The problem presents a proportion in the form a : b :: c : d. This notation means that the ratio a to b is equal to the ratio c to d. We can write this as a fraction equality: \[\frac{a}{b} = \frac{c}{d}\]. To find the unknown value, we will set up this equation and solve for the variable.
The given proportion is 6 : 18 :: 18 : y.
Convert the proportion into an equation using fractions:
\[ \frac{6}{18} = \frac{18}{y} \]To solve for \(y\), we can cross-multiply:
\[ 6 \times y = 18 \times 18 \]Calculate the product on the right side:
\[ 6y = 324 \]Divide both sides by 6 to isolate \(y\):
\[ y = \frac{324}{6} \]Perform the division:
\[ y = 54 \]Thus, the value of \(y\) is 54.
Correct Option: C) 54
Find the circumference (in m) of the largest circle that can be inscribed in a rectangle whose dimensions are given as 12 m and 7 m.
Take π=\( \frac{22}{7} \)
To find the circumference of the largest circle that can be inscribed in a rectangle, we first need to determine the diameter of this circle. The largest circle that can be inscribed in a rectangle will have a diameter equal to the shorter side of the rectangle. Once the diameter is known, the circumference can be calculated using the formula \(C = \pi d\).
The given dimensions of the rectangle are 12 m and 7 m.
For a circle to be inscribed in a rectangle, its diameter must be equal to the shorter side of the rectangle. If the diameter were equal to the longer side, the circle would extend beyond the shorter dimension of the rectangle.
Comparing the given dimensions, 7 m is shorter than 12 m.
Therefore, the diameter \((d)\) of the largest inscribed circle is 7 m.
The formula for the circumference \((C)\) of a circle is \(C = \pi d\).
We are given \(\pi = \frac{22}{7}\) and we found \(d = 7 \text{ m}\).
Substitute these values into the formula:
\[ C = \frac{22}{7} \times 7 \]\[ C = 22 \text{ m} \]So, the circumference of the largest circle that can be inscribed in the rectangle is 22 m.
Correct Option: C) 22
The question asks to identify the cell component where chlorophyll pigment is located. This requires knowledge of basic cell biology, specifically the organelles involved in photosynthesis.
Correct Option: C) Chloroplasts
Chloroplasts are the primary sites of photosynthesis in plant cells and some protists. They contain chlorophyll, a green pigment that absorbs sunlight, initiating the process of converting light energy into chemical energy (sugars).
A man sells two articles for Rs 20,000 each. He gets a profit of 15% on one article and loses 15% on the other article. What is his overall profit or loss (to the nearest integer) in this transaction?
When two articles are sold at the same selling price, and one is sold at a profit of \(x\%\) and the other at a loss of \(x\%\), there will always be an overall loss. The overall loss percentage can be calculated using the formula: \(\left(\frac{x}{10}\right)^2\%\). We will first calculate the cost price of each article and then find the total cost price and total selling price to determine the overall profit or loss.
Calculate the Selling Price (SP) for each article:
Given, each article is sold for Rs 20,000.
\[ \text{SP}_1 = \text{Rs } 20,000 \]\[ \text{SP}_2 = \text{Rs } 20,000 \]Calculate the Cost Price (CP) for the first article (15% profit):
If there is a profit of 15%, then the selling price is 115% of the cost price.
\[ \text{SP}_1 = \text{CP}_1 \times \left(1 + \frac{15}{100}\right) \]\[ 20,000 = \text{CP}_1 \times \frac{115}{100} \]\[ \text{CP}_1 = 20,000 \times \frac{100}{115} \]\[ \text{CP}_1 = \frac{2,000,000}{115} \approx \text{Rs } 17,391.30 \]Calculate the Cost Price (CP) for the second article (15% loss):
If there is a loss of 15%, then the selling price is 85% of the cost price.
\[ \text{SP}_2 = \text{CP}_2 \times \left(1 - \frac{15}{100}\right) \]\[ 20,000 = \text{CP}_2 \times \frac{85}{100} \]\[ \text{CP}_2 = 20,000 \times \frac{100}{85} \]\[ \text{CP}_2 = \frac{2,000,000}{85} \approx \text{Rs } 23,529.41 \]Calculate the Total Selling Price (Total SP):
\[ \text{Total SP} = \text{SP}_1 + \text{SP}_2 \]\[ \text{Total SP} = 20,000 + 20,000 = \text{Rs } 40,000 \]Calculate the Total Cost Price (Total CP):
\[ \text{Total CP} = \text{CP}_1 + \text{CP}_2 \]\[ \text{Total CP} = \frac{2,000,000}{115} + \frac{2,000,000}{85} \]\[ \text{Total CP} = 2,000,000 \times \left(\frac{1}{115} + \frac{1}{85}\right) \]\[ \text{Total CP} = 2,000,000 \times \left(\frac{85 + 115}{115 \times 85}\right) \]\[ \text{Total CP} = 2,000,000 \times \frac{200}{9775} \]\[ \text{Total CP} = \frac{400,000,000}{9775} \approx \text{Rs } 40,910.485 \]Calculate the Overall Profit or Loss:
Since Total CP > Total SP, there is an overall loss.
\[ \text{Overall Loss} = \text{Total CP} - \text{Total SP} \]\[ \text{Overall Loss} = 40,910.485 - 40,000 \]\[ \text{Overall Loss} = \text{Rs } 910.485 \]Round to the nearest integer:
The overall loss is approximately Rs 910.
Alternatively, using the direct formula for loss percentage when SP is same and profit/loss percentage is same:
Loss percentage \( = \left(\frac{x}{10}\right)^2 \% \)
Here, \(x = 15\).
Loss percentage \( = \left(\frac{15}{10}\right)^2 \% = (1.5)^2 \% = 2.25 \% \)
Total SP \( = \text{Rs } 40,000 \)
Let Total CP be \(C\).
Loss percentage \( = \frac{\text{Loss}}{\text{Total CP}} \times 100 \)
Loss \( = \text{Total CP} - \text{Total SP} \)
So, \( 2.25 = \frac{C - 40,000}{C} \times 100 \)
This method calculates loss percentage on CP. However, a more straightforward way to use the loss percentage is to find the total CP from the total SP and loss percentage:
If there is a loss of 2.25%, then Total SP is \((100 - 2.25)\%\) of Total CP.
\[ \text{Total SP} = \text{Total CP} \times \left(1 - \frac{2.25}{100}\right) \]\[ 40,000 = \text{Total CP} \times \frac{97.75}{100} \]\[ \text{Total CP} = 40,000 \times \frac{100}{97.75} \]\[ \text{Total CP} = \frac{4,000,000}{97.75} \approx \text{Rs } 40,910.485 \]\[ \text{Overall Loss} = \text{Total CP} - \text{Total SP} \]\[ \text{Overall Loss} = 40,910.485 - 40,000 = \text{Rs } 910.485 \]Rounding to the nearest integer:
The overall loss is Rs 910.
Comparing this with the options, the closest is Rs 921 loss.
Let's recheck the calculation of \(\frac{400,000,000}{9775}\).
\[\frac{400,000,000}{9775} = \frac{80,000,000}{1955} = \frac{16,000,000}{391}\]
\(16,000,000 \div 391 \approx 40920.716\)
So, \(\text{Total CP} \approx \text{Rs } 40,920.716\)
\(\text{Overall Loss} = 40,920.716 - 40,000 = \text{Rs } 920.716\)
Rounding to the nearest integer, the loss is Rs 921.
B) 921 loss
This question tests general knowledge about recent scientific developments in India, specifically related to agricultural technology and research by Indian Institutes of Technology (IITs). The key is to recall which IIT was in the news in February 2022 for developing a biodegradable nanoparticle for crop protection.
Correct Option: D) IIT Kanpur
The question asks about the material used for the heating element of a bulb. This refers to the filament in incandescent light bulbs. The choice of material for a heating element is crucial and depends on its electrical and thermal properties, particularly its melting point and resistance to oxidation at high temperatures.
D) tungsten — Tungsten is chosen for the filament of incandescent bulbs because it has the highest melting point of all metals, allowing it to be heated to incandescence (glowing white hot) without melting. It also has high electrical resistance, which helps in generating heat and light efficiently.
Two pipes can fill a tank in 25 hours and 35 hours, respectively. The time (in hours) required to fill the tank when both pipes are opened simultaneously is:
\(\frac{165}{12}\)
\(\frac{175}{14}\)
\(\frac{175}{12}\)
\(\frac{165}{14}\)
This problem involves calculating the combined work rate of two pipes to fill a tank. We will determine the fraction of the tank each pipe can fill in one hour and then sum these fractions to find their combined filling rate. Finally, we will take the reciprocal of the combined rate to find the total time required to fill the tank.
The first pipe can fill the tank in 25 hours. Therefore, in one hour, it fills \(\frac{1}{25}\) of the tank.
The second pipe can fill the tank in 35 hours. Therefore, in one hour, it fills \(\frac{1}{35}\) of the tank.
When both pipes are opened simultaneously, their individual rates add up. The combined rate in one hour is:
\[ \text{Combined Rate} = \frac{1}{25} + \frac{1}{35} \]The least common multiple (LCM) of 25 and 35 is 175.
\[ \text{Combined Rate} = \frac{1 \times 7}{25 \times 7} + \frac{1 \times 5}{35 \times 5} \]\[ \text{Combined Rate} = \frac{7}{175} + \frac{5}{175} \]\[ \text{Combined Rate} = \frac{7 + 5}{175} \]\[ \text{Combined Rate} = \frac{12}{175} \text{ (of the tank per hour)} \]If the pipes fill \(\frac{12}{175}\) of the tank in one hour, then the total time required to fill the entire tank (which is 1 unit of work) is the reciprocal of this rate.
\[ \text{Time} = \frac{1}{\text{Combined Rate}} \]\[ \text{Time} = \frac{1}{\frac{12}{175}} \]\[ \text{Time} = \frac{175}{12} \text{ hours} \]Correct Option: C) \(\frac{175}{12}\)
The problem requires us to identify specific digits from a given set of three-digit numbers based on their magnitude (greatest, smallest) and then perform an addition operation. We need to first list the numbers, arrange them to find the third greatest and the smallest, identify the required digits, and finally add them.
Correct Option: B) 10
The question asks to identify the Indian state that hosts the oldest operating oil refinery, established in 1901. This requires knowledge of the history of oil exploration and refining in India.
C) Assam — The Digboi Refinery in Assam, established in 1901, is indeed the oldest operating oil refinery in India. It was set up following the discovery of oil in the region in 1889.
To determine which element does not belong to Period 3, we need to recall the definition of a period in the Modern Periodic Table and identify the atomic number or electron configuration of each given element to place it correctly.
B) Ge (Germanium) has an atomic number of 32 and is located in Period 4 of the Modern Periodic Table, as its outermost electrons are in the 4th principal energy shell.