We are given three numbers in a specific ratio and their Least Common Multiple (LCM). We need to find their Highest Common Factor (HCF). The key concept here is the relationship between numbers, their HCF, and their LCM. If numbers are in a given ratio, we can express them as multiples of their HCF. Then, we can use the formula relating LCM, HCF, and the numbers.
Let the three numbers be \(N_1\), \(N_2\), and \(N_3\). They are in the ratio \(1 : 15 : 7\).
Let the HCF of these three numbers be \(H\). Then, the numbers can be expressed as multiples of \(H\):
\[ N_1 = 1 \times H = H \]\[ N_2 = 15 \times H = 15H \]\[ N_3 = 7 \times H = 7H \]The LCM of three numbers \(aH\), \(bH\), and \(cH\) (where \(a, b, c\) are coprime) is given by \(H \times \text{LCM}(a, b, c)\). In our case, the numbers are \(H, 15H, 7H\). Here, \(1, 15, 7\) are the coprime parts (since \(15 = 3 \times 5\) and \(7\) is prime, and \(1\) is coprime to all).
Calculate the LCM of the ratio parts \(1, 15, 7\):
The LCM of \(1, 15, 7\) is \(1 \times 3 \times 5 \times 7 = 105\).
Now, we can express the LCM of the three numbers \(N_1, N_2, N_3\) using \(H\):
\[ \text{LCM}(N_1, N_2, N_3) = H \times \text{LCM}(1, 15, 7) \]\[ \text{LCM}(N_1, N_2, N_3) = H \times 105 \]We are given that the LCM of the numbers is \(4830\). So, we can set up the equation:
\[ 105H = 4830 \]Solve for \(H\):
\[ H = \frac{4830}{105} \]\[ H = 46 \]Therefore, the HCF of the three numbers is \(46\).
D) 46
To solve this problem, we need to identify the highest and lowest numbers from the given set, then extract specific digits from these identified numbers as per the question's instructions, and finally perform the requested arithmetic operation.
Correct Option: A) 14
A conical vessel has base radius 6 cm and height 20 cm. Water is poured into the vessel until it is \( \frac{3}{4} \) full. Find the volume (in cm³) of water in the vessel.
To find the volume of water in the conical vessel, we first need to calculate the total volume of the conical vessel. Then, we will find three-fourths of this total volume, as the vessel is filled to that extent.
The volume \(V\) of a cone with base radius \(r\) and height \(h\) is given by:
\[ V = \frac{1}{3} \pi r^2 h \]Substitute the given values into the volume formula:
\[ V_{\text{total}} = \frac{1}{3} \pi (6 \text{ cm})^2 (20 \text{ cm}) \] \[ V_{\text{total}} = \frac{1}{3} \pi (36 \text{ cm}^2) (20 \text{ cm}) \] \[ V_{\text{total}} = \frac{1}{3} \pi (720 \text{ cm}^3) \] \[ V_{\text{total}} = 240 \pi \text{ cm}^3 \]The vessel is \( \frac{3}{4} \) full of water. So, the volume of water is \( \frac{3}{4} \) of the total volume:
\[ V_{\text{water}} = \frac{3}{4} \times V_{\text{total}} \] \[ V_{\text{water}} = \frac{3}{4} \times (240 \pi \text{ cm}^3) \] \[ V_{\text{water}} = 3 \times (60 \pi \text{ cm}^3) \] \[ V_{\text{water}} = 180 \pi \text{ cm}^3 \]Correct Option: C) 180π
The problem states that Vipul spends 55% of his income and saves Rs. 18,000. We need to find his total income. We can determine the percentage of income saved and then use this percentage along with the saved amount to calculate the total income.
If Vipul spends 55% of his income, then the remaining percentage is saved. The total income is considered 100%.
\[ \text{Percentage Saved} = 100\% - \text{Percentage Spent} \]\[ \text{Percentage Saved} = 100\% - 55\% = 45\% \]We are given that Vipul saves Rs. 18,000. This amount corresponds to 45% of his total income.
\[ 45\% \text{ of Income} = \text{Rs. } 18,000 \]Let Vipul's total income be \(I\). We can set up the equation:
\[ 0.45 \times I = 18,000 \]\[ I = \frac{18,000}{0.45} \]\[ I = \frac{18,000}{\frac{45}{100}} \]\[ I = \frac{18,000 \times 100}{45} \]\[ I = \frac{1,800,000}{45} \]\[ I = 40,000 \]Thus, Vipul's income is Rs. 40,000.
C) 40,000 — As calculated, if Vipul saves 45% of his income, and that amount is Rs. 18,000, then his total income is Rs. 40,000.
The question asks about the industrial uses of Freon. Freons are a group of chlorofluorocarbons (CFCs) that were widely used in various applications due to their chemical stability and non-toxic nature. However, their use has been largely phased out due to their detrimental effect on the ozone layer.
Correct Option: A) It is used in refrigeration and aerosol production.
Freons, particularly dichlorodifluoromethane (Freon-12 or R-12) and chlorodifluoromethane (Freon-22 or R-22), were extensively used as refrigerants in air conditioning systems and refrigerators. They were also widely used as propellants in aerosol sprays for products like deodorants, hairsprays, and insecticides due to their low boiling points and non-flammable nature. Although their use has been largely discontinued due to their ozone-depleting potential, this was their primary industrial application.
The problem requires us to modify a given number based on specific rules for its odd and even digits. Then, we need to find the sum of the highest and lowest digits in the newly formed number.
Correct Option: D) 8. The sum of the highest (8) and lowest (0) digits in the new number is \(8 + 0 = 8\).
The question asks to identify the winner of the Best Directorial Debut award for the film 'Madgaon Express' at the IIFA Awards, 2025. This requires knowledge of recent film awards and their recipients.
D) Kunal Kemmu — Kunal Kemmu made his directorial debut with the film 'Madgaon Express' and won the Best Directorial Debut award at the IIFA Awards, 2025 for this film.
The question asks to identify which option represents a collection of eight poetic anthologies from ancient Tamil literature. This requires knowledge of the classification and names of prominent works in Sangam literature.
Correct Option: A) Ettuttokai is a collection of eight poetic anthologies from ancient Tamil literature. The name itself means "Eight Collections" (எட்டு - eight, தொகை - collection).