The question asks to identify the Article of the Indian Constitution that vests the executive powers of the State in the Governor. This requires knowledge of the specific Articles related to the Governor's powers and functions.
B) Article 154 — This Article explicitly vests the executive power of the State in the Governor, making him the constitutional head of the state executive.
This problem involves finding two numbers given their LCM, HCF, and their sum. We will use the fundamental relationship between two numbers, their LCM, and HCF, along with the given conditions, to set up and solve a system of equations.
Let the two numbers be \(a\) and \(b\). Let their HCF be \(H\) and their LCM be \(L\).
From the problem statement, we are given two conditions:
Substitute Equation 1 into Equation 2:
\[ 10H + H = 451 \] \[ 11H = 451 \] \[ H = \frac{451}{11} \] \[ H = 41 \]Now, find \(L\) using Equation 1:
\[ L = 10H = 10 \times 41 \] \[ L = 410 \]We know that for any two numbers \(a\) and \(b\), the product of the numbers is equal to the product of their HCF and LCM:
\[ a \times b = H \times L \] \[ a \times b = 41 \times 410 \] \[ a \times b = 16810 \quad \text{(Equation 3)} \]We are also given that the sum of the numbers is 287:
\[ a + b = 287 \quad \text{(Equation 4)} \]Now we have a system of two equations with two variables \(a\) and \(b\). We can express \(b\) in terms of \(a\) from Equation 4:
\[ b = 287 - a \]Substitute this expression for \(b\) into Equation 3:
\[ a \times (287 - a) = 16810 \] \[ 287a - a^2 = 16810 \] \[ a^2 - 287a + 16810 = 0 \]This is a quadratic equation. We can solve it using the quadratic formula \(a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\) or by factorization. Let's try to factorize it. We need two numbers whose sum is 287 and product is 16810.
Notice that \(H = 41\). This means both numbers \(a\) and \(b\) must be multiples of 41. Let \(a = 41x\) and \(b = 41y\), where \(x\) and \(y\) are co-prime integers.
From \(a + b = 287\):
\[ 41x + 41y = 287 \] \[ 41(x + y) = 287 \] \[ x + y = \frac{287}{41} \] \[ x + y = 7 \]From \(a \times b = 16810\):
\[ (41x) \times (41y) = 16810 \] \[ 1681xy = 16810 \] \[ xy = \frac{16810}{1681} \] \[ xy = 10 \]Now we need to find two co-prime integers \(x\) and \(y\) such that their sum is 7 and their product is 10. The pairs of factors of 10 are (1, 10) and (2, 5).
So, \(x = 2\) and \(y = 5\) (or vice versa).
Now, find the numbers \(a\) and \(b\):
If \(x = 2\), then \(a = 41 \times 2 = 82\).
If \(y = 5\), then \(b = 41 \times 5 = 205\).
The two numbers are 82 and 205.
Check the sum: \(82 + 205 = 287\). This matches the given condition.
One of the two numbers is 82. This matches option A.
A) 82 is one of the two numbers found (the other being 205).
This is a logical reasoning problem involving arrangement on different floors. We need to deduce the position of each person based on the given clues and then identify the floor number of Z.
Establish the floors: There are six floors, numbered 1 to 6 from bottom to top.
Place N: "N lives on floor number 2."
| Floor | Person |
|---|---|
| 6 | |
| 5 | |
| 4 | |
| 3 | |
| 2 | N |
| 1 |
Place R: "No one lives between N and R." Since N is on floor 2, R must be on floor 1 or floor 3.
| Floor | Person |
|---|---|
| 6 | |
| 5 | |
| 4 | |
| 3 | |
| 2 | N |
| 1 | R |
| Floor | Person |
|---|---|
| 6 | |
| 5 | |
| 4 | |
| 3 | R |
| 2 | N |
| 1 |
Place S: "Only three people live between R and S."
| Floor | Person |
|---|---|
| 6 | |
| 5 | S |
| 4 | |
| 3 | |
| 2 | N |
| 1 | R |
So, Case 1 is the only valid arrangement so far.
Place W and X: "W lives immediately above X." We have floors 3, 4, and 6 remaining. The only consecutive floors available are 3 and 4, or 5 and 6 (but 5 is taken by S). So, W must be on floor 4 and X on floor 3.
| Floor | Person |
|---|---|
| 6 | |
| 5 | S |
| 4 | W |
| 3 | X |
| 2 | N |
| 1 | R |
Place Z: The only remaining person is Z and the only remaining floor is 6. Therefore, Z lives on floor 6.
| Floor | Person |
|---|---|
| 6 | Z |
| 5 | S |
| 4 | W |
| 3 | X |
| 2 | N |
| 1 | R |
B) Z lives on floor number 6.
Which two numbers from amongst the given options should be interchanged to make the given equation correct?
382 + 9 ÷ 4 – 39 × 172 = 74
The problem requires us to find which two numbers, when interchanged in the given equation, will make the equation mathematically correct. We need to test each option by interchanging the specified numbers and then evaluating the expression using the order of operations (BODMAS/PEMDAS).
The given equation is:
\[ 382 + 9 \div 4 - 39 \times 172 = 74 \]If we interchange 9 and 4, the equation becomes:
\[ 382 + 4 \div 9 - 39 \times 172 \]\[ 382 + 0.444... - 6708 \]\[ 382.444... - 6708 = -6325.555... \]This is not equal to 74.
If we interchange 4 and 172, the equation becomes:
\[ 382 + 9 \div 172 - 39 \times 4 \]\[ 382 + 0.052... - 156 \]\[ 382.052... - 156 = 226.052... \]This is not equal to 74.
If we interchange 382 and 39, the equation becomes:
\[ 39 + 9 \div 4 - 382 \times 172 \]\[ 39 + 2.25 - 65624 \]\[ 41.25 - 65624 = -65582.75 \]This is not equal to 74.
If we interchange 9 and 172, the equation becomes:
\[ 382 + 172 \div 4 - 39 \times 9 \]Now, let's evaluate this expression using the order of operations (division and multiplication first, then addition and subtraction):
Since \(74 = 74\), the equation is correct after interchanging 9 and 172.
D) 9 and 172
The median of the following data is 28.5. The total frequency is 60. The values of x and y, respectively, are
| Age in years | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
| No. of patients | 5 | x | 20 | 15 | y | 5 | 60 |
x = 7, y = 8
x = 6, y = 9
x = 8, y = 7
x = 9, y = 6
To find the missing frequencies x and y, we will use the given total frequency and the median value. First, we will construct a cumulative frequency table. Then, we will identify the median class based on the median value. Using the formula for the median of grouped data, we will set up an equation involving x and y. A second equation will come from the total frequency. Solving these two linear equations will give us the values of x and y.
Construct the cumulative frequency table:
| Age in years (Class Interval) | No. of patients (Frequency, \(f\)) | Cumulative Frequency (cf) |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | x | \(5 + x\) |
| 20-30 | 20 | \(5 + x + 20 = 25 + x\) |
| 30-40 | 15 | \(25 + x + 15 = 40 + x\) |
| 40-50 | y | \(40 + x + y\) |
| 50-60 | 5 | \(40 + x + y + 5 = 45 + x + y\) |
Use the total frequency:
The total frequency is given as 60. From the cumulative frequency table, the last cumulative frequency is \(45 + x + y\).
\[ 45 + x + y = 60 \] \[ x + y = 60 - 45 \] \[ x + y = 15 \quad \text{(Equation 1)} \]Identify the median class:
The median is given as 28.5. This value lies in the class interval 20-30. Therefore, the median class is 20-30.
From the median class:
Apply the median formula:
The formula for the median of grouped data is:
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]Substitute the known values:
\[ 28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10 \] \[ 28.5 - 20 = \left( \frac{30 - 5 - x}{20} \right) \times 10 \] \[ 8.5 = \left( \frac{25 - x}{20} \right) \times 10 \] \[ 8.5 = \frac{25 - x}{2} \] \[ 8.5 \times 2 = 25 - x \] \[ 17 = 25 - x \] \[ x = 25 - 17 \] \[ x = 8 \quad \text{(Equation 2)} \]Solve for y:
Substitute the value of \(x = 8\) into Equation 1:
\[ x + y = 15 \] \[ 8 + y = 15 \] \[ y = 15 - 8 \] \[ y = 7 \]Thus, the values of x and y are 8 and 7, respectively.
C) x = 8, y = 7. This matches our calculated values for x and y.
Refer to the following number and symbol series and answer the question.
(Left) 8 # 7 * 3 $ 9 6 2 @ 7 & 9 % 4 3 7 % & 2 5 ^ 6 (Right)
How many such symbols are there in the series which are immediately preceded by an odd number and also immediately followed by an even number?
The problem asks us to identify symbols in a given series that satisfy two conditions: they must be immediately preceded by an odd number and immediately followed by an even number. We will systematically scan the series, looking for this specific pattern: Odd Number - Symbol - Even Number.
C) Two
Wait, upon re-checking the series carefully, let's re-evaluate the symbols one by one:
Series: 8 # 7 * 3 \) 9 6 2 @ 7 & 9 % 4 3 7 % & 2 5 ^ 6
My apologies for the previous error in counting. There are only two such symbols: '%' (9 % 4) and '^' (5 ^ 6).
Therefore, the correct option is C.
The perimeter of a regular pentagon whose each side is 2a cm long is:
To find the perimeter of a regular polygon, we multiply the number of sides by the length of one side. A regular pentagon has 5 equal sides.
Correct Option: D) 10a cm
The question asks to identify which of the given options is NOT a scheme or initiative run by the Ministry of Youth Affairs and Sports. We need to evaluate each option to determine its administrative ministry.
Based on this analysis, the Shyama Prasad Mukherjee National Sports Welfare Fund does not align with the known schemes of the Ministry of Youth Affairs and Sports.
Correct Option: B) Shyama Prasad Mukherjee National Sports Welfare Fund
The question asks for the SI unit of electric power. We need to recall the definitions and SI units of various physical quantities to identify the correct option.
Therefore, the SI unit of electric power is Watt.
Correct Option: B) Watt
A group of numbers/symbols is coded using letters as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
| Number/symbol | 1 | & | 2 | 9 | * | 7 | 3 | $ | 8 | 5 | 6 | 4 | @ | # |
| Code | B | F | I | K | A | V | S | U | E | Q | C | P | X | T |
Conditions:
(i) If the first element is a symbol and the second element is an even number, the first and the second elements are to be coded as E.
(ii) If both the first and the last elements are numbers, the codes for these two (the first and the last elements) are to be interchanged.
(iii) If the last element is a symbol, the element is to be coded as B.
5 @ & 2 7