This is a logical reasoning problem involving the arrangement of boxes. We need to deduce the exact position of each of the seven boxes based on the given clues and then identify the box at the lowermost position. We will build the arrangement step-by-step, eliminating possibilities as we go.
Clue 1: Only three boxes are kept above B.
This means B is at the 4th position from the top (or 4th from the bottom, if we count from 1 to 7 from top to bottom).
Position: 7 Position: 6 Position: 5 Position: 4 - B Position: 3 Position: 2 Position: 1
Clue 2: Only one box is kept between C and B.
Since B is at position 4, C can be at position 2 (below B) or position 6 (above B).
Case 1: C is above B
Position: 7 Position: 6 - C Position: 5 Position: 4 - B Position: 3 Position: 2 Position: 1
Case 2: C is below B
Position: 7 Position: 6 Position: 5 Position: 4 - B Position: 3 Position: 2 - C Position: 1
Clue 3: Only three boxes are kept between C and Y.
Let's check Case 1 (C at position 6): If C is at 6, then Y must be at position 2 (6 - 3 - 1 = 2). This is possible.
Position: 7 Position: 6 - C Position: 5 Position: 4 - B Position: 3 Position: 2 - Y Position: 1
Let's check Case 2 (C at position 2): If C is at 2, then Y must be at position 6 (2 + 3 + 1 = 6). This is also possible.
Position: 7 Position: 6 - Y Position: 5 Position: 4 - B Position: 3 Position: 2 - C Position: 1
Clue 4: Y is kept at some place above B.
This clue eliminates Case 1 where Y is at position 2 (below B). So, Case 2 is the correct one.
Current arrangement:
Position: 7 Position: 6 - Y Position: 5 Position: 4 - B Position: 3 Position: 2 - C Position: 1
Clue 5: Z is kept immediately below Y.
Since Y is at position 6, Z must be at position 5.
Position: 7 Position: 6 - Y Position: 5 - Z Position: 4 - B Position: 3 Position: 2 - C Position: 1
Clue 6: A is kept at one of the places above X.
The remaining positions are 7, 3, and 1. The remaining boxes are A, D, and X. For A to be above X, A must be at position 7, and X can be at position 3 or 1. D will take the remaining spot.
Clue 7: D is not kept immediately above or below C.
C is at position 2. So, D cannot be at position 3 or position 1.
From Clue 6, A is at position 7. The remaining boxes are D and X, and the remaining positions are 3 and 1. Since D cannot be at position 3 or 1 (because C is at position 2), this creates a contradiction. Let's re-evaluate Clue 7.
Ah, the phrasing "D is not kept immediately above or below C" means D cannot be at position 3 (immediately above C) AND D cannot be at position 1 (immediately below C). This means D must be at position 7.
If D is at position 7, then A and X must occupy positions 3 and 1. Since A is above X, A must be at position 3 and X must be at position 1.
Let's re-construct the final arrangement:
Position: 7 - D Position: 6 - Y Position: 5 - Z Position: 4 - B Position: 3 - A Position: 2 - C Position: 1 - X
Final Check:
- Three boxes (D, Y, Z) are above B. (Correct)
- One box (A) is between C and B. (Correct)
- Three boxes (Z, B, A) are between C and Y. (Correct)
- Y is above B. (Correct)
- Z is immediately below Y. (Correct)
- A is above X. (Correct)
- D is not immediately above (position 3) or below (position 1) C. (Correct, D is at position 7)
The arrangement is consistent with all clues.
Identify the lowermost box:
From the final arrangement, the box at the lowermost position (Position 1) is X.
C) X is kept at the lowermost position.
Which state hosted the 61st edition of the Indian National Chess Championship in 2024?
The question asks about the host state for a specific edition of the Indian National Chess Championship. This requires knowledge of recent events in Indian sports, particularly chess.
Correct Option: B) Haryana
The 61st edition of the Indian National Chess Championship was hosted by Rohtak, Haryana, in 2024. This event is a significant tournament in the Indian chess calendar, determining the national champion.
The question asks about the substance used in antacids to treat acidity problems. Acidity is caused by an excess of stomach acid (hydrochloric acid). Therefore, antacids must contain a basic substance to neutralize this acid.
B) Milk of magnesia is a suspension of magnesium hydroxide (\(\text{Mg(OH)}_2\)), which is a mild base. It reacts with the excess hydrochloric acid in the stomach to form water and magnesium chloride, thereby neutralizing the acid and relieving acidity symptoms. The reaction is: \(\text{Mg(OH)}_2\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + 2\text{H}_2\text{O(l)}\).
A, B and C invest sums in the ratio 76 : 37 : 52 respectively. If they earned a total profit of βΉ3,850 at the end of the year, then what is the difference between the shares of B and C?
The profit earned by partners in a business is distributed in the ratio of their investments, assuming the investment period is the same for all. We will first find the sum of the ratios, then calculate the individual shares of B and C from the total profit, and finally find the difference between their shares.
The investment ratio of A, B, and C is given as \(76 : 37 : 52\).
Sum of ratios \( = 76 + 37 + 52 = 165\).
The total profit earned is βΉ\(3,850\).
B's share \[= \left( \frac{\text{B's ratio}}{\text{Sum of ratios}} \right) \times \text{Total profit}\]
B's share \( = \frac{37}{165} \times 3850\)
To simplify, we can divide \(3850\) by \(165\). Both are divisible by \(5\):
\(165 \div 5 = 33\)
\(3850 \div 5 = 770\)
So, B's share \( = \frac{37}{33} \times 770\)
Now, \(770\) is divisible by \(33\). \(770 = 33 \times 23.33...\) This indicates a potential calculation error or that \(770\) is not directly divisible by \(33\). Let's recheck the division of \(3850\) by \(165\).
Let's simplify \(3850/165\) by dividing both by \(5\): \(770/33\). Now, \(770 = 77 \times 10 = 7 \times 11 \times 10\). And \(33 = 3 \times 11\).
So, \[\frac{770}{33} = \frac{7 \times 11 \times 10}{3 \times 11} = \frac{70}{3}\].
B's share \[= 37 \times \frac{70}{3} = \frac{2590}{3} \approx 863.33\].
This suggests that the numbers might be designed to be simpler. Let's re-evaluate the division of \(3850\) by \(165\).
Notice that \(3850 = 165 \times 23.33...\) This implies that the total profit might be a multiple of \(165\). Let's check if \(3850\) is divisible by \(165\).
\(3850 / 165 = 23.33...\)
Let's re-examine the problem statement or my calculation. The problem implies exact values for the shares. Let's check if \(3850\) is a multiple of \(165\). \(165 \times 10 = 1650\), \(165 \times 20 = 3300\). \(3850 - 3300 = 550\). \(165 \times 3 = 495\). So \(3850/165\) is not an integer. This is unusual for competitive exams unless the options are also decimals, which they are not.
Let's assume the question intends for the shares to be exact. Let's recheck the sum of ratios: \(76 + 37 + 52 = 165\). This is correct.
Perhaps there is a common factor between \(3850\) and \(165\). We already found \(5\). \(3850/5 = 770\), \(165/5 = 33\). So we have \( \frac{770}{33} \). \(770 = 70 \times 11\), \(33 = 3 \times 11\). So \[\frac{770}{33} = \frac{70}{3}\].
So B's share \[= 37 \times \frac{70}{3} = \frac{2590}{3}\].
And C's share \[= 52 \times \frac{70}{3} = \frac{3640}{3}\].
Difference \[= \frac{3640}{3} - \frac{2590}{3} = \frac{1050}{3} = 350\].
This works out perfectly! My initial hesitation about non-integer division was unnecessary as the final difference is an integer.
C's share \[= \left( \frac{\text{C's ratio}}{\text{Sum of ratios}} \right) \times \text{Total profit}\]
C's share \( = \frac{52}{165} \times 3850\)
C's share \[= 52 \times \frac{70}{3} = \frac{3640}{3}\].
Difference \( = \text{C's share} - \text{B's share}\)
Difference \[= \frac{3640}{3} - \frac{2590}{3}\]
Difference \( = \frac{3640 - 2590}{3}\)
Difference \( = \frac{1050}{3}\)
Difference \( = 350\).
Difference in ratios \( = 52 - 37 = 15\).
Difference in shares \[= \left( \frac{\text{Difference in ratios}}{\text{Sum of ratios}} \right) \times \text{Total profit}\]
Difference in shares \( = \frac{15}{165} \times 3850\)
Simplify the fraction \( \frac{15}{165} \). Both are divisible by \(15\):
\(15 \div 15 = 1\)
\(165 \div 15 = 11\)
So, Difference in shares \( = \frac{1}{11} \times 3850\)
Now, divide \(3850\) by \(11\):
\(3850 \div 11 = 350\).
The difference between the shares of B and C is βΉ\(350\).
Correct Option: D) βΉ350
To find the median age, we first need to arrange the given ages in ascending order. Then, we identify the middle value(s). If the number of data points is odd, the median is the single middle value. If the number of data points is even, the median is the average of the two middle values.
A) 24.5 is the correct median age, calculated by averaging the two middle values after arranging the data in ascending order.
The question asks for the modern term for what Gregor Mendel referred to as 'factors' in his studies of inheritance. This requires knowledge of basic genetics and the historical development of genetic terminology.
Mendel's 'factors' are indeed what we now call genes. Genes are the fundamental units of heredity, carrying instructions for building and maintaining an organism, and are passed from one generation to the next.
The problem asks us to identify symbols in a given series that are immediately preceded by an odd number and immediately followed by an even number. We need to scan the series from left to right and check each symbol for this specific pattern.
\(7 \& 4 \ 2 \& \% 4 \ 6 \@ 8 \ \Omega 9 \# 6 \* \pounds 5 \= 2\)
Correct Option: C) Three
To determine the nature of the solution formed when sodium carbonate dissolves in water, we need to analyze the hydrolysis of its constituent ions. Sodium carbonate (\(\text{Na}_2\text{CO}_3\)) is a salt formed from a strong base (\(\text{NaOH}\)) and a weak acid (\(\text{H}_2\text{CO}_3\)).
Dissociation of Sodium Carbonate: When sodium carbonate dissolves in water, it dissociates into its ions:
\[ \text{Na}_2\text{CO}_3 \text{(s)} \xrightarrow{\text{H}_2\text{O}} 2\text{Na}^+ \text{(aq)} + \text{CO}_3^{2-} \text{(aq)} \]Hydrolysis of Ions: We then consider the hydrolysis of these ions:
The sodium ion (\(\text{Na}^+\)) is the conjugate acid of a strong base (\(\text{NaOH}\)). Therefore, it does not hydrolyze significantly in water and does not affect the pH.
The carbonate ion (\(\text{CO}_3^{2-}\)) is the conjugate base of a weak acid (\(\text{HCO}_3^-\), which itself comes from \(\text{H}_2\text{CO}_3\)). Conjugate bases of weak acids are strong enough to react with water (hydrolyze) and produce hydroxide ions (\(\text{OH}^-\)).
The hydrolysis reaction is:
\[ \text{CO}_3^{2-} \text{(aq)} + \text{H}_2\text{O} \text{(l)} \rightleftharpoons \text{HCO}_3^- \text{(aq)} + \text{OH}^- \text{(aq)} \]Effect on pH: The production of hydroxide ions (\(\text{OH}^-\)) increases the concentration of \(\text{OH}^-\), making the solution basic (pH > 7).
C) basic β The carbonate ion (\(\text{CO}_3^{2-}\)) hydrolyzes in water to produce hydroxide ions (\(\text{OH}^-\)), which makes the solution basic.
A) highly reactive with sodium chloride β Sodium carbonate and sodium chloride are both salts containing sodium ions. They do not typically react with each other in a way that would be described as "highly reactive" in solution. Sodium chloride solution is neutral.
B) acidic β An acidic solution would require the production of hydrogen ions (\(\text{H}^+\)) or hydronium ions (\(\text{H}_3\text{O}^+\)). Since \(\text{CO}_3^{2-}\) produces \(\text{OH}^-\), the solution is not acidic.
D) neutral β A neutral solution would have a pH of 7, meaning equal concentrations of \(\text{H}^+\) and \(\text{OH}^-\). Due to the hydrolysis of the carbonate ion producing \(\text{OH}^-\), the solution is not neutral.
The problem asks us to find a relationship between two numbers based on given examples and then apply that relationship to a new number. We need to identify the mathematical operation(s) that transform the first number into the second number in the given pairs.
Analyze the first pair: 5 is related to 26.
Let's try common mathematical operations:
The relationship \(n^2 + 1\) seems promising because it involves squaring the number, which is a common pattern in such problems.
Analyze the second pair: 12 is related to 145.
Let's test the \(n^2 + 1\) relationship with the number 12:
\[ 12^2 + 1 = 144 + 1 = 145 \]This matches the given relationship perfectly. Therefore, the logic is that the second number is obtained by squaring the first number and then adding 1.
Apply the logic to the third number: 13.
Following the same logic, we need to find the number related to 13 by squaring 13 and adding 1:
\[ 13^2 + 1 = 169 + 1 = 170 \]Correct Option: C) 170
The logic established from the first two pairs is that if a number is \(n\), the related number is \(n^2 + 1\). Applying this to 13, we get \(13^2 + 1 = 169 + 1 = 170\).