To find the amount to be paid at the end of two years, we need to calculate the simple interest first and then add it to the principal amount. The formula for simple interest is \(I = \frac{P \times R \times T}{100}\), where \(P\) is the principal, \(R\) is the rate of interest, and \(T\) is the time period. The amount is then \(A = P + I\).
Using the formula \(I = \frac{P \times R \times T}{100}\):
\[ I = \frac{10000 \times 15 \times 2}{100} \]\[ I = \frac{300000}{100} \]\[ I = 3000 \]The total amount (\(A\)) is the sum of the principal and the simple interest:
\[ A = P + I \]\[ A = 10000 + 3000 \]\[ A = 13000 \]Therefore, the amount to be paid at the end of two years is ₹13,000.
Correct Option: A) ₹13,000 is the correct amount calculated by adding the simple interest to the principal.
Let the ages of the two colleagues be \(x\) and \(y\). We will translate the given conditions into algebraic equations and solve them to find the values of \(x\) and \(y\). We will then check the options against our derived ages.
Formulate the first equation:
The sum of the reciprocals of their ages is nine times the difference of the reciprocals of their ages.
Simplify the first equation:
Multiply both sides by \(xy\):
Formulate the second equation:
The ratio of the product of their ages to the sum of their ages is 100 : 9.
Solve the system of equations:
From the first simplified equation, we have \(x = \frac{4}{5}y\). Substitute this into the second equation:
Find the value of \(x\):
Substitute \(y = 25\) back into \(x = \frac{4}{5}y\):
Verify the solution:
The ages are 20 years and 25 years. Let's check the original conditions:
Sum of reciprocals: \[\frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}\]
Difference of reciprocals: \[\frac{1}{20} - \frac{1}{25} = \frac{5-4}{100} = \frac{1}{100}\]
Is \[\frac{9}{100} = 9 \times \frac{1}{100}\]? Yes, it is.
Product to sum ratio: \[\frac{20 \times 25}{20 + 25} = \frac{500}{45} = \frac{100}{9}\]. This also matches.
Both conditions are satisfied.
Correct Option: A) 20 years, 25 years
To find the largest 4-digit number divisible by 66, we first identify the largest 4-digit number. Then, we divide this number by 66 and use the remainder to find the desired number. If the remainder is 0, the largest 4-digit number itself is divisible by 66. Otherwise, we subtract the remainder from the largest 4-digit number to get the largest 4-digit number perfectly divisible by 66.
Correct Option: D) 9966 is the largest 4-digit number that is completely divisible by 66, as calculated by subtracting the remainder from the largest 4-digit number (9999).
The question asks for the year when the Government of India instituted the Bharat Ratna and Padma Vibhushan awards. This is a direct factual question related to Indian history and polity, specifically concerning national awards.
Correct Option: D) 1954
If cot(\(\frac{\pi}{2}\) − θ) = \(\frac{1}{\sqrt{3}}\), then find the value of cos θ.
\( \frac{1}{2} \)
\(\frac{\sqrt{3}}{2}\)
\(\sqrt{3}\)
\(\frac{1}{\sqrt{3}}\)
The problem requires us to find the value of \(\cos \theta\) given an equation involving \(\cot\) and \(\theta\). We will use trigonometric identities to simplify the given equation and find the value of \(\theta\), then substitute it into \(\cos \theta\).
The given equation is:
\[ \cot\left(\frac{\pi}{2} - \theta\right) = \frac{1}{\sqrt{3}} \]Recall the trigonometric identity: \(\cot\left(\frac{\pi}{2} - x\right) = \tan(x)\). Applying this identity to our equation, we get:
\[ \tan(\theta) = \frac{1}{\sqrt{3}} \]We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). In radians, \(30^\circ = \frac{\pi}{6}\). Therefore,
\[ \tan(\theta) = \tan\left(\frac{\pi}{6}\right) \]From this, we can deduce that \(\theta = \frac{\pi}{6}\).
Now, we need to find the value of \(\cos \theta\). Substitute the value of \(\theta\) we found:
\[ \cos(\theta) = \cos\left(\frac{\pi}{6}\right) \]We know that \[\cos\left(\frac{\pi}{6}\right) = \cos(30^\circ) = \frac{\sqrt{3}}{2}\].
Thus, \(\cos \theta = \frac{\sqrt{3}}{2}\).
B) The value of \(\cos \theta\) is \(\frac{\sqrt{3}}{2}\).
To solve this problem, we need to set up a system of two linear equations based on the given information. We will define two variables for the two numbers, express one in terms of the other, and then use their sum to find their individual values. Finally, we will calculate their product.
Define the variables: Let the smaller number be \(x\). Let the bigger number be \(y\).
Formulate the first equation: "The big number is obtained by adding 5 to the other." This can be written as:
\[ y = x + 5 \quad \text{(Equation 1)} \]Formulate the second equation: "The total of the two numbers is 19." This means their sum is 19:
\[ x + y = 19 \quad \text{(Equation 2)} \]Solve the system of equations: Substitute Equation 1 into Equation 2:
\[ x + (x + 5) = 19 \]Combine like terms:
\[ 2x + 5 = 19 \]Subtract 5 from both sides:
\[ 2x = 19 - 5 \] \[ 2x = 14 \]Divide by 2 to find \(x\):
\[ x = \frac{14}{2} \] \[ x = 7 \]Find the value of \(y\): Substitute \(x = 7\) back into Equation 1:
\[ y = 7 + 5 \] \[ y = 12 \]So, the two numbers are 7 and 12.
Calculate the product of the numbers: Product \( = x \times y \) Product \( = 7 \times 12 \) Product \( = 84 \)
A) 84 is the correct product of the two numbers, 7 and 12.
If 2 L of water is evaporated on boiling from 8 L of sugar solution containing 4% sugar, find the percentage of sugar in the remaining solution.
The problem involves calculating the new percentage concentration of sugar in a solution after some water has evaporated. The key principle here is that the amount of sugar remains constant, while the total volume of the solution decreases due to evaporation of water. We will first calculate the initial amount of sugar, then find the new volume of the solution, and finally determine the new percentage concentration.
Calculate the initial amount of sugar:
The initial volume of sugar solution is \(8 \, \text{L}\).
The initial percentage of sugar is \(4\%\).
Amount of sugar \( = \text{Percentage} \times \text{Total Volume}\)
\[ \text{Amount of sugar} = 4\% \times 8 \, \text{L} = \frac{4}{100} \times 8 \, \text{L} = \frac{32}{100} \, \text{L} = 0.32 \, \text{L} \]So, there is \(0.32 \, \text{L}\) of sugar in the solution.
Calculate the volume of the remaining solution:
Initial volume of solution \( = 8 \, \text{L}\).
Volume of water evaporated \( = 2 \, \text{L}\).
Remaining volume of solution \( = \text{Initial volume} - \text{Volume evaporated}\)
\[ \text{Remaining volume} = 8 \, \text{L} - 2 \, \text{L} = 6 \, \text{L} \]Calculate the new percentage of sugar:
The amount of sugar remains constant, which is \(0.32 \, \text{L}\).
The new total volume of the solution is \(6 \, \text{L}\).
New percentage of sugar \[= \frac{\text{Amount of sugar}}{\text{Remaining volume}} \times 100\%\]
\[ \text{New percentage of sugar} = \frac{0.32 \, \text{L}}{6 \, \text{L}} \times 100\% = \frac{32}{100 \times 6} \times 100\% = \frac{32}{6}\% \]Simplify the fraction:
\[ \frac{32}{6}\% = \frac{16}{3}\% \]Correct Option: B) \( \frac{16}{3} \) %
The percentage of sugar in the remaining solution is \( \frac{16}{3} \) %.
The question asks to identify the odd word out from a given set of terms related to the Indian Constitution. We need to understand the nature of each term to find the one that differs from the others.
Correct Option: C) Ordinance is the odd one out because it is a temporary legislative instrument, whereas Preamble, Articles, and Schedules are permanent and integral parts of the Constitution of India.
The problem involves decoding a pattern in a given word and applying the same pattern to another word. We need to identify the transformation rule applied to each letter of "CRICKET" to get "ZOFZHBQ" and then use that rule for "FOOTBALL".
Let's write down the letters of CRICKET and their corresponding coded letters:
We'll find the positional difference in the alphabet for each pair.
The pattern is consistently subtracting 3 from the alphabetical position of each letter (or moving 3 steps backward in the alphabet).
Now, we apply the "-3" rule to each letter of FOOTBALL:
Combining these, FOOTBALL is coded as CLLQYXII.
Correct Option: B) CLLQYXII
We are given the sum of two numbers and their HCF. We need to find a possible pair of such numbers from the given options. If the HCF of two numbers is \(H\), then the numbers can be expressed as \(Hx\) and \(Hy\), where \(x\) and \(y\) are coprime (i.e., their HCF is 1).
A) 225, 625