To find the percentage discount, we first need to calculate the actual discount amount by subtracting the selling price from the marked price. Then, we divide this discount amount by the marked price and multiply by 100 to express it as a percentage.
Identify the Marked Price (MP) and Selling Price (SP):
Calculate the Discount Amount:
The discount is the difference between the Marked Price and the Selling Price.
\[ \text{Discount} = \text{MP} - \text{SP} \\ \text{Discount} = \text{₹}500 - \text{₹}425 \\ \text{Discount} = \text{₹}75 \]Calculate the Percentage Discount:
The percentage discount is calculated by dividing the discount amount by the Marked Price and multiplying by 100.
\[ \text{Percentage Discount} = \left( \frac{\text{Discount}}{\text{MP}} \right) \times 100\% \\ \text{Percentage Discount} = \left( \frac{75}{500} \right) \times 100\% \]Simplify the Calculation:
\[ \text{Percentage Discount} = \left( \frac{75}{5} \right)\% \\ \text{Percentage Discount} = 15\% \]Correct Option: B) 15% is the calculated percentage discount.
Pipe A can fill a tank in 15 hours, pipe B can fill the same tank in 27 hours and pipe C can fill the same tank in 10 hours. The time taken by them to fill the same tank if they operate together is:
8 \( \frac{10}{11} \) hours
9 \( \frac{10}{11} \) hours
4 \( \frac{10}{11} \) hours
12 \( \frac{10}{11} \) 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 to fill the tank together is the reciprocal of the combined rate.
Calculate the filling rate of Pipe A:
Pipe A fills the tank in 15 hours. So, in 1 hour, Pipe A fills:
\[ \text{Rate of A} = \frac{1}{15} \text{ of the tank} \]Calculate the filling rate of Pipe B:
Pipe B fills the tank in 27 hours. So, in 1 hour, Pipe B fills:
\[ \text{Rate of B} = \frac{1}{27} \text{ of the tank} \]Calculate the filling rate of Pipe C:
Pipe C fills the tank in 10 hours. So, in 1 hour, Pipe C fills:
\[ \text{Rate of C} = \frac{1}{10} \text{ of the tank} \]Calculate the combined filling rate of A, B, and C:
When all three pipes operate together, their rates add up. In 1 hour, they fill:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C} \] \[ \text{Combined Rate} = \frac{1}{15} + \frac{1}{27} + \frac{1}{10} \]To add these fractions, find the Least Common Multiple (LCM) of 15, 27, and 10.
LCM \((15, 27, 10) = 2 \times 3^3 \times 5 = 2 \times 27 \times 5 = 270\)
Now, rewrite the fractions with the common denominator:
\[ \text{Combined Rate} = \frac{1 \times 18}{15 \times 18} + \frac{1 \times 10}{27 \times 10} + \frac{1 \times 27}{10 \times 27} \] \[ \text{Combined Rate} = \frac{18}{270} + \frac{10}{270} + \frac{27}{270} \] \[ \text{Combined Rate} = \frac{18 + 10 + 27}{270} \] \[ \text{Combined Rate} = \frac{55}{270} \]This fraction can be simplified by dividing the numerator and denominator by 5:
\[ \text{Combined Rate} = \frac{55 \div 5}{270 \div 5} = \frac{11}{54} \text{ of the tank per hour} \]Calculate the total time taken to fill the tank together:
The time taken is the reciprocal of the combined rate:
\[ \text{Time} = \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{11}{54}} = \frac{54}{11} \text{ hours} \]Convert the improper fraction to a mixed number:
Divide 54 by 11:
\[ 54 \div 11 = 4 \text{ with a remainder of } 10 \]So, the time taken is \(4 \frac{10}{11}\) hours.
Correct Option: C) \(4 \frac{10}{11}\) hours
This problem involves calculating the actual profit percentage when a vendor claims a loss but also uses a faulty weight. We need to consider two factors: the claimed loss percentage and the cheating percentage due to incorrect weighing. We can assume a base quantity and cost to simplify calculations.
Let's assume the vendor claims to sell 1000 grams of wheat. Let the cost price (CP) of 1 gram of wheat be ₹1.
So, the actual cost price for 1000 grams is \(1000 \times ₹1 = ₹1000\).
The vendor claims to sell at a loss of 20%. This means the selling price is 20% less than the cost price.
\[ \text{Claimed SP} = \text{CP} - (0.20 \times \text{CP}) = ₹1000 - (0.20 \times ₹1000) = ₹1000 - ₹200 = ₹800 \]The vendor cheats by using weights that are 35% less than the stated weight. When a customer asks for 1000 grams, the vendor actually gives 35% less.
\[ \text{Actual weight delivered} = 1000 \text{ g} - (0.35 \times 1000 \text{ g}) = 1000 \text{ g} - 350 \text{ g} = 650 \text{ g} \]The vendor's actual cost for the 650 grams of wheat delivered is:
\[ \text{Actual CP} = 650 \text{ g} \times ₹1/\text{g} = ₹650 \]The vendor sells the 650 grams (which cost him ₹650) for ₹800 (the claimed SP for 1000 grams).
\[ \text{Actual Profit} = \text{Selling Price} - \text{Actual Cost Price} = ₹800 - ₹650 = ₹150 \]Profit percentage is calculated on the actual cost price.
\[ \text{Profit Percentage} = \left( \frac{\text{Actual Profit}}{\text{Actual CP}} \right) \times 100 \]\[ \text{Profit Percentage} = \left( \frac{₹150}{₹650} \right) \times 100 \]\[ \text{Profit Percentage} = \frac{150}{650} \times 100 = \frac{15}{65} \times 100 = \frac{3}{13} \times 100 \]\[ \text{Profit Percentage} = \frac{300}{13} \approx 23.0769\% \]Rounding 23.0769% to two decimal places gives 23.08%.
Correct Option: C) 23.08%
To find the area of a triangle given its three side lengths, we can first check if it's a right-angled triangle using the Pythagorean theorem. If it is, the area can be calculated as half the product of its perpendicular sides. If not, Heron's formula can be used.
Check for a right-angled triangle: Let the sides be \(a = 9 \text{ cm}\), \(b = 12 \text{ cm}\), and \(c = 15 \text{ cm}\). According to the Pythagorean theorem, for a right-angled triangle, the square of the longest side (hypotenuse) must be equal to the sum of the squares of the other two sides. So, we check if \(a^2 + b^2 = c^2\).
\[ 9^2 + 12^2 = 81 + 144 = 225 \]\[ 15^2 = 225 \]Since \(9^2 + 12^2 = 15^2\), the triangle is a right-angled triangle with the sides 9 cm and 12 cm forming the right angle.
Calculate the area: The area of a right-angled triangle is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]In a right-angled triangle, the two shorter sides can be considered as the base and height. So, we use 9 cm and 12 cm.
\[ \text{Area} = \frac{1}{2} \times 9 \text{ cm} \times 12 \text{ cm} \]\[ \text{Area} = \frac{1}{2} \times 108 \text{ cm}^2 \]\[ \text{Area} = 54 \text{ cm}^2 \]B) 54 cm² is the correct area of the triangle.
A) 60 cm² — This value might be obtained if an incorrect formula or calculation was used, for example, if the sum of two sides was mistakenly used in some part of the calculation or if the sides were incorrectly identified for a different type of triangle area calculation.
C) 42 cm² — This value is incorrect. It might result from an arithmetic error during calculation or an incorrect application of a formula.
D) 48 cm² — This value is incorrect. It could be a result of a calculation mistake, such as using incorrect dimensions or an erroneous formula application.
Simplify: 2((\( \frac{6}{2} \))x² − 22x + 15) − 6(x² + 5x − 18).
To simplify the given algebraic expression, we need to first distribute the constants into the parentheses and then combine like terms. This involves careful application of the distributive property and attention to signs.
Write down the given expression:
\[ 2\left(\left( \frac{6}{2} \right)x^2 - 22x + 15\right) - 6(x^2 + 5x - 18) \]Simplify the fraction inside the first parenthesis:
\[ \frac{6}{2} = 3 \]Substitute this back into the expression:
\[ 2(3x^2 - 22x + 15) - 6(x^2 + 5x - 18) \]Distribute the \(2\) into the first set of parentheses:
\[ 2 \times 3x^2 = 6x^2 \] \[ 2 \times (-22x) = -44x \] \[ 2 \times 15 = 30 \]So the first part becomes:
\[ 6x^2 - 44x + 30 \]Distribute the \(-6\) into the second set of parentheses:
\[ -6 \times x^2 = -6x^2 \] \[ -6 \times 5x = -30x \] \[ -6 \times (-18) = 108 \]So the second part becomes:
\[ -6x^2 - 30x + 108 \]Combine the results from step 3 and step 4:
\[ (6x^2 - 44x + 30) + (-6x^2 - 30x + 108) \]Group like terms:
For \(x^2\) terms:
\[ 6x^2 - 6x^2 = 0x^2 = 0 \]For \(x\) terms:
\[ -44x - 30x = -74x \]For constant terms:
\[ 30 + 108 = 138 \]Combine the simplified terms to get the final expression:
\[ 0 - 74x + 138 = -74x + 138 \]Correct Option: D) \(-74x + 138\)
To find the profit, we need to calculate the total cost price (CP) and the total selling price (SP). The cost price is given directly. For the selling price, we first determine the number of light bulbs remaining after some broke and then multiply by the selling price per bulb.
The problem states that 37 light bulbs were purchased for ₹37.
\[ \text{CP of 37 light bulbs} = \text{₹}37 \]Initially, there were 37 light bulbs. 19 light bulbs broke in transit.
\[ \text{Remaining light bulbs} = \text{Total light bulbs} - \text{Broken light bulbs} \]\[ \text{Remaining light bulbs} = 37 - 19 = 18 \]The trader sold the remaining 18 light bulbs at ₹3.90 each.
\[ \text{SP of 18 light bulbs} = \text{Number of remaining bulbs} \times \text{SP per bulb} \]\[ \text{SP of 18 light bulbs} = 18 \times \text{₹}3.90 \]\[ \text{SP of 18 light bulbs} = \text{₹}70.20 \]Profit is the difference between the total selling price and the total cost price.
\[ \text{Profit} = \text{SP} - \text{CP} \]\[ \text{Profit} = \text{₹}70.20 - \text{₹}37 \]\[ \text{Profit} = \text{₹}33.20 \]B) ₹33.20 — The calculated profit is ₹33.20, which matches this option.
This problem involves setting up and solving a system of linear equations based on the given information about the ages of P and Q. We will define variables for their present ages, translate the given statements into equations, and then solve for the unknown age of Q.
D) The present age of Q is 4 years, which matches our calculation.
We need to find the measure of \(\angle AEB\) using the given information about \(\triangle ABC\). We will use the properties of triangles, specifically the sum of angles in a triangle and the exterior angle theorem.
Identify angles in \(\triangle BDC\):
Given that \(BD \perp AC\) at \(D\), this means \(\angle BDC = 90^\circ\).
We are also given \(\angle DBC = 44^\circ\).
The sum of angles in \(\triangle BDC\) is \(180^\circ\). So,
\[ \angle BCD + \angle BDC + \angle DBC = 180^\circ \]Substitute the known values:
\[ \angle BCD + 90^\circ + 44^\circ = 180^\circ \]\[ \angle BCD + 134^\circ = 180^\circ \]\[ \angle BCD = 180^\circ - 134^\circ \]\[ \angle BCD = 46^\circ \]Note that \(\angle BCD\) is the same as \(\angle C\) of \(\triangle ABC\).
Identify angles in \(\triangle AEC\):
We know \(\angle C = 46^\circ\) (from step 1).
We are given \(\angle CAE = 23^\circ\).
The sum of angles in \(\triangle AEC\) is \(180^\circ\). So,
\[ \angle AEC + \angle ECA + \angle CAE = 180^\circ \]Substitute the known values:
\[ \angle AEC + 46^\circ + 23^\circ = 180^\circ \]\[ \angle AEC + 69^\circ = 180^\circ \]\[ \angle AEC = 180^\circ - 69^\circ \]\[ \angle AEC = 111^\circ \]Find \(\angle AEB\) using angles on a straight line:
Angles \(\angle AEB\) and \(\angle AEC\) form a linear pair on the line \(BC\).
Therefore, their sum is \(180^\circ\).
\[ \angle AEB + \angle AEC = 180^\circ \]Substitute the value of \(\angle AEC\) from step 2:
\[ \angle AEB + 111^\circ = 180^\circ \]\[ \angle AEB = 180^\circ - 111^\circ \]\[ \angle AEB = 69^\circ \]B) \(69^\circ\) is the correct measure for \(\angle AEB\).
The problem involves calculating the price of the cheapest item given the average price of three items and their price ratio. We will first use the average price to find the total price of the three items. Then, we will use the given ratio to express the individual prices in terms of a common variable and solve for that variable. Finally, we will determine the price of the cheapest item.
The average price of three items is ₹5,300.
\[ \text{Total Price} = \text{Average Price} \times \text{Number of Items} \]\[ \text{Total Price} = 5300 \times 3 = \text{₹}15,900 \]The prices of the three items are in the ratio 3 : 5 : 7. Let the prices of the three items be \(3x\), \(5x\), and \(7x\) respectively, where \(x\) is a common multiplier.
The sum of the prices must equal the total price calculated in Step 1.
\[ 3x + 5x + 7x = 15900 \]\[ 15x = 15900 \]\[ x = \frac{15900}{15} \]\[ x = 1060 \]The cheapest item corresponds to the smallest ratio part, which is \(3x\).
\[ \text{Price of Cheapest Item} = 3x \]\[ \text{Price of Cheapest Item} = 3 \times 1060 \]\[ \text{Price of Cheapest Item} = \text{₹}3,180 \]B) 3,180 is the price of the cheapest item, as calculated in the step-by-step solution.
To determine which of the given options divides the number 541096943, we can use the division algorithm. We will divide the given number by each option and check for a zero remainder. Alternatively, for larger numbers, divisibility rules or modular arithmetic can be applied, but direct division is often the most straightforward method when options are provided.
Let's perform long division of 541096943 by 97.
\[ 541096943 \div 97 \]\[ \begin{array}{r} 5578319 \ 97 \overline{\smash) 541096943} \ -485 \downarrow \ \hline 560 \downarrow \ -485 \downarrow \ \hline 759 \downarrow \ -679 \downarrow \ \hline 806 \downarrow \ -776 \downarrow \ \hline 309 \downarrow \ -291 \downarrow \ \hline 184 \downarrow \ -97 \downarrow \ \hline 873 \ -873 \ \hline 0 \end{array} \]Since the remainder is 0, 541096943 is divisible by 97.
A) 97 — As shown by the long division, 541096943 divided by 97 yields a quotient of 5578319 with a remainder of 0. Therefore, 541096943 is divisible by 97.