To find the ratio of two quantities, they must be expressed in the same units. We will convert kilograms to grams and then simplify the ratio.
Identify the given weights and their units:
First weight = \(27 \text{ kg}\)
Second weight = \(108 \text{ g}\)
Convert the units to be consistent:
Since \(1 \text{ kg} = 1000 \text{ g}\), we convert the first weight from kilograms to grams:
\[ 27 \text{ kg} = 27 \times 1000 \text{ g} = 27000 \text{ g} \]Form the ratio:
Now, both weights are in grams. The ratio of the first weight to the second weight is:
\[ \text{Ratio} = \frac{27000 \text{ g}}{108 \text{ g}} \]Simplify the ratio:
We need to divide 27000 by 108.
We can simplify this step-by-step:
So, the ratio is \(250 : 1\).
C) 250 : 1 is the correct ratio after converting units and simplifying.
This problem involves calculating the passing marks based on percentages and a given deficit. We will first determine the percentage difference between the required passing marks and the marks secured. This percentage difference corresponds to the marks by which the student failed. From this, we can find the total marks for the exam and then calculate the passing marks.
Identify the given information:
Calculate the percentage difference: The difference in percentage between passing marks and secured marks is:
\[ \text{Percentage Difference} = \text{Required Percentage} - \text{Secured Percentage} \] \[ \text{Percentage Difference} = 20\% - 10\% = 10\% \]Relate percentage difference to marks: This \(10\%\) difference corresponds to the \(20\) marks by which the student failed.
\[ 10\% \text{ of Total Marks} = 20 \text{ marks} \]Calculate the Total Marks: If \(10\%\) of the total marks is \(20\), then \(1\%\) of the total marks is \(\frac{20}{10} = 2\). Therefore, \(100\%\) (the total marks) will be:
\[ \text{Total Marks} = \frac{20}{10\%} = \frac{20}{0.10} = 200 \text{ marks} \]Calculate the Passing Marks: Passing marks are \(20\%\) of the total marks.
\[ \text{Passing Marks} = 20\% \text{ of } 200 \] \[ \text{Passing Marks} = \frac{20}{100} \times 200 \] \[ \text{Passing Marks} = 20 \times 2 = 40 \text{ marks} \]Correct Option: A) 40
A shopkeeper incurs a loss of 20% after selling a machine for ₹4,800. In order to gain a profit of 20%, at what price should the shopkeeper sell the same machine?
To solve this problem, we first need to determine the Cost Price (CP) of the machine. We are given the Selling Price (SP) and the percentage loss. Once the CP is known, we can calculate the new selling price required to achieve a 20% profit.
The shopkeeper incurs a loss of 20% when selling the machine for ₹4,800. This means the selling price is 80% of the cost price (100% - 20% loss).
\[ \text{SP} = \text{CP} \times (1 - \text{Loss Percentage}) \]\[ 4800 = \text{CP} \times (1 - 0.20) \]\[ 4800 = \text{CP} \times 0.80 \]\[ \text{CP} = \frac{4800}{0.80} \]\[ \text{CP} = \text{₹}6000 \]Now that we know the Cost Price is ₹6,000, we need to find the selling price that will yield a 20% profit. This means the new selling price will be 120% of the cost price (100% + 20% profit).
\[ \text{New SP} = \text{CP} \times (1 + \text{Profit Percentage}) \]\[ \text{New SP} = 6000 \times (1 + 0.20) \]\[ \text{New SP} = 6000 \times 1.20 \]\[ \text{New SP} = \text{₹}7200 \]Correct Option: A) ₹7,200
The problem involves calculating the original cost price of an item given a selling price and a loss percentage, and then determining a new selling price to achieve a desired profit percentage. We will use the formulas relating cost price, selling price, loss percentage, and profit percentage.
C) ₹660. This is the price at which Shakila should sell the part to achieve a 10% profit, calculated based on the original cost price of ₹600.
The question asks for another name for Tamil New Year. This requires knowledge of the traditional names for New Year celebrations in different regions of India, particularly those associated with the Tamil community.
A) Varusha Pirappu — This is the traditional name for the Tamil New Year, celebrated by Tamils in Tamil Nadu and Puducherry in India, and by the Tamil diaspora worldwide. It falls on the first day of the Tamil month of Chithirai, usually around April 14th.
To find the value of the given trigonometric expression, we will use the complementary angle identity for trigonometric functions, specifically \(\cos(90^\circ - \theta) = \sin(\theta)\) or \(\sin(90^\circ - \theta) = \cos(\theta)\).
Correct Option: C) 0
The question asks to identify the Raja of Burdwan at the time of the imposition of the Permanent Settlement. This requires knowledge of specific historical figures and events related to British colonial administration in India.
Correct Option: B) Tejchand was the Raja of Burdwan when the Permanent Settlement was imposed in 1793. He was a prominent zamindar whose estate was significantly affected by the new land revenue system.
To keep the expenditure on an item unchanged when its price increases, the consumption must be reduced. This problem involves calculating the percentage reduction in consumption required to offset a percentage increase in price, maintaining a constant expenditure. We will use the formula: Expenditure = Price × Consumption.
Correct Option: A) 20% is the correct percentage by which consumption should be reduced to keep the expenditure unchanged.
The question asks about the protective mechanism for the stomach's inner lining against its own acidic environment. We need to identify the component responsible for this defense.
Correct Option: C) Mucus
Mucus is secreted by goblet cells and mucous glands in the stomach lining. It forms a protective barrier that shields the stomach wall from the highly acidic gastric juice and digestive enzymes like pepsin. This barrier is crucial for preventing ulcers and other damage to the stomach.
The question asks to identify an electrical safety device that operates based on the heating effect of electric current. We need to examine the working principle of each given option.
A) fuse — A fuse is specifically designed to melt and break an electrical circuit when the current flowing through it exceeds a safe limit. This melting action is a direct result of the heat generated by the excessive current, which is an application of the heating effect of electric current (Joule's law).