This problem involves partnership and profit sharing. The profit share of each partner is directly proportional to the product of their investment and the time period for which the investment was made. We will set up equations based on the given information about investments, time, and profit distribution, and then solve for the unknown capital contributed by M.
Correct Option: A) ₹42,500 is the capital contributed by M.
To find the simple interest, we use the formula \(SI = \frac{P \times R \times T}{100}\), where \(P\) is the principal amount, \(R\) is the rate of interest per period, and \(T\) is the time period. It's crucial to ensure that the rate and time are consistent in their units (e.g., both per month or both per year).
C) ₹480 is the correct simple interest calculated using the given principal, rate, and time.
To simplify the given algebraic expression, we need to perform the multiplication operations first, then combine like terms. This involves applying the distributive property and then adding or subtracting coefficients of terms with the same variable and exponent.
First, distribute \(x\) into the first parenthesis and \(3\) into the second parenthesis:
\[ x(7x - 5) + 3(x^2 - 2) + 14 \]\[ = (x \cdot 7x) - (x \cdot 5) + (3 \cdot x^2) - (3 \cdot 2) + 14 \]\[ = 7x^2 - 5x + 3x^2 - 6 + 14 \]Next, group the terms with the same powers of \(x\) and the constant terms together.
Now, perform the addition/subtraction for each group:
Combine the simplified terms to get the final expression:
\[ 10x^2 - 5x + 8 \]Correct Option: C) \(10x^2 - 5x + 8\)
Pipe P can fill \( \frac{4}{5} \) part of a tank in 12 hours and pipe Q can fill \( \frac{2}{7} \) part of the same tank in 30 hours.
Both P and Q were kept open for 5 hours, then both were closed. Pipe R alone was then opened and it emptied the water in the tank in 8 hours.
Pipes P, Q and R together can fill the empty tank in:
This problem involves calculating the work rates of pipes and then combining them to find the total time taken to fill the tank. We will first find the individual rates of pipes P and Q. Then, calculate the amount of tank filled by P and Q together in 5 hours. Next, we will determine the emptying rate of pipe R based on the remaining water. Finally, we will combine the rates of P, Q, and R to find the time taken to fill the empty tank.
Calculate the time taken by pipe P to fill the entire tank:
Pipe P fills \( \frac{4}{5} \) of the tank in 12 hours.
\[ \text{Time to fill 1 tank by P} = 12 \text{ hours} \times \frac{5}{4} = 15 \text{ hours} \]Calculate the time taken by pipe Q to fill the entire tank:
Pipe Q fills \( \frac{2}{7} \) of the tank in 30 hours.
\[ \text{Time to fill 1 tank by Q} = 30 \text{ hours} \times \frac{7}{2} = 105 \text{ hours} \]Calculate the filling rate of pipes P and Q:
Rate of P = \( \frac{1}{15} \) tank per hour.
Rate of Q = \( \frac{1}{105} \) tank per hour.
Calculate the amount of tank filled by P and Q together in 5 hours:
Combined rate of P and Q = \[\frac{1}{15} + \frac{1}{105}\]
\[ \frac{1}{15} + \frac{1}{105} = \frac{7}{105} + \frac{1}{105} = \frac{8}{105} \text{ tank per hour} \]Amount filled in 5 hours = \[5 \times \frac{8}{105} = \frac{40}{105} = \frac{8}{21}\] of the tank.
Calculate the emptying rate of pipe R:
Pipe R empties the \( \frac{8}{21} \) part of the tank in 8 hours.
Rate of R (emptying) = \[\frac{8}{21} \div 8 = \frac{8}{21} \times \frac{1}{8} = \frac{1}{21}\] tank per hour.
Calculate the combined rate of P, Q, and R when working together:
P and Q fill, R empties. So, the effective rate is (Rate of P + Rate of Q) - Rate of R.
\[ \text{Combined rate} = \left( \frac{1}{15} + \frac{1}{105} \right) - \frac{1}{21} \]\[ \text{Combined rate} = \frac{8}{105} - \frac{1}{21} \]To subtract, find a common denominator, which is 105.
\[ \frac{8}{105} - \frac{5}{105} = \frac{3}{105} = \frac{1}{35} \text{ tank per hour} \]Calculate the time taken by P, Q, and R together to fill the empty tank:
Time = \[\frac{1}{\text{Combined rate}}\]
\[ \text{Time} = \frac{1}{\frac{1}{35}} = 35 \text{ hours} \]Correct Option: B) 35 hours
To find the average of the first 'n' even numbers, we can use the formula for the sum of the first 'n' even numbers and then divide by 'n'. Alternatively, we can recognize that the first 'n' even numbers form an arithmetic progression, and the average of an arithmetic progression is the average of its first and last terms.
Both methods yield the same result.
Correct Option: D) 164
The value of \( \frac{5}{3} \) + \(\left(\frac{1}{1+\frac{4}{9}}\right)\) − \( \frac{2}{5} \) is
\(\frac{384}{191}\)
\(\frac{382}{195}\)
\(\frac{379}{201}\)
\(\frac{380}{197}\)
To solve this problem, we need to simplify the given expression by performing the operations in the correct order. We will first simplify the term inside the parenthesis, then perform the addition and subtraction.
The term inside the parenthesis is \[\left(\frac{1}{1+\frac{4}{9}}\right)\]. First, simplify the denominator \(1+\frac{4}{9}\).
\[ 1 + \frac{4}{9} = \frac{9}{9} + \frac{4}{9} = \frac{9+4}{9} = \frac{13}{9} \]Therefore, the term becomes:\[ \frac{1}{1+\frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13} \]The original expression is \[\frac{5}{3} + \left(\frac{1}{1+\frac{4}{9}}\right) - \frac{2}{5}\]. Substituting \(\frac{9}{13}\) for the parenthesis term, we get:
\[ \frac{5}{3} + \frac{9}{13} - \frac{2}{5} \]To add and subtract these fractions, we need to find a common denominator for 3, 13, and 5. The least common multiple (LCM) of 3, 13, and 5 is \(3 \times 13 \times 5 = 195\).
Convert each fraction to have the denominator 195:
\[ \frac{5}{3} = \frac{5 \times 65}{3 \times 65} = \frac{325}{195} \]\[ \frac{9}{13} = \frac{9 \times 15}{13 \times 15} = \frac{135}{195} \]\[ \frac{2}{5} = \frac{2 \times 39}{5 \times 39} = \frac{78}{195} \]Now, perform the operations:\[ \frac{325}{195} + \frac{135}{195} - \frac{78}{195} = \frac{325 + 135 - 78}{195} \]\[ = \frac{460 - 78}{195} = \frac{382}{195} \]B) The calculated value of the expression is \(\frac{382}{195}\).
To find the greater of two consecutive natural numbers whose product is 110, we can either set up an algebraic equation or test the options provided. Since the numbers are consecutive, if one is \(x\), the other is \(x+1\).
We can also test the options. If the given option is the greater number, the smaller number would be one less than it. Then we multiply them to see if the product is 110.
Thus, the greater number is 11.
Correct Option: C) 11
The value of \(\sqrt{121}\) + \(\sqrt{0.0225}\) − \(\sqrt{5.76}\) =
To solve this problem, we need to evaluate each square root term individually and then perform the indicated addition and subtraction. We will find the square root of whole numbers and decimal numbers.
We know that \(11 \times 11 = 121\).
\[ \sqrt{121} = 11 \]First, consider the number without the decimal point, which is \(225\). We know that \(15 \times 15 = 225\).
Since there are four decimal places in \(0.0225\), its square root will have half that number of decimal places, which is two.
\[ \sqrt{0.0225} = 0.15 \]First, consider the number without the decimal point, which is \(576\). We know that \(24 \times 24 = 576\).
Since there are two decimal places in \(5.76\), its square root will have half that number of decimal places, which is one.
\[ \sqrt{5.76} = 2.4 \]The original expression is \(\sqrt{121} + \sqrt{0.0225} - \sqrt{5.76}\).
Substitute the calculated values:
\[ 11 + 0.15 - 2.4 \]First, add \(11\) and \(0.15\):
\[ 11 + 0.15 = 11.15 \]Next, subtract \(2.4\) from \(11.15\):
\[ 11.15 - 2.4 = 8.75 \]C) 8.75 is the correct answer, as derived from the step-by-step calculation.
Simplify: \(\frac{\tan^2\beta\,\mathrm{cosec}^2\beta}{1+\tan^2\beta}\)
To simplify the given trigonometric expression, we will use fundamental trigonometric identities to convert all terms into sine and cosine, or use other standard identities to simplify directly.
The given expression is:
\[ \frac{\tan^2\beta\,\mathrm{cosec}^2\beta}{1+\tan^2\beta} \]Recall the fundamental trigonometric identities:
\(\tan\beta = \frac{\sin\beta}{\cos\beta}\)
\(\mathrm{cosec}\beta = \frac{1}{\sin\beta}\)
\(1+\tan^2\beta = \sec^2\beta\)
\(\sec\beta = \frac{1}{\cos\beta}\)
Substitute \(1+\tan^2\beta = \sec^2\beta\) into the denominator:
\[ \frac{\tan^2\beta\,\mathrm{cosec}^2\beta}{\sec^2\beta} \]Now, express \(\tan^2\beta\), \(\mathrm{cosec}^2\beta\), and \(\sec^2\beta\) in terms of \(\sin\beta\) and \(\cos\beta\):
\(\tan^2\beta = \frac{\sin^2\beta}{\cos^2\beta}\)
\(\mathrm{cosec}^2\beta = \frac{1}{\sin^2\beta}\)
\(\sec^2\beta = \frac{1}{\cos^2\beta}\)
Substitute these into the expression:
\[ \frac{\left(\frac{\sin^2\beta}{\cos^2\beta}\right) \left(\frac{1}{\sin^2\beta}\right)}{\frac{1}{\cos^2\beta}} \]Simplify the numerator:
\[ \left(\frac{\sin^2\beta}{\cos^2\beta}\right) \left(\frac{1}{\sin^2\beta}\right) = \frac{1}{\cos^2\beta} \]Substitute the simplified numerator back into the main expression:
\[ \frac{\frac{1}{\cos^2\beta}}{\frac{1}{\cos^2\beta}} \]This simplifies to:
\[ 1 \]C) 1
The problem involves finding an unknown number given the HCF (Highest Common Factor) and LCM (Least Common Multiple) of two numbers, and one of the numbers. The fundamental relationship between HCF, LCM, and two numbers is that the product of the two numbers is equal to the product of their HCF and LCM.
A) 216 is the correct answer. The other number is 216.