There are three groups G1, G2, G3 in a college with 30, 60 and 30 students respectively. The average marks obtained by groups G1, G2, G3 are 50%, 60% and 70% respectively. Find the average marks of the entire college?
To find the average marks of the entire college, we need to calculate the total marks obtained by all students and divide it by the total number of students. This is a weighted average problem, where the number of students in each group acts as the weight for their respective average marks.
Calculate total marks for Group G1:
Number of students in G1 = 30
Average marks of G1 = 50%
Total marks for G1 = \(30 \times 50 = 1500\)
Calculate total marks for Group G2:
Number of students in G2 = 60
Average marks of G2 = 60%
Total marks for G2 = \(60 \times 60 = 3600\)
Calculate total marks for Group G3:
Number of students in G3 = 30
Average marks of G3 = 70%
Total marks for G3 = \(30 \times 70 = 2100\)
Calculate the total marks for the entire college:
Total marks = Total marks (G1) + Total marks (G2) + Total marks (G3)
Total marks = \(1500 + 3600 + 2100 = 7200\)
Calculate the total number of students in the college:
Total students = Students in G1 + Students in G2 + Students in G3
Total students = \(30 + 60 + 30 = 120\)
Calculate the average marks of the entire college:
\[ \text{Average Marks} = \frac{\text{Total Marks}}{\text{Total Students}} \] \[ \text{Average Marks} = \frac{7200}{120} = 60 \]So, the average marks of the entire college is 60%.
Correct Option: A) 60%
The question asks to identify the element whose chemical symbol consists of only one letter. This requires knowledge of the chemical symbols for common elements.
Based on this analysis, Carbon is the only element among the options whose symbol consists of only one letter.
Correct Option: C) Carbon
The chemical symbol for Carbon is 'C', which is a single letter.
The question asks for the specific year when Chandragupta II extended the Gupta Kingdom to Gujarat. This requires knowledge of the chronology of the Gupta Empire and the reign of Chandragupta II.
A) 390 AD is the correct year when Chandragupta II extended the Gupta Kingdom to Gujarat by defeating the Western Satraps.
The question asks to identify the language for which Mahabaleshwar Sail received the 2016 Saraswati Samman. This requires knowledge of prominent literary awards and their recipients.
Correct Option: D) Konkani — Mahabaleshwar Sail was awarded the 2016 Saraswati Samman for his Konkani novel 'Hawthan'. The Saraswati Samman is an annual literary award given by the K. K. Birla Foundation for outstanding prose or poetry literary works in any of the 22 languages of India listed in Schedule VIII of the Constitution of India.
Let the three prime numbers be \(p_1, p_2, p_3\). We are given two conditions: the product of the first two numbers is 1147, and the product of the last two numbers is 1517. We need to find these three prime numbers. The common prime number between the two products will be a common factor of 1147 and 1517. We can find this common factor using the HCF method or by factoring the given products.
Let the three prime numbers be \(p_1, p_2, p_3\).
Given:
Notice that \(p_2\) is a common factor of both 1147 and 1517. Therefore, \(p_2\) must be the Highest Common Factor (HCF) of 1147 and 1517, or a factor of the HCF.
Let's find the HCF of 1147 and 1517 using the Euclidean algorithm:
\[ 1517 = 1147 \times 1 + 370 \]
\[ 1147 = 370 \times 3 + 37 \]
\[ 370 = 37 \times 10 + 0 \]
The HCF of 1147 and 1517 is 37.
Since \(p_2\) is a common prime factor, and 37 is a prime number, \(p_2 = 37\).
Now, we can find \(p_1\) and \(p_3\):
The three prime numbers are 31, 37, and 41. We can verify that all three are prime numbers.
D) 31, 37, 41. This matches our calculated prime numbers.
The question asks us to find the word that relates to "Bell" in the same way that "Play" relates to "Instrument". This is an analogy question, where we need to identify the relationship between the first pair of words and apply the same relationship to the third word to find the fourth.
Correct Option: A) Ring. The relationship is that the second word is the action performed on the first word. An instrument is played, and a bell is rung.
This is a question based on syllogism. We need to evaluate the given conclusions based on the provided statements. We can use Venn diagrams to visually represent the statements and then check the validity of each conclusion.
Analyze the Statements:
From these statements, we know that both 'Buses' and 'Ships' are subsets of 'Cars'. However, there is no direct relationship given between 'Buses' and 'Ships'. They could be separate, or they could overlap, or one could be a subset of the other, as long as they are both within 'Cars'.
Evaluate Conclusion I: "All cars are ships."
Evaluate Conclusion II: "All buses are cars."
Only Conclusion II logically follows from the statements because it is a direct restatement of one of the given premises.
To find the value of one-fourth of G, we first need to calculate the value of G. We will use the order of operations (BODMAS/PEMDAS) to correctly evaluate the expression for G. BODMAS stands for Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction. We perform operations in this specific order.
First, evaluate the expressions inside the brackets:
\[ G = (96 \div 12) + 14 \times (12 + 8) \div 2 \]\[ G = (8) + 14 \times (20) \div 2 \]Next, perform division and multiplication from left to right. There are two multiplication/division operations: \(14 \times 20\) and the subsequent division by 2.
\[ G = 8 + (14 \times 20) \div 2 \]\[ G = 8 + 280 \div 2 \]Now, perform the division:
\[ G = 8 + 140 \]Finally, perform the addition:
\[ G = 148 \]The question asks for one-fourth of G. So, we need to calculate \(\frac{1}{4} \times G\):
\[ \text{One-fourth of G} = \frac{1}{4} \times 148 \]\[ \text{One-fourth of G} = 37 \]Correct Option: D) The value of one-fourth of G is 37.
To find the height at which the object is located, we need to use the formula for gravitational potential energy. This formula relates the potential energy to the mass of the object, the acceleration due to gravity, and the height.
The formula for gravitational potential energy (\(PE\)) is given by:
\[ PE = mgh \]where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height.To find the height, we can rearrange the formula as follows:
\[ h = \frac{PE}{mg} \]Now, plug in the values:
\[ h = \frac{480 \text{ J}}{12 \text{ kg} \times 10 \text{ ms}^{-2}} \]Perform the multiplication in the denominator:
\[ h = \frac{480 \text{ J}}{120 \text{ kg ms}^{-2}} \]\[ h = \frac{480}{120} \text{ m} \]\[ h = 4 \text{ m} \]Thus, the height at which the object is located is 4 meters.
Correct Option: C) 4 m
In the Fibonacci series each number is defined as Fn = Fn−1 + Fn−2. If F0 = 0 and F1 = 1, then find the value of F10.
The problem asks us to find the 10th term (\(F_{10}\)) of the Fibonacci series, given the recursive definition \(F_n = F_{n-1} + F_{n-2}\) and the initial conditions \(F_0 = 0\) and \(F_1 = 1\). We will generate the terms of the Fibonacci series sequentially until we reach \(F_{10}\).
Correct Option: B) 55