The question asks for the scientific name of baking soda. This requires knowledge of common chemical compounds and their scientific nomenclature.
Correct Option: A) Sodium bicarbonate (\( \text{NaHCO}_3 \)) is the scientific name for baking soda. It is a white crystalline powder commonly used in baking as a leavening agent and also as an antacid.
Simplify the following.
1 – 2 × [ { –2(2 + 3) × 20 } ÷ 2 ]
To simplify the given expression, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)).
The simplified value of the expression is \(201\).
D) 201 is the correct answer, as derived by following the order of operations.
Shyam and Sundar can do a piece of work in 16 days and 24 days, respectively.
Shyam started the work alone and after 10 days, Sundar joined him and the duo worked together till the completion of the work. What was the total time taken for the work to be completed?
13\( \frac{3}{5} \) days
12\( \frac{3}{5} \) days
15\( \frac{2}{5} \) days
10\( \frac{3}{5} \) days
This problem involves calculating the total time taken to complete a work when two individuals, Shyam and Sundar, work at different rates and for different durations. We will first determine their individual work rates, then calculate the amount of work done by Shyam alone, and finally find the time taken for both to complete the remaining work.
Shyam can do a piece of work in 16 days. So, Shyam's 1-day work rate is:
\[ \text{Shyam's 1-day work} = \frac{1}{16} \text{ of the work} \]Sundar can do the same piece of work in 24 days. So, Sundar's 1-day work rate is:
\[ \text{Sundar's 1-day work} = \frac{1}{24} \text{ of the work} \]Shyam started the work alone and worked for 10 days.
\[ \text{Work done by Shyam in 10 days} = 10 \times \frac{1}{16} = \frac{10}{16} = \frac{5}{8} \text{ of the work} \]The total work is 1. So, the remaining work after Shyam worked for 10 days is:
\[ \text{Remaining work} = 1 - \frac{5}{8} = \frac{3}{8} \text{ of the work} \]When Shyam and Sundar work together, their combined 1-day work rate is the sum of their individual rates:
\[ \text{Combined 1-day work} = \frac{1}{16} + \frac{1}{24} \]To add these fractions, find a common denominator, which is 48.
\[ \text{Combined 1-day work} = \frac{3}{48} + \frac{2}{48} = \frac{5}{48} \text{ of the work} \]The remaining work is \( \frac{3}{8} \) and their combined rate is \( \frac{5}{48} \) per day. Time taken to complete the remaining work is:
\[ \text{Time} = \frac{\text{Remaining Work}}{\text{Combined 1-day work}} = \frac{\frac{3}{8}}{\frac{5}{48}} \]\[ \text{Time} = \frac{3}{8} \times \frac{48}{5} = \frac{3 \times 6}{5} = \frac{18}{5} \text{ days} \]This can be written as a mixed fraction:
\[ \frac{18}{5} = 3\frac{3}{5} \text{ days} \]Total time = Time Shyam worked alone + Time Shyam and Sundar worked together
\[ \text{Total time} = 10 \text{ days} + 3\frac{3}{5} \text{ days} = 13\frac{3}{5} \text{ days} \]Correct Option: A) 13\( \frac{3}{5} \) days
A displacement reaction is a chemical reaction in which a more reactive element displaces a less reactive element from its compound. We need to identify the reaction among the given options that fits this definition.
Correct Option: B) \(\text{Zn (s)} + \text{CuSO}_4\text{ (aq)} \to \text{ZnSO}_4\text{ (aq)} + \text{Cu (s)}\)
This is a single displacement reaction where zinc, being more reactive than copper, displaces copper from its salt solution (\(\text{CuSO}_4\)).
If ‘+’ means ‘×’, ‘−’ means ‘÷’, ‘×’ means ‘−’, and ‘÷’ means ‘+’, what will come in place of the question mark (?) in the following equation?
9 + 7 ÷ 8 − 4 × 2 =?
This problem involves substituting given symbol meanings into a mathematical expression and then evaluating it using the order of operations (BODMAS/PEMDAS).
The original equation is: \(9 + 7 \div 8 - 4 \times 2 = ?\)
Applying the new meanings:
\[ 9 \times 7 + 8 \div 4 - 2 = ? \]BODMAS/PEMDAS stands for Brackets/Parentheses, Orders/Exponents, Division, Multiplication, Addition, Subtraction. Operations are performed from left to right for Division/Multiplication and Addition/Subtraction.
The value of the expression is \(63\).
A) 63 — After correctly substituting the symbols and following the order of operations, the expression evaluates to 63.
The present age of Arun is \( \frac{1}{3} \) of his uncle’s age.
After 5 years, the sum of their ages will be 86 years.
The present age of Arun is:
The problem involves setting up and solving a system of two linear equations based on the given age relationships. We will use variables to represent the present ages of Arun and his uncle, form equations from the given conditions, and then solve for Arun's present age.
Define variables:
Let Arun's present age be \(A\) years.
Let his uncle's present age be \(U\) years.
Formulate the first equation from the first condition:
"The present age of Arun is \( \frac{1}{3} \) of his uncle's age."
This translates to:
\[ A = \frac{1}{3} U \]Formulate the second equation from the second condition:
"After 5 years, the sum of their ages will be 86 years."
After 5 years, Arun's age will be \(A + 5\).
After 5 years, his uncle's age will be \(U + 5\).
The sum of their ages after 5 years is:
\[ (A + 5) + (U + 5) = 86 \]\[ A + U + 10 = 86 \]\[ A + U = 86 - 10 \]\[ A + U = 76 \]Solve the system of equations:
We have two equations:
1) \( A = \frac{1}{3} U \)
2) \( A + U = 76 \)
Substitute the expression for \(A\) from equation (1) into equation (2):
\[ \left(\frac{1}{3} U\right) + U = 76 \]\[ \frac{U}{3} + \frac{3U}{3} = 76 \]\[ \frac{4U}{3} = 76 \]Calculate the uncle's present age (\(U\)):
\[ 4U = 76 \times 3 \]\[ 4U = 228 \]\[ U = \frac{228}{4} \]\[ U = 57 \text{ years} \]Calculate Arun's present age (\(A\)):
Using equation (1):
Verify the answer:
Arun's present age = 19 years
Uncle's present age = 57 years
After 5 years:
Arun's age = \(19 + 5 = 24\) years
Uncle's age = \(57 + 5 = 62\) years
Sum of their ages after 5 years = \(24 + 62 = 86\) years. This matches the given condition.
B) 19 years
The question asks for the year when the Centre for Excellence of Classical Tamil was relocated from Mysore to Chennai. This is a factual question related to the history and establishment of a significant cultural institution in India.
Correct Option: B) The Centre for Excellence of Classical Tamil was shifted from Mysore to Chennai in 2008. This is a well-documented fact regarding the establishment and relocation of the Central Institute of Classical Tamil (CICT).
In this question, a group of letters is coded using numbers/symbols as per the table given below and the conditions which follow.
The correct combination of codes following the conditions is your answer.
| Letters | E | O | L | Y | A | W | N | S |
| Numbers/Symbols | 5 | % | 7 | # | 1 | 4 | @ | $ |
Conditions:
(i) If a word has a vowel at the end, the vowel present at the end will be coded as 6.
(ii) If a word has more than 2 vowels, all the vowels will be coded as 8.
(iii) If a letter is present exactly twice in a word, it will be coded as 2 in both such places.
What will be the correct code for YELLOW?
To code the word "YELLOW", we need to apply the given conditions in the specified order. First, we'll find the base codes for each letter from the table. Then, we'll check and apply the conditions one by one. Condition (i) checks for a vowel at the end. Condition (ii) checks for more than 2 vowels. Condition (iii) checks for letters appearing exactly twice.
The word YELLOW ends with 'W', which is a consonant. Therefore, Condition (i) does not apply.
The vowels in YELLOW are E and O. There are exactly 2 vowels. Since the condition states "more than 2 vowels", Condition (ii) does not apply.
In the word YELLOW, the letter 'L' appears exactly twice. According to this condition, both 'L's should be coded as 2.
Replacing the 'L's in the initial code #577%4:
The final code becomes #522%4.
D) #522%4 — This matches our derived code after applying all conditions correctly.
To find the effective resistance of resistors connected in parallel, we use the formula for parallel resistance. This formula states that the reciprocal of the total effective resistance is equal to the sum of the reciprocals of individual resistances.
Correct Option: D) 1.2 Ω
This problem involves analyzing a statement with inequalities and equalities to determine which of the given conclusions logically follows. We need to break down the given statement into smaller relationships and then check each conclusion against these relationships.
The given statement is: \(L \le H = P > O > D \ge S = U\)
Let's analyze each conclusion:
A) 1 — Conclusion 1 (\(L \le P\)) is derived directly from \(L \le H\) and \(H = P\), making it true.