Which among the following States topped the 4th Khelo India Youth Games 2021 medals tally?
The question asks to identify the state that topped the medals tally in the 4th Khelo India Youth Games 2021. This requires knowledge of the results of this specific sporting event.
Correct Option: C) Haryana
To find the mode of the data, we first need to determine the value of \(x\). We are given the mean of the data set. We will use the formula for the mean to set up an equation and solve for \(x\). Once \(x\) is known, we will substitute it back into the data points to get the actual numerical values. Finally, we will identify the mode, which is the value that appears most frequently in the data set.
The data points are \(11, 17, x + 1, 3x, 19, 2x - 4, x + 5\).
There are 7 data points.
\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \]
\[ 21 = \frac{(11 + 17 + (x + 1) + 3x + 19 + (2x - 4) + (x + 5))}{7} \]
Combine the constant terms and the terms with \(x\):
Constant terms: \(11 + 17 + 1 + 19 - 4 + 5 = 49\)
\(x\) terms: \(x + 3x + 2x + x = 7x\)
So, the numerator becomes \(7x + 49\).
\[ 21 = \frac{7x + 49}{7} \]
Multiply both sides by 7:
\[ 21 \times 7 = 7x + 49 \]
\[ 147 = 7x + 49 \]
Subtract 49 from both sides:
\[ 147 - 49 = 7x \]
\[ 98 = 7x \]
Divide by 7:
\[ x = \frac{98}{7} \]
\[ x = 14 \]
The data set now becomes: \(11, 17, 15, 42, 19, 24, 19\).
The mode is the value that appears most frequently in the data set.
In the data set \(\{11, 17, 15, 42, 19, 24, 19\}\), the number 19 appears twice, while all other numbers appear only once.
Therefore, the mode is 19.
C) 19 — After calculating \(x=14\) and substituting it back into the data set, the values are \(11, 17, 15, 42, 19, 24, 19\). The number 19 appears twice, which is more frequent than any other number. Thus, 19 is the mode.
In a row of students facing north, Krish is 19th from the extreme left while Maya is 30th from the extreme right. When both of them interchange their positions, Krish becomes 36th from the extreme left end. How many students are there in the row?
This problem involves finding the total number of students in a row using the concept of ranking and position interchange. We will use the formula for total number of persons in a row when the position of one person from both ends is known.
When Krish moves to Maya's original position, his new position from the left is given as 36th. Since Maya was 30th from the extreme right, Krish, now at Maya's position, is also 30th from the extreme right.
So, Krish's position from the left = 36
Krish's position from the right (which was Maya's original position) = 30
The formula to find the total number of persons in a row when the position of one person from both ends is known is:
\[ \text{Total persons} = (\text{Position from left} + \text{Position from right}) - 1 \]Using Krish's new position:
\[ \text{Total students} = (36 + 30) - 1 \]\[ \text{Total students} = 66 - 1 \]\[ \text{Total students} = 65 \]D) 65 — By using the formula for total persons in a row after Krish's new position from both ends is determined, we find the total number of students to be 65.
To determine the value of the absolute refractive index, we need to understand its definition and the physical properties of light in different media. The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium.
Correct Option: B) more than 1
The absolute refractive index of any material medium is always greater than 1 because the speed of light in a vacuum is the maximum possible speed, and light slows down when it enters any material medium.
The question asks to identify the state where the Mahendragiri Hills are located. This is a factual question related to Indian geography. We need to recall the geographical distribution of major hill ranges in India.
Correct Option: B) Mahendragiri Hills are located in the state of Odisha. It is a significant peak in the Eastern Ghats and is known for its ecological and cultural importance.
Six frogs, P, Q, R, S, T and U, were sitting around a circular pond, facing the centre. T was second to the left of S. P was second to the right of U. P is not seated three places to the right of Q. There are exactly two frogs between Q and S.
Which frog was sitting to the immediate right of P?
This question is based on a circular seating arrangement involving six frogs, P, Q, R, S, T, and U, facing the centre. We need to arrange them according to the given conditions and identify the frog sitting immediately to the right of P.
Since all the frogs face the centre:
We can fix S at any position because rotating a circular arrangement does not change the relative positions.
Fix S at the first position.
T sits second to the left of S.
Count two positions clockwise from S:
Therefore, T occupies the third position in the clockwise arrangement.
Exactly two frogs sit between Q and S.
Since there are six positions, Q must sit opposite S.
Therefore, Q occupies the fourth position.
The partial clockwise arrangement is:
S → Vacant → T → Q → Vacant → Vacant
P sits second to the right of U.
Since everyone faces the centre, P must be two positions anticlockwise from U.
The available positions are 2, 5, and 6.
The only possible placement satisfying this condition is:
Therefore, the arrangement becomes:
S → U → T → Q → Vacant → P
The only remaining vacant position is position 5.
Therefore, R occupies this position.
The complete clockwise arrangement is:
S → U → T → Q → R → P
Equivalently, starting from P:
P → S → U → T → Q → R
All the given conditions are satisfied.
The clockwise arrangement is:
P → S → U → T → Q → R
Since all frogs face the centre, move one position anticlockwise from P.
The frog immediately anticlockwise from P is R.
Therefore, R sits immediately to the right of P.
Correct Option: C) R
The complete clockwise arrangement is P, S, U, T, Q, and R. Since all frogs face the centre, the immediate right of P is the adjacent anticlockwise position, occupied by R. Therefore, the correct option is C) R.
Six teams named P, Q, R, S, T and U won tournaments in different months of the same year, viz. January, February, March, April, May and June.
R won the tournament in January. Q won in the month immediately before U. P won the tournament in the month immediately before S. Exactly two teams won the tournaments between the months of R and S.
In which month did T win the tournament?
This is a logical reasoning problem involving ordering and sequencing. We need to deduce the month in which each team won a tournament based on the given clues. We will create a table or a sequence of months and fill in the teams as we process each clue.
D) From our step-by-step deduction, T won the tournament in February.
The product for the following reaction will be:
CH₃ – CHâ‚‚ – OH + 2[O] → (Ethanol) (Oxygen from the oxidizing agent)
Conditions:
Alkaline KMnOâ‚„ + Heat
or
Acidified K₂Cr₂O₇ + Heat
The question asks for the product of the oxidation of ethanol under strong oxidizing conditions (alkaline KMnO₄ + heat or acidified K₂Cr₂O₇ + heat). This is a classic organic chemistry reaction involving the oxidation of a primary alcohol.
C) CH₃COOH — Ethanol is a primary alcohol. Strong oxidizing agents like alkaline KMnO₄ or acidified K₂Cr₂O₇ oxidize primary alcohols completely to carboxylic acids. Therefore, ethanol is oxidized to ethanoic acid (CH₃COOH).
This problem can be solved using Venn diagrams or by applying the rules of syllogism. We will draw Venn diagrams for each statement and then check the validity of each conclusion.
This means there is an overlap between the set of monkeys and the set of pandas. We can represent this with two overlapping circles, one for 'Monkeys' and one for 'Pandas'.
This means there is an overlap between the set of pandas and the set of giraffes. The 'Giraffes' circle will overlap with the 'Pandas' circle. It may or may not overlap with the 'Monkeys' circle based on this statement alone.
This means the entire set of 'Giraffes' is contained within the set of 'Lions'. So, the 'Giraffes' circle will be completely inside the 'Lions' circle.
We have Monkeys overlapping Pandas. Pandas overlapping Giraffes. Giraffes completely inside Lions. This implies that anything that is a Giraffe is also a Lion.
From Statement II, some pandas are giraffes. From Statement III, all giraffes are lions. Since some pandas are giraffes, and all giraffes are lions, it logically follows that those pandas that are giraffes are also lions. Therefore, some pandas are lions, or conversely, some lions are pandas. This conclusion follows.
We know some monkeys are pandas (Statement I) and some pandas are giraffes (Statement II). We also know all giraffes are lions (Statement III). However, there is no direct or indirect definite link established between 'Monkeys' and 'Lions'. The overlap between Monkeys and Pandas does not necessarily extend to Giraffes or Lions. It is possible that the monkeys that are pandas are not the same pandas that are giraffes. Thus, we cannot conclude that some lions are monkeys. This conclusion does not follow.
We know some monkeys are pandas (Statement I) and some pandas are giraffes (Statement II). Similar to Conclusion II, there is no definite link established between 'Giraffes' and 'Monkeys'. The overlap between Monkeys and Pandas does not guarantee an overlap between Monkeys and Giraffes. This conclusion does not follow.
Correct Option: B) Only conclusion I follows
If 1 + \(\sqrt{3}\) and 1 − \(\sqrt{3}\) are the roots of a quadratic equation, then the quadratic equation is:
x² − 2x + 3 = 0
x² − 2x − 3 = 0
x² − 2x − 2 = 0
x² − 2x + 2 = 0
To find the quadratic equation given its roots, we use the standard form of a quadratic equation: \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\). We will first calculate the sum and product of the given roots and then substitute these values into the standard form.
Identify the roots:
Let the two roots be \(\alpha\) and \(\beta\).
\(\alpha = 1 + \sqrt{3}\)
\(\beta = 1 - \sqrt{3}\)
Calculate the sum of the roots:
Sum of roots \( = \alpha + \beta \)
\[ = (1 + \sqrt{3}) + (1 - \sqrt{3}) \]\[ = 1 + \sqrt{3} + 1 - \sqrt{3} \]\[ = 1 + 1 \]\[ = 2 \]
Calculate the product of the roots:
Product of roots \( = \alpha \cdot \beta \)
\[ = (1 + \sqrt{3})(1 - \sqrt{3}) \]This is in the form \((a+b)(a-b) = a^2 - b^2\).
Here, \(a=1\) and \(b=\sqrt{3}\).
\[ = (1)^2 - (\sqrt{3})^2 \]\[ = 1 - 3 \]\[ = -2 \]
Form the quadratic equation:
The general form of a quadratic equation with roots \(\alpha\) and \(\beta\) is:
\[ x^2 - (\alpha + \beta)x + (\alpha \cdot \beta) = 0 \]
Substitute the calculated sum and product of roots:
\[ x^2 - (2)x + (-2) = 0 \]\[ x^2 - 2x - 2 = 0 \]
Correct Option: C) The calculated quadratic equation is \(x^2 - 2x - 2 = 0\), which matches option C.