This problem can be solved by working backward from the known number of passengers at the third station. We will define variables for the number of passengers at each stage and set up equations based on the information given about dropping and taking in passengers at each stop.
Let \(P\) be the initial number of passengers when the train started.
At the first stop:
At the second stop:
We are given that \(P_2 = 248\). Now we can work backward to find \(P\).
Substitute \(P_2 = 248\) into the equation for the second stop:
\[ 248 = \frac{1}{2}P_1 + 12 \]Solve for \(P_1\):
\[ 248 - 12 = \frac{1}{2}P_1 \]\[ 236 = \frac{1}{2}P_1 \]\[ P_1 = 236 \times 2 \]\[ P_1 = 472 \]Now substitute \(P_1 = 472\) into the equation for the first stop:
\[ 472 = \frac{2}{3}P + 96 \]Solve for \(P\):
\[ 472 - 96 = \frac{2}{3}P \]\[ 376 = \frac{2}{3}P \]\[ P = \frac{376 \times 3}{2} \]\[ P = 188 \times 3 \]\[ P = 564 \]Thus, the train started with 564 passengers.
A) 564 is the correct answer, as calculated by working backward through the given conditions.
A person sells his table at a profit of 12\( \frac{1}{2} \)% and chair at a loss of 8\( \frac{1}{3} \)% but on the whole he gains ₹25. On the other hand, if he sells the table at a loss of 8\( \frac{1}{3} \)% and the chair at a profit of 12\( \frac{1}{2} \)% then he neither gains nor loses. Find the cost price of the table.
We are given two scenarios involving the profit/loss percentages on a table and a chair, and the overall gain/loss. We need to find the cost price of the table. We will set up a system of linear equations based on the given information and solve for the cost prices.
Let \(C_T\) be the cost price of the table and \(C_C\) be the cost price of the chair.
Profit on table: \[12\frac{1}{2}\% = \frac{25}{2}\% = \frac{25}{200} = \frac{1}{8}\]
Loss on chair: \[8\frac{1}{3}\% = \frac{25}{3}\% = \frac{25}{300} = \frac{1}{12}\]
In the first scenario, the person sells the table at a profit of \(12\frac{1}{2}\%\) and the chair at a loss of \(8\frac{1}{3}\%\), gaining ₹25 overall.
\[ \frac{1}{8} C_T - \frac{1}{12} C_C = 25 \quad \text{(Equation 1)} \]In the second scenario, if the table is sold at a loss of \(8\frac{1}{3}\%\) and the chair at a profit of \(12\frac{1}{2}\%\), there is neither gain nor loss.
\[ -\frac{1}{12} C_T + \frac{1}{8} C_C = 0 \quad \text{(Equation 2)} \]From Equation 2, we can express \(C_C\) in terms of \(C_T\):
\[ \frac{1}{8} C_C = \frac{1}{12} C_T \]\[ C_C = \frac{8}{12} C_T \]\[ C_C = \frac{2}{3} C_T \]Substitute \(C_C = \frac{2}{3} C_T\) into Equation 1:
\[ \frac{1}{8} C_T - \frac{1}{12} \left(\frac{2}{3} C_T\right) = 25 \]\[ \frac{1}{8} C_T - \frac{2}{36} C_T = 25 \]\[ \frac{1}{8} C_T - \frac{1}{18} C_T = 25 \]Find a common denominator for 8 and 18, which is 72.
\[ \frac{9}{72} C_T - \frac{4}{72} C_T = 25 \]\[ \frac{5}{72} C_T = 25 \]\[ C_T = \frac{25 \times 72}{5} \]\[ C_T = 5 \times 72 \]\[ C_T = 360 \]The cost price of the table is ₹360.
B) ₹360 is the correct cost price of the table, as calculated from the system of equations.
This problem can be solved using the concept of 'Work and Time', specifically the principle that the total work done is constant. The total work is proportional to the number of men multiplied by the number of days they work. We can set up an equation based on the initial plan and the actual scenario to find the original number of men.
C) 30 men
To simplify the given expression, we will convert all decimal numbers into fractions to make the division and multiplication operations easier. We will then perform the operations in the correct order (division first, then multiplication from left to right).
Write the given expression:
\[ 0.25 \div 0.0025 \times 0.025 \times 2.5 \]Convert each decimal to a fraction:
Substitute these fractions back into the expression:
\[ \frac{25}{100} \div \frac{25}{10000} \times \frac{25}{1000} \times \frac{25}{10} \]Perform the division first. Dividing by a fraction is the same as multiplying by its reciprocal:
\[ \frac{25}{100} \times \frac{10000}{25} \times \frac{25}{1000} \times \frac{25}{10} \]Cancel out the common term \(25\) from the first two fractions:
\[ \frac{\cancel{25}}{100} \times \frac{10000}{\cancel{25}} = \frac{10000}{100} = 100 \]Now the expression becomes:
\[ 100 \times \frac{25}{1000} \times \frac{25}{10} \]Perform the multiplication from left to right. First, multiply \(100\) by \(\frac{25}{1000}\):
\[ 100 \times \frac{25}{1000} = \frac{100 \times 25}{1000} = \frac{2500}{1000} = \frac{25}{10} = 2.5 \]Now the expression is:
\[ 2.5 \times \frac{25}{10} \]We know \(\frac{25}{10} = 2.5\). So, the final multiplication is:
\[ 2.5 \times 2.5 = 6.25 \]B) The simplified value of the expression is \(6.25\).
The question asks to identify which of the given options is NOT a flagship programme of India. Flagship programmes are major government initiatives designed to address specific national priorities and often involve significant funding and widespread implementation. We need to evaluate each option to determine its status as a flagship programme.
Correct Option: C) Desert Development Programme is NOT considered a flagship programme in the same category as the others listed, which are broad national initiatives.
The problem asks for the unit's place digit of a positive number \(N\) that leaves a remainder of 3 when divided by 5. We can use the definition of division with remainder to express \(N\) and then determine the possible unit's digits.
Correct Option: B) 3 or 8
The question asks for the year in which the Maulana Abul Kalam Azad (MAKA) Trophy was instituted. This is a factual question related to sports awards in India, requiring knowledge of specific historical dates.
The Maulana Abul Kalam Azad (MAKA) Trophy is a prestigious award given by the Ministry of Youth Affairs and Sports, Government of India, to the university that has shown overall top performance in sports. It was instituted in the year 1956-57. This trophy is named after India's first Education Minister, Maulana Abul Kalam Azad, and is one of the oldest national sports awards in India.
This problem involves direct proportionality. The amount of paper required is directly proportional to the number of copies and the number of sheets per book. We can set up a ratio or find the paper required per unit (per copy per sheet) and then scale it up for the new requirements.
The number of reams of paper (\(R\)) is directly proportional to the number of copies (\(C\)) and the number of sheets per book (\(S\)).
\[ R \propto C \times S \]\[ R = k \times C \times S \]where \(k\) is the constant of proportionality.We can simplify the multiplication:
\[ R_2 = \frac{26}{1500} \times 750 \times 13 \]\[ R_2 = 26 \times \frac{750}{1500} \times 13 \]\[ R_2 = 26 \times \frac{1}{2} \times 13 \]\[ R_2 = 13 \times 13 \]\[ R_2 = 169 \]A) 169 reams. Based on the calculations, 169 reams of paper are required.
The problem requires us to substitute the given symbols with their corresponding mathematical operators and then evaluate each equation to find the true one. We need to follow the order of operations (BODMAS/PEMDAS) while evaluating.
First, let's list the symbol-to-operator mappings:
Now, we will evaluate each option by substituting the symbols and performing the calculations.
Substitute the symbols:
\[ (81 - 4) \times 3 = 8 \div 15 \times 7 \]\[ 77 \times 3 = \frac{8}{15} \times 7 \]\[ 231 = \frac{56}{15} \]\[ 231 = 3.733... \]This is false.Substitute the symbols:
\[ (81 - 4) \times 3 = 4 + 16 - 3 \]\[ 77 \times 3 = 20 - 3 \]\[ 231 = 17 \]This is false.Substitute the symbols:
\[ (15 - 4) \times 3 = (60 \div 2) + 3 \]\[ 11 \times 3 = 30 + 3 \]\[ 33 = 33 \]This is true.Substitute the symbols:
\[ (6 + 2) \div 2 = (2 \times 8) - 1 \]\[ 8 \div 2 = 16 - 1 \]\[ 4 = 15 \]This is false.C) After substituting the symbols and evaluating both sides of the equation, we found that \(33 = 33\), which is a true statement.
The question asks for the unit used to measure the intensity of sound. We need to identify the correct unit from the given options and understand what each of the other units measures.
C) Decibel is the unit used to measure the intensity of sound. It is a logarithmic unit that expresses the ratio of a sound's intensity to a reference intensity, reflecting how humans perceive loudness.