The problem involves calculating the marked price of an article given its selling price after a certain percentage discount. We can use the formula relating marked price, selling price, and discount percentage to find the unknown marked price.
The selling price is obtained by subtracting the discount from the marked price. The discount is calculated as a percentage of the marked price.
\[ \text{Discount} = \text{Discount Percentage} \times \text{Marked Price} \]\[ \text{Selling Price} = \text{Marked Price} - \text{Discount} \]Substitute the discount formula into the selling price formula:
\[ \text{Selling Price} = \text{Marked Price} - (\text{Discount Percentage} \times \text{Marked Price}) \]\[ \text{Selling Price} = \text{Marked Price} (1 - \text{Discount Percentage}) \]Let MP be the Marked Price.
\[ 3240 = \text{MP} (1 - 0.28) \]Thus, the marked price of the article is ₹4,500.
D) ₹4,500 is the correct answer. This is obtained by calculating the marked price using the given selling price and discount percentage.
To solve this problem, we need to use the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
The given 6-digit number is N84M69.
The digits are:
According to the rule, \( S_{\text{odd}} - S_{\text{even}} \) must be 0 or a multiple of 11.
\[ (17 + M) - (10 + N) = 0 \text{ or } \pm 11, \pm 22, \dots \]Since M and N are single digits (0-9), the value of \( M - N \) will be between \( 0 - 9 = -9 \) and \( 9 - 0 = 9 \).
Therefore, \( 7 + M - N \) will be between \( 7 + (-9) = -2 \) and \( 7 + 9 = 16 \).
The only multiple of 11 that falls within the range \([-2, 16]\) is 0 or 11.
We look for an option that matches either \( M - N = -7 \) or \( M - N = 4 \).
Option D is \( M - N = -7 \).
Correct Option: D) M - N = -7
Pipe A can fill a tank in 12 hours, pipe B can fill the same tank in 22 hours and pipe C can fill the same tank in 8 hours. The time taken by them to fill the same tank if they operate together is:
4 \( \frac{63}{67} \) hours
3 \( \frac{63}{67} \) hours
2 \( \frac{63}{67} \) hours
5 \( \frac{63}{67} \) hours
To find the time taken by multiple pipes working together to fill a tank, we first calculate the individual work rate of each pipe (the fraction of the tank filled per hour). Then, we sum these individual rates to get the combined work rate. Finally, the total time taken is the reciprocal of the combined work rate.
When all three pipes work together, their combined work rate per hour is the sum of their individual rates:
\[ \text{Combined work rate} = \frac{1}{12} + \frac{1}{22} + \frac{1}{8} \]To add these fractions, find the Least Common Multiple (LCM) of 12, 22, and 8.
LCM \((12, 22, 8) = 2^3 \times 3 \times 11 = 8 \times 3 \times 11 = 264\)
Now, rewrite each fraction with the common denominator:
\[ \frac{1}{12} = \frac{1 \times 22}{12 \times 22} = \frac{22}{264} \]Add the fractions:
\[ \text{Combined work rate} = \frac{22}{264} + \frac{12}{264} + \frac{33}{264} = \frac{22 + 12 + 33}{264} = \frac{67}{264} \]So, together, the pipes fill \(\frac{67}{264}\) of the tank in 1 hour.
The time taken to fill the tank is the reciprocal of the combined work rate:
\[ \text{Time taken} = \frac{1}{\text{Combined work rate}} = \frac{1}{\frac{67}{264}} = \frac{264}{67} \text{ hours} \]Convert this improper fraction to a mixed number:
\[ 264 \div 67 \]We know that \(67 \times 3 = 201\) and \(67 \times 4 = 268\). So, 264 divided by 67 is 3 with a remainder.
\[ 264 = 3 \times 67 + 63 \]Therefore,
\[ \frac{264}{67} = 3 \frac{63}{67} \text{ hours} \]Correct Option: B) 3 \( \frac{63}{67} \) hours
This problem involves relative speed concepts. We need to first determine the speed of the first train using the information about it overtaking a man. Then, we will use this train's speed to calculate the time it takes to cross another train moving in the opposite direction.
The speeds are given in km/hr, and lengths/times are in meters/seconds. We need to convert km/hr to m/s by multiplying by \(\frac{5}{18}\).
When a train overtakes a man moving in the same direction, the relative speed is the difference between their speeds. The distance covered is the length of the train.
When two trains cross each other moving in opposite directions, the relative speed is the sum of their speeds. The total distance covered is the sum of their lengths.
The time taken for the first train to completely cross the second train is 18 seconds.
D) 18
This problem involves compound interest. We are given the principal amount, the final amount, and the time period. We need to find the annual rate of interest. The formula for compound interest will be used to set up an equation and solve for the rate.
Identify the given values:
Recall the compound interest formula:
The formula for the amount \(A\) after \(n\) years with compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]Substitute the given values into the formula:
\[ 24389 = 15625 \left(1 + \frac{R}{100}\right)^3 \]Isolate the term containing \(R\):
\[ \frac{24389}{15625} = \left(1 + \frac{R}{100}\right)^3 \]Simplify the fraction:
To find the cube root easily, we can recognize that \(15625 = 25^3\). Let's check if \(24389\) is a perfect cube. We can try dividing \(24389\) by small prime numbers or look for cubes of numbers close to 25. For example, \(29^3 = 24389\).
So, the equation becomes:
\[ \left(\frac{29}{25}\right)^3 = \left(1 + \frac{R}{100}\right)^3 \]Take the cube root of both sides:
\[ \frac{29}{25} = 1 + \frac{R}{100} \]Solve for \(R\):
\[ \frac{29}{25} - 1 = \frac{R}{100} \]\[ \frac{29 - 25}{25} = \frac{R}{100} \]\[ \frac{4}{25} = \frac{R}{100} \]\[ R = \frac{4}{25} \times 100 \]\[ R = 4 \times 4 \]\[ R = 16\% \]B) The rate of interest per annum is 16%.
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), and Addition and Subtraction (from left to right).
The value of the expression is 434.
D) 434 is the correct answer, as calculated by following the BODMAS rule.
The problem involves simple interest calculations. We are given the total investment, two different interest rates, and the condition that the interest earned from both parts is equal after a certain period. We need to find the amount invested at the lower interest rate. We will set up equations based on the simple interest formula and solve for the unknown amounts.
The sum invested at 3% per annum is ₹17,360.
Correct Option: A) 17,360
When a number divides several other numbers and leaves the same remainder each time, that number must be a common divisor of the differences between those numbers. To find the greatest such number, we need to find the HCF (Highest Common Factor) of the absolute differences between all pairs of the given numbers.
B) 28 — The HCF of the differences (56, 28, 84) is 28. This means 28 is the largest number that will divide 183, 127, and 211, leaving the same remainder. Let's verify:
Simplify: \(\frac{\sin A\tan A}{1-\cos A}\)
To simplify the given trigonometric expression, we will convert all terms into sine and cosine functions. Then, we will use algebraic manipulation and fundamental trigonometric identities to simplify the expression to one of the given options.
Start with the given expression:
\[ \frac{\sin A\tan A}{1-\cos A} \]Replace \(\tan A\) with \(\frac{\sin A}{\cos A}\):
\[ \frac{\sin A \left(\frac{\sin A}{\cos A}\right)}{1-\cos A} \]Multiply the terms in the numerator:
\[ \frac{\frac{\sin^2 A}{\cos A}}{1-\cos A} \]Rewrite the expression by moving \(\cos A\) from the numerator's denominator to the main denominator:
\[ \frac{\sin^2 A}{\cos A (1-\cos A)} \]Use the trigonometric identity \(\sin^2 A = 1 - \cos^2 A\):
\[ \frac{1-\cos^2 A}{\cos A (1-\cos A)} \]Factor the numerator using the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a=1\) and \(b=\cos A\):
\[ \frac{(1-\cos A)(1+\cos A)}{\cos A (1-\cos A)} \]Cancel out the common term \((1-\cos A)\) from the numerator and denominator (assuming \(1-\cos A \neq 0\), which means \(A \neq 2n\pi\)):
\[ \frac{1+\cos A}{\cos A} \]Separate the terms in the numerator:
\[ \frac{1}{\cos A} + \frac{\cos A}{\cos A} \]Simplify the terms:
\[ \sec A + 1 \]Rearrange the terms to match the options:
\[ 1 + \sec A \]A) \(1 + \sec A\) is the simplified form of the given expression, as derived in the step-by-step solution.
The problem asks us to find the greater of two consecutive natural numbers whose product is 182. We can solve this by setting up an algebraic equation or by testing the given options.
Correct Option: A) 14
The greater of the two consecutive natural numbers is 14, as \(13 \times 14 = 182\).