The HCF and the LCM of two numbers are 1080 and 30240, respectively. If one of the numbers is 4320, then the other number is ________.
This problem involves the relationship between the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two numbers. A fundamental property states that for any two positive integers, the product of the numbers is equal to the product of their HCF and LCM.
Let the two numbers be \(N_1\) and \(N_2\).
We are given the HCF and LCM of these two numbers:
\(\text{HCF} = 1080\)
\(\text{LCM} = 30240\)
One of the numbers is given as \(N_1 = 4320\).
We need to find the other number, \(N_2\).
The key property relating HCF, LCM, and the two numbers is:
\[ N_1 \times N_2 = \text{HCF} \times \text{LCM} \]Substitute the given values into the formula:
\[ 4320 \times N_2 = 1080 \times 30240 \]Now, solve for \(N_2\):
\[ N_2 = \frac{1080 \times 30240}{4320} \]Simplify the expression. We can cancel out common factors. Notice that \(4320 = 4 \times 1080\).
\[ N_2 = \frac{1080 \times 30240}{4 \times 1080} \]Cancel \(1080\) from the numerator and denominator:
\[ N_2 = \frac{30240}{4} \]Perform the division:
\[ N_2 = 7560 \]D) 7560 — The calculation shows that the other number is 7560, which matches option D.
A) 1080 — This is the HCF of the two numbers, not the other number itself. If the other number were 1080, then \(4320 \times 1080 \neq 1080 \times 30240\).
B) 8640 — This value is obtained if there's an error in calculation, perhaps by multiplying 4320 by 2 or by an incorrect division. It does not satisfy the HCF-LCM property.
C) 30240 — This is the LCM of the two numbers, not the other number itself. If the other number were 30240, then \(4320 \times 30240 \neq 1080 \times 30240\).
The problem is a coding-decoding question where words are coded based on a specific rule. We need to identify the rule by analyzing the given examples and then apply it to the target word.
Correct Option: A) OPRST — The letters of the word 'SPORT' when arranged in alphabetical order give 'OPRST'.
To solve this calendar problem, we need to calculate the total number of odd days between the given date (7 September 2012) and the target date (17 December 2014). Odd days are the remainder when the total number of days is divided by 7. Each odd day shifts the day of the week by one. We must account for leap years.
September: \(30 - 7 = 23\) days
October: \(31\) days
November: \(30\) days
December: \(31\) days
Total days in 2012 after Sep 7: \(23 + 31 + 30 + 31 = 115\) days.
Odd days in 2012: \(115 \div 7 = 16\) weeks and \(3\) odd days.
2013 is a normal year (not a leap year), so it has \(365\) days.
Odd days in 2013: \(365 \div 7 = 52\) weeks and \(1\) odd day.
January: \(31\) days
February: \(28\) days (2014 is not a leap year)
March: \(31\) days
April: \(30\) days
May: \(31\) days
June: \(30\) days
July: \(31\) days
August: \(31\) days
September: \(30\) days
October: \(31\) days
November: \(30\) days
December: \(17\) days
Total days in 2014: \(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 17 = 352\) days.
Odd days in 2014: \(352 \div 7 = 50\) weeks and \(2\) odd days.
Total odd days = (Odd days in 2012) + (Odd days in 2013) + (Odd days in 2014)
Total odd days = \(3 + 1 + 2 = 6\) odd days.
Given 7 September 2012 is a Friday.
Adding 6 odd days to Friday:
Friday + 1 day = Saturday
Friday + 2 days = Sunday
Friday + 3 days = Monday
Friday + 4 days = Tuesday
Friday + 5 days = Wednesday
Friday + 6 days = Thursday
Alternatively, adding 6 days is equivalent to subtracting 1 day (since \(7-6=1\)). So, Friday - 1 day = Thursday.
Let's recheck the calculation of leap years. 2012 is a leap year. The leap day (Feb 29) has already passed by September 7, 2012. So, the remaining days of 2012 do not include the leap day. 2013 is not a leap year. 2014 is not a leap year. The calculations for odd days in each segment are correct.
Let's re-evaluate the total odd days from 7 September 2012 to 17 December 2014.
From 7 September 2012 to 7 September 2013:
This period includes the leap day of 2012 (Feb 29, 2012) if we count from a date before Feb 29, 2012. However, our start date is 7 September 2012. So, the leap day of 2012 is not relevant for the period *after* 7 September 2012.
Number of days from 7 September 2012 to 7 September 2013:
2012 (remaining): Sep (23), Oct (31), Nov (30), Dec (31) = 115 days
2013 (full year): Jan (31), Feb (28), Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug (31), Sep (7) = 250 days
Total days = \(115 + 250 = 365\) days. This is a normal year, so 1 odd day.
So, 7 September 2013 will be Friday + 1 = Saturday.
From 7 September 2013 to 7 September 2014:
2014 is not a leap year. So, 365 days, which is 1 odd day.
So, 7 September 2014 will be Saturday + 1 = Sunday.
Now, we need to find the day for 17 December 2014, starting from 7 September 2014 (Sunday).
Remaining days in 2014 from 7 September:
September: \(30 - 7 = 23\) days
October: \(31\) days
November: \(30\) days
December: \(17\) days
Total days = \(23 + 31 + 30 + 17 = 101\) days.
Odd days = \(101 \div 7 = 14\) weeks and \(3\) odd days.
So, 17 December 2014 will be Sunday + 3 days.
Sunday + 1 = Monday
Sunday + 2 = Tuesday
Sunday + 3 = Wednesday
A) Wednesday is the correct day of the week.
This is a syllogism problem. We will use Venn diagrams to represent the given statements and then evaluate the conclusions based on these diagrams.
Since both conclusions logically follow from the given statements, the correct option is C.
C) Both conclusions I and II follow
The question asks to identify the specific term for the angle formed between the incident ray and the emergent ray when light passes through a prism. This is a fundamental concept in optics related to the behavior of light during refraction through a prism.
A) angle of deviation — This is the correct definition. The angle of deviation (\(\delta\)) is the angle between the direction of the incident ray and the direction of the emergent ray. It represents the total change in the direction of the light ray caused by the prism.
The question asks about the number of children who received the Pradhan Mantri Rashtriya Bal Puraskar (PMRBP) award in 2022. This is a factual question requiring knowledge of current affairs related to national awards.
Correct Option: B) 29 children received the Pradhan Mantri Rashtriya Bal Puraskar 2022 award.
The question asks about the broad classification of government budget expenditure. Understanding the different ways government spending is categorized is crucial for public finance and economics. The Indian government's budget classifies expenditure based on various criteria to provide a clear picture of its financial operations.
Government budget expenditure is broadly classified in multiple ways. The most common and comprehensive classification, particularly in the context of Indian competitive exams, considers five main categories:
While there are various ways to classify expenditure (e.g., revenue vs. capital, plan vs. non-plan, developmental vs. non-developmental), the question asks for the broadest classification. When considering the comprehensive view, the five categories often referred to are Revenue, Capital, Plan, Non-Plan, and a further distinction into Developmental and Non-Developmental. However, the most commonly accepted broad classification for competitive exams that covers the primary distinctions is often considered to be five major types, which can be derived from combining the pairs (Revenue/Capital, Plan/Non-Plan, Developmental/Non-Developmental) or by focusing on the distinct nature of spending.
A more precise and widely accepted classification in the context of Indian public finance often refers to the following five categories:
Alternatively, some classifications consider Revenue, Capital, Developmental, Non-Developmental, and Transfer Payments as five categories. Given the options, '5' is the most appropriate answer reflecting a comprehensive classification.
The budget expenditure of the government is broadly classified into 5 categories: Revenue Expenditure, Capital Expenditure, Plan Expenditure, Non-Plan Expenditure, and further sub-classification into Developmental and Non-Developmental Expenditure. While some classifications might simplify it to two (Revenue and Capital) or four (Revenue, Capital, Plan, Non-Plan), the most comprehensive and broadly accepted classification in the context of public finance often extends to five or more distinct types when considering all aspects.
To find the correct balanced chemical equation, we need to ensure that the number of atoms of each element on the reactant side is equal to the number of atoms of the same element on the product side. We will start by writing the unbalanced equation and then balance it step-by-step.
| Element | Reactant Side | Product Side |
|---|---|---|
| Ca | 1 | 1 |
| Cl | 2 | 1 |
| H | 2 | 1 |
| S | 1 | 1 |
| O | 4 | 4 |
| Element | Reactant Side | Product Side |
|---|---|---|
| Ca | 1 | 1 |
| Cl | 2 | \(1 \times 2 = 2\) |
| H | 2 | \(1 \times 2 = 2\) |
| S | 1 | 1 |
| O | 4 | 4 |
B) CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl is the correct balanced equation as it satisfies the law of conservation of mass.
Which of the following describes the nature of the roots of the quadratic equation
4x² − 7\(\sqrt{3}\) x + 12 = 0?
One real root and one imaginary root
Two distinct real roots
No real roots
Two equal roots
To determine the nature of the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\), we use the discriminant, \(\Delta = b^2 - 4ac\). The value of the discriminant tells us whether the roots are real, imaginary, distinct, or equal.
Correct Option: C) No real roots
The discriminant \(\Delta = -45\), which is less than zero. Therefore, the roots of the quadratic equation are not real; they are complex conjugates.
A train travels 20 km in 24 minutes. If its speed is increased by 5 km/h, the time taken by it to cover the same distance will be ______ minutes.
29 \( \frac{9}{11} \)
27 \( \frac{9}{11} \)
21 \( \frac{9}{11} \)
26 \( \frac{9}{11} \)
To solve this problem, we first need to calculate the initial speed of the train using the given distance and time. Then, we increase the speed by 5 km/h. Finally, we calculate the new time taken to cover the same distance with the increased speed and convert it to minutes.
Given distance \(D = 20 \text{ km}\).
Given time \(T = 24 \text{ minutes}\).
Convert time to hours: \[T = \frac{24}{60} \text{ hours} = \frac{2}{5} \text{ hours}\].
The formula for speed is \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\).
\[ \text{Initial Speed} = \frac{20 \text{ km}}{\frac{2}{5} \text{ hours}} = 20 \times \frac{5}{2} \text{ km/h} = 50 \text{ km/h} \]The speed is increased by 5 km/h.
\[ \text{New Speed} = \text{Initial Speed} + 5 \text{ km/h} = 50 \text{ km/h} + 5 \text{ km/h} = 55 \text{ km/h} \]The distance remains \(D = 20 \text{ km}\).
The formula for time is \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
\[ \text{New Time} = \frac{20 \text{ km}}{55 \text{ km/h}} = \frac{20}{55} \text{ hours} = \frac{4}{11} \text{ hours} \]To convert hours to minutes, multiply by 60.
\[ \text{New Time in minutes} = \frac{4}{11} \times 60 \text{ minutes} = \frac{240}{11} \text{ minutes} \]Divide 240 by 11:
\[ 240 \div 11 = 21 \text{ with a remainder of } 9 \]\[ \text{New Time} = 21 \frac{9}{11} \text{ minutes} \]Correct Option: C) 21 \( \frac{9}{11} \)