To find the measure of \(\angle AEB\), we will use the properties of triangles, specifically the sum of angles in a triangle and the concept of exterior angles. We will first find the angles in \(\triangle BDC\) and then use them to find angles in \(\triangle ABC\) and finally \(\triangle AEC\) or \(\triangle ABE\).
Consider \(\triangle BDC\).
Given that \(BD \perp AC\) at \(D\), so \(\angle BDC = 90^\circ\).
Given \(\angle DBC = 60^\circ\).
The sum of angles in \(\triangle BDC\) is \(180^\circ\).
\[ \angle BCD + \angle BDC + \angle DBC = 180^\circ \]
\[ \angle BCD + 90^\circ + 60^\circ = 180^\circ \]
\[ \angle BCD + 150^\circ = 180^\circ \]
\[ \angle BCD = 180^\circ - 150^\circ = 30^\circ \]
So, \(\angle C = 30^\circ\).
Now consider \(\triangle AEC\).
We know \(\angle C = 30^\circ\).
Given \(\angle CAE = 49^\circ\).
The angle \(\angle AEB\) is an exterior angle to \(\triangle AEC\) at vertex \(E\).
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Therefore, \(\angle AEB = \angle CAE + \angle ACE\).
Here, \(\angle ACE\) is the same as \(\angle C\) in \(\triangle ABC\), which we found to be \(30^\circ\).
\[ \angle AEB = 49^\circ + 30^\circ \]
\[ \angle AEB = 79^\circ \]
D) \(79^\circ\)
This problem involves calculating individual weights using given average weights. We will use the formula for average weight (sum of weights divided by the number of individuals) to set up a system of equations and then solve for the unknown weight of Mandar.
Let R, M, and K be the weights of Rupesh, Mandar, and Ketan, respectively.
Given the average weight of Rupesh, Mandar, and Ketan is 45 kg:
\[ \frac{R + M + K}{3} = 45 \]Multiplying both sides by 3, we get the total weight:
\[ R + M + K = 45 \times 3 = 135 \quad \text{(Equation 1)} \]Given the average weight of Rupesh and Mandar is 43 kg:
\[ \frac{R + M}{2} = 43 \]Multiplying both sides by 2, we get the total weight of Rupesh and Mandar:
\[ R + M = 43 \times 2 = 86 \quad \text{(Equation 2)} \]Given the average weight of Mandar and Ketan is 49 kg:
\[ \frac{M + K}{2} = 49 \]Multiplying both sides by 2, we get the total weight of Mandar and Ketan:
\[ M + K = 49 \times 2 = 98 \quad \text{(Equation 3)} \]Now we have a system of three equations. We can find the weight of Mandar by adding Equation 2 and Equation 3, and then subtracting Equation 1.
Add Equation 2 and Equation 3:
\[ (R + M) + (M + K) = 86 + 98 \] \[ R + 2M + K = 184 \quad \text{(Equation 4)} \]Subtract Equation 1 from Equation 4:
\[ (R + 2M + K) - (R + M + K) = 184 - 135 \] \[ R + 2M + K - R - M - K = 49 \] \[ M = 49 \]So, the weight of Mandar is 49 kg.
Correct Option: B) 49
Simplify the following.
2((\( \frac{5}{2} \))x² − 28x + 17) − 5(x² + 6x − 18)
To simplify the given algebraic expression, we need to distribute the constants into the parentheses and then combine like terms. This involves multiplying each term inside the parentheses by the factor outside and then adding or subtracting terms with the same variable and power.
Let's perform the operations:
The given expression is:
\[ 2\left( \frac{5}{2}x^2 - 28x + 17 \right) - 5(x^2 + 6x - 18) \]Step 1: Distribute \(2\) into the first parenthesis.
\[ 2 \times \frac{5}{2}x^2 - 2 \times 28x + 2 \times 17 \]\[ 5x^2 - 56x + 34 \]Step 2: Distribute \(-5\) into the second parenthesis.
\[ -5 \times x^2 - 5 \times 6x - 5 \times (-18) \]\[ -5x^2 - 30x + 90 \]Step 3: Combine the results from Step 1 and Step 2.
\[ (5x^2 - 56x + 34) + (-5x^2 - 30x + 90) \]Group the like terms:
\[ (5x^2 - 5x^2) + (-56x - 30x) + (34 + 90) \]Perform the operations for each group:
So, the simplified expression is:
\[ 0 - 86x + 124 \]\[ -86x + 124 \]A) \(-86x + 124\) is the correct simplified expression after performing the distribution and combining like terms.
Pipe A can fill a tank in 17 hours, pipe B can fill the same tank in 24 hours and pipe C can fill the same tank in 18 hours. The time taken by them to fill the same tank if they operate together is:
10 \( \frac{78}{191} \) hours
6 \( \frac{78}{191} \) hours
16 \( \frac{78}{191} \) hours
13 \( \frac{78}{191} \) hours
To find the combined time taken by multiple pipes to fill a tank, we first determine the individual filling rates of each pipe. The rate is the reciprocal of the time taken to fill the tank. Then, we sum these individual rates to get the combined filling rate. Finally, the total time taken when all pipes operate together is the reciprocal of the combined rate.
The rate at which a pipe fills a tank is \[\frac{1}{\text{Time taken}}\].
When all pipes operate together, their rates add up.
\[ \text{Combined Rate} = \frac{1}{17} + \frac{1}{24} + \frac{1}{18} \]Prime factorization:
LCM \( = 17 \times 2^3 \times 3^2 = 17 \times 8 \times 9 = 17 \times 72 = 1224 \).
The total time taken is the reciprocal of the combined rate.
\[ \text{Time taken} = \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{191}{1224}} = \frac{1224}{191} \text{ hours} \]Divide 1224 by 191:
\( 1224 \div 191 \)
\( 191 \times 6 = 1146 \)
\( 1224 - 1146 = 78 \)
So, \[\frac{1224}{191} = 6 \frac{78}{191}\] hours.
B) \( 6 \frac{78}{191} \) hours is the correct time taken for all three pipes to fill the tank together.
When partners invest in a business, their share of the profit is directly proportional to the amount of capital they invest. We need to find the ratio of their investments, then use this ratio to distribute the total profit among them.
First, write down the investments made by Neha, Rajat, and Avinash.
The ratio of their investments is:
\[ \text{Neha} : \text{Rajat} : \text{Avinash} = 1430 : 1680 : 1340 \]Divide each number by the greatest common divisor (GCD) to simplify the ratio. In this case, all numbers are divisible by 10.
\[ 143 : 168 : 134 \]Add the simplified ratio parts:
\[ 143 + 168 + 134 = 445 \]The total profit is ₹1,780. Avinash's share will be his ratio part divided by the total ratio parts, multiplied by the total profit.
\[ \text{Avinash's share} = \frac{\text{Avinash's ratio part}}{\text{Sum of ratio parts}} \times \text{Total Profit} \]\[ \text{Avinash's share} = \frac{134}{445} \times 1780 \]\[ \text{Avinash's share} = 134 \times \frac{1780}{445} \]We can observe that \( 445 \times 4 = 1780 \). So, \( \frac{1780}{445} = 4 \).\[ \text{Avinash's share} = 134 \times 4 \]\[ \text{Avinash's share} = \text{₹}536 \]By how much is 65% of 75 greater than \( \frac{3}{5} \) of 25?
To solve this problem, we need to calculate two values: first, 65% of 75, and second, \( \frac{3}{5} \) of 25. After calculating both values, we will find the difference between the first value and the second value to determine how much greater the first is than the second.
Calculate 65% of 75:
To find a percentage of a number, convert the percentage to a decimal or a fraction and multiply it by the number.
\[ 65\% \text{ of } 75 = \frac{65}{100} \times 75 \] \[ = \frac{13 \times 5}{20 \times 5} \times 75 \] \[ = \frac{13}{20} \times 75 \]We can simplify this by dividing 75 by 5 and 20 by 5:
\[ = \frac{13}{4} \times 15 \] \[ = \frac{195}{4} \] \[ = 48.75 \]So, 65% of 75 is 48.75.
Calculate \( \frac{3}{5} \) of 25:
To find a fraction of a number, multiply the fraction by the number.
\[ \frac{3}{5} \text{ of } 25 = \frac{3}{5} \times 25 \]We can simplify this by dividing 25 by 5:
\[ = 3 \times 5 \] \[ = 15 \]So, \( \frac{3}{5} \) of 25 is 15.
Find the difference:
Now, we need to find how much 65% of 75 (which is 48.75) is greater than \( \frac{3}{5} \) of 25 (which is 15).
\[ \text{Difference} = 48.75 - 15 \] \[ = 33.75 \]D) 33.75 is the correct answer, as calculated by finding the difference between 65% of 75 (48.75) and \( \frac{3}{5} \) of 25 (15).
To find the LCM of decimal numbers, it is best to convert them into fractions. Then, we can use the formula for the LCM of fractions: \[\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}\].
The given numbers are 4.7 and 0.04.
\[ 4.7 = \frac{47}{10} \] \[ 0.04 = \frac{4}{100} \]The fraction \(\frac{47}{10}\) is already in its simplest form.
The fraction \(\frac{4}{100}\) can be simplified by dividing both numerator and denominator by 4:
\[ \frac{4}{100} = \frac{4 \div 4}{100 \div 4} = \frac{1}{25} \]So, we need to find the LCM of \(\frac{47}{10}\) and \(\frac{1}{25}\).
The formula is \[\text{LCM}(a/b, c/d) = \frac{\text{LCM}(a, c)}{\text{HCF}(b, d)}\].
Here, \(a=47\), \(b=10\), \(c=1\), \(d=25\).
The LCM of any number and 1 is the number itself.
\[ \text{LCM}(47, 1) = 47 \]Factors of 10 are 1, 2, 5, 10.
Factors of 25 are 1, 5, 25.
The highest common factor (HCF) is 5.
\[ \text{HCF}(10, 25) = 5 \]Correct Option: A) 9.4
To find the average speed for the whole journey, we need to calculate the total distance traveled and the total time taken. The average speed is then the total distance divided by the total time.
Correct Option: D) 58.3
The problem asks us to find the third proportional of two given numbers. The concept of the third proportional states that for two numbers, say 'a' and 'b', their third proportional 'x' satisfies the relationship that the ratio of the first to the second is equal to the ratio of the second to the third. This can be expressed as \(a : b :: b : x\), or \[\frac{a}{b} = \frac{b}{x}\].
Thus, the value of \(x\) is 18.
Correct Option: D) 18
We are given two statements relating the present ages of Mary and Dia. We can form a system of two linear equations with two variables representing their ages. By solving this system, we can find their individual ages and then calculate the difference between them.
Let Mary's present age be \(M\) years and Dia's present age be \(D\) years.
Translate the first statement into an equation:
"3 times Mary's present age is 22 years more than 5 times Dia's present age"
\[ 3M = 5D + 22 \quad \text{(Equation 1)} \]Translate the second statement into an equation:
"3 times Dia's present age is 2 years less than Mary's present age"
\[ 3D = M - 2 \]Rearrange Equation 2 to express \(M\) in terms of \(D\):
\[ M = 3D + 2 \quad \text{(Equation 2 revised)} \]Substitute the expression for \(M\) from Equation 2 revised into Equation 1:
\[ 3(3D + 2) = 5D + 22 \]Solve for \(D\):
\[ 9D + 6 = 5D + 22 \]\[ 9D - 5D = 22 - 6 \]\[ 4D = 16 \]\[ D = \frac{16}{4} \]\[ D = 4 \]Now that we have Dia's age (\(D=4\)), substitute this value back into Equation 2 revised to find Mary's age (\(M\)):
\[ M = 3(4) + 2 \]\[ M = 12 + 2 \]\[ M = 14 \]Mary's age is 14 years and Dia's age is 4 years.
Calculate the difference between their ages:
\[ \text{Difference} = M - D \]\[ \text{Difference} = 14 - 4 \]\[ \text{Difference} = 10 \]C) 10
The difference between Mary's age (14 years) and Dia's age (4 years) is 10 years.