In a partnership business, if the investments are made for the same duration, the profit is distributed among the partners in the ratio of their investments. We will first find the ratio of investments of Vikram, Komal, and Rishi, and then use this ratio to determine Rishi's share of the total profit.
Identify the investments:
Determine the ratio of investments:
The ratio of investments for Vikram : Komal : Rishi is:
Simplify the ratio:
We can divide each number by 10 to simplify the ratio:
Check for further simplification:
Let's check if there's a common factor for 154, 176, and 117.
There are no common prime factors for all three numbers. So, the simplified ratio is \(154 : 176 : 117\).
Calculate the sum of the ratio parts:
\[ \text{Sum of ratio parts} = 154 + 176 + 117 = 447 \]Identify the total profit:
Total profit = Rs. 1,490
Calculate Rishi's share of the profit:
Rishi's share is the total profit multiplied by Rishi's ratio part divided by the sum of the ratio parts.
So, Rishi's share in the profit is Rs. 390.
B) Rs.390 is the correct share of profit for Rishi, calculated based on the ratio of investments.
This problem involves calculating the principal amount when the future amount, time, and simple interest rate are given. We will use the formula for simple interest and the relationship between Amount, Principal, and Simple Interest.
The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]where \(P\) is the Principal amount.The Amount is the sum of the Principal and the Simple Interest:
\[ A = P + SI \]Substitute the expression for \(SI\) from step 2 into the equation from step 3:
\[ A = P + \frac{P \times R \times T}{100} \]Factor \(P\) from the right side of the equation:
\[ A = P \left(1 + \frac{R \times T}{100}\right) \]Now, plug in the known values for \(A\), \(R\), and \(T\):
\[ 2280 = P \left(1 + \frac{5 \times 4}{100}\right) \]Divide both sides by 1.20 to find the Principal \(P\):
\[ P = \frac{2280}{1.20} \]\[ P = \frac{2280}{\frac{120}{100}} \]\[ P = \frac{2280 \times 100}{120} \]\[ P = \frac{228000}{120} \]\[ P = \frac{22800}{12} \]\[ P = 1900 \]Thus, the sum of money (Principal) is Rs. 1,900.
A) Rs.1,900 is the correct principal amount that will amount to Rs. 2,280 in 4 years at 5% simple interest.
To solve this problem, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS. This means we perform operations in the following order: Brackets (Parentheses), Orders (Exponents), Division and Multiplication (from left to right), Addition and Subtraction (from left to right).
The innermost bracket is \((4 - 8)\).
\[ 4 - 8 = -4 \]The expression becomes:
\[ 147 \div \{29 + 2 \times (-4)\} \]Next, we evaluate \(2 \times (-4)\).
\[ 2 \times (-4) = -8 \]The expression now is:
\[ 147 \div \{29 + (-8)\} \]Now, we evaluate \(29 + (-8)\).
\[ 29 + (-8) = 29 - 8 = 21 \]The expression simplifies to:
\[ 147 \div 21 \]Finally, we divide 147 by 21.
\[ 147 \div 21 = 7 \]D) 7 — Following the BODMAS rule, the value of the expression is 7.
To find the average speed for a journey where the distance traveled in each direction is the same but the speeds are different, we use the formula for average speed: Total Distance / Total Time. Alternatively, for two equal distances covered at different speeds, a specific formula can be used.
Define variables:
Calculate time for each leg of the journey:
Calculate total distance and total time:
Calculate average speed:
Alternative Method (Harmonic Mean for equal distances):
When an object covers the same distance at two different speeds, the average speed can be calculated using the formula:
\[ \text{Average Speed} = \frac{2 S_1 S_2}{S_1 + S_2} \]
D) The average speed for the entire journey is 8 km/hr.
Simplify: \(\frac{1}{\tan^2\theta}\) + \(\frac{\tan^2\theta}{\sin^2\theta}\) × cos²θ
To simplify the given trigonometric expression, we will use fundamental trigonometric identities to convert all terms into sine and cosine functions. Then, we will perform algebraic operations to combine and simplify the expression.
The given expression is:
\[ \frac{1}{\tan^2\theta} + \frac{\tan^2\theta}{\sin^2\theta} \times \cos^2\theta \]Recall the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \). Therefore, \( \tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} \).
Substitute this identity into the first term:
\[ \frac{1}{\tan^2\theta} = \frac{1}{\frac{\sin^2\theta}{\cos^2\theta}} = \frac{\cos^2\theta}{\sin^2\theta} \]Substitute the identity into the second term:
\[ \frac{\tan^2\theta}{\sin^2\theta} \times \cos^2\theta = \frac{\frac{\sin^2\theta}{\cos^2\theta}}{\sin^2\theta} \times \cos^2\theta \]Simplify the second term:
\[ \frac{\sin^2\theta}{\cos^2\theta \cdot \sin^2\theta} \times \cos^2\theta \]\[ = \frac{1}{\cos^2\theta} \times \cos^2\theta \]\[ = 1 \]Now, substitute the simplified terms back into the original expression:
\[ \frac{\cos^2\theta}{\sin^2\theta} + 1 \]Combine the terms by finding a common denominator:
\[ \frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\sin^2\theta} \]\[ = \frac{\cos^2\theta + \sin^2\theta}{\sin^2\theta} \]Recall the Pythagorean identity: \( \sin^2\theta + \cos^2\theta = 1 \).
Substitute this into the expression:
\[ \frac{1}{\sin^2\theta} \]Recall the reciprocal identity: \( \csc\theta = \frac{1}{\sin\theta} \). Therefore, \( \csc^2\theta = \frac{1}{\sin^2\theta} \).
So, the simplified expression is \( \csc^2\theta \).
Correct Option: C) cosec²θ
The problem involves finding an angle in a trapezium with parallel sides and an isosceles triangle formed by a diagonal and a non-parallel side. We will use properties of triangles (sum of angles, isosceles triangle properties) and parallel lines (alternate interior angles) to solve this.
D) 118° is the correct answer, as calculated above.
Find the value of 69-4 ÷ 6910 × 69-17.
The problem involves simplifying an expression with exponents. We will use the rules of exponents, specifically the product rule \((a^m \times a^n = a^{m+n})\) and the quotient rule \((a^m \div a^n = a^{m-n})\).
B) \(69^{-31}\) is the correct answer, as derived by applying the rules of exponents.
To find the Least Common Multiple (LCM) of numbers expressed in their prime factorized form, we identify all unique prime factors present in any of the numbers. For each unique prime factor, we take the highest power (exponent) that appears in any of the given numbers. The product of these highest powers will be the LCM.
Note that \(9 = 3^2\) and \(14 = 2 \times 7\). Let's rewrite the numbers using only prime bases:
However, the options are given with 9 and 14 as bases, so it's easier to work with the original composite bases (9 and 14) and prime bases (5 and 19) as presented in the question, treating 9 and 14 as distinct factors for the purpose of finding LCM, and then combining their highest powers.
Let's re-examine the numbers as given:
Correct Option: A) \(5^3 \times 9^3 \times 14^2 \times 19^2\)
This option incorrectly uses \(9^2\) instead of \(9^3\) and \(19^1\) instead of \(19^2\). The highest power of 9 is \(9^3\) and the highest power of 19 is \(19^2\).
This option incorrectly uses \(5^2\) instead of \(5^3\), \(9^2\) instead of \(9^3\), and \(19^3\) is not the highest power of 19 (it should be \(19^2\)).
This option incorrectly uses \(5^2\) instead of \(5^3\) and \(14^1\) instead of \(14^2\). The highest power of 5 is \(5^3\) and the highest power of 14 is \(14^2\).
This problem involves setting up and solving a system of linear equations based on the given information about the cost of pens and pencils under two different scenarios. We will define variables for the original cost of a pen and a pencil, form two equations, solve them to find the original costs, and then calculate the cost of 12 pens and 11 pencils.
Define variables:
Let \(P\) be the original cost of one pen (in Rs.).
Let \(C\) be the original cost of one pencil (in Rs.).
Formulate the first equation:
The first statement says "13 pens and 3 pencils cost Rs.152".
\[ 13P + 3C = 152 \quad \text{(Equation 1)} \]Formulate the second equation:
The second statement describes a change in prices: "If the price of pens decreases by Rs.1 per pen and the price of pencils increases by Rs.7 per pencil, then 18 pens and 3 pencils cost Rs.165."
New price of one pen \( = P - 1 \)
New price of one pencil \( = C + 7 \)
The cost of 18 pens and 3 pencils with these new prices is Rs.165:
\[ 18(P - 1) + 3(C + 7) = 165 \]\[ 18P - 18 + 3C + 21 = 165 \]\[ 18P + 3C + 3 = 165 \]\[ 18P + 3C = 165 - 3 \]\[ 18P + 3C = 162 \quad \text{(Equation 2)} \]Solve the system of equations:
We have two equations:
Equation 1: \( 13P + 3C = 152 \)
Equation 2: \( 18P + 3C = 162 \)
Subtract Equation 1 from Equation 2 to eliminate \(3C\):
\[ (18P + 3C) - (13P + 3C) = 162 - 152 \]\[ 18P - 13P = 10 \]\[ 5P = 10 \]\[ P = \frac{10}{5} \]\[ P = 2 \]Find the value of \(C\):
Substitute \(P = 2\) into Equation 1:
\[ 13(2) + 3C = 152 \]\[ 26 + 3C = 152 \]\[ 3C = 152 - 26 \]\[ 3C = 126 \]\[ C = \frac{126}{3} \]\[ C = 42 \]Calculate the original cost of 12 pens and 11 pencils:
Original cost \( = 12P + 11C \)
Substitute the values of \(P = 2\) and \(C = 42\):
\[ \text{Cost} = 12(2) + 11(42) \]\[ \text{Cost} = 24 + 462 \]\[ \text{Cost} = 486 \]The original cost of 12 pens and 11 pencils is Rs. 486.
Correct Option: B) Rs.486
To find the number of prime numbers between 402 and 409 (inclusive), we need to list all integers in this range and then test each one for primality. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Therefore, there is 1 prime number between 402 and 409, both included.
Correct Option: A) 1