This problem involves calculating distances and times based on the speed of sound and the speed of drones. We need to work backward from the reception times of the drone pictures to determine the explosion time. The key is to understand that the first drone starts flying when the sound reaches the camp, and the second drone starts flying immediately after the first picture is received at the camp.
Let \(D\) be the distance from the camp to the explosion spot.
Let \(t_E\) be the time of explosion.
Let \(v_S\) be the speed of sound, \(v_S = 300 \text{ m/s}\).
Let \(v_D\) be the speed of the drones, \(v_D = 30 \text{ m/s}\).
Time when sound reaches the camp:
The sound travels distance \(D\) at speed \(v_S\).
Time taken for sound to reach camp \[t_{\text{sound travel}} = \frac{D}{v_S}\].
The sensor receives the sound at \(t_E + \frac{D}{v_S}\). At this moment, the first drone starts flying.
First Drone's Journey:
The first drone flies from the camp to the explosion spot (distance \(D\)) at speed \(v_D\).
Time taken for the first drone to reach the spot \[t_{\text{drone1 travel}} = \frac{D}{v_D}\].
The drone clicks a picture as soon as it reaches the spot and sends it back. The problem states the camp receives it at the same time. This implies the picture transmission is instantaneous or negligible.
So, the time the first picture is received at the camp is \[t_E + \frac{D}{v_S} + \frac{D}{v_D}\].
We are given this time is 5:02 PM.
\[ t_E + \frac{D}{300} + \frac{D}{30} = 5:02 \text{ PM} \quad (1) \]
Second Drone's Journey:
The second drone starts flying immediately after the first picture is received at the camp (5:02 PM).
It flies to the spot (distance \(D\)) at speed \(v_D\).
Time taken for the second drone to reach the spot \[t_{\text{drone2 travel}} = \frac{D}{v_D}\].
It sends a picture as soon as it reaches the spot. This picture is received at 5:05 PM.
So, the time the second picture is received at the camp is \(5:02 \text{ PM} + \frac{D}{v_D}\).
\[ 5:02 \text{ PM} + \frac{D}{30} = 5:05 \text{ PM} \quad (2) \]
Calculate Distance \(D\):
From equation (2):
\[ \frac{D}{30} = 5:05 \text{ PM} - 5:02 \text{ PM} \]\[ \frac{D}{30} = 3 \text{ minutes} \]\[ \frac{D}{30} = 3 \times 60 \text{ seconds} \]\[ \frac{D}{30} = 180 \text{ seconds} \]\[ D = 180 \times 30 \]\[ D = 5400 \text{ meters} \]
Calculate Explosion Time \(t_E\):
Substitute \(D = 5400 \text{ m}\) into equation (1):
\[ t_E + \frac{5400}{300} + \frac{5400}{30} = 5:02 \text{ PM} \]\[ t_E + 18 \text{ seconds} + 180 \text{ seconds} = 5:02 \text{ PM} \]\[ t_E + 198 \text{ seconds} = 5:02 \text{ PM} \]\[ t_E = 5:02 \text{ PM} - 198 \text{ seconds} \]
Convert 5:02 PM to seconds from a reference point (e.g., 5:00 PM):
\(5:02 \text{ PM} = 5 \text{ hours } 2 \text{ minutes}\)
\(2 \text{ minutes} = 120 \text{ seconds}\)
So, \(t_E = 5:00 \text{ PM} + 120 \text{ seconds} - 198 \text{ seconds}\)
\(t_E = 5:00 \text{ PM} - 78 \text{ seconds}\)
Subtract 78 seconds from 5:00 PM:
\(5:00 \text{ PM} - 78 \text{ seconds} = 4:59 \text{ PM} - 18 \text{ seconds}\)
\(4:59 \text{ PM} - 18 \text{ seconds} = 4:58 \text{ PM } 42 \text{ seconds}\)
So, \(t_E = 4:58:42 \text{ PM}\).
C) 4:58:42 PM is the calculated time of the explosion.
The digit in the unit place of the number 6129 x 7307 is
To find the digit in the unit place of a product of two numbers, we only need to consider the unit digits of the numbers being multiplied. The unit digit of the product will be the unit digit of the product of their unit digits.
C) 8. Assuming the question intended to ask for the unit digit of \(6^{129} \times 7^{307}\), the calculation is as follows:
To solve this problem, we will assume an initial salary and then calculate the initial savings and expenditure. We will then calculate the new salary and new expenditure based on the given percentage increases. Finally, we will determine the new savings and calculate the percentage change in savings.
Let the initial salary be \( \text{Rs. } 100 \).
A person saves 10% of his salary. So, initial savings will be:
\[ \text{Initial Savings} = 10\% \text{ of } 100 = \frac{10}{100} \times 100 = \text{Rs. } 10 \]Expenditure is Salary - Savings. So, initial expenditure will be:
\[ \text{Initial Expenditure} = \text{Initial Salary} - \text{Initial Savings} = 100 - 10 = \text{Rs. } 90 \]His salary increases by 12%. So, the new salary will be:
\[ \text{New Salary} = \text{Initial Salary} + 12\% \text{ of Initial Salary} \] \[ \text{New Salary} = 100 + \frac{12}{100} \times 100 = 100 + 12 = \text{Rs. } 112 \]His expenditure increases by 10%. So, the new expenditure will be:
\[ \text{New Expenditure} = \text{Initial Expenditure} + 10\% \text{ of Initial Expenditure} \] \[ \text{New Expenditure} = 90 + \frac{10}{100} \times 90 = 90 + 9 = \text{Rs. } 99 \]New Savings = New Salary - New Expenditure.
\[ \text{New Savings} = 112 - 99 = \text{Rs. } 13 \]Change in Savings = New Savings - Initial Savings.
\[ \text{Change in Savings} = 13 - 10 = \text{Rs. } 3 \]Percentage Change in Savings is calculated as:
\[ \text{Percentage Change} = \frac{\text{Change in Savings}}{\text{Initial Savings}} \times 100\% \] \[ \text{Percentage Change} = \frac{3}{10} \times 100\% = 30\% \]Since the change is positive, it is an increase.
B) 30% increase
This is a blood relation problem. We need to deduce the relationship between Q and S based on the given statements. We can represent the family tree to visualize the connections.
Correct Option: D) Q is a sister of S's husband
To find the number of three-digit numbers that can be expressed as an integral power of 2, we need to list the powers of 2 and identify which of them fall within the range of three-digit numbers (100 to 999).
Understand the range of three-digit numbers: A three-digit number is any integer from 100 to 999, inclusive.
List the integral powers of 2: We will start listing powers of 2 until we exceed the range of three-digit numbers.
Count the identified numbers: The powers of 2 that are three-digit numbers are 128, 256, and 512.
Determine the total count: There are 3 such numbers.
Correct Option: C) 3
The task is to identify correct inferences based solely on the provided passage. Each statement must be evaluated against the text to determine its validity.
Based on this evaluation, statements 1 and 3 are correct inferences.
Correct Option: C) 1 and 3
The question asks to identify the correct statements regarding measures taken in the National Capital Region (NCR) concerning pollution and environmental regulations. We need to evaluate each statement based on known environmental policies and Supreme Court directives.
Statement 2 is factually correct. The assurance of CNG supplies was a necessary step to support the transition away from polluting diesel vehicles, especially in public transport, following various bans and restrictions on diesel vehicles in the NCR.
This is a syllogism problem where we need to deduce conclusions from given statements. We can visualize these relationships using Venn diagrams or by following the logical flow of the statements. The core idea is that if A is part of B, and B is part of C, then A must also be part of C.
Analyze the given statements:
Evaluate statement P: Every blue is yellow.
Evaluate statement Q: Every red is green.
Evaluate statement R: Every red is yellow.
Since statements P, Q, and R are all correct, the option that includes all three is the answer.
D) P, Q and R
All three statements P, Q, and R are logically derivable from the given premises. The relationships form a chain: Red \(\rightarrow\) Blue \(\rightarrow\) Green \(\rightarrow\) Yellow. Therefore, any item earlier in the chain is also part of all subsequent categories.
To find out how many times the digit '5' appears in all two-digit positive integers, we need to consider two cases: when '5' appears in the units place and when '5' appears in the tens place. We will list out the numbers for each case and sum the occurrences.
Numbers where '5' appears in the units place:
These are two-digit numbers ending in 5. The tens digit can be any digit from 1 to 9 (since it's a two-digit number, the tens digit cannot be 0). The numbers are:
\(15, 25, 35, 45, 55, 65, 75, 85, 95\)
There are 9 such numbers. In each of these numbers, the digit '5' appears once in the units place. So, the count of '5's from this set is 9.
Numbers where '5' appears in the tens place:
These are two-digit numbers starting with 5. The units digit can be any digit from 0 to 9.
\(50, 51, 52, 53, 54, 55, 56, 57, 58, 59\)
There are 10 such numbers. In each of these numbers, the digit '5' appears once in the tens place. So, the count of '5's from this set is 10.
Total occurrences:
We need to sum the occurrences from both cases. However, we must be careful not to double-count the number 55. In our counting, 55 was included in both lists (as a number ending in 5 and as a number starting with 5). We counted the '5' in the units place of 55 once and the '5' in the tens place of 55 once. This means we have correctly counted each '5' individually.
Total number of times '5' appears = (Occurrences from units place) + (Occurrences from tens place)
Total = \(9 + 10 = 19\)
Correct Option: B) 19
The digit '5' appears 9 times in the units place (15, 25, ..., 95) and 10 times in the tens place (50, 51, ..., 59). The number 55 contributes two '5's, and our method correctly counts both instances. Therefore, the total number of times '5' appears is \(9 + 10 = 19\).
To determine how many minutes early Y completes the journey compared to X, we need to calculate the total time taken by each person. The total journey is 6 km, divided into three segments with different speeds. We will calculate the time for each segment using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), sum these times to get the total journey time for X and Y, and then find the difference in minutes.
D) 6 — The calculations show that X takes 23/20 hours and Y takes 21/20 hours. The difference is 2/20 hours or 1/10 hours, which is 6 minutes. Therefore, Y completes the journey 6 minutes early.