We need to find the population 3 years ago and 2 years ago. Since the population is increasing, we can work backward from the current population using the given growth rate. The formula for population growth is \(P_t = P_0 (1 + r)^t\), where \(P_t\) is the population after \(t\) years, \(P_0\) is the initial population, and \(r\) is the growth rate. To find the population in the past, we can rearrange this formula or apply the inverse operation (division by \(1+r\)).
Let \(P_{-1}\) be the population 1 year ago. The current population is \(P_{-1} \times (1 + r)\).
\[ P_{\text{current}} = P_{-1} \times (1 + 0.25) \]\[ 2,50,000 = P_{-1} \times 1.25 \]\[ P_{-1} = \frac{2,50,000}{1.25} = 2,00,000 \]Let \(P_{-2}\) be the population 2 years ago. The population 1 year ago is \(P_{-2} \times (1 + r)\).
\[ P_{-1} = P_{-2} \times (1 + 0.25) \]\[ 2,00,000 = P_{-2} \times 1.25 \]\[ P_{-2} = \frac{2,00,000}{1.25} = 1,60,000 \]Let \(P_{-3}\) be the population 3 years ago. The population 2 years ago is \(P_{-3} \times (1 + r)\).
\[ P_{-2} = P_{-3} \times (1 + 0.25) \]\[ 1,60,000 = P_{-3} \times 1.25 \]\[ P_{-3} = \frac{1,60,000}{1.25} = 1,28,000 \]Difference = \(P_{-2} - P_{-3}\)
\[ \text{Difference} = 1,60,000 - 1,28,000 = 32,000 \]Correct Option: D) 32,000
To find the smallest number by which 6627 must be multiplied to make it a perfect square, we need to perform prime factorization of 6627. A number is a perfect square if all the exponents in its prime factorization are even. We will identify the prime factors with odd exponents and multiply 6627 by the product of those factors to make all exponents even.
D) 3 — Multiplying 6627 by 3 makes its prime factorization \(3^2 \times 47^2\), which is a perfect square.
The question asks about the authority responsible for administering the oath of office and secrecy to Union Ministers as per Article 75(4) of the Indian Constitution. This is a direct constitutional provision, so recalling the relevant article is key.
A) President of India is the correct answer. Article 75(4) of the Constitution of India clearly states that the President shall administer the oaths of office and secrecy to the Union Ministers.
To find the positive difference in simple interest, we need to calculate the simple interest paid to each bank separately and then find the absolute difference between these two amounts. The formula for simple interest is \(SI = \frac{P \times R \times T}{100}\), where \(P\) is the principal amount, \(R\) is the rate of interest per annum, and \(T\) is the time in years.
Given:
Simple Interest for Bank A (\(SI_A\)) is:
\[ SI_A = \frac{P_A \times R_A \times T}{100} = \frac{5,00,000 \times 3.5 \times 4}{100} \] \[ SI_A = 5,000 \times 3.5 \times 4 = 5,000 \times 14 = \text{₹}70,000 \]Given:
Simple Interest for Bank B (\(SI_B\)) is:
\[ SI_B = \frac{P_B \times R_B \times T}{100} = \frac{5,00,000 \times 6 \times 4}{100} \] \[ SI_B = 5,000 \times 6 \times 4 = 5,000 \times 24 = \text{₹}1,20,000 \]The positive difference between the amounts of simple interest paid is:
\[ \text{Difference} = |SI_B - SI_A| = |1,20,000 - 70,000| = \text{₹}50,000 \]Alternatively, we can calculate the difference in interest rates first:
Difference in rates \(= R_B - R_A = 6\% - 3.5\% = 2.5\%\)
Then, the difference in simple interest can be calculated directly:
\[ \text{Difference} = \frac{P \times (R_B - R_A) \times T}{100} = \frac{5,00,000 \times 2.5 \times 4}{100} \] \[ \text{Difference} = 5,000 \times 2.5 \times 4 = 5,000 \times 10 = \text{₹}50,000 \]The question asks for the launch year of the Pradhan Mantri Awas Yojana (PMAY). This requires knowledge of major government schemes and their inception dates, which is a common topic in competitive exams.
Correct Option: D) 2015 — The Pradhan Mantri Awas Yojana (PMAY) was launched by the Government of India on June 25, 2015, with the objective of providing "Housing for All by 2022".
To solve this problem, we need to process each digit of the given number according to the specified rules: add 1 to odd digits and subtract 2 from even digits. After forming the new number, we identify its highest and lowest digits and then calculate their sum.
D) 8 — The highest digit in the new number is 8 and the lowest digit is 0. Their sum is \(8 + 0 = 8\).
To determine which number is divisible by 64, we need to perform division for each option or use divisibility rules. Since 64 is a power of 2 (\(2^6\)), a number is divisible by 64 if it can be expressed as \(64 \times k\) for some integer \(k\). We will divide each given option by 64 and check for a remainder of zero.
Divide 5129 by 64.
\[ \frac{5129}{64} \]\[ 5129 = 64 \times 80 + 9 \]The remainder is 9. So, 5129 is not divisible by 64.Divide 4797 by 64.
\[ \frac{4797}{64} \]\[ 4797 = 64 \times 74 + 61 \]The remainder is 61. So, 4797 is not divisible by 64.Divide 5760 by 64.
\[ \frac{5760}{64} \]\[ 5760 = 64 \times 90 + 0 \]The remainder is 0. So, 5760 is divisible by 64.Divide 5327 by 64.
\[ \frac{5327}{64} \]\[ 5327 = 64 \times 83 + 15 \]The remainder is 15. So, 5327 is not divisible by 64.C) 5760 is the only number among the given options that, when divided by 64, leaves a remainder of 0. Specifically, \(5760 \div 64 = 90\).
Harry ranked 34th from the top and 25th from the bottom in his class. How many students are there in his class?
This problem involves finding the total number of students in a class given a person's rank from both the top and the bottom. The key is to understand that when a person's rank is counted from both ends, that person is counted twice. Therefore, to find the total number of students, we sum the ranks from the top and bottom and then subtract 1 (to account for counting the person twice).
Therefore, there are 58 students in the class.
Correct Option: C) 58
To find the profit or loss per pen on average, we need to calculate the total cost price (CP) of all pens and the total selling price (SP) of all pens. Then, we can find the average cost price per pen and the average selling price per pen to determine the profit or loss per pen.
The shopkeeper sells all 50 pens at ₹14 each.
\[ \text{Total SP} = 50 \times \text{₹}14 = \text{₹}700 \]The selling price per pen is given as ₹14.
\[ \text{Average SP per pen} = \text{₹}14 \]Since the average selling price per pen (₹14) is greater than the average cost price per pen (₹12), there is a profit.
\[ \text{Profit per pen} = \text{Average SP per pen} - \text{Average CP per pen} \] \[ \text{Profit per pen} = \text{₹}14 - \text{₹}12 = \text{₹}2 \]Correct Option: D) ₹2 profit
The problem asks us to identify the number of symbols in a given series that satisfy two conditions simultaneously: they must be immediately preceded by a number and immediately followed by another symbol. We need to scan the series from left to right and check each symbol against these conditions.
A) 2 — There are two such symbols ('£' and '#') in the series that are immediately preceded by a number and immediately followed by a symbol.