To solve this blood relation problem, we will systematically build a family tree or list relationships based on the given statements. We will then trace the connection between A and B.
Now, let's trace the relationship between A and B:
Therefore, A is B's mother's sister's husband.
Correct Option: A) Mother's sister's husband
The question asks us to identify the correct statements regarding Exercise Tiger Triumph – 25. We need to evaluate each statement based on general knowledge about military exercises, particularly those involving India.
C) Only 2 and 4
Based on our evaluation, statements 2 and 4 are correct, while statements 1 and 3 are incorrect.
The question asks for the MS Excel feature that allows users to print only a specific part of a spreadsheet. We need to identify the option that directly controls the printable area.
Correct Option: B) Print Area is the feature in MS Excel that allows users to select a specific range of cells to be printed, excluding other parts of the worksheet.
To find a single equivalent discount for consecutive discounts, we can assume an initial marked price (e.g., 100 units) and calculate the price after each discount. Alternatively, we can use the formula for equivalent discount for two consecutive discounts repeatedly.
Assume an initial Marked Price (MP): Let the Marked Price be \( \\(100 \).
Calculate price after the first discount: The first discount is 20%.
Price after 1st discount \( = 100 - (20\% \text{ of } 100) = 100 - 20 = \\)80 \).
Calculate price after the second discount: The second discount is 13%. This discount is applied to the price after the first discount, which is \( \\(80 \).
Discount amount \[= 13\% \text{ of } 80 = \frac{13}{100} \times 80 = 13 \times 0.8 = \\]10.40 \).
Price after 2nd discount \( = 80 - 10.40 = \\(69.60 \).
Calculate price after the third discount: The third discount is 20%. This discount is applied to the price after the second discount, which is \( \\)69.60 \).
Discount amount \[= 20\% \text{ of } 69.60 = \frac{20}{100} \times 69.60 = 0.20 \times 69.60 = \\(13.92\].
Price after 3rd discount \( = 69.60 - 13.92 = \\)55.68 \).
Calculate the single equivalent discount: The final selling price is \( \\(55.68 \) when the initial marked price was \( \\)100 \).
Total discount \( = \text{Marked Price} - \text{Final Selling Price} = 100 - 55.68 = \$44.32 \).
Equivalent single discount percentage \[= \frac{\text{Total Discount}}{\text{Marked Price}} \times 100 = \frac{44.32}{100} \times 100 = 44.32\%\].
For two consecutive discounts \( d_1 \) and \( d_2 \), the equivalent single discount \( D \) is given by:
\[ D = d_1 + d_2 - \frac{d_1 d_2}{100} \]First two discounts: \( d_1 = 20\% \), \( d_2 = 13\% \).
\[ D_{1,2} = 20 + 13 - \frac{20 \times 13}{100} = 33 - \frac{260}{100} = 33 - 2.6 = 30.4\% \]Combine \( D_{1,2} \) with the third discount: Now we have two discounts: \( D_{1,2} = 30.4\% \) and \( d_3 = 20\% \).
\[ D_{\text{total}} = 30.4 + 20 - \frac{30.4 \times 20}{100} \]\[ D_{\text{total}} = 50.4 - \frac{608}{100} \]\[ D_{\text{total}} = 50.4 - 6.08 \]\[ D_{\text{total}} = 44.32\% \]D) 44.32% is the single equivalent discount for the three consecutive discounts of 20%, 13%, and 20%.
The problem asks us to identify numbers in a given series that are immediately preceded by a number and immediately followed by a number. We need to scan the series from left to right and apply this condition to each number.
B) 6 is the correct answer because there are 6 numbers (3, 5, 6, 2, 7, 5) in the series that are immediately preceded by a number and immediately followed by a number.
The problem asks us to identify symbols that are immediately preceded by a number and immediately followed by a number in the given series. We need to scan the series from left to right and look for the pattern: Number - Symbol - Number.
@ 9 7 8 2 @ @ 9 3 * 2 6 2 © # 9 & 7 8 9 @ 4
D) 3. There are three symbols (*, &, @) that are immediately preceded by a number and immediately followed by a number.
What positive value of X satisfies the equation \( \frac{X}{176} \) = \( \frac{99}{X} \)?
To find the positive value of X that satisfies the given equation, we need to cross-multiply the terms and then solve for X. This will involve isolating \(X^2\) and then taking the square root of both sides.
Write down the given equation:
\[ \frac{X}{176} = \frac{99}{X} \]Cross-multiply the terms to eliminate the denominators:
Multiply \(X\) by \(X\) on the left side and \(176\) by \(99\) on the right side.
\[ X \times X = 176 \times 99 \] \[ X^2 = 176 \times 99 \]Calculate the product \(176 \times 99\):
It's often easier to write \(99\) as \((100 - 1)\):
\[ X^2 = 176 \times (100 - 1) \] \[ X^2 = 176 \times 100 - 176 \times 1 \] \[ X^2 = 17600 - 176 \] \[ X^2 = 17424 \]Take the square root of both sides to find X. Since the question asks for a positive value of X, we only consider the positive root:
\[ X = \sqrt{17424} \]To find the square root of \(17424\), we can use prime factorization or estimation. Let's try to estimate. We know that \(100^2 = 10000\) and \(150^2 = 22500\). So X should be between 100 and 150. The last digit of \(17424\) is 4, which means the last digit of X must be 2 or 8.
Let's check the options:
So, only 132 is a potential candidate among the options.
Let's calculate \(132^2\):
\[ 132^2 = 132 \times 132 \] \[ 132 \times 132 = 17424 \]Thus, \(X = 132\).
A) 132 is the correct answer because it satisfies the equation \(X^2 = 17424\).
What is the median of the following distribution?
| Class | 140-190 | 190-240 | 240-290 | 290-340 | 340-390 |
| Frequency | 23 | 13 | 24 | 25 | 35 |
To find the median of a grouped frequency distribution, we first need to calculate the cumulative frequencies. Then, we identify the median class, which is the class containing the \( (N/2)^{th} \) observation. Finally, we use the median formula for grouped data.
Let's create a table with class intervals, frequencies (\(f\)), and cumulative frequencies (\(cf\)).
| Class Interval | Frequency (\(f\)) | Cumulative Frequency (\(cf\)) |
|---|---|---|
| 140-190 | 23 | 23 |
| 190-240 | 13 | \(23 + 13 = 36\) |
| 240-290 | 24 | \(36 + 24 = 60\) |
| 290-340 | 25 | \(60 + 25 = 85\) |
| 340-390 | 35 | \(85 + 35 = 120\) |
The total number of observations is the sum of all frequencies, which is the last cumulative frequency.
\[ N = \sum f = 120 \]The median is the \( (N/2)^{th} \) observation.
\[ \frac{N}{2} = \frac{120}{2} = 60 \]The 60th observation falls in the class where the cumulative frequency first exceeds or is equal to 60. From our table, the cumulative frequency of 60 corresponds to the class interval 240-290. So, the median class is 240-290.
The formula for the median of grouped data is:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]Where:
Substitute these values into the formula:
\[ \text{Median} = 240 + \left( \frac{\frac{120}{2} - 36}{24} \right) \times 50 \] \[ \text{Median} = 240 + \left( \frac{60 - 36}{24} \right) \times 50 \] \[ \text{Median} = 240 + \left( \frac{24}{24} \right) \times 50 \] \[ \text{Median} = 240 + 1 \times 50 \] \[ \text{Median} = 240 + 50 \] \[ \text{Median} = 290 \]Correct Option: C) 290