To find the value of the given expression, we can multiply the numerical parts and handle the decimal places separately. The expression involves several decimal numbers and integers. We will first multiply all the non-decimal parts and then count the total number of decimal places to correctly place the decimal point in the final product.
The given expression is:
\[ 7 \times 0.7 \times 0.07 \times 0.007 \times 70 \]We can write each term as a product of an integer and a power of 10, or simply multiply the integers first and then adjust for decimal places.
Let's multiply the integer parts: \(7 \times 7 \times 7 \times 7 \times 70\).
This can be simplified as \(7^4 \times 7 \times 10 = 7^5 \times 10\).
Calculate \(7^5\):
So, the product of the integer parts is \(16807 \times 10 = 168070\).
The total number of decimal places is \(1 + 2 + 3 = 6\).
We have the numerical product \(168070\). We need to place the decimal point such that there are 6 digits after it. Starting from the right of \(168070\), move the decimal point 6 places to the left.
Original number: \(168070.\)
Moving 6 places to the left:
So, the final value is \(0.16807\).
D) 0.16807 — This matches our calculated value after multiplying the numbers and correctly placing the decimal point.
Calculate the compound interest on ₹15,000 in one year at 4% per annum, if the interest is compounded half yearly.
To calculate the compound interest when interest is compounded half-yearly, we need to adjust the annual interest rate and the time period. The annual rate is divided by 2, and the number of years is multiplied by 2 to get the number of compounding periods. Then, we can use the compound interest formula or calculate it step-by-step for each half-year period.
Since the interest is compounded half-yearly, the annual rate needs to be divided by 2, and the time period needs to be multiplied by 2.
Interest for the 1st half-year:
\[ I_1 = P \times \frac{r}{100} = 15000 \times \frac{2}{100} = 150 \times 2 = \text{₹}300 \]Amount after 1st half-year:
\[ A_1 = P + I_1 = 15000 + 300 = \text{₹}15,300 \]For the 2nd half-year, the principal becomes \(A_1\).
Interest for the 2nd half-year:
\[ I_2 = A_1 \times \frac{r}{100} = 15300 \times \frac{2}{100} = 153 \times 2 = \text{₹}306 \]Amount after 2nd half-year (Total Amount):
\[ A_2 = A_1 + I_2 = 15300 + 306 = \text{₹}15,606 \]Compound Interest = Total Amount - Principal
\[ \text{CI} = A_2 - P = 15606 - 15000 = \text{₹}606 \]B) ₹606 is the correct compound interest calculated after one year with half-yearly compounding.
The question asks for the term that describes the persistence of sound in an auditorium due to repeated reflections. We need to identify the correct physics term from the given options.
D) Reverberation — Reverberation is precisely defined as the persistence of sound in an enclosed space, such as an auditorium, due to multiple reflections of sound waves from the surfaces of the enclosure. This phenomenon contributes to the overall acoustic quality of a space.
At what rate percent per annum will the simple interest in 15 years on a sum of money be \( \frac{3}{4} \) of the sum invested?
This problem involves calculating the rate of simple interest. We are given the time period and the relationship between the simple interest and the principal amount. We will use the standard formula for simple interest to find the unknown rate.
Let the principal sum invested be \(P\).
The simple interest (\(SI\)) is given as \( \frac{3}{4} \) of the sum invested, so \( SI = \frac{3}{4} P \).
The time period (\(T\)) is 15 years.
We need to find the rate of interest per annum (\(R\)).
The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]Substitute \( SI = \frac{3}{4} P \) and \( T = 15 \) into the formula:
\[ \frac{3}{4} P = \frac{P \times R \times 15}{100} \]We can cancel \(P\) from both sides of the equation (assuming \(P \neq 0\)):
\[ \frac{3}{4} = \frac{R \times 15}{100} \]Multiply both sides by 100:
\[ \frac{3}{4} \times 100 = R \times 15 \]\[ 3 \times 25 = R \times 15 \]\[ 75 = R \times 15 \]Divide by 15:
\[ R = \frac{75}{15} \]\[ R = 5 \]The rate of interest is 5% per annum.
Correct Option: B) 5% is the correct rate percent per annum at which the simple interest in 15 years will be \( \frac{3}{4} \) of the sum invested.
The problem asks us to evaluate two courses of action based on a given statement. We need to determine which course(s) of action logically follow from the statement that "Road accidents are going to increase day by day." The goal of any course of action in such a scenario should be to mitigate or address the problem stated.
Correct Option: B) Only I follows
To solve this problem, we need to test each option by interchanging the specified mathematical operators in the given equation and then evaluating the expression using the BODMAS/PEMDAS rule (Brackets, Orders/Exponents, Division, Multiplication, Addition, Subtraction). The goal is to find the interchange that makes the equation equal to 19.
The original equation is: \(56 - 8 + 42 \div 6 \times 5 = 19\)
The equation becomes: \(56 + 8 - 42 \div 6 \times 5\)
Since \(29 \neq 19\), Option A is incorrect.
The equation becomes: \(56 - 8 \div 42 + 6 \times 5\)
Since \(\frac{1802}{21} \neq 19\), Option B is incorrect.
The equation becomes: \(56 \times 8 + 42 \div 6 - 5\)
Since \(450 \neq 19\), Option C is incorrect.
The equation becomes: \(56 \div 8 + 42 - 6 \times 5\)
Since \(19 = 19\), Option D is correct.
D) ÷ and − — When division and subtraction are interchanged, the equation evaluates to 19, making the statement correct.
The question asks to identify India's first indigenously made film in color. This requires knowledge of Indian cinematic history, specifically milestones related to film technology and production.
The film Kisan Kanya, released in 1937, is widely recognized as India's first indigenously produced color film. It was produced by Ardeshir Irani, who also produced India's first talkie, Alam Ara. The film used the Cinecolor process.
B) Kisan Kanya — Kisan Kanya, released in 1937, was India's first indigenously made film in color. It was produced by Ardeshir Irani and directed by Moti B. Gidwani.
To find the value of \(k\), we will use the relationship between the coefficients of a quadratic polynomial and its zeros. For a quadratic polynomial \(ax^2 + bx + c\), if \(\alpha\) and \(\beta\) are its zeros, then the sum of the zeros is \(\alpha + \beta = -b/a\) and the product of the zeros is \(\alpha \beta = c/a\). We are given the sum of the squares of the zeros, \(\alpha^2 + \beta^2 = 40\). We can express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha \beta\).
Identify coefficients: The given quadratic polynomial is \(f(x) = x^2 - 8x + k\). Comparing this with the standard form \(ax^2 + bx + c\), we have:
Find sum and product of zeros: Let \(\alpha\) and \(\beta\) be the zeros of the polynomial.
Use the given condition: We are given that the sum of the squares of the zeros is 40, i.e., \( \alpha^2 + \beta^2 = 40 \).
Relate sum of squares to sum and product of zeros: We know the algebraic identity \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \). Rearranging this, we get \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
Substitute values and solve for k: Substitute the values we found for \( \alpha + \beta \) and \( \alpha \beta \) into the rearranged identity:
\[ 40 = (8)^2 - 2(k) \]\[ 40 = 64 - 2k \]Isolate k:
\[ 2k = 64 - 40 \]\[ 2k = 24 \]\[ k = \frac{24}{2} \]\[ k = 12 \]Thus, the value of \(k\) is 12.
Correct Option: B) 12
The question asks to identify the author of a famous quote about the United Nations. This requires knowledge of prominent figures associated with the UN and their notable statements. The quote emphasizes the UN's role in preventing catastrophe rather than achieving an ideal state.
Correct Option: C) Dag Hammarskjold. This quote is widely attributed to Dag Hammarskjöld, who served as the second Secretary-General of the United Nations from 1953 until his death in 1961. His tenure was marked by significant efforts in peacekeeping and international diplomacy, often in challenging circumstances, reflecting the sentiment of the quote.
The question asks to identify the category of elements that are electronegative and form negatively charged ions by gaining electrons. We need to recall the fundamental properties of different types of substances.
A) Non-metals are known as electronegative elements because they form negatively charged ions by the gain of electrons. This is a defining characteristic of non-metallic elements.