The question asks about the bud development process in Hydra. Hydra reproduces asexually through budding. We need to identify the fundamental cellular event that initiates this process.
A) Cell divisions at one site — This is the correct description of the initiation of bud development in Hydra. The localized, rapid cell divisions lead to an outgrowth that eventually forms a new individual.
This question is about the formation of an ionic bond between sodium (Na) and chlorine (Cl) to form sodium chloride (NaCl). We need to understand the electron configuration of these atoms and how they achieve a stable octet (or duplet) by gaining or losing electrons.
Therefore, each sodium atom loses one electron during this reaction.
C) loses one electron. Sodium has one valence electron, and it readily loses this electron to achieve a stable electron configuration, forming a positive ion (\(\text{Na}^+\)).
This problem involves comparing simple interest (SI) and compound interest (CI) over two years at the same rate. We can use the relationship between SI and CI for two years to find the rate of interest and then the principal amount.
Let the principal be \(P\), the rate of interest be \(R\) per annum, and the time be \(T = 2\) years.
The simple interest (SI) for 2 years is given as ₹56.
The formula for simple interest is \(SI = \frac{P \times R \times T}{100}\).
So, \(56 = \frac{P \times R \times 2}{100}\)
This simplifies to \(PR = \frac{56 \times 100}{2} = 2800\). (Equation 1)
The compound interest (CI) for 2 years is given as ₹72.
The formula for compound interest is \[CI = P \left[ \left(1 + \frac{R}{100}\right)^T - 1 \right]\].
So, \[72 = P \left[ \left(1 + \frac{R}{100}\right)^2 - 1 \right]\]
Expanding the term: \[72 = P \left[ 1 + \frac{2R}{100} + \frac{R^2}{100^2} - 1 \right]\]
\[72 = P \left[ \frac{2R}{100} + \frac{R^2}{10000} \right]\]
\[72 = \frac{2PR}{100} + \frac{PR^2}{10000}\] (Equation 2)
Substitute \(PR = 2800\) from Equation 1 into Equation 2:
\[72 = \frac{2 \times 2800}{100} + \frac{2800 \times R}{10000}\]
\(72 = 2 \times 28 + \frac{28R}{100}\)
\(72 = 56 + \frac{7R}{25}\)
Solve for \(R\):
\(72 - 56 = \frac{7R}{25}\)
\(16 = \frac{7R}{25}\)
\[R = \frac{16 \times 25}{7} = \frac{400}{7}\] %
Now, substitute the value of \(R\) back into Equation 1 (\(PR = 2800\)) to find the principal \(P\):
\(P \times \frac{400}{7} = 2800\)
\(P = \frac{2800 \times 7}{400}\)
\(P = 7 \times 7\)
\(P = 49\)
For 2 years, the difference between CI and SI is given by the formula:
\(CI - SI = P \left( \frac{R}{100} \right)^2\)
Given \(SI = 56\) and \(CI = 72\).
\(CI - SI = 72 - 56 = 16\)
So, \(16 = P \left( \frac{R}{100} \right)^2\). (Equation 3)
The simple interest for 1 year is \[SI_1 = \frac{SI}{2} = \frac{56}{2} = 28\].
We know that the difference between CI and SI for 2 years is the interest on the first year's simple interest. That is, \(CI - SI = SI_1 \times \frac{R}{100}\).
So, \(16 = 28 \times \frac{R}{100}\)
\[R = \frac{16 \times 100}{28} = \frac{4 \times 100}{7} = \frac{400}{7}\] %
Now, use the simple interest formula for 2 years: \(SI = \frac{P \times R \times T}{100}\)
\[56 = \frac{P \times \frac{400}{7} \times 2}{100}\]
\(56 = \frac{P \times 800}{700}\)
\(56 = \frac{8P}{7}\)
\(P = \frac{56 \times 7}{8}\)
\(P = 7 \times 7\)
\(P = 49\)
The principal amount is ₹49.
Correct Option: C) 49
To solve this problem, we need to apply two rules to the digits of the given number: add 1 to each odd digit and subtract 2 from each even digit. After transforming the number, we identify the second digit from the left and the second digit from the right and then calculate their sum.
Identify the original number: The given number is 4835176.
Apply the rules to each digit:
Let's process each digit:
The new number formed by these transformed digits is 2646284.
Identify the second digit from the left: In the new number 2646284, the second digit from the left is 6.
Identify the second digit from the right: In the new number 2646284, the second digit from the right is 8.
Calculate the sum: Add the two identified digits.
\[ \text{Sum} = 6 + 8 = 14 \]D) 14. The sum of the second digit from the left (6) and the second digit from the right (8) in the new number is 14.
A balanced chemical equation adheres to the Law of Conservation of Mass, stating that matter cannot be created or destroyed. This means the number of atoms of each element on the reactant side (left side) must be equal to the number of atoms of the same element on the product side (right side) of the equation.
This equation represents the formation of hydrogen peroxide from water and oxygen. Let's check the balance:
A) \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\)
This equation has \(4\) hydrogen atoms and \(2\) oxygen atoms on both the reactant and product sides, satisfying the Law of Conservation of Mass.
If x varies inversely as y³ - 1 and is equal to 7 when y = 2, find x when y = 6.
\(\frac{49}{215}\)
\(\frac{49}{216}\)
\(\frac{50}{216}\)
\(\frac{51}{215}\)
The problem describes an inverse variation relationship between two variables, \(x\) and \(y\). We are given an initial condition (\(x\) when \(y=2\)) and asked to find \(x\) for a new value of \(y\). The key is to first establish the constant of proportionality using the given values and then use this constant to find the unknown value.
A question is followed by two statements numbered (I) and (II). You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and select the most appropriate answer.
Question: Five friends, P, Q, R, S and T, are sitting around a circular table facing the centre. Who is sitting second to the left of T?
Statements:
(I) S sits second to the right of P. T sits to the immediate left of P.
(II) Only one person sits between Q and R. P sits to the immediate left of R.
Data in statement I alone is sufficient to answer the question while data in statement II is not.
Data in statements I and II together are not sufficient to answer the question.
Data in statement II alone is sufficient to answer the question while data in statement I is not.
Data in statements I and II together are sufficient to answer the question.
This problem involves setting up equations based on the given average ages and the relationship between Mohan's age and the average age of his friends. We will use the formula for average: Average = (Sum of quantities) / (Number of quantities).
Let Mohan's age be \(M\) and the ages of his four friends be \(F_1, F_2, F_3, F_4\).
The average age of Mohan and his four friends (5 people) is 36 years. So, the sum of their ages is:
\[ M + F_1 + F_2 + F_3 + F_4 = 5 \times 36 = 180 \quad \text{(Equation 1)} \]Let the average age of his four friends be \(A_F\). Then, the sum of the ages of his four friends is:
\[ F_1 + F_2 + F_3 + F_4 = 4 \times A_F \quad \text{(Equation 2)} \]Substitute Equation 2 into Equation 1:
\[ M + 4 A_F = 180 \quad \text{(Equation 3)} \]We are given that Mohan is 5 years older than the average age of his four friends. So,
\[ M = A_F + 5 \quad \text{(Equation 4)} \]Now, substitute Equation 4 into Equation 3:
\[ (A_F + 5) + 4 A_F = 180 \]Combine like terms:
\[ 5 A_F + 5 = 180 \]Subtract 5 from both sides:
\[ 5 A_F = 180 - 5 \]\[ 5 A_F = 175 \]Divide by 5 to find \(A_F\):
\[ A_F = \frac{175}{5} \]\[ A_F = 35 \text{ years} \]Finally, substitute the value of \(A_F\) back into Equation 4 to find Mohan's age \(M\):
\[ M = A_F + 5 \]\[ M = 35 + 5 \]\[ M = 40 \text{ years} \]B) Mohan's age is 40 years.
This problem involves calculating the combined work rate of multiple taps (two filling and one emptying) to determine the total time required to fill a cistern. We will find the individual rates of each tap and then combine them to find the net filling rate.
Correct Option: D) 2
To find the smallest perfect square divisible by 12, 15, and 18, we first need to find the Least Common Multiple (LCM) of these numbers. Then, we will make the LCM a perfect square by multiplying it by the necessary factors. Finally, we will divide this perfect square by 25 and find the sum of the digits of the quotient.
The LCM is found by taking the highest power of all prime factors present in the numbers.
\[ \text{LCM}(12, 15, 18) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180 \]For a number to be a perfect square, all the exponents in its prime factorization must be even. The prime factorization of the LCM is \(2^2 \times 3^2 \times 5^1\). Here, the exponent of 5 is 1 (odd). To make it a perfect square, we need to multiply it by \(5^1\).
\[ M = (2^2 \times 3^2 \times 5^1) \times 5^1 = 2^2 \times 3^2 \times 5^2 \]\[ M = 4 \times 9 \times 25 = 36 \times 25 = 900 \]The quotient is 36. The sum of its digits is \(3 + 6 = 9\).
B) 9 — The smallest perfect square divisible by 12, 15, and 18 is 900. When 900 is divided by 25, the quotient is 36. The sum of the digits of 36 is \(3 + 6 = 9\).