If a, b are the roots of the equation
x2 − 5x + 6 = 0, then a2 + b2 − ab is:
The problem asks us to find the value of an expression involving the roots of a quadratic equation. We can use Vieta's formulas to relate the coefficients of the quadratic equation to the sum and product of its roots. Then, we can express the given expression in terms of the sum and product of the roots.
The given quadratic equation is \(x^2 - 5x + 6 = 0\). Let its roots be \(a\) and \(b\).
For a quadratic equation of the form \(Ax^2 + Bx + C = 0\), the sum of the roots is \(-\frac{B}{A}\) and the product of the roots is \(\frac{C}{A}\).
In our equation, \(A=1\), \(B=-5\), and \(C=6\).
We need to find the value of \(a^2 + b^2 - ab\).
We know the identity: \((a+b)^2 = a^2 + b^2 + 2ab\).
From this, we can write: \(a^2 + b^2 = (a+b)^2 - 2ab\).
Substitute this into the expression we need to evaluate:
\[ a^2 + b^2 - ab = ((a+b)^2 - 2ab) - ab \]\[ a^2 + b^2 - ab = (a+b)^2 - 3ab \]We found \(a+b = 5\) and \(ab = 6\).
\[ (a+b)^2 - 3ab = (5)^2 - 3(6) \]\[ = 25 - 18 \]\[ = 7 \]C) The value of \(a^2 + b^2 - ab\) is 7.
This problem involves setting up equations based on the given information about incomes, savings, and expenditures of two individuals, A and B. We will use variables to represent their annual income, savings, and expenditures, and then solve the system of equations to find the annual income.
Let the annual income of A and B be \(I\). Since their annual incomes are equal, both A and B earn \(I\) per year.
A saves one-fifth of his income every year. So, A's annual savings (\(S_A\)) and annual expenditure (\(E_A\)) are:
\[ S_A = \frac{1}{5} I \]\[ E_A = I - S_A = I - \frac{1}{5} I = \frac{4}{5} I \]B spends Rs 80,000 more than A every year. So, B's annual expenditure (\(E_B\)) is:
\[ E_B = E_A + 80000 \]\[ E_B = \frac{4}{5} I + 80000 \]B's annual savings (\(S_B\)) can be calculated as income minus expenditure:
\[ S_B = I - E_B \]\[ S_B = I - \left( \frac{4}{5} I + 80000 \right) \]\[ S_B = I - \frac{4}{5} I - 80000 \]\[ S_B = \frac{1}{5} I - 80000 \]In 4 years, B runs into a debt of Rs 2,20,000. This means B's total savings over 4 years is \(-\text{Rs } 2,20,000\). So, 4 times B's annual savings equals this debt:
\[ 4 \times S_B = -220000 \]Substitute the expression for \(S_B\) from step 4 into the equation from step 5:
\[ 4 \times \left( \frac{1}{5} I - 80000 \right) = -220000 \]Solve for \(I\):
\[ \frac{4}{5} I - 4 \times 80000 = -220000 \]\[ \frac{4}{5} I - 320000 = -220000 \]\[ \frac{4}{5} I = -220000 + 320000 \]\[ \frac{4}{5} I = 100000 \]\[ I = \frac{100000 \times 5}{4} \]\[ I = 25000 \times 5 \]\[ I = 125000 \]Thus, the annual income of each of A and B is Rs 1,25,000.
Correct Option: D) Rs 1,25,000
The question asks about the number of elements in specific periods of the modern periodic table. This can be determined by understanding the electron shell filling rules and the structure of the periodic table.
Understanding Periods: Each period in the modern periodic table corresponds to the principal quantum number (\(n\)) of the outermost electron shell being filled.
Period 3: In the third period, the 3s and 3p subshells are filled. The 3s subshell can hold 2 electrons and the 3p subshell can hold 6 electrons. Thus, the total number of elements is \(2 + 6 = 8\).
Period 4: In the fourth period, the 4s, 3d, and 4p subshells are filled. The 4s subshell holds 2 electrons, the 3d subshell holds 10 electrons, and the 4p subshell holds 6 electrons. Thus, the total number of elements is \(2 + 10 + 6 = 18\).
Period 5: In the fifth period, the 5s, 4d, and 5p subshells are filled. The 5s subshell holds 2 electrons, the 4d subshell holds 10 electrons, and the 5p subshell holds 6 electrons. Thus, the total number of elements is \(2 + 10 + 6 = 18\).
Period 6: In the sixth period, the 6s, 4f, 5d, and 6p subshells are filled. The 6s subshell holds 2 electrons, the 4f subshell holds 14 electrons (lanthanides), the 5d subshell holds 10 electrons, and the 6p subshell holds 6 electrons. Thus, the total number of elements is \(2 + 14 + 10 + 6 = 32\).
Period 7: In the seventh period, the 7s, 5f, 6d, and 7p subshells are filled. The 7s subshell holds 2 electrons, the 5f subshell holds 14 electrons (actinides), the 6d subshell holds 10 electrons, and the 7p subshell holds 6 electrons. Thus, the total number of elements is \(2 + 14 + 10 + 6 = 32\).
Summary: The number of elements in the third, fourth, fifth, sixth, and seventh periods are 8, 18, 18, 32, and 32, respectively.
Correct Option: C) 8, 18, 18, 32 and 32, respectively
Abaz and Aman can complete painting a building in 16 days and 12 days, respectively. If Abaz starts the work and they work on alternate days, then the building painting will be completed in how many days?
This problem involves calculating the time taken to complete a task when two individuals work on alternate days. We will first determine the individual work rates, then calculate the work done in a 2-day cycle, and finally find the total number of days required to complete the entire work.
Abaz can complete the work in 16 days. So, Abaz's 1-day work is:
\[ \text{Abaz's 1-day work} = \frac{1}{16} \]Aman can complete the work in 12 days. So, Aman's 1-day work is:
\[ \text{Aman's 1-day work} = \frac{1}{12} \]Since Abaz starts the work and they work on alternate days:
Work done in 2 days (1 cycle) = Abaz's 1-day work + Aman's 1-day work
\[ \text{Work in 2 days} = \frac{1}{16} + \frac{1}{12} \]To add these fractions, find a common denominator, which is 48.
\[ \text{Work in 2 days} = \frac{3}{48} + \frac{4}{48} = \frac{7}{48} \]Let \( N \) be the number of 2-day cycles. We need to find how many cycles are needed to complete most of the work without exceeding the total work.
We want to find \( N \) such that \( N \times \frac{7}{48} \le 1 \).
Multiply both sides by 48:
\[ 7N \le 48 \]\[ N \le \frac{48}{7} \]\[ N \le 6.85... \]So, we consider 6 full cycles. After 6 cycles (12 days), the work completed will be:
\[ \text{Work completed after 6 cycles} = 6 \times \frac{7}{48} = \frac{42}{48} = \frac{7}{8} \]Remaining work = Total work - Work completed after 6 cycles
\[ \text{Remaining work} = 1 - \frac{7}{8} = \frac{1}{8} \]After 12 days (6 cycles), it's the start of the 13th day. On the 13th day, Abaz works.
Abaz's 1-day work = \( \frac{1}{16} \).
Since the remaining work is \( \frac{1}{8} \), and Abaz can do \( \frac{1}{16} \) of the work in one day, Abaz will take:
\[ \text{Time taken by Abaz} = \frac{\text{Remaining work}}{\text{Abaz's 1-day work}} = \frac{\frac{1}{8}}{\frac{1}{16}} = \frac{1}{8} \times 16 = 2 \text{ days} \]However, Abaz only needs to complete \( \frac{1}{8} \) of the work. His daily rate is \( \frac{1}{16} \). So, the fraction of a day Abaz needs is:
\[ \text{Fraction of day for Abaz} = \frac{\text{Remaining work}}{\text{Abaz's 1-day work}} = \frac{1/8}{1/16} = \frac{1}{8} \times 16 = 2 \]This calculation is incorrect. Let's re-evaluate. Abaz's 1-day work is \( \frac{1}{16} \). The remaining work is \( \frac{1}{8} \). Since \[\frac{1}{8} = \frac{2}{16}\], Abaz needs to do work equivalent to 2 days of his work. This means Abaz cannot finish the remaining work in one day. This implies Aman will also have to work.
Let's re-check the number of cycles. We need to find the largest number of cycles such that the work done is less than 1, and the next person can finish it.
Work done in 6 cycles (12 days) = \( \frac{7}{8} \). Remaining work = \( \frac{1}{8} \).
On the 13th day, Abaz works. Abaz's 1-day work = \( \frac{1}{16} \).
Work remaining after Abaz works on Day 13:
\[ \text{Work remaining} = \frac{1}{8} - \frac{1}{16} = \frac{2}{16} - \frac{1}{16} = \frac{1}{16} \]So, after 13 days, \[\frac{7}{8} + \frac{1}{16} = \frac{14}{16} + \frac{1}{16} = \frac{15}{16}\] of the work is completed.
The remaining work is \[1 - \frac{15}{16} = \frac{1}{16}\].
It is now the 14th day, and Aman's turn to work. Aman's 1-day work = \( \frac{1}{12} \).
Time taken by Aman to complete the remaining \( \frac{1}{16} \) work:
\[ \text{Time} = \frac{\text{Remaining work}}{\text{Aman's 1-day work}} = \frac{1/16}{1/12} = \frac{1}{16} \times 12 = \frac{12}{16} = \frac{3}{4} \text{ days} \]Total days = 13 days (12 days for 6 cycles + 1 day for Abaz) + \( \frac{3}{4} \) days for Aman
\[ \text{Total days} = 13 + \frac{3}{4} = 13\frac{3}{4} \text{ days} \]C) \( 13\frac{3}{4} \)
The question asks about the behavior of light when it passes from one medium (air) to another (water) along the normal. This involves understanding the principles of refraction and Snell's Law.
Correct Option: A) not be changed. When light is incident along the normal, both the angle of incidence and the angle of refraction are \(0^\circ\), meaning the light ray passes straight through without any change in direction.
The question asks to identify the Kalaripayattu exponent who received the Padma Shri award in 2022 at the age of 93. This requires knowledge of recent Padma awardees, specifically those associated with Kalaripayattu.
Correct Option: A) Sankaranarayana Menon Chundayil was a celebrated Kalaripayattu guru who received the Padma Shri in 2022 at the age of 93. He is known for his lifelong dedication to preserving and promoting this ancient martial art form.
A, B, C, D, E, F, G and H are sitting around a square table facing the centre of the table (but not necessarily in the same order). Some of them are sitting at the corners, while some are sitting at the exact centre of the sides. A is not sitting at any of the corners and is second to the right of H. G is sitting at one of the corners and is sitting fourth to the left of C. E and B are the immediate neighbours of C. F is sitting at one of the corners and is to the immediate left of B.
Who among the following are the immediate neighbours of H?
Suppose a pin is placed in front of a concave mirror and a real image that is thrice the size of the pin is formed on a screen. The pin and the screen are then moved until the image is six times the size of the object. If the shift of the screen is 24 cm, then the shift in the object is:
\( \frac{7}{3} \) cm
\( \frac{4}{3} \) cm
\( \frac{2}{3} \) cm
\( \frac{5}{3} \) cm
This problem involves the magnification formula for a concave mirror and the mirror formula. We will use the given magnification values and the shift in the screen (image position) to find the shift in the object position. Since a real image is formed, both the object distance (\(u\)) and image distance (\(v\)) will be negative according to the sign convention, and magnification (\(m\)) will be negative.
First Case: Magnification \(m_1 = -3\)
For a concave mirror, the magnification \(m\) is given by \(m = - \frac{v}{u}\). Since the image is real and inverted, \(m\) is negative.
So, \(m_1 = -3\)
\[ -3 = - \frac{v_1}{u_1} \implies v_1 = 3u_1 \]Using the mirror formula: \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
\[ \frac{1}{f} = \frac{1}{3u_1} + \frac{1}{u_1} = \frac{1+3}{3u_1} = \frac{4}{3u_1} \] \[ f = \frac{3u_1}{4} \quad \text{or} \quad u_1 = \frac{4f}{3} \]Since \(u_1\) is an object distance for a real image, it should be negative. Let's consider magnitudes for now and apply signs consistently. If \(f\) is negative for a concave mirror, then \(u_1\) will be negative. Let's use the magnitude of \(u_1\) and \(v_1\).
So, \(|u_1| = \frac{4|f|}{3}\) and \(|v_1| = 3|u_1| = 4|f|\).
Second Case: Magnification \(m_2 = -6\)
Similarly, \(m_2 = -6\)
\[ -6 = - \frac{v_2}{u_2} \implies v_2 = 6u_2 \]Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{6u_2} + \frac{1}{u_2} = \frac{1+6}{6u_2} = \frac{7}{6u_2} \] \[ f = \frac{6u_2}{7} \quad \text{or} \quad u_2 = \frac{7f}{6} \]So, \(|u_2| = \frac{7|f|}{6}\) and \(|v_2| = 6|u_2| = 7|f|\).
Calculate the shift in screen (image) position
The shift of the screen is given as 24 cm. This is the difference in image distances.
Since magnification increased from 3 to 6, the object must have moved closer to the mirror, and the image must have moved further away from the mirror (for a real image formed by a concave mirror beyond F, as the object moves from infinity to F, the image moves from F to infinity). So \(|v_2| > |v_1|\).
\[ |v_2| - |v_1| = 24 \text{ cm} \] \[ 7|f| - 4|f| = 24 \] \[ 3|f| = 24 \] \[ |f| = 8 \text{ cm} \]Calculate the object distances
Now substitute the value of \(|f|\) back into the expressions for \(|u_1|\) and \(|u_2|\):
\[ |u_1| = \frac{4|f|}{3} = \frac{4 \times 8}{3} = \frac{32}{3} \text{ cm} \] \[ |u_2| = \frac{7|f|}{6} = \frac{7 \times 8}{6} = \frac{56}{6} = \frac{28}{3} \text{ cm} \]Calculate the shift in object position
The shift in the object is \(|u_1| - |u_2|\) (since the object moves closer to the mirror, \(|u_1| > |u_2|\)).
\[ \text{Shift in object} = |u_1| - |u_2| = \frac{32}{3} - \frac{28}{3} = \frac{32-28}{3} = \frac{4}{3} \text{ cm} \]Correct Option: B) \( \frac{4}{3} \) cm
To find the sum of deviations from the mean, we first need to calculate the mean of the given data set. Then, for each data point, we subtract the mean to find its deviation. Finally, we sum all these deviations.
General Property: A fundamental property of the arithmetic mean is that the sum of the deviations of all individual values from the mean is always zero. This is because the mean is the balancing point of the data set.
D) 0. As calculated, the sum of the deviations of the values from their mean is 0.
This question tests knowledge about the administrative structure of Municipal Corporations in India, specifically identifying the political head. Understanding the roles of various functionaries in urban local bodies is crucial.
Correct Option: D) Mayor is the political head of a Municipal Corporation. The Mayor is elected by the Councillors and serves as the ceremonial head and the first citizen of the city, presiding over the corporation's meetings.