This question is based on letter coding and decoding using reverse alphabetical positions. In this coding system, each letter is assigned a numerical value by counting backward from the end of the English alphabet, where A = 26, B = 25, C = 24, and Z = 1. The code for a word is obtained by adding the reverse alphabetical values of all its letters. We must verify this rule using BOTTLE = 88 and then apply it to GREEN.
In the normal alphabetical order, A = 1, B = 2, C = 3, and Z = 26.
In reverse alphabetical order, the values are assigned as follows:
The reverse alphabetical value of a letter is calculated using:
\(\text{Reverse alphabetical value} = 27 - \text{Normal alphabetical position}\)
The word BOTTLE contains six letters: B, O, T, T, L, and E.
The reverse alphabetical values are:
Add the reverse alphabetical values of all six letters:
\(25+12+7+7+15+22=88\)
Therefore:
BOTTLE = 88
This matches the given code and confirms the intended coding rule.
The word GREEN contains five letters: G, R, E, E, and N.
The reverse alphabetical values are:
\(20+9+22+22+13=86\)
Therefore:
GREEN = 86
GREEN contains five letters. The sum of its normal alphabetical positions is:
\(7+18+5+5+14=49\)
Since each reverse alphabetical value equals 27 minus the normal alphabetical position, the total reverse alphabetical value is:
\((27\times5)-49=135-49=86\)
This confirms that GREEN is coded as 86.
Correct Option: D) 86
The coding rule assigns reverse alphabetical values to the letters, where A = 26 and Z = 1. The values of the letters in GREEN are G = 20, R = 9, E = 22, E = 22, and N = 13. Adding these values gives \(20+9+22+22+13=86\). Therefore, option D is correct.
If tan θ = \( \frac{12}{13} \), then find the value of \(\frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}\).
\( \frac{13}{12} \)
\( \frac{413}{25} \)
\( \frac{312}{25} \)
\( \frac{512}{25} \)
The problem provides the value of \(\tan \theta\) and asks for the value of an expression involving \(\sin \theta\) and \(\cos \theta\). We can solve this by either constructing a right-angled triangle to find \(\sin \theta\) and \(\cos \theta\) or by manipulating the given expression to relate it to \(\tan \theta\).
Identify the given information:
We are given \(\tan \theta = \frac{12}{13}\).
Identify the expression to evaluate:
We need to find the value of \[\frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}\].
Manipulate the expression to use \(\tan \theta\):
Divide both the numerator and the denominator by \(\cos^2\theta\). This is a common technique to convert expressions involving \(\sin \theta\) and \(\cos \theta\) into expressions involving \(\tan \theta\).
\[ \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} = \frac{\frac{2\sin\theta\cos\theta}{\cos^2\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}} \]Simplify the numerator:
\[ \frac{2\sin\theta\cos\theta}{\cos^2\theta} = \frac{2\sin\theta}{\cos\theta} = 2\tan\theta \]Simplify the denominator:
\[ \frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta} = \frac{\cos^2\theta}{\cos^2\theta} - \frac{\sin^2\theta}{\cos^2\theta} = 1 - \left(\frac{\sin\theta}{\cos\theta}\right)^2 = 1 - \tan^2\theta \]Substitute the simplified terms back into the expression:
\[ \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} = \frac{2\tan\theta}{1-\tan^2\theta} \]This is also the formula for \(\tan(2\theta)\).
Substitute the given value of \(\tan \theta\):
Given \(\tan \theta = \frac{12}{13}\).
\[ \frac{2\left(\frac{12}{13}\right)}{1-\left(\frac{12}{13}\right)^2} = \frac{\frac{24}{13}}{1-\frac{144}{169}} \]Calculate the denominator:
\[ 1-\frac{144}{169} = \frac{169-144}{169} = \frac{25}{169} \]Complete the calculation:
\[ \frac{\frac{24}{13}}{\frac{25}{169}} = \frac{24}{13} \times \frac{169}{25} = \frac{24 \times 13}{25} = \frac{312}{25} \]Correct Option: C) \(\frac{312}{25}\)
To find the compound interest, we can calculate the amount after 2 years using the compound interest formula and then subtract the principal amount. Alternatively, we can calculate the interest for each year separately, adding the previous year's interest to the principal for the next year's calculation.
The formula for the amount (A) compounded annually is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
Substitute the given values:
\[ A = 8000 \left(1 + \frac{5}{100}\right)^2 \]
\[ A = 8000 \left(1 + 0.05\right)^2 \]
\[ A = 8000 \left(1.05\right)^2 \]
\[ A = 8000 \times 1.1025 \]
\[ A = 8820 \]
The amount after 2 years is ₹8,820.
Now, calculate the Compound Interest (CI):
\[ \text{CI} = A - P \]
\[ \text{CI} = 8820 - 8000 \]
\[ \text{CI} = 820 \]
For the 1st year:
Simple Interest for 1st year = \(\frac{P \times R \times T}{100}\)
\[ \text{Interest}_1 = \frac{8000 \times 5 \times 1}{100} = \frac{40000}{100} = 400 \]
Amount at the end of 1st year = \(8000 + 400 = 8400\)
For the 2nd year:
The principal for the 2nd year becomes ₹8,400.
\[ \text{Interest}_2 = \frac{8400 \times 5 \times 1}{100} = \frac{42000}{100} = 420 \]
Total Compound Interest = Interest for 1st year + Interest for 2nd year
\[ \text{CI} = 400 + 420 = 820 \]
Correct Option: B) ₹820
The question asks to identify the primary energy currency used by cells for most of their processes. This requires knowledge of fundamental cellular biochemistry and metabolism.
A) ATP — ATP (Adenosine Triphosphate) is universally recognized as the primary energy currency of the cell. It stores chemical energy in its phosphate bonds, which is released upon hydrolysis to power various cellular activities.
The question asks about two specific resolutions piloted by India and adopted by the United Nations Environment Assembly (UNEA) in 2019. This requires knowledge of India's environmental diplomacy and key outcomes of UNEA sessions.
Correct Option: A) Single-use plastic and sustainable nitrogen management. These were indeed the two resolutions piloted by India and adopted by the United Nations Environment Assembly (UNEA) in 2019 (specifically, UNEA-4).
The problem asks for the rise in water level in a cuboidal pond when a certain number of people displace water. The key idea is that the total volume of water displaced by all persons will be equal to the volume of the raised water level in the pond. We will calculate the total displaced volume and then use the pond's dimensions to find the rise in height.
Given that 500 persons are taking a dip, and the average displacement of water by one person is \(0.04 \text{ m}^3\).
\[\text{Total displaced volume} = \text{Number of persons} \times \text{Displacement per person}\] \[\text{Total displaced volume} = 500 \times 0.04 \text{ m}^3\] \[\text{Total displaced volume} = 20 \text{ m}^3\]The pond is cuboidal with a length of \(80 \text{ m}\) and a breadth of \(50 \text{ m}\). Let the rise in water level be \(h\).
The volume of the raised water in the pond can be calculated as:
\[\text{Volume of raised water} = \text{Length} \times \text{Breadth} \times \text{Height (rise)}\] \[\text{Volume of raised water} = 80 \text{ m} \times 50 \text{ m} \times h\] \[\text{Volume of raised water} = 4000h \text{ m}^3\]The total displaced volume is equal to the volume of the raised water.
\[20 \text{ m}^3 = 4000h \text{ m}^3\] \[h = \frac{20}{4000} \text{ m}\] \[h = \frac{1}{200} \text{ m}\] \[h = 0.005 \text{ m}\]Since \(1 \text{ m} = 100 \text{ cm}\), we convert the height from meters to centimeters.
\[h = 0.005 \times 100 \text{ cm}\] \[h = 0.5 \text{ cm}\]Correct Option: A) 0.5 cm
The question asks about the fundamental language a computer uses to process data. This requires understanding how computers internally represent and manipulate information.
Correct Option: D) Binary — Computers operate using electrical signals that are either on or off. This two-state system is represented by the binary digits 0 and 1. All data and instructions processed by a computer are ultimately converted into and understood as binary code.
If x = asinθ, and y = btanθ, then find the value of \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\).
The problem asks us to find the value of an expression involving \(x\) and \(y\), given their definitions in terms of trigonometric functions. We will substitute the given relations for \(x\) and \(y\) into the expression and simplify it using fundamental trigonometric identities.
Given \(x = a \sin\theta\), we can write:
\[ \frac{a}{x} = \frac{a}{a \sin\theta} = \frac{1}{\sin\theta} = \csc\theta \]Given \(y = b \tan\theta\), we can write:
\[ \frac{b}{y} = \frac{b}{b \tan\theta} = \frac{1}{\tan\theta} = \cot\theta \]The expression we need to evaluate is \[\frac{a^2}{x^2} - \frac{b^2}{y^2}\].
We can rewrite this as:
\[ \left(\frac{a}{x}\right)^2 - \left(\frac{b}{y}\right)^2 \]Substitute the values from Step 1:
\[ (\csc\theta)^2 - (\cot\theta)^2 = \csc^2\theta - \cot^2\theta \]Recall the fundamental trigonometric identity:
\[ \csc^2\theta - \cot^2\theta = 1 \]Therefore, the value of the expression is \(1\).
A) 1 is the correct answer because, after substituting the given relations and simplifying, the expression reduces to the fundamental trigonometric identity \(\csc^2\theta - \cot^2\theta = 1\).
The problem asks us to find the number that has the same relationship with 9 as 54 has with 7. We need to identify the pattern or mathematical operation connecting the first pair of numbers (7 and 54) and then apply the same pattern to the third number (9) to find the missing number.
Let the first number be \(N_1 = 7\) and the second number be \(N_2 = 54\).
We need to find a mathematical operation that transforms 7 into 54. Let's try common operations:
Let's test the pattern \(N_1^2 + 5\):
\[ 7^2 + 5 = 49 + 5 = 54 \]This pattern works for the first pair.
Let the third number be \(N_3 = 9\). We need to find \(N_4\) using the same pattern.
Using the pattern \(N^2 + 5\):
\[ 9^2 + 5 = 81 + 5 = 86 \]So, the missing number is 86.
Correct Option: D) 86
The question asks for the launch date of the Samagra Shiksha Abhiyan, a significant educational scheme in India. To answer this, one needs to recall the specific timeline of major government initiatives related to education.
Correct Option: B) May 2018 is the correct launch date for the Samagra Shiksha Abhiyan. This scheme integrates various educational initiatives to improve the quality of education and leverage technology for students and teachers.