This problem can be solved using the Remainder Theorem. The Remainder Theorem states that if a polynomial \( P(x) \) is divided by \( (x - a) \), then the remainder is \( P(a) \). We are given a polynomial and a divisor, along with the remainder in terms of \( p \). We will substitute the value from the divisor into the polynomial and equate it to the given remainder to find the value of \( p \).
The given polynomial is \( P(x) = x^{4} - px^{3} + 2x^{2} - 5x + 8 \).
The divisor is \( (x - 1) \).
According to the Remainder Theorem, when \( P(x) \) is divided by \( (x - a) \), the remainder is \( P(a) \). Here, \( a = 1 \).
So, the remainder is \( P(1) \).
Substitute \( x = 1 \) into the polynomial \( P(x) \):
\[ P(1) = (1)^{4} - p(1)^{3} + 2(1)^{2} - 5(1) + 8 \]Simplify the expression:
\[ P(1) = 1 - p + 2 - 5 + 8 \]\[ P(1) = (1 + 2 - 5 + 8) - p \]\[ P(1) = (11 - 5) - p \]\[ P(1) = 6 - p \]We are given that the remainder is \( 2p \).
Therefore, we set our calculated remainder equal to \( 2p \):
\[ 6 - p = 2p \]Add \( p \) to both sides of the equation:
\[ 6 = 2p + p \]\[ 6 = 3p \]Divide by 3:
\[ p = \frac{6}{3} \]\[ p = 2 \]Correct Option: C) 2
The question asks to identify the person responsible for inviting Mahatma Gandhi to Champaran in 1917. This requires knowledge of the historical events leading up to the Champaran Satyagraha.
A) Raj Kumar Shukla was the peasant leader who played a crucial role in bringing Mahatma Gandhi to Champaran in 1917, thereby initiating the Champaran Satyagraha.
The question asks to identify the organization that jointly developed the 'SAMYUKTA' electronic warfare program with the Indian Army. This requires knowledge of India's defense research and development organizations and their projects.
B) DRDO — The Defence Research and Development Organisation (DRDO) is the premier agency of the Government of India, charged with the military's research and development. It is well-known for developing various indigenous defense systems, including the 'SAMYUKTA' electronic warfare program, in collaboration with the Indian Army and other partners.
The problem involves calculating the total amount paid for a washing machine, considering an initial down payment and the remaining balance paid after one year with simple interest. We need to find the principal amount on which interest is charged, calculate the simple interest for one year, and then add this interest to the remaining balance to find the final payment. Finally, sum the initial payment and the final payment to get the total amount.
Identify the total cost of the washing machine:
Total cost \( = \text{₹}12,000 \)
Identify the initial payment (down payment):
Initial payment \( = \text{₹}2,000 \)
Calculate the remaining balance to be paid:
Remaining balance \( = \text{Total cost} - \text{Initial payment} \)
Remaining balance \( = \text{₹}12,000 - \text{₹}2,000 = \text{₹}10,000 \)
This remaining balance is the principal amount on which simple interest will be charged.
Identify the rate of simple interest and time period:
Rate of interest \( (R) = 10\% \) per year
Time period \( (T) = 1 \) year
Principal amount \( (P) = \text{₹}10,000 \)
Calculate the simple interest (SI) on the remaining balance:
The formula for simple interest is:
\[\text{SI} = \frac{P \times R \times T}{100}\]Substitute the values:
\[\text{SI} = \frac{10,000 \times 10 \times 1}{100}\] \[\text{SI} = \frac{100,000}{100}\] \[\text{SI} = \text{₹}1,000\]Calculate the amount paid after 1 year:
Amount paid after 1 year \( = \text{Remaining balance} + \text{Simple Interest} \)
Amount paid after 1 year \( = \text{₹}10,000 + \text{₹}1,000 = \text{₹}11,000 \)
Calculate the total amount paid for the washing machine:
Total amount paid \( = \text{Initial payment} + \text{Amount paid after 1 year} \)
Total amount paid \( = \text{₹}2,000 + \text{₹}11,000 = \text{₹}13,000 \)
A) ₹13,000 is the correct total amount paid by the man for the washing machine.
The question asks to identify the device in automobiles that measures the distance travelled. We need to recall the functions of common automobile instruments.
Correct Option: A) Odometer
An odometer is an instrument used for measuring the distance travelled by a vehicle, such as a bicycle or car. It is typically found on the dashboard of an automobile.
To solve this problem, we need to first determine the cost price of one mango. We are given the selling price and the loss percentage. Once we have the cost price, we can calculate the selling price required to achieve a 20% gain. Finally, we can find out how many mangoes should be sold for ₹28 at this new selling price.
Calculate the selling price of one mango:
Given that 50 mangoes are sold for ₹30, the selling price (SP) of one mango is:
\[ \text{SP per mango} = \frac{\text{Total Selling Price}}{\text{Number of Mangoes}} = \frac{30}{50} = \text{₹}0.60 \]Calculate the cost price of one mango:
The person loses 10% by selling at ₹0.60 per mango. This means the selling price is 90% of the cost price (CP).
\[ \text{SP} = \text{CP} \times \left(1 - \frac{\text{Loss Percentage}}{100}\right) \]\[ 0.60 = \text{CP} \times \left(1 - \frac{10}{100}\right) \]\[ 0.60 = \text{CP} \times \frac{90}{100} \]\[ \text{CP} = \frac{0.60 \times 100}{90} = \frac{60}{90} = \frac{2}{3} \text{₹} \]Calculate the new selling price per mango for a 20% gain:
To gain 20%, the new selling price (New SP) should be 120% of the cost price.
\[ \text{New SP} = \text{CP} \times \left(1 + \frac{\text{Gain Percentage}}{100}\right) \]\[ \text{New SP} = \frac{2}{3} \times \left(1 + \frac{20}{100}\right) \]\[ \text{New SP} = \frac{2}{3} \times \frac{120}{100} \]\[ \text{New SP} = \frac{2}{3} \times \frac{6}{5} = \frac{12}{15} = \frac{4}{5} = \text{₹}0.80 \]Calculate the number of mangoes to sell for ₹28 to gain 20%:
If each mango is sold for ₹0.80, then for ₹28, the number of mangoes sold will be:
\[ \text{Number of Mangoes} = \frac{\text{Total Amount}}{\text{New SP per mango}} = \frac{28}{0.80} \]\[ \text{Number of Mangoes} = \frac{28}{\frac{4}{5}} = 28 \times \frac{5}{4} = 7 \times 5 = 35 \]A) 35
The question asks to identify the constitutional head of the Executive of the Union of India. This requires knowledge of the fundamental structure of the Indian government as outlined in the Constitution.
B) President is the correct answer. The President of India is the constitutional head of the Executive of the Union, symbolizing the nation and exercising powers in a formal capacity, though generally on the aid and advice of the Council of Ministers.
The value of 2 sin 30° + 5 cos 60° + 3 sin 180° is:
\( \frac{1}{2} \)
\( \frac{7}{2} \)
-\( \frac{1}{2} \)
\( \frac{9}{2} \)
To solve this problem, we need to recall the standard trigonometric values for specific angles: \(\sin 30^\circ\), \(\cos 60^\circ\), and \(\sin 180^\circ\). Once these values are known, we substitute them into the given expression and perform the arithmetic operations.
The given expression is \(2 \sin 30^\circ + 5 \cos 60^\circ + 3 \sin 180^\circ\).
Substitute the known values:
\[ 2 \left(\frac{1}{2}\right) + 5 \left(\frac{1}{2}\right) + 3 (0) \]To add \(1\) and \(\frac{5}{2}\), we can write \(1\) as \(\frac{2}{2}\):
\[ \frac{2}{2} + \frac{5}{2} = \frac{2+5}{2} = \frac{7}{2} \]Thus, the value of the expression is \(\frac{7}{2}\).
Correct Option: B) \( \frac{7}{2} \)
The adjacent sides of a parallelogram are 4a and 3a. If the angle between them is 60°, then one of the diagonals of the parallelogram will be:
\(\sqrt{13}\) a
2\(\sqrt{3}\) a
5\(\sqrt{3}\) a
3\(\sqrt{3}\)a
To find the length of the diagonals of a parallelogram, we can use the Law of Cosines. A parallelogram has two pairs of equal adjacent sides and two pairs of equal opposite angles. If one angle is given, the other angle can be determined. We will apply the Law of Cosines to the triangles formed by the diagonals and the sides of the parallelogram.
Let the adjacent sides of the parallelogram be \(a_1 = 4a\) and \(a_2 = 3a\).
Let the angle between these sides be \(\theta = 60^\circ\). In a parallelogram, consecutive angles are supplementary, so the other angle between the sides will be \(180^\circ - 60^\circ = 120^\circ\).
Let \(d_1\) and \(d_2\) be the lengths of the two diagonals.
Consider the triangle formed by sides \(4a\), \(3a\), and one diagonal \(d_1\). The angle opposite to \(d_1\) is \(\theta = 60^\circ\). Using the Law of Cosines:
\[ d_1^2 = (4a)^2 + (3a)^2 - 2(4a)(3a)\cos(60^\circ) \]\[ d_1^2 = 16a^2 + 9a^2 - 24a^2 \left(\frac{1}{2}\right) \]\[ d_1^2 = 25a^2 - 12a^2 \]\[ d_1^2 = 13a^2 \]\[ d_1 = \sqrt{13a^2} = \sqrt{13}a \]Now, consider the triangle formed by sides \(4a\), \(3a\), and the other diagonal \(d_2\). The angle opposite to \(d_2\) is \(180^\circ - 60^\circ = 120^\circ\). Using the Law of Cosines:
\[ d_2^2 = (4a)^2 + (3a)^2 - 2(4a)(3a)\cos(120^\circ) \]\[ d_2^2 = 16a^2 + 9a^2 - 24a^2 \left(-\frac{1}{2}\right) \]\[ d_2^2 = 25a^2 + 12a^2 \]\[ d_2^2 = 37a^2 \]\[ d_2 = \sqrt{37a^2} = \sqrt{37}a \]The question asks for "one of the diagonals". Comparing our calculated diagonal lengths with the given options, \(\sqrt{13}a\) matches option A.
A) \(\sqrt{13}a\) is one of the calculated diagonal lengths, matching the result from applying the Law of Cosines to the triangle with the \(60^\circ\) angle.
The numerator of a fraction is increased by 20% and the denominator is decreased by 50%. If the resultant fraction is \( \frac{5}{6} \), then what will be the original fraction?
\( \frac{25}{84} \)
\( \frac{25}{72} \)
\( \frac{72}{25} \)
\( \frac{25}{12} \)
We need to find the original fraction given the changes to its numerator and denominator and the resulting fraction. We can represent the original fraction algebraically and then set up an equation based on the given information.
Let the original fraction be \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.
The numerator is increased by 20%. An increase of 20% means the new numerator will be \( x + 0.20x = 1.20x \).
The denominator is decreased by 50%. A decrease of 50% means the new denominator will be \( y - 0.50y = 0.50y \).
The resultant fraction is given as \( \frac{5}{6} \). So, we can set up the equation:
\[ \frac{1.20x}{0.50y} = \frac{5}{6} \]Simplify the left side of the equation:
\[ \frac{1.20}{0.50} \cdot \frac{x}{y} = \frac{5}{6} \] \[ \frac{120}{50} \cdot \frac{x}{y} = \frac{5}{6} \] \[ \frac{12}{5} \cdot \frac{x}{y} = \frac{5}{6} \]Now, solve for the original fraction \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{5}{6} \div \frac{12}{5} \] \[ \frac{x}{y} = \frac{5}{6} \times \frac{5}{12} \] \[ \frac{x}{y} = \frac{25}{72} \]Correct Option: B) \( \frac{25}{72} \)