The problem asks us to find the cost price of a gold ring given its selling price and the percentage loss incurred. We can use the formula relating selling price, cost price, and loss percentage to solve this problem.
When there is a loss, the Selling Price is less than the Cost Price. The loss percentage is calculated on the Cost Price.
If the loss is 15%, it means the selling price is \(100\% - 15\% = 85\%\) of the Cost Price.
Let CP be the Cost Price.
\[ \text{SP} = \text{CP} - (\text{Loss Percentage of CP}) \]\[ \text{SP} = \text{CP} - \left(\frac{15}{100} \times \text{CP}\right) \]\[ \text{SP} = \text{CP} \left(1 - \frac{15}{100}\right) \]\[ \text{SP} = \text{CP} \left(\frac{100 - 15}{100}\right) \]\[ \text{SP} = \text{CP} \left(\frac{85}{100}\right) \]We have \(\text{SP} = \text{₹}19,125\).
\[ 19,125 = \text{CP} \times \frac{85}{100} \]\[ \text{CP} = 19,125 \times \frac{100}{85} \]To simplify, we can divide 19,125 by 85.
Divide 19125 by 85:
\[ \frac{19125}{85} = \frac{3825 \times 5}{17 \times 5} = \frac{3825}{17} \]\[ 3825 \div 17 \]\[ 17 \times 2 = 34 \]\[ 38 - 34 = 4 \]\[ \text{Bring down 2, so } 42 \]\[ 17 \times 2 = 34 \]\[ 42 - 34 = 8 \]\[ \text{Bring down 5, so } 85 \]\[ 17 \times 5 = 85 \]\[ 85 - 85 = 0 \]So, \(19125 \div 85 = 225\).Now, multiply by 100:
\[ \text{CP} = 225 \times 100 \]\[ \text{CP} = \text{₹}22,500 \]B) ₹22,500 is the correct cost price of the gold ring.
To find the ratio of the perimeters of two squares given the ratio of their areas, we first need to establish the relationship between the side length, area, and perimeter of a square. We will use the given area ratio to find the ratio of their side lengths, and then use the side length ratio to find the ratio of their perimeters.
Let the side lengths of the two squares be \(s_1\) and \(s_2\) respectively.
The area of a square with side length \(s\) is given by \(A = s^2\). Therefore, the areas of the two squares are \(A_1 = s_1^2\) and \(A_2 = s_2^2\).
The problem states that the ratio of their areas is 16 : 9. So, we have:
\[ \frac{A_1}{A_2} = \frac{s_1^2}{s_2^2} = \frac{16}{9} \]To find the ratio of their side lengths, we take the square root of both sides of the equation:
\[ \sqrt{\frac{s_1^2}{s_2^2}} = \sqrt{\frac{16}{9}} \] \[ \frac{s_1}{s_2} = \frac{\sqrt{16}}{\sqrt{9}} = \frac{4}{3} \]So, the ratio of their side lengths is 4 : 3.
The perimeter of a square with side length \(s\) is given by \(P = 4s\). Therefore, the perimeters of the two squares are \(P_1 = 4s_1\) and \(P_2 = 4s_2\).
Now, we find the ratio of their perimeters:
\[ \frac{P_1}{P_2} = \frac{4s_1}{4s_2} = \frac{s_1}{s_2} \]Since we found \[\frac{s_1}{s_2} = \frac{4}{3}\], the ratio of their perimeters is also 4 : 3.
The ratio of the perimeters is 4 : 3, which matches option A.
The question asks to identify the captain of the Indian cricket team when they won the Cricket World Cup in 1983. This is a factual question related to Indian sports history.
D) Kapil Dev was the captain of the Indian cricket team when they won the Cricket World Cup in 1983. His leadership was instrumental in this historic achievement.
When two objects move in opposite directions from the same point, their relative speed is the sum of their individual speeds. We can use this relative speed to calculate the total distance between them after a given time.
They are moving in opposite directions from the same point.
Since they are moving in opposite directions, their speeds add up to find the rate at which the distance between them increases.
\[ \text{Relative Speed} (S_{\text{rel}}) = S_R + S_M \]\[ S_{\text{rel}} = 80 \text{ km/h} + 40 \text{ km/h} \]\[ S_{\text{rel}} = 120 \text{ km/h} \]The given time is 15 minutes. To use it with speed in km/h, convert minutes to hours.
\[ \text{Time} (T) = 15 \text{ min} \]\[ T = \frac{15}{60} \text{ hours} \]\[ T = \frac{1}{4} \text{ hours} \]The distance between them will be the relative speed multiplied by the time.
\[ \text{Distance} (D) = S_{\text{rel}} \times T \]\[ D = 120 \text{ km/h} \times \frac{1}{4} \text{ hours} \]\[ D = 30 \text{ km} \]Correct Option: B) 30 km
The sum of two angles is 155° and their difference is \(\frac{\pi}{2}\). The value of the greater angle (in radians) is:
\(\frac{53\pi}{72}\)
\(\frac{49\pi}{72}\)
\(\frac{47\pi}{72}\)
\(\frac{51\pi}{72}\)
We are given the sum and difference of two angles, one in degrees and the other in radians. The goal is to find the greater angle in radians. First, we will convert all given angle measures to a consistent unit (either degrees or radians). Then, we will set up a system of two linear equations with two variables representing the angles and solve for them. Finally, we will identify the greater angle and convert it to radians if necessary.
Let the two angles be \(A\) and \(B\), where \(A > B\).
From the problem statement, the sum of the two angles is 155°:
\[ A + B = 155^\circ \quad \text{(Equation 1)} \]The difference between the two angles is \(\frac{\pi}{2}\). We need to convert this to degrees for consistency with Equation 1. We know that \(\pi \text{ radians} = 180^\circ\).
\[ \frac{\pi}{2} \text{ radians} = \frac{180^\circ}{2} = 90^\circ \]So, the difference between the two angles is 90°:
\[ A - B = 90^\circ \quad \text{(Equation 2)} \]Now we have a system of two linear equations:
Equation 1: \(A + B = 155^\circ\)
Equation 2: \(A - B = 90^\circ\)
Add Equation 1 and Equation 2:
\[ (A + B) + (A - B) = 155^\circ + 90^\circ \]\[ 2A = 245^\circ \]\[ A = \frac{245^\circ}{2} = 122.5^\circ \]Substitute the value of \(A\) into Equation 1 to find \(B\):
\[ 122.5^\circ + B = 155^\circ \]\[ B = 155^\circ - 122.5^\circ \]\[ B = 32.5^\circ \]The two angles are \(122.5^\circ\) and \(32.5^\circ\). The greater angle is \(A = 122.5^\circ\).
We need to express the greater angle in radians. To convert degrees to radians, we use the conversion factor \(\frac{\pi}{180^\circ}\).
\[ A \text{ (in radians)} = 122.5^\circ \times \frac{\pi}{180^\circ} \]\[ A = \frac{122.5}{180} \pi \]To simplify the fraction, we can multiply the numerator and denominator by 10 to remove the decimal:
\[ A = \frac{1225}{1800} \pi \]Now, simplify the fraction by dividing by common factors. Both are divisible by 25:
\[ 1225 \div 25 = 49 \]\[ 1800 \div 25 = 72 \]So, the greater angle in radians is:
\[ A = \frac{49}{72} \pi \text{ radians} \]B) The greater angle is \(\frac{49\pi}{72}\) radians, which matches our calculated value.
The question asks to identify the two steel plants in India that were established with the collaboration of Russia (former USSR). This requires knowledge of the foreign collaborations behind India's major steel plants.
Correct Option: A) Bokaro Steel Plant and Bhilai Steel Plant were both set up with the collaboration of the former USSR (Russia).
The question asks to identify the Indian state associated with Madhubani art. This requires knowledge of traditional Indian art forms and their geographical origins.
Correct Option: D) Bihar is the correct answer. Madhubani art, also known as Mithila painting, originated in the Mithila region of Bihar, India. It is a significant cultural heritage of the state.
This problem involves finding two positive numbers given the product of their LCM and HCF, and their difference. We will use the fundamental property that for any two positive numbers, the product of the numbers is equal to the product of their LCM and HCF. Then, we will use the given difference to set up a system of equations or test the options.
Let the two positive numbers be \(a\) and \(b\).
We are given that the product of their LCM and HCF is 36. A fundamental property of numbers states that for any two positive integers \(a\) and \(b\):
\[ a \times b = \text{LCM}(a, b) \times \text{HCF}(a, b) \]Given \(\text{LCM}(a, b) \times \text{HCF}(a, b) = 36\), we can write:
\[ a \times b = 36 \]We are also given that the difference of the two numbers is 5. Assuming \(a > b\), we have:
\[ a - b = 5 \]Now we have a system of two equations:
From equation (ii), we can express \(a\) in terms of \(b\):
\[ a = b + 5 \]Substitute this expression for \(a\) into equation (i):
\[ (b + 5) \times b = 36 \] \[ b^2 + 5b = 36 \] \[ b^2 + 5b - 36 = 0 \]Solve this quadratic equation for \(b\). We can factor the quadratic equation:
\[ b^2 + 9b - 4b - 36 = 0 \] \[ b(b + 9) - 4(b + 9) = 0 \] \[ (b - 4)(b + 9) = 0 \]This gives two possible values for \(b\): \(b = 4\) or \(b = -9\).
Since the numbers must be positive, we take \(b = 4\).
Substitute \(b = 4\) back into the equation \(a = b + 5\):
\[ a = 4 + 5 \] \[ a = 9 \]So, the two numbers are 4 and 9.
Let's verify these numbers with the given conditions:
C) 4 and 9
The question asks to identify the space shuttle in which Indian-American astronaut Kalpana Chawla died in 2003. This is a factual recall question related to significant events in space exploration history.
Correct Option: D) Columbia was the space shuttle in which Kalpana Chawla and her six crewmates perished during re-entry on February 1, 2003.
The question asks to identify the nutrient whose deficiency leads to kwashiorkor. This requires knowledge of common nutritional deficiency diseases and their causes.
Correct Option: D) Protein — Kwashiorkor is a severe form of protein-energy malnutrition, specifically characterized by a severe deficiency of protein, often with adequate calorie intake from carbohydrates. This leads to symptoms like edema, stunted growth, and a distended abdomen.