This problem involves simple interest calculations. We need to divide the total investment into two parts, calculate the simple interest for each part, and then equate them as per the problem statement. Finally, we solve for the unknown investment amount.
The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]According to the problem, she earns equal interests from the two investments, so \(SI_1 = SI_2\):
\[ \frac{55 P_1}{100} = \frac{15 (20300 - P_1)}{100} \]Multiply both sides by 100 to clear the denominators:
\[ 55 P_1 = 15 (20300 - P_1) \]Divide both sides by 5 to simplify:
\[ 11 P_1 = 3 (20300 - P_1) \]Distribute 3 on the right side:
\[ 11 P_1 = 60900 - 3 P_1 \]Add \(3 P_1\) to both sides:
\[ 11 P_1 + 3 P_1 = 60900 \] \[ 14 P_1 = 60900 \]Divide by 14:
\[ P_1 = \frac{60900}{14} \] \[ P_1 = 4350 \]A) 4,350 is the sum invested at 11% per annum.
To find the number of prime numbers between 430 and 443 (inclusive), we need to list all integers in this range and then identify which ones are prime. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We will use the trial division method to check for primality.
Correct Option: A) 4
By how much is 65% of 65 greater than \( \frac{1}{5} \) of 25?
To solve this problem, we need to calculate two values: first, 65% of 65, and second, \(\frac{1}{5}\) of 25. Once we have these two values, we will find the difference between the first value and the second value.
To find a percentage of a number, we convert the percentage to a decimal or a fraction and then multiply it by the number.
\[ 65\% \text{ of } 65 = \frac{65}{100} \times 65 \]\[ = 0.65 \times 65 \]\[ = 42.25 \]To find a fraction of a number, we multiply the fraction by the number.
\[ \frac{1}{5} \text{ of } 25 = \frac{1}{5} \times 25 \]\[ = \frac{25}{5} \]\[ = 5 \]The question asks by how much the first value is greater than the second value. So, we subtract the second value from the first value.
\[ \text{Difference} = (\text{65% of 65}) - (\frac{1}{5} \text{ of } 25) \]\[ = 42.25 - 5 \]\[ = 37.25 \]B) 37.25 is the correct answer, as calculated by finding 65% of 65 and subtracting \(\frac{1}{5}\) of 25 from it.
This problem involves setting up and solving a system of two linear equations based on the given information about the ages of P and Q. We will define variables for their present ages and then translate the word problem into algebraic equations. Finally, we will solve these equations simultaneously to find the present age of Q.
Define variables: Let the present age of P be \(P\) years. Let the present age of Q be \(Q\) years.
Formulate the first equation: "Two times the present age of P exceeds the present age of Q by 41 years."
\[ 2P = Q + 41 \quad \text{(Equation 1)} \]Formulate the second equation: "After 7 years, nine times Q's age will be 24 years less than six times P's age at that time."
So, the statement translates to:
\[ 9(Q + 7) = 6(P + 7) - 24 \]Simplify the second equation:
\[ 9Q + 63 = 6P + 42 - 24 \] \[ 9Q + 63 = 6P + 18 \] \[ 9Q - 6P = 18 - 63 \] \[ 9Q - 6P = -45 \]Divide the entire equation by 3 to simplify:
\[ 3Q - 2P = -15 \quad \text{(Equation 2)} \]Solve the system of equations: We have two equations:
Equation 1: \(2P = Q + 41 \implies 2P - Q = 41\)
Equation 2: \(-2P + 3Q = -15\)
Add Equation 1 and Equation 2 to eliminate \(P\):
\[ (2P - Q) + (-2P + 3Q) = 41 + (-15) \] \[ 2P - Q - 2P + 3Q = 26 \] \[ 2Q = 26 \] \[ Q = \frac{26}{2} \] \[ Q = 13 \]Verify (optional): Substitute \(Q = 13\) into Equation 1:
\[ 2P = 13 + 41 \] \[ 2P = 54 \] \[ P = 27 \]Now check with Equation 2:
\[ 3(13) - 2(27) = 39 - 54 = -15 \]The values satisfy both equations.
Correct Option: D) 13
To find the LCM of decimal numbers, we first convert them into fractions. Then, we use the formula for the LCM of fractions: \[\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}\]. Finally, we convert the resulting fraction back to a decimal.
The given numbers are 14.2 and 0.05.
\[ 14.2 = \frac{142}{10} \] \[ 0.05 = \frac{5}{100} \]For \( \frac{142}{10} \), both numerator and denominator are divisible by 2:
\[ \frac{142}{10} = \frac{71}{5} \]For \( \frac{5}{100} \), both numerator and denominator are divisible by 5:
\[ \frac{5}{100} = \frac{1}{20} \]So, we need to find the LCM of \( \frac{71}{5} \) and \( \frac{1}{20} \).
The formula for the LCM of fractions is:
\[ \text{LCM} \left( \frac{a}{b}, \frac{c}{d} \right) = \frac{\text{LCM}(a, c)}{\text{HCF}(b, d)} \]Here, \(a = 71\), \(c = 1\), \(b = 5\), \(d = 20\).
The LCM of any number and 1 is the number itself.
\[ \text{LCM}(71, 1) = 71 \]The factors of 5 are 1, 5.
The factors of 20 are 1, 2, 4, 5, 10, 20.
The highest common factor is 5.
\[ \text{HCF}(5, 20) = 5 \]Alternatively, one can make the denominators of the fractions equal before finding the LCM. This is often easier for decimals.
The numbers are 14.2 and 0.05.
\[ 14.2 = \frac{142}{10} = \frac{142 \times 10}{10 \times 10} = \frac{1420}{100} \] \[ 0.05 = \frac{5}{100} \]When fractions have the same denominator, the LCM is simply the LCM of the numerators divided by the common denominator.
We need to find LCM of \( \frac{1420}{100} \) and \( \frac{5}{100} \).
\[ \text{LCM} = \frac{\text{LCM}(1420, 5)}{100} \]Since 1420 is a multiple of 5 (\(1420 = 5 \times 284\)), the LCM of 1420 and 5 is 1420.
\[ \text{LCM}(1420, 5) = 1420 \]Correct Option: A) 14.2
To find the mean proportional of two numbers, say \(a\) and \(b\), we need to find a number \(x\) such that \(a:x = x:b\). This relationship implies that the product of the extremes equals the product of the means, i.e., \(x^2 = ab\), which means \(x = \sqrt{ab}\).
Thus, the mean proportional of 400 and 4 is 40.
Correct Option: C) 40
The problem involves calculating the list price of a fan given its cost price markup, discount percentage, and the total profit earned. We will use the relationships between Cost Price (CP), List Price (LP), Selling Price (SP), discount, and profit to solve this problem. We can assume a variable for the Cost Price and then express the List Price and Selling Price in terms of this variable, finally using the given profit to find the actual values.
Let the Cost Price (CP) of the fan be \(\text{₹}x\).
The shopkeeper lists the price at 44% above its cost price. So, the List Price (LP) is:
\[ \text{LP} = \text{CP} + 44\% \text{ of CP} = x + 0.44x = 1.44x \]A 25% discount is offered on the List Price. So, the Selling Price (SP) is:
\[ \text{SP} = \text{LP} - 25\% \text{ of LP} = \text{LP}(1 - 0.25) = 0.75 \times \text{LP} \]Substitute the expression for LP into the SP equation:
\[ \text{SP} = 0.75 \times (1.44x) = 1.08x \]The profit earned is given as ₹126. Profit is the difference between Selling Price and Cost Price:
\[ \text{Profit} = \text{SP} - \text{CP} \]\[ 126 = 1.08x - x \]\[ 126 = 0.08x \]Now, solve for \(x\) (the Cost Price):
\[ x = \frac{126}{0.08} = \frac{126}{\frac{8}{100}} = \frac{126 \times 100}{8} \]\[ x = \frac{12600}{8} = 1575 \]So, the Cost Price (CP) is ₹1575.The question asks for the List Price (LP). We found that \(\text{LP} = 1.44x\). Substitute the value of \(x\):
\[ \text{LP} = 1.44 \times 1575 \]\[ \text{LP} = 2268 \]Thus, the list price of the fan is ₹2268.
D) 2,268
The problem involves finding an angle in a trapezium given some angles and a side equality. We will use properties of triangles (sum of angles, isosceles triangle properties) and parallel lines (alternate interior angles) to solve this.
Analyze the given information:
Find \(\angle ACB\) in \(\triangle ABC\):
The sum of angles in a triangle is \(180^\circ\). In \(\triangle ABC\):
Use the parallel lines property:
Since \(BC \parallel AD\) and \(AC\) is a transversal, the alternate interior angles are equal:
Use the isosceles triangle property in \(\triangle ACD\):
Given \(AC = CD\), the angles opposite to these sides must be equal:
Find \(\angle ACD\) in \(\triangle ACD\):
The sum of angles in \(\triangle ACD\) is \(180^\circ\):
C) 80° is the correct measure for \(\angle ACD\).
This problem involves the relationship between speed, distance, and time. The key is that the distance covered is the same in both scenarios. We can first calculate the total distance traveled using the initial speed and time. Then, using this calculated distance and the new time, we can find the required new speed.
The formula relating speed, distance, and time is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]Given:
Substitute these values into the formula:
\[ \text{Distance} = 84 \text{ km/hr} \times 20 \text{ hours} \]\[ \text{Distance} = 1680 \text{ km} \]The bus needs to cover the same distance (1680 km) in a new time.
Given:
The formula for speed is:
\[ \text{New Speed} = \frac{\text{Distance}}{\text{New Time}} \]Substitute the values:
\[ \text{New Speed} = \frac{1680 \text{ km}}{10 \text{ hours}} \]\[ \text{New Speed} = 168 \text{ km/hr} \]Therefore, the bus's speed should be 168 km/hr to cover the same distance in 10 hours.
Correct Option: D) 168 km/hr