The problem involves simple interest. We are given the principal amount, the amount after 3 years, and then asked to find the new amount if the interest rate is increased by 4%. First, we need to find the original simple interest rate. Then, we calculate the new interest rate and finally the new amount.
Given Principal \((P) = \text{₹}800\)
Amount \((A) = \text{₹}920\)
Time \((T) = 3\) years
Simple Interest \((SI) = A - P = \text{₹}920 - \text{₹}800 = \text{₹}120\)
The formula for Simple Interest is \(SI = \frac{P \times R \times T}{100}\).
Rearranging to find \(R\):
\[ R = \frac{SI \times 100}{P \times T} \]\[ R = \frac{120 \times 100}{800 \times 3} \]\[ R = \frac{12000}{2400} \]\[ R = 5\% \]The interest is increased by 4%.
New Rate \((R') = R + 4\% = 5\% + 4\% = 9\%\)
Using the new rate \(R' = 9\%\), Principal \(P = \text{₹}800\), and Time \(T = 3\) years:
\[ SI' = \frac{P \times R' \times T}{100} \]\[ SI' = \frac{800 \times 9 \times 3}{100} \]\[ SI' = 8 \times 9 \times 3 \]\[ SI' = 72 \times 3 \]\[ SI' = \text{₹}216 \]New Amount \((A') = P + SI'\)
\[ A' = 800 + 216 \]\[ A' = \text{₹}1016 \]Correct Option: D) ₹1,016
First, we need to find the rate of interest from the simple interest information provided. Once the rate is determined, we can use it to calculate the compound interest for the given principal amount and time period.
Calculate the Rate of Interest (R) from Simple Interest:
Let the principal amount be \(P\). The increase in amount is 60% of \(P\), which means the simple interest (SI) is \(0.60P\).
Given: Time (\(T\)) = 6 years.
The formula for Simple Interest is:
\[ SI = \frac{P \times R \times T}{100} \]Substitute the given values:
\[ 0.60P = \frac{P \times R \times 6}{100} \]Divide both sides by \(P\) (assuming \(P \neq 0\)):
\[ 0.60 = \frac{R \times 6}{100} \] \[ R = \frac{0.60 \times 100}{6} \] \[ R = \frac{60}{6} \] \[ R = 10\% \text{ per annum} \]Calculate the Compound Interest (CI):
Given: Principal (\(P\)) = ₹10,000
Rate (\(R\)) = 10% per annum
Time (\(n\)) = 3 years
The formula for the Amount (\(A\)) in compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]Substitute the values:
\[ A = 10000 \left(1 + \frac{10}{100}\right)^3 \] \[ A = 10000 \left(1 + 0.1\right)^3 \] \[ A = 10000 \left(1.1\right)^3 \] \[ A = 10000 \times 1.331 \] \[ A = \text{₹}13310 \]Now, calculate the Compound Interest (CI):
\[ CI = A - P \] \[ CI = 13310 - 10000 \] \[ CI = \text{₹}3310 \]Correct Option: D) ₹3,310
The question asks to identify the largest producer of bauxite in India as of 2021. This requires knowledge of the geographical distribution of mineral resources and their production statistics in India.
D) Odisha — Odisha has consistently been the largest producer of bauxite in India, contributing over 50% of the country's total production. Its vast reserves and active mining operations make it the leading state in this sector.
The question asks to identify the Governor-General of Bengal responsible for introducing the Permanent Settlement in 1793. This requires knowledge of key figures and reforms during British colonial rule in India.
Correct Option: D) Charles Cornwallis
Lord Charles Cornwallis introduced the Permanent Settlement in Bengal in 1793. This system fixed the land revenue payable by the zamindars to the Company permanently. It aimed to ensure a stable revenue income for the Company and create a loyal class of zamindars.
If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of its first five terms to the sum of its first 10 terms.
We are given the ratio of the 11th term to the 18th term of an Arithmetic Progression (AP). We need to find the ratio of the sum of its first five terms to the sum of its first 10 terms. We will use the standard formulas for the \(n\)-th term and the sum of the first \(n\) terms of an AP.
Let the first term of the AP be \(a\) and the common difference be \(d\).
The \(n\)-th term of an AP is given by the formula: \(a_n = a + (n-1)d\).
So, the 11th term is \(a_{11} = a + (11-1)d = a + 10d\).
And the 18th term is \(a_{18} = a + (18-1)d = a + 17d\).
We are given that the ratio of the 11th term to the 18th term is 2 : 3.
\[ \frac{a_{11}}{a_{18}} = \frac{a + 10d}{a + 17d} = \frac{2}{3} \]Cross-multiply to find a relationship between \(a\) and \(d\):
\[ 3(a + 10d) = 2(a + 17d) \]\[ 3a + 30d = 2a + 34d \]\[ 3a - 2a = 34d - 30d \]\[ a = 4d \]Now, we need to find the ratio of the sum of its first five terms to the sum of its first 10 terms. The sum of the first \(n\) terms of an AP is given by the formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\).
The sum of the first five terms is:
\[ S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}[2a + 4d] \]The sum of the first 10 terms is:
\[ S_{10} = \frac{10}{2}[2a + (10-1)d] = 5[2a + 9d] \]Now, substitute \(a = 4d\) into the expressions for \(S_5\) and \(S_{10}\):
For \(S_5\):
\[ S_5 = \frac{5}{2}[2(4d) + 4d] = \frac{5}{2}[8d + 4d] = \frac{5}{2}[12d] = 5 \times 6d = 30d \]For \(S_{10}\):
\[ S_{10} = 5[2(4d) + 9d] = 5[8d + 9d] = 5[17d] = 85d \]Finally, find the ratio of \(S_5\) to \(S_{10}\):
\[ \frac{S_5}{S_{10}} = \frac{30d}{85d} \]\[ \frac{S_5}{S_{10}} = \frac{30}{85} \]Simplify the ratio by dividing both numerator and denominator by their greatest common divisor, which is 5:
\[ \frac{30 \div 5}{85 \div 5} = \frac{6}{17} \]The ratio of the sum of its first five terms to the sum of its first 10 terms is 6 : 17.
D) 6 : 17
The question asks to identify the book for which Arundhati Roy won The Booker Prize. This requires knowledge of famous literary awards and the works of prominent authors, a common type of question in general knowledge and literature sections of competitive exams.
Correct Option: B) The God of Small Things
Arundhati Roy won The Booker Prize in 1997 for her novel The God of Small Things. This novel is her most famous work and brought her international recognition.
The problem provides a set of observations arranged in ascending order and states the median of this data. We need to find the value of 'x'. The median of a data set depends on whether the number of observations is odd or even. For an even number of observations, the median is the average of the two middle terms.
Correct Option: A) 13
The question asks for the establishment date of the Border Roads Organisation (BRO) in India. This is a factual question requiring knowledge of Indian history and government organizations.
Correct Option: D) May 1960
The Border Roads Organisation (BRO) was established on May 7, 1960, with the primary objective of developing and maintaining road networks in India's border areas and friendly neighboring countries. This was a crucial step for India's defense and economic development, especially in strategically important and difficult terrains.
To solve this problem, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)). We will start by simplifying the innermost parentheses and work our way outwards.
First, simplify the expression inside the innermost parentheses: \((45 + 10 \div 2 \times 3 - 50)\).
Within these parentheses, perform division first:
\[ 10 \div 2 = 5 \]Next, perform multiplication:
\[ 5 \times 3 = 15 \]Now, substitute this value back into the innermost parentheses:
\[ 45 + 15 - 50 \]Perform addition and subtraction from left to right:
\[ 45 + 15 = 60 \]\[ 60 - 50 = 10 \]So, the expression inside the innermost parentheses simplifies to \(10\).
Substitute this back into the main expression:
\[ 3[35 + (10) + 5] \]Now, simplify the expression inside the square brackets:
\[ 35 + 10 + 5 \]Perform the additions from left to right:
\[ 35 + 10 = 45 \]\[ 45 + 5 = 50 \]Finally, multiply the result by \(3\):
\[ 3 \times 50 = 150 \]B) 150 is the correct answer, obtained by strictly following the order of operations.
The question asks about the geographical popularity of the folk song 'Kajari'. To answer this, one needs to recall the cultural and geographical distribution of Indian folk music forms.
B) Uttar Pradesh. Kajari is a popular folk song form predominantly sung in Uttar Pradesh, especially in the eastern parts (Purvanchal) and parts of Bihar. It is traditionally associated with the monsoon season and expresses themes of separation, longing, and the beauty of nature during the rains.