This problem involves calculating the annual rate of increase, which is analogous to finding the rate of interest in compound interest problems. We can use the compound interest formula to solve this, where the initial production is the principal amount, and the final production is the amount after a certain period.
Identify the given values:
Recall the compound interest formula:
The formula for compound interest (or compounded growth) is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]Substitute the given values into the formula:
\[ 7986 = 6600 \left(1 + \frac{R}{100}\right)^2 \]Isolate the term containing \(R\):
\[ \frac{7986}{6600} = \left(1 + \frac{R}{100}\right)^2 \]Simplify the fraction:
Divide both the numerator and denominator by common factors. Both are divisible by 6:
\[ \frac{7986 \div 6}{6600 \div 6} = \frac{1331}{1100} \]So, we have:
\[ \frac{1331}{1100} = \left(1 + \frac{R}{100}\right)^2 \]Wait, let's recheck the division. \(7986 / 6600 = 1.21\). This is a more direct approach.
\[ 1.21 = \left(1 + \frac{R}{100}\right)^2 \]Take the square root of both sides:
\[ \sqrt{1.21} = 1 + \frac{R}{100} \] \[ 1.1 = 1 + \frac{R}{100} \]Solve for \(R\):
\[ 1.1 - 1 = \frac{R}{100} \] \[ 0.1 = \frac{R}{100} \] \[ R = 0.1 \times 100 \] \[ R = 10 \]So, the rate of increase is 10%.
Correct Option: D) 10%
To simplify the given expression, we will convert all decimal numbers into fractions to make the division and multiplication operations easier. Then, we will perform the division first, followed by the multiplications, adhering to the order of operations.
Convert decimals to fractions:
Substitute the fractions into the expression:
\[ \frac{25}{100} \div \frac{25}{10000} \times \frac{25}{1000} \times \frac{25}{10} \]
Perform the division: Dividing by a fraction is equivalent to multiplying by its reciprocal.
\[ \frac{25}{100} \times \frac{10000}{25} \times \frac{25}{1000} \times \frac{25}{10} \]
Simplify the expression: Cancel out common terms.
\[ \frac{\cancel{25}}{100} \times \frac{10000}{\cancel{25}} \times \frac{25}{1000} \times \frac{25}{10} \]
\[ \frac{1}{100} \times 10000 \times \frac{25}{1000} \times \frac{25}{10} \]
\[ \frac{10000}{100} \times \frac{25}{1000} \times \frac{25}{10} \]
\[ 100 \times \frac{25}{1000} \times \frac{25}{10} \]
\[ \frac{100 \times 25}{1000} \times \frac{25}{10} \]
\[ \frac{2500}{1000} \times \frac{25}{10} \]
\[ 2.5 \times 2.5 \]
\[ 6.25 \]
Alternatively, after step 3:
\[ \frac{25}{100} \times \frac{10000}{25} \times \frac{25}{1000} \times \frac{25}{10} \]
Cancel the first 25 with the second 25:
\[ \frac{1}{100} \times 10000 \times \frac{25}{1000} \times \frac{25}{10} \]
Simplify \( \frac{10000}{100} = 100 \):
\[ 100 \times \frac{25}{1000} \times \frac{25}{10} \]
Simplify \[\frac{100}{1000} = \frac{1}{10}\]:
\[ \frac{1}{1} \times \frac{25}{10} \times \frac{25}{10} \]
\[ \frac{25}{10} \times \frac{25}{10} \]
\[ 2.5 \times 2.5 = 6.25 \]
B) The simplified value of the expression is \(6.25\).
The question asks to identify which of the given options is NOT a public sector unit (PSU). A public sector unit is a company owned by the government (either central or state government) by holding the majority of its shares (typically more than 50%). We need to examine each option to determine its ownership structure.
B) ICICI — ICICI Bank Limited is a private sector bank. While it originated as a public financial institution, it was privatized and is now predominantly owned by private shareholders.
We will set up equations based on the given profit percentages and changes in cost price (CP) and selling price (SP). Let the original CP be \(C\) and the original SP be \(S\). We will use the profit percentage formula to form two equations and solve them simultaneously to find the value of \(C\).
Define initial conditions:
Let the original Cost Price be \(C\).
The article is sold at a profit of 20%. So, the original Selling Price \(S\) can be expressed as:
Define new conditions:
If both the cost price and selling price are ₹100 less, then:
New Cost Price \((C') = C - 100\)
New Selling Price \((S') = S - 100\)
Calculate the new profit percentage:
The original profit percentage is 20%. If the profit will be 4% more, the new profit percentage is \(20\% + 4\% = 24\%\).
Formulate the second equation:
Using the new CP and SP, and the new profit percentage:
Substitute the expressions for \(S'\) and \(C'\):
\[ S - 100 = 1.24(C - 100) \]Substitute \(S\) from the first equation into the second equation:
We know \(S = 1.20C\). Substitute this into the equation from Step 4:
Solve for \(C\):
\[ 1.20C - 100 = 1.24C - 1.24 \times 100 \] \[ 1.20C - 100 = 1.24C - 124 \]Rearrange the terms to isolate \(C\):
\[ 124 - 100 = 1.24C - 1.20C \] \[ 24 = 0.04C \] \[ C = \frac{24}{0.04} \] \[ C = \frac{24}{\frac{4}{100}} \] \[ C = \frac{24 \times 100}{4} \] \[ C = 6 \times 100 \] \[ C = 600 \]The cost price is ₹600.
Correct Option: D) ₹600
The question asks to identify what Kamet, Namcha Barwa, and Gurla Mandhata are, specifically in the context of the Himalayas. This requires general knowledge about the geography of the Himalayan region.
Correct Option: B) mountain peaks in Himalayas
If \(\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\) = x + y\(\sqrt{14}\), find the value of y.
\( \frac{6}{5} \)
\( \frac{4}{5} \)
\( \frac{3}{5} \)
\( \frac{2}{5} \)
To find the value of \(y\), we need to simplify the given expression \(\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\) by rationalizing the denominator. After simplification, we will equate the result with \(x + y\sqrt{14}\) and compare the coefficients to find \(y\).
Rationalize the denominator: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{7}+\sqrt{2}\).
\[ \frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}} = \frac{(\sqrt{7}+\sqrt{2})(\sqrt{7}+\sqrt{2})}{(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2})} \]Apply algebraic identities: Use the identities \((a+b)^2 = a^2 + b^2 + 2ab\) for the numerator and \((a-b)(a+b) = a^2 - b^2\) for the denominator.
Numerator: \((\sqrt{7}+\sqrt{2})^2 = (\sqrt{7})^2 + (\sqrt{2})^2 + 2(\sqrt{7})(\sqrt{2}) = 7 + 2 + 2\sqrt{14} = 9 + 2\sqrt{14}\)
Denominator: \((\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2}) = (\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5\)
Substitute the simplified terms back into the expression:
\[ \frac{9 + 2\sqrt{14}}{5} \]Separate the terms:
\[ \frac{9}{5} + \frac{2\sqrt{14}}{5} = \frac{9}{5} + \frac{2}{5}\sqrt{14} \]Equate with the given expression: We are given that \[\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}} = x + y\sqrt{14}\].
So, \[\frac{9}{5} + \frac{2}{5}\sqrt{14} = x + y\sqrt{14}\].
Compare the coefficients: By comparing the rational and irrational parts on both sides of the equation, we get:
\(x = \frac{9}{5}\)
\(y = \frac{2}{5}\)
Correct Option: D) \(\frac{2}{5}\)
The question asks to identify the option that does NOT support rural development. We need to evaluate each option based on its direct or indirect impact on the economic and social well-being of rural areas.
Correct Option: A) Urban credit does NOT directly support rural development. While there might be indirect linkages, its primary focus is on urban economies.
Of the three given equations, the first two are solved on the basis of a certain system. Find the correct answer for the unsolved third equation on the same basis.
32 × 34 = 96 25 × 14 = 29 18 × 51 = ?
The question asks to identify the two freedom fighters responsible for throwing smoke bombs in the Delhi Central Legislative Assembly in 1929. This is a direct factual recall question from Indian history, specifically related to the Indian independence movement.
Correct Option: B) Bhagat Singh and Batukeshwar Dutt
This problem involves calculating a value based on a percentage increase. We are given the marks obtained in the second term and the percentage increase from the first term. We need to find the marks obtained in the first term. We can set up an equation where the first term's marks, increased by 16%, equal the second term's marks.
Let the marks obtained in the first term be \(M_1\).
The marks obtained in the second term are 16% more than the marks in the first term. This can be expressed as:
\[ M_2 = M_1 + 16\% \text{ of } M_1 \]Or, equivalently:
\[ M_2 = M_1 \left(1 + \frac{16}{100}\right) \] \[ M_2 = M_1 \left(1 + 0.16\right) \] \[ M_2 = 1.16 \times M_1 \]We are given that the marks in the second term (\(M_2\)) are 87.
\[ 87 = 1.16 \times M_1 \]Now, we need to solve for \(M_1\):
\[ M_1 = \frac{87}{1.16} \]To simplify the division, we can multiply the numerator and denominator by 100:
\[ M_1 = \frac{87 \times 100}{1.16 \times 100} \] \[ M_1 = \frac{8700}{116} \]Perform the division:
\[ 8700 \div 116 = 75 \]So, \(M_1 = 75\).
Therefore, the marks in the first term were 75.
C) 75 is the correct answer. This is derived by setting up the equation where the first term's marks, increased by 16%, equal the given second term's marks of 87.
A) 78: If the first term marks were 78, then 16% more would be \(78 \times 1.16 = 90.48\), which is not 87.
B) 60: If the first term marks were 60, then 16% more would be \(60 \times 1.16 = 69.6\), which is not 87.
D) 80: If the first term marks were 80, then 16% more would be \(80 \times 1.16 = 92.8\), which is not 87.