Compute: 10584 ÷ 168 − 63
To solve this problem, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS. This means performing division before subtraction.
Let's calculate the division:
\[ 10584 \div 168 \]We can perform long division or simplify by finding common factors. Let's try to estimate or perform the division directly.
If we divide \(10584\) by \(168\):
\[ \begin{array}{r} 63 \\ 168 \overline{) 10584} \\ -1008 \downarrow \\ \hline 504 \\ -504 \\ \hline 0 \end{array} \]So, \(10584 \div 168 = 63\).
Now, substitute this value back into the original expression:
\[ 63 - 63 \]\[ 63 - 63 = 0 \]B) \(0\) is the correct answer, as \(10584 \div 168 - 63 = 63 - 63 = 0\).
Which is the correct ascending order of the given numbers?
\( \frac{4}{9} \), 0.3, \( \frac{2}{7} \)
\( \frac{4}{9} \), 0.3, \( \frac{2}{7} \)
0.3, \( \frac{4}{9} \),\( \frac{2}{7} \)
\( \frac{2}{7} \), 0.3, \( \frac{4}{9} \)
\( \frac{2}{7} \), \( \frac{4}{9} \), 0.3
To arrange the given numbers in ascending order, we need to convert all of them into a common format, preferably decimal form, to facilitate easy comparison. Then, we will order them from smallest to largest.
First, convert the given fractions to their decimal equivalents.
Now we have all numbers in decimal form:
Compare the decimal values:
So, the ascending order of the decimal values is \( 0.2857... \), \( 0.3 \), \( 0.444... \).
Replace the decimal values with their original forms:
Therefore, the correct ascending order is \( \frac{2}{7} \), \( 0.3 \), \( \frac{4}{9} \).
C) \( \frac{2}{7} \), \( 0.3 \), \( \frac{4}{9} \) is the correct ascending order as determined by converting all numbers to decimal form and comparing them.
The question asks to identify the missing river that, along with the Penganga and Wainganga rivers, forms the Pranhita River, which is the largest tributary of the Godavari River. This requires knowledge of the tributaries of the Godavari River system.
A) Wardha. The Pranhita River is indeed formed by the combined waters of the Penganga, Wardha, and Wainganga rivers, making it the largest tributary of the Godavari River by discharge volume.
There are total 200 students in a school, of which \( \frac{2}{5} \) are boys. Find the number of girls in the school.
To find the number of girls, we first need to determine the number of boys. We are given the total number of students and the fraction of students who are boys. Once we have the number of boys, we can subtract it from the total number of students to find the number of girls.
Identify the total number of students: Total students = \(200\)
Identify the fraction of boys: Fraction of boys = \( \frac{2}{5} \)
Calculate the number of boys: Number of boys = Fraction of boys \( \times \) Total students
\[ \text{Number of boys} = \frac{2}{5} \times 200 \] \[ \text{Number of boys} = 2 \times \frac{200}{5} \] \[ \text{Number of boys} = 2 \times 40 \] \[ \text{Number of boys} = 80 \]Calculate the number of girls: Number of girls = Total students - Number of boys
\[ \text{Number of girls} = 200 - 80 \] \[ \text{Number of girls} = 120 \]Correct Option: C) 120
The question asks for the location of the Meenakshi Temple. This is a factual question related to Indian geography and cultural heritage, specifically important religious sites.
Correct Option: A) Tamil Nadu
The Meenakshi Temple is famously located in Madurai, a city in the state of Tamil Nadu, India. It is one of the most prominent and architecturally significant temples in South India.
If Sin x = \( \frac{4}{5} \), then Sec x + Tan x = ?
\( \frac{37}{20} \)
\( \frac{31}{12} \)
3
\( \frac{1}{3} \)
To find the value of \(\text{Sec } x + \text{Tan } x\) given \(\text{Sin } x = \frac{4}{5}\), we first need to determine the values of \(\text{Sec } x\) and \(\text{Tan } x\). We can do this by using the fundamental trigonometric identity \(\text{Sin}^2 x + \text{Cos}^2 x = 1\) to find \(\text{Cos } x\), and then derive \(\text{Sec } x = \frac{1}{\text{Cos } x}\) and \(\text{Tan } x = \frac{\text{Sin } x}{\text{Cos } x}\).
Given:
\[ \text{Sin } x = \frac{4}{5} \]Find \(\text{Cos } x\) using the identity \(\text{Sin}^2 x + \text{Cos}^2 x = 1\):
\[ \left(\frac{4}{5}\right)^2 + \text{Cos}^2 x = 1 \]Find \(\text{Sec } x\):
\[ \text{Sec } x = \frac{1}{\text{Cos } x} \]Find \(\text{Tan } x\):
\[ \text{Tan } x = \frac{\text{Sin } x}{\text{Cos } x} \]Calculate \(\text{Sec } x + \text{Tan } x\):
\[ \text{Sec } x + \text{Tan } x = \frac{5}{3} + \frac{4}{3} \]Correct Option: C) 3
The question asks to identify the sport associated with Korada Ramana. This requires general knowledge about prominent Indian athletes and their respective sports.
C) Weightlifting — Korada Ramana is an Indian weightlifter who has achieved recognition in the sport.
This problem can be solved using the principle of inclusion-exclusion or by using Venn diagrams. We are given the total number of people, the number of people who do not subscribe to any newspaper, and the number of subscribers for two specific newspapers. We need to find the number of people who subscribe to 'The Indian Express' only.
Total people = 5000
People who do not subscribe to any newspaper = 1200
Number of people who subscribe to at least one newspaper = Total people - People who do not subscribe to any newspaper
\[ \text{Subscribers (at least one)} = 5000 - 1200 = 3800 \]Let \(IE\) be the set of people who subscribe to 'The Indian Express' and \(H\) be the set of people who subscribe to 'The Hindu'.
We know:
The formula for the union of two sets is:
\[ |IE \cup H| = |IE| + |H| - |IE \cap H| \]Substitute the known values into the formula:
\[ 3800 = 2700 + 1800 - |IE \cap H| \]\[ 3800 = 4500 - |IE \cap H| \]\[ |IE \cap H| = 4500 - 3800 \]\[ |IE \cap H| = 700 \]The number of people who subscribe to 'The Indian Express' only is the total number of 'The Indian Express' subscribers minus those who also subscribe to 'The Hindu'.
\[ \text{IE only} = |IE| - |IE \cap H| \]\[ \text{IE only} = 2700 - 700 \]\[ \text{IE only} = 2000 \]B) 2000 — As calculated in the step-by-step solution, 2000 people subscribe to 'The Indian Express' only.
This problem can be solved using the principle of inclusion-exclusion for two sets. We are given the total number of people, the number of people who subscribe to newspaper A, and the number of people who subscribe to newspaper B. We also know the number of people who do not subscribe to either newspaper. We need to find the number of people who subscribe to both newspapers.
Let \(N\) be the total number of people surveyed.
Let \(A\) be the set of people who subscribe to newspaper A.
Let \(B\) be the set of people who subscribe to newspaper B.
From the problem statement, we have:
The number of people who subscribe to at least one newspaper is the total number of people minus those who subscribe to neither. Using De Morgan's laws, \(|A' \cap B'| = |(A \cup B)'|\).
\[ |A \cup B| = N - |(A \cup B)'| \]\[ |A \cup B| = 5000 - 1000 \]\[ |A \cup B| = 4000 \]The principle of inclusion-exclusion for two sets states:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]We want to find \(|A \cap B|\), which is the number of people who subscribe to both newspapers. Rearranging the formula:
\[ |A \cap B| = |A| + |B| - |A \cup B| \]Substitute the known values into the equation:
\[ |A \cap B| = 2500 + 2200 - 4000 \]\[ |A \cap B| = 4700 - 4000 \]\[ |A \cap B| = 700 \]Thus, 700 people subscribe to both newspapers.
C) 700 is the correct answer, as calculated using the principle of inclusion-exclusion.
This is an analogy question. The goal is to identify the relationship between the first pair of words ("When : Where") and then apply the same relationship to the second word ("Time") to find the missing word.
C) Place — "Time" is a concept related to the temporal dimension, and "Place" is a concept related to the spatial dimension. This directly mirrors the relationship between "When" (asking about time) and "Where" (asking about place).