This problem involves compound interest calculated half-yearly. We are given the principal amount, the final amount, and the time period. We need to find the annual rate of interest. The key is to adjust the time period and the rate for half-yearly compounding.
Since the interest is compounded half-yearly, we need to convert the time into half-year periods and the annual rate into a half-yearly rate.
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]Where \(A\) is the final amount, \(P\) is the principal, \(r\) is the rate per compounding period, and \(n\) is the number of compounding periods.
Simplify the fraction:
\[ \frac{15625}{8000} = \frac{3125}{1600} = \frac{625}{320} = \frac{125}{64} \]So,
\[ \frac{125}{64} = \left(1 + \frac{R}{200}\right)^3 \]We know that \(5^3 = 125\) and \(4^3 = 64\).
\[ \left(\frac{5}{4}\right)^3 = \left(1 + \frac{R}{200}\right)^3 \]\[ \frac{5}{4} = 1 + \frac{R}{200} \]The annual rate of interest is 50%.
B) 50% is the correct annual rate of interest.
Find the value of \(\sqrt{125+\sqrt{344+\sqrt{289}}}\).
To solve this problem, we need to evaluate the expression by working from the innermost square root outwards. This involves calculating the square root of perfect squares at each step and then performing the addition.
Start with the innermost square root: \(\sqrt{289}\).
\[ \sqrt{289} = 17 \]Substitute this value back into the expression:
\[ \sqrt{125+\sqrt{344+17}} \]Perform the addition inside the next square root:
\[ 344 + 17 = 361 \]Now, evaluate the next square root: \(\sqrt{361}\).
\[ \sqrt{361} = 19 \]Substitute this value back into the expression:
\[ \sqrt{125+19} \]Perform the final addition:
\[ 125 + 19 = 144 \]Finally, evaluate the outermost square root: \(\sqrt{144}\).
\[ \sqrt{144} = 12 \]Thus, the value of the given expression is 12.
B) 12 is the correct answer, as derived from the step-by-step calculation.
This problem involves calculating displacement based on a series of movements in different directions. We can solve this by breaking down Arpan's journey into components along the East-West and North-South axes. We'll track his final position relative to his starting point (Point A) and then determine the shortest distance and direction needed to return to Point A.
Therefore, Arpan needs to drive 4 km to the South to reach Point A.
A) 4 km to the south - Based on the step-by-step calculation, Arpan's final position is 4 km North of Point A. To return to Point A, he must drive 4 km South.
To find the greatest number that divides two given numbers leaving specific remainders, we first subtract the respective remainders from the numbers. The problem then reduces to finding the Greatest Common Divisor (GCD) or Highest Common Factor (HCF) of the resulting numbers. The GCD of these new numbers will be the required greatest number.
We are given that the greatest number divides 152 leaving a remainder of 2. This means that if we subtract 2 from 152, the resulting number will be perfectly divisible by the greatest number.
\[ 152 - 2 = 150 \]Similarly, the greatest number divides 181 leaving a remainder of 6. So, if we subtract 6 from 181, the resulting number will also be perfectly divisible by the greatest number.
\[ 181 - 6 = 175 \]Now, we need to find the greatest number that divides both 150 and 175 exactly. This is equivalent to finding the Greatest Common Divisor (GCD) or Highest Common Factor (HCF) of 150 and 175.
Let's find the prime factorization of 150:
\[ 150 = 2 \times 75 = 2 \times 3 \times 25 = 2 \times 3 \times 5^2 \]Now, let's find the prime factorization of 175:
\[ 175 = 5 \times 35 = 5 \times 5 \times 7 = 5^2 \times 7 \]To find the GCD, we take the common prime factors raised to the lowest power they appear in either factorization.
The common prime factor is 5, and the lowest power is \(5^2\).
\[ \text{GCD}(150, 175) = 5^2 = 25 \]Thus, the greatest number that divides 152 and 181 leaving remainders 2 and 6, respectively, is 25.
Correct Option: C) 25
The question pertains to the powers of the President of India regarding the reconsideration of advice given by the Council of Ministers. This is a specific provision under the Indian Constitution, amended to clarify the President's role.
Correct Option: D) The President can ask the Prime Minister to have an executive recommendation reconsidered by the Cabinet only once. This is a direct consequence of the 44th Constitutional Amendment Act, which allows for a single instance of reconsideration, after which the President is bound by the reconsidered advice.
To solve this problem, we need to track Ganesh's movements step-by-step on a coordinate plane or by drawing a path. We will determine his final position relative to his starting point A and then calculate the shortest distance and direction needed to return to A.
Therefore, Ganesh needs to drive 4 km to the west to reach point A again.
Correct Option: A) 4 km to the west
The question asks for the names of the venues where the Khelo India Para Games (KIPG) 2025 were held. This requires specific knowledge about the event's locations.
D) Jawaharlal Nehru Indoor Stadium, Indira Gandhi Indoor Stadium, and Dr Karni Singh Shooting Range. These three venues were indeed the hosts for the inaugural Khelo India Para Games, which were held in Delhi.
To find the median of a given set of data, we first need to arrange the data in ascending or descending order. Once ordered, the median is the middle value if the number of data points is odd. If the number of data points is even, the median is the average of the two middle values.
List the given data:
\(72, 47, 98, 52, 20, 86, 85, 69, 59, 81, 53\)
Count the number of data points (n):
There are \(11\) data points.
Arrange the data in ascending order:
\(20, 47, 52, 53, 59, 69, 72, 81, 85, 86, 98\)
Determine the position of the median:
Since \(n = 11\) (an odd number), the median is the \(\left(\frac{n+1}{2}\right)^{\text{th}}\) term.
\[\text{Median position} = \left(\frac{11+1}{2}\right)^{\text{th}} = \left(\frac{12}{2}\right)^{\text{th}} = 6^{\text{th}}\text{ term}\]Identify the median value:
Looking at the ordered list, the \(6^{\text{th}}\) term is \(69\).
\(20, 47, 52, 53, 59, \mathbf{69}, 72, 81, 85, 86, 98\)
Therefore, the median of the given data is \(69\).
B) 69
This is a scheduling puzzle. We need to determine the exam day for each person (P, Q, R, S, T, U, V) based on the given clues, starting from Monday to Sunday. We will use a table to systematically fill in the information.
| Day | Person |
| Monday | |
| Tuesday | |
| Wednesday | |
| Thursday | |
| Friday | |
| Saturday | |
| Sunday |
| Day | Person |
| Monday | |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | |
| Sunday |
This means S and U cannot be on Sunday. If 4 people are between S and U, they must be at the extreme ends of the available days (excluding Sunday). The only way to have 4 people between them is if one is on Monday and the other on Saturday, or vice versa.
Let's consider Case 1 first: S on Monday, U on Saturday.
| Day | Person |
| Monday | S |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | U |
| Sunday |
This means R and P must be on consecutive days. From the current table, the only consecutive empty slots are Tuesday and Wednesday.
| Day | Person |
| Monday | S |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | |
| Saturday | U |
| Sunday |
The remaining empty slots are Friday and Sunday. If T cannot be on Friday, T must be on Sunday.
| Day | Person |
| Monday | S |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | |
| Saturday | U |
| Sunday | T |
The last remaining person is Q, who must have their exam on Friday.
| Day | Person |
| Monday | S |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | Q |
| Saturday | U |
| Sunday | T |
In our current arrangement, S is on Monday and P is on Wednesday. So, S (Monday) is not after P (Wednesday). This arrangement is consistent with all clues.
Let's check the other case from Step 3 (U on Monday, S on Saturday) to ensure it doesn't work:
| Day | Person |
| Monday | U |
| Tuesday | |
| Wednesday | |
| Thursday | V |
| Friday | |
| Saturday | S |
| Sunday |
Now, R immediately before P. The only consecutive slots are Tuesday and Wednesday. So, R on Tuesday, P on Wednesday.
| Day | Person |
| Monday | U |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | |
| Saturday | S |
| Sunday |
Remaining slots are Friday and Sunday. T cannot be on Friday, so T must be on Sunday. Q on Friday.
| Day | Person |
| Monday | U |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | Q |
| Saturday | S |
| Sunday | T |
Now check Clue 5: S does not have an exam after P. In this case, S is on Saturday and P is on Wednesday. S (Saturday) is after P (Wednesday). This contradicts the clue. Therefore, this case is invalid.
The first arrangement is the correct one.
| Day | Person |
| Monday | S |
| Tuesday | R |
| Wednesday | P |
| Thursday | V |
| Friday | Q |
| Saturday | U |
| Sunday | T |
The question asks who has an exam on Tuesday. From the final arrangement, R has an exam on Tuesday.
C) R — Based on the step-by-step deduction, R has an exam on Tuesday.