The question asks to identify the tree whose twig is traditionally used for cleaning teeth. This is a common practice in many parts of India and other countries, relying on the natural properties of certain plants for oral hygiene.
A) Neem. Neem twigs (datun) have been used for centuries in India and other parts of the world for oral hygiene. They possess strong antimicrobial and antiseptic properties that help in cleaning teeth, preventing gum diseases, and combating bad breath.
The question asks for the condition under which a chemical equation is considered balanced. This relates directly to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, the total number of atoms of each element must remain constant from reactants to products.
A) Atoms are the same in both sides — This is the correct definition of a balanced chemical equation. For example, in the reaction \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\), there are 4 hydrogen atoms and 2 oxygen atoms on both the reactant and product sides.
The problem presents a series of equations where a custom operation (represented by '*') is defined. We need to identify the pattern or rule governing this operation and then apply it to the final expression to find the result.
Let's consider the numbers involved in each equation. A common approach for such problems is to perform basic arithmetic operations (addition, subtraction, multiplication, division) on the digits or the numbers themselves.
If we add the two numbers: \(16 + 24 = 40\). If we divide by 2: \(40 / 2 = 20\).
If we add the two numbers: \(71 + 19 = 90\). If we divide by 2: \(90 / 2 = 45\).
If we add the two numbers: \(8 + 52 = 60\). If we divide by 2: \(60 / 2 = 30\).
The pattern is consistent across all given examples. The operation '*' means to add the two numbers and then divide the sum by 2.
We need to find the value of \(35 * 27\).
Therefore, \(35 * 27 = 31\).
D) 31 — The pattern identified is to sum the two numbers and then divide the result by 2. Applying this to \(35 * 27\) gives \((35 + 27) / 2 = 62 / 2 = 31\).
This problem involves concepts of speed, distance, and time in the context of boats and streams. We will use the formulas for speed upstream (against current) and speed downstream (along current) to set up two equations and then solve them simultaneously to find the speed of the water current.
Let the speed of the boat in still water be \(v_b\) km/h and the speed of the water current be \(v_c\) km/h.
When the boat travels against the current (upstream), its effective speed is the speed of the boat minus the speed of the current:
\[ \text{Speed upstream} = v_b - v_c \]When the boat travels along the current (downstream), its effective speed is the speed of the boat plus the speed of the current:
\[ \text{Speed downstream} = v_b + v_c \]We are given that the boat covers 24 km against the current in 6 hours. Using the formula \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\), we can write:
\[ v_b - v_c = \frac{24 \text{ km}}{6 \text{ h}} \] \[ v_b - v_c = 4 \text{ km/h} \quad \text{(Equation 1)} \]We are also given that the boat covers 36 km along the current in 6 hours. Similarly, we can write:
\[ v_b + v_c = \frac{36 \text{ km}}{6 \text{ h}} \] \[ v_b + v_c = 6 \text{ km/h} \quad \text{(Equation 2)} \]Now we have a system of two linear equations:
Equation 1: \(v_b - v_c = 4\)
Equation 2: \(v_b + v_c = 6\)
To find the speed of the water current (\(v_c\)), we can subtract Equation 1 from Equation 2:
\[ (v_b + v_c) - (v_b - v_c) = 6 - 4 \] \[ v_b + v_c - v_b + v_c = 2 \] \[ 2v_c = 2 \] \[ v_c = \frac{2}{2} \] \[ v_c = 1 \text{ km/h} \]Correct Option: C) 1 km/h
The question asks to identify the type of organism that can survive without oxygen. This requires knowledge of basic biological classifications based on oxygen requirements for respiration.
Correct Option: D) Anaerobe
The problem presents an alphabet series with groups of two letters. To find the missing group, we need to identify the pattern in how the letters within each group are related and how consecutive groups are related.
Each group consists of two consecutive letters in reverse alphabetical order.
This means the missing group will also have two consecutive letters, with the first letter being alphabetically after the second.
Let's look at the first letter of each group and the last letter of each group.
Consider the alphabetical positions (A=1, B=2, ..., Z=26):
So, the first letter of the missing group is Q.
So, the second letter of the missing group is P.
The missing group is QP.
A) QP is the correct answer. The pattern involves a decreasing sequence of letters, with the gap between the first letters of consecutive groups increasing by one each time (1, 2, 3, 4 letters skipped). The same pattern applies to the second letters of consecutive groups. Within each group, the letters are consecutive in reverse alphabetical order (e.g., ZY, XW).
The problem asks us to find the missing number in a given series. We need to identify the pattern or rule governing the sequence of numbers.
B) 120 — The series follows a pattern where each term is multiplied by a decreasing integer: \(1 \times 6 = 6\), \(6 \times 5 = 30\), \(30 \times 4 = 120\), \(120 \times 3 = 360\), \(360 \times 2 = 720\). Thus, the missing number is 120.
We need to find a number that satisfies two conditions: first, when divided by 5, 6, 7, and 8, it leaves a remainder of 3; second, it must be a multiple of 9. We will use the concept of LCM (Least Common Multiple) and the remainder theorem, followed by checking divisibility by 9.
A) 1683 — As calculated in the step-by-step explanation, 1683 is the least number that leaves a remainder of 3 when divided by 5, 6, 7, and 8, and is also a multiple of 9.
A man invested \( \frac{1}{3} \) of his capital at 7%, \( \frac{1}{4} \) at 8% and the rest at 10% thereby getting an annual income of Rs.561. What was his capital?
To find the total capital, we need to set up an equation based on the annual income received from different investments. We will calculate the interest earned from each part of the capital and sum them up to equal the total annual income given.
Let the total capital be \(P\).
The capital is divided into three parts:
Calculate the fraction of the capital for the third part:
\[ \text{Fraction of rest} = 1 - \left( \frac{1}{3} + \frac{1}{4} \right) \]\[ \text{Fraction of rest} = 1 - \left( \frac{4+3}{12} \right) \]\[ \text{Fraction of rest} = 1 - \frac{7}{12} \]\[ \text{Fraction of rest} = \frac{12-7}{12} = \frac{5}{12} \]Now, calculate the simple interest (annual income) from each part:
The total annual income is Rs. 561. So, the sum of these interests must be equal to 561:
\[ I_1 + I_2 + I_3 = 561 \]\[ \frac{7P}{300} + \frac{2P}{100} + \frac{5P}{120} = 561 \]Find a common denominator for 300, 100, and 120. The least common multiple (LCM) is 600.
\[ \frac{7P \times 2}{600} + \frac{2P \times 6}{600} + \frac{5P \times 5}{600} = 561 \]\[ \frac{14P}{600} + \frac{12P}{600} + \frac{25P}{600} = 561 \]\[ \frac{14P + 12P + 25P}{600} = 561 \]\[ \frac{51P}{600} = 561 \]Solve for \(P\):
\[ P = \frac{561 \times 600}{51} \]\[ P = 11 \times 600 \]\[ P = 6600 \]Thus, the total capital was Rs. 6,600.
D) Rs.6,600 — The calculation shows that the total capital is Rs. 6,600, which yields an annual income of Rs. 561 based on the given investment distribution and interest rates.