these fractions, find a common denominator, which is 6:
\[ E = \frac{2}{6} S + \frac{3}{6} S + \frac{6}{6} S \]\[ E = \left( \frac{2+3+6}{6} \right) S \]\[ E = \frac{11}{6} S \]The fraction is \[\frac{\text{Yusuf's payment}}{\text{Total expenditure}}\]:
\[ \text{Fraction by Yusuf} = \frac{Y}{E} \]Substitute the expressions for \(Y\) and \(E\) in terms of \(S\):
\[ \text{Fraction by Yusuf} = \frac{\frac{1}{2} S}{\frac{11}{6} S} \]Cancel \(S\) from the numerator and denominator:
\[ \text{Fraction by Yusuf} = \frac{\frac{1}{2}}{\frac{11}{6}} \]To divide by a fraction, multiply by its reciprocal:
\[ \text{Fraction by Yusuf} = \frac{1}{2} \times \frac{6}{11} \]\[ \text{Fraction by Yusuf} = \frac{6}{22} \]\[ \text{Fraction by Yusuf} = \frac{3}{11} \]B) \( \frac{3}{11} \) is the correct fraction of the total expenditure paid by Yusuf.
This problem involves concepts of speed, time, and distance, specifically in the context of boats and streams. We need to calculate the upstream and downstream speeds, set up an equation based on the given time difference for two round trips, and then solve for the distance.
Downstream speed is the sum of the boat's speed and the current's speed.
\[ V_d = V_b + V_c = 15 + 3 = 18 \text{ km/h} \]Upstream speed is the difference between the boat's speed and the current's speed.
\[ V_u = V_b - V_c = 15 - 3 = 12 \text{ km/h} \]Let \(D\) be the distance between the two ends (in km).
Time = Distance / Speed
\[ T_d = \frac{D}{V_d} = \frac{D}{18} \text{ hours} \]The time taken for one upstream journey and one downstream journey is \(T_u - T_d\).
\[ T_u - T_d = \frac{D}{12} - \frac{D}{18} \]\[ T_u - T_d = \frac{3D - 2D}{36} = \frac{D}{36} \text{ hours} \]The problem states the difference between the time taken for upstream and downstream to complete two trips. This means two round trips. So, the total time difference is twice the difference for one round trip.
\[ \text{Total time difference} = 2 \times \left( \frac{D}{12} - \frac{D}{18} \right) = 2 \times \frac{D}{36} = \frac{D}{18} \text{ hours} \]10 minutes = \(\frac{10}{60}\) hours = \(\frac{1}{6}\) hours.
Equate the total time difference calculated in step 8 to the given time difference in hours.
\[ \frac{D}{18} = \frac{1}{6} \]\[ D = \frac{18}{6} \]\[ D = 3 \text{ km} \]D) 3 km is the correct distance between the two ends, as calculated by setting up the equation for the time difference over two round trips.
Simplify: \( 1.8 + 2 \times \frac{3}{2} \times \frac{1}{2}\left\{ \frac{2}{5} \div \left( \frac{3}{5} \times \frac{2}{3} \right) \times \frac{3}{2} - 1 \right\} \)
To simplify the given expression, we will follow the order of operations (BODMAS/PEMDAS): Brackets (Parentheses), Orders (Exponents), Division and Multiplication (from left to right), Addition and Subtraction (from left to right). We will start by simplifying the innermost brackets and work our way outwards.
First, simplify the expression inside the innermost parentheses: \[\left( \frac{3}{5} \times \frac{2}{3} \right)\]
\[ \frac{3}{5} \times \frac{2}{3} = \frac{3 \times 2}{5 \times 3} = \frac{6}{15} = \frac{2}{5} \]Next, substitute this value back into the curly braces and simplify the division:
\[ \frac{2}{5} \div \left( \frac{2}{5} \right) = \frac{2}{5} \times \frac{5}{2} = 1 \]Now, the expression inside the curly braces becomes:
\[ \left\{ 1 \times \frac{3}{2} - 1 \right\} \]Perform the multiplication inside the curly braces:
\[ 1 \times \frac{3}{2} = \frac{3}{2} \]Perform the subtraction inside the curly braces:
\[ \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]Substitute this value back into the main expression:
\[ 1.8 + 2 \times \frac{3}{2} \times \frac{1}{2} \left( \frac{1}{2} \right) \]Perform the multiplications from left to right:
\[ 2 \times \frac{3}{2} = 3 \]\[ 3 \times \frac{1}{2} = \frac{3}{2} \]\[ \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \]Finally, perform the addition:
\[ 1.8 + \frac{3}{4} \]Convert \(\frac{3}{4}\) to a decimal:
\[ \frac{3}{4} = 0.75 \]Add the decimal numbers:
\[ 1.8 + 0.75 = 2.55 \]D) The simplified value of the expression is \(2.55\).
The question asks about the route of the first passenger train in India. This is a factual question related to Indian history, specifically the history of railways.
Correct Option: D) Bombay and Thane
The first passenger train in India ran on April 16, 1853, from Bori Bunder (Bombay) to Thane, covering a distance of 34 km. It was hauled by three locomotives: Sindh, Sultan, and Sahib.
The question asks to identify the Chief Justice of India as of November 2020. This requires knowledge of the individuals who held this position around that time.
Correct Option: A) S Arvind Bobde was the 47th Chief Justice of India, serving from November 18, 2019, to April 23, 2021. Thus, he was the Chief Justice of India in November 2020.
The question asks for the location of INTERPOL's headquarters. This is a factual question about international organizations and their administrative centers. Knowledge of major international bodies and their headquarters is essential for competitive exams.
Correct Option: B) France
INTERPOL, the world's largest international police organization, has its headquarters in Lyon, France. It facilitates international police cooperation and crime control.
The question asks about a specific initiative launched by the Reserve Bank of India (RBI) to enhance the safety and security of e-payment options. This requires knowledge of RBI's policy initiatives related to digital payments.
Correct Option: B) Vision 2021
The Reserve Bank of India (RBI) launched the "Payment and Settlement Systems in India: Vision 2019-2021" document. This vision aimed to make India a cash-lite society by ensuring a safe, secure, convenient, and affordable e-payment system for everyone. Therefore, 'Vision 2021' (referring to the end year of the vision period) is the correct answer.
To find the length of the wire, we can visualize the situation as a right-angled triangle. The difference in heights of the poles will form one side of the triangle, and the wire will be the hypotenuse. We can use trigonometric ratios, specifically the sine function, as we are given an angle and need to find the hypotenuse.
Let the height of the first pole be \(h_1 = 20\) meters and the height of the second pole be \(h_2 = 14\) meters.
The wire connects the tops of the two poles. Imagine a horizontal line drawn from the top of the shorter pole to the taller pole. This creates a right-angled triangle.
The vertical side of this right-angled triangle will be the difference in the heights of the poles:
\[ \text{Difference in height} = h_1 - h_2 = 20 - 14 = 6 \text{ meters} \]Let the length of the wire be \(L\). The wire makes an angle of \(30^\circ\) with the horizontal. In the right-angled triangle formed, the difference in height (6 meters) is the side opposite to the \(30^\circ\) angle, and the length of the wire \(L\) is the hypotenuse.
We can use the sine trigonometric ratio, which is defined as:
\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]Substitute the known values:
\[ \sin(30^\circ) = \frac{6}{L} \]We know that \(\sin(30^\circ) = \frac{1}{2}\).
So, the equation becomes:
\[ \frac{1}{2} = \frac{6}{L} \]Now, solve for \(L\):
\[ L = 6 \times 2 \]\[ L = 12 \text{ meters} \]Thus, the length of the wire is 12 meters.
The length of the wire is calculated to be 12 meters using the sine function in the right-angled triangle formed by the poles and the wire.
We need to find the greater of two numbers. We are given two conditions: the difference between the numbers is 5, and a ratio involving modified versions of these numbers is 1:2. We will set up a system of two linear equations based on these conditions and solve for the numbers.
Let the two numbers be \(x\) and \(y\). Assume \(x\) is the greater number and \(y\) is the smaller number.
From the first condition, "The difference between two numbers is 5", we can write the equation:
\[ x - y = 5 \quad \text{(Equation 1)} \]From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = x - 5 \]From the second condition, "If 25 is subtracted from the smaller number and 20 is added to the greater number the ratio becomes 1:2".
The ratio of these modified numbers is 1:2. So, we can write:
\[ \frac{y - 25}{x + 20} = \frac{1}{2} \]Substitute \(y = x - 5\) into the ratio equation:
\[ \frac{(x - 5) - 25}{x + 20} = \frac{1}{2} \]\[ \frac{x - 30}{x + 20} = \frac{1}{2} \]Cross-multiply to solve for \(x\):
\[ 2(x - 30) = 1(x + 20) \]\[ 2x - 60 = x + 20 \]Rearrange the terms to find \(x\):
\[ 2x - x = 20 + 60 \]\[ x = 80 \]Since \(x\) is the greater number, the greater number is 80.
To verify, we can find the smaller number \(y\):
\[ y = x - 5 = 80 - 5 = 75 \]Check the ratio condition:
The ratio is \[\frac{50}{100} = \frac{1}{2}\], which matches the given condition.
C) 80 is the greater number.