The question asks for the location of a specific two-day Chintan Shivir held in March 2025 to discuss preparations for the 2028 and 2036 Olympics. This is a factual question related to current affairs and government initiatives concerning sports.
Correct Option: B) Hyderabad
The two-day Chintan Shivir, organized by the Ministry of Youth Affairs and Sports to discuss preparations for the 2028 Los Angeles Olympics and the 2036 Olympic bid, was held in Hyderabad on March 7-8, 2025. This event brought together various stakeholders to strategize India's Olympic aspirations.
To solve this problem, we need to track Uday's movement in a coordinate system or by drawing a path. We'll determine his final position relative to his starting point (Point A) and then calculate the shortest distance and direction needed to return to Point A.
Therefore, Uday needs to drive 5 km towards East to reach Point A.
D) 5 km; East is the correct answer. Uday's final position is 5 km to the west of his starting point. To return to the starting point, he must travel 5 km to the east.
The problem requires us to evaluate a given mathematical expression after interchanging certain arithmetic operators. We need to replace '+' with '−', '−' with '+', '×' with '÷', and '÷' with '×'. After performing these substitutions, we will evaluate the new expression following the order of operations (BODMAS/PEMDAS).
\[ 63 \div 12 - 98 \times 7 + 35 = ? \]
\[ 63 \times 12 + 98 \div 7 - 35 = ? \]
First, perform multiplication and division from left to right.
\[ 63 \times 12 = 756 \]
\[ 98 \div 7 = 14 \]
Substitute these values back into the expression:
\[ 756 + 14 - 35 = ? \]
Next, perform addition and subtraction from left to right.
\[ 756 + 14 = 770 \]
\[ 770 - 35 = 735 \]
Correct Option: B) 735
Solve: 70 + \( \frac{72}{6} \) + 29 − 30
To solve this problem, 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 given expression is:
\[ 70 + \frac{72}{6} + 29 - 30 \]B) 81
The question asks about the target percentage for rural households to receive tap water under the Jal Jeevan Mission by 2024. This requires knowledge of the mission's objectives and timelines.
Correct Option: A) 100%
To find the height of the cone, we will first use the given curved surface area and diameter to calculate the slant height. Once the slant height and radius are known, we can use the Pythagorean theorem to find the height of the cone.
The radius (\(r\)) is half of the diameter.
\[ r = \frac{\text{Diameter}}{2} = \frac{100 \text{ cm}}{2} = 50 \text{ cm} \]The formula for the curved surface area of a cone is \( \text{CSA} = \pi r l \), where \(l\) is the slant height.
\[ 6500\pi = \pi \times 50 \times l \]Divide both sides by \( \pi \):
\[ 6500 = 50 \times l \]Solve for \(l\):
\[ l = \frac{6500}{50} = 130 \text{ cm} \]For a right circular cone, the radius (\(r\)), height (\(h\)), and slant height (\(l\)) form a right-angled triangle, with \(l\) as the hypotenuse. According to the Pythagorean theorem:
\[ l^2 = r^2 + h^2 \]Substitute the values of \(l\) and \(r\):
\[ (130)^2 = (50)^2 + h^2 \] \[ 16900 = 2500 + h^2 \]Solve for \(h^2\):
\[ h^2 = 16900 - 2500 \] \[ h^2 = 14400 \]Take the square root to find \(h\):
\[ h = \sqrt{14400} = 120 \text{ cm} \]Correct Option: B) 120
Two pipes, A and B, can fill a tank of 3700 litres in 9 hours and 6 hours, respectively. If they are opened together, how many hours will they take to fill an empty tank of 6400 litres?
\( \frac{1152}{185} \) hours
\( \frac{1149}{185} \) hours
\( \frac{1159}{185} \) hours
\( \frac{1153}{185} \) hours
This problem involves calculating the combined work rate of two pipes and then using that rate to find the time required to fill a different tank capacity. The key is to first determine the rate at which each pipe fills the tank and then sum these rates to find the combined rate. Finally, divide the new tank volume by the combined rate to get the time.
Calculate the filling rate of Pipe A:
Pipe A fills a 3700-litre tank in 9 hours. So, its rate of filling is:
\[ \text{Rate of A} = \frac{\text{Volume}}{\text{Time}} = \frac{3700 \text{ litres}}{9 \text{ hours}} = \frac{3700}{9} \text{ litres/hour} \]Calculate the filling rate of Pipe B:
Pipe B fills a 3700-litre tank in 6 hours. So, its rate of filling is:
\[ \text{Rate of B} = \frac{\text{Volume}}{\text{Time}} = \frac{3700 \text{ litres}}{6 \text{ hours}} = \frac{3700}{6} \text{ litres/hour} \]Calculate the combined filling rate of Pipe A and Pipe B:
When opened together, their rates add up:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} \] \[ \text{Combined Rate} = \frac{3700}{9} + \frac{3700}{6} \]To add these fractions, find a common denominator, which is 18:
\[ \text{Combined Rate} = \frac{3700 \times 2}{9 \times 2} + \frac{3700 \times 3}{6 \times 3} \] \[ \text{Combined Rate} = \frac{7400}{18} + \frac{11100}{18} \] \[ \text{Combined Rate} = \frac{7400 + 11100}{18} = \frac{18500}{18} \text{ litres/hour} \]Simplify the combined rate:
\[ \text{Combined Rate} = \frac{9250}{9} \text{ litres/hour} \]Calculate the time taken to fill a 6400-litre tank:
Now, we need to find the time \(T\) to fill a new tank of 6400 litres with the combined rate:
\[ \text{Time} = \frac{\text{New Tank Volume}}{\text{Combined Rate}} \] \[ T = \frac{6400 \text{ litres}}{\frac{9250}{9} \text{ litres/hour}} \] \[ T = 6400 \times \frac{9}{9250} \] \[ T = \frac{6400 \times 9}{9250} \] \[ T = \frac{57600}{9250} \]Simplify the fraction by dividing the numerator and denominator by 50:
\[ T = \frac{57600 \div 50}{9250 \div 50} \] \[ T = \frac{1152}{185} \text{ hours} \]A) \( \frac{1152}{185} \) hours is the correct time calculated.
To find the difference between compound interest (CI) and simple interest (SI) for 2 years, we first calculate the simple interest and compound interest separately. Alternatively, we can use a direct formula for the difference between CI and SI for 2 years.
The formula for Simple Interest is:
\[ \text{SI} = \frac{P \times R \times T}{100} \]\[ \text{SI} = \frac{47100 \times 9 \times 2}{100} \]\[ \text{SI} = 471 \times 18 \]\[ \text{SI} = \text{₹}8478 \]The formula for Amount (\(A\)) compounded annually is:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]\[ A = 47100 \left(1 + \frac{9}{100}\right)^2 \]\[ A = 47100 \left(\frac{109}{100}\right)^2 \]\[ A = 47100 \times \frac{109 \times 109}{100 \times 100} \]\[ A = 47100 \times \frac{11881}{10000} \]\[ A = 471 \times \frac{11881}{100} \]\[ A = \frac{5595951}{100} \]\[ A = \text{₹}55959.51 \]Now, calculate Compound Interest (CI):
\[ \text{CI} = A - P \]\[ \text{CI} = 55959.51 - 47100 \]\[ \text{CI} = \text{₹}8859.51 \]For 2 years, the difference between CI and SI is given by:
\[ \text{Difference} = P \left(\frac{R}{100}\right)^2 \]\[ \text{Difference} = 47100 \left(\frac{9}{100}\right)^2 \]\[ \text{Difference} = 47100 \times \frac{81}{10000} \]\[ \text{Difference} = \frac{471 \times 81}{100} \]\[ \text{Difference} = \frac{38151}{100} \]\[ \text{Difference} = \text{₹}381.51 \]C) ₹381.51 is the calculated difference between the compound interest and simple interest.
The question asks to identify who among the given options was NOT part of the Santhal Rebellion. This requires knowledge of the key figures and movements associated with the Santhal Rebellion and other related tribal movements in India.
Correct Option: D) Tana Bhagats were not part of the Santhal Rebellion. They were associated with a later movement (Tana Bhagat Movement) that started in 1914 among the Oraon tribe.
The arithmetic mean of 25 real numbers is 110. If 10 numbers are increased by 12\( \frac{1}{2} \) each, another 10 numbers are increased by 15 each and the remaining 5 numbers are unaltered, then the new value of the arithmetic mean is:
To find the new arithmetic mean, we first calculate the total sum of the 25 numbers. Then, we determine the total increase in the sum due to the changes. Finally, we add this increase to the original sum and divide by the total number of items (which remains 25) to get the new mean.
Calculate the original total sum:
The arithmetic mean of 25 numbers is 110. The sum of numbers is given by \(\text{Mean} \times \text{Number of items}\).
Calculate the total increase from the first set of numbers:
10 numbers are increased by \(12 \frac{1}{2}\) each. \(12 \frac{1}{2}\) can be written as 12.5.
Calculate the total increase from the second set of numbers:
Another 10 numbers are increased by 15 each.
Calculate the total increase in the sum:
The remaining 5 numbers are unaltered, so they contribute no increase.
Calculate the new total sum:
\[ \text{New Sum} = \text{Original Sum} + \text{Total Increase} = 2750 + 275 = 3025 \]Calculate the new arithmetic mean:
The total number of items remains 25.
B) 121 is the new value of the arithmetic mean.