This is a blood relation problem. We need to establish the relationships between the individuals mentioned and then determine how P is related to K. We can use a family tree diagram or simply deduce the relationships step-by-step.
Correct Option: D) Brother's son
P is the son of S. K is the sister of S. Therefore, P is the son of K's brother (S).
This problem involves calculating the total marks based on given averages and then finding the difference to determine the marks in a specific subject. We will use the formula: \(\text{Total Marks} = \text{Average Marks} \times \text{Number of Subjects}\).
Given that the average marks in 7 subjects is 76.
\[ \text{Total marks in 7 subjects} = \text{Average} \times \text{Number of subjects} = 76 \times 7 \] \[ \text{Total marks in 7 subjects} = 532 \]Given that the average marks in 6 subjects (excluding English) is 70.
\[ \text{Total marks in 6 subjects} = \text{Average} \times \text{Number of subjects} = 70 \times 6 \] \[ \text{Total marks in 6 subjects} = 420 \]The marks in English can be found by subtracting the total marks in 6 subjects from the total marks in 7 subjects.
\[ \text{Marks in English} = \text{Total marks in 7 subjects} - \text{Total marks in 6 subjects} \] \[ \text{Marks in English} = 532 - 420 \] \[ \text{Marks in English} = 112 \]Correct Option: C) 112
The question asks to identify the RBI window that allows banks to borrow overnight funds at a rate higher than the repo rate. This mechanism is crucial for banks facing sudden liquidity shortages.
The Marginal Standing Facility (MSF) is a window for scheduled commercial banks to borrow overnight funds from the RBI in emergency situations when interbank liquidity dries up. The interest rate for MSF is typically higher than the repo rate, acting as a penal rate to discourage frequent use and to serve as a ceiling for the overnight interbank rate.
B) Marginal standing facility is the correct answer. It allows banks to borrow overnight funds from the RBI at a rate higher than the repo rate, typically used in emergency liquidity situations.
We are given a trapezium ABCD with BC || AD. We are also given that AC = CD, which means triangle ACD is an isosceles triangle. We need to find the measure of \(\angle ACD\). We will use properties of triangles (sum of angles, isosceles triangle properties) and parallel lines (alternate interior angles, consecutive interior angles).
The sum of angles in a triangle is \(180^\circ\). In \(\triangle ABC\), we have:
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]Given \(\angle ABC = 13^\circ\) and \(\angle BAC = 164^\circ\).
\[ 164^\circ + 13^\circ + \angle ACB = 180^\circ \] \[ 177^\circ + \angle ACB = 180^\circ \] \[ \angle ACB = 180^\circ - 177^\circ = 3^\circ \]Since BC || AD, AC is a transversal. Therefore, alternate interior angles are equal:
\[ \angle CAD = \angle ACB \] \[ \angle CAD = 3^\circ \]In \(\triangle ACD\), it is given that AC = CD. This means \(\triangle ACD\) is an isosceles triangle, and the angles opposite the equal sides are equal.
\[ \angle CAD = \angle CDA \] \[ \angle CDA = 3^\circ \]The sum of angles in \(\triangle ACD\) is \(180^\circ\).
\[ \angle CAD + \angle CDA + \angle ACD = 180^\circ \] \[ 3^\circ + 3^\circ + \angle ACD = 180^\circ \] \[ 6^\circ + \angle ACD = 180^\circ \] \[ \angle ACD = 180^\circ - 6^\circ = 174^\circ \]Correct Option: D) 174°
The question asks why a touchscreen is considered both an input and an output device. We need to understand the definitions of input and output devices in the context of computer hardware and then apply them to a touchscreen.
Correct Option: B) It allows users to interact with visual data and sends touch signals to the computer.
This option accurately describes both the output (allows users to interact with visual data) and input (sends touch signals to the computer) functions of a touchscreen. Visual data is displayed on the screen, making it an output device. When the user touches the screen, these touch actions are converted into signals that are sent to the computer for processing, making it an input device.
This is a logical reasoning problem involving the arrangement of boxes. We need to deduce the correct order of the seven boxes (D, E, F, G, I, J, K) based on the given clues and then identify the box kept second below J.
"No box is kept above F." This means F is the topmost box.
| Position | Box |
|---|---|
| 7 | F |
| 6 | |
| 5 | |
| 4 | |
| 3 | |
| 2 | |
| 1 |
"Only three boxes are kept between F and K." Since F is at position 7, K must be at position 3 (7 - 3 - 1 = 3).
| Position | Box |
|---|---|
| 7 | F |
| 6 | |
| 5 | |
| 4 | |
| 3 | K |
| 2 | |
| 1 |
"E is kept immediately above K." K is at position 3, so E is at position 4.
| Position | Box |
|---|---|
| 7 | F |
| 6 | |
| 5 | |
| 4 | E |
| 3 | K |
| 2 | |
| 1 |
"Only one box is kept between D and E." E is at position 4. This means D can be either at position 2 (below E) or position 6 (above E). Let's consider both possibilities.
Case 1: D is at position 2.
| Position | Box |
|---|---|
| 7 | F |
| 6 | |
| 5 | |
| 4 | E |
| 3 | K |
| 2 | D |
| 1 |
Case 2: D is at position 6.
| Position | Box |
|---|---|
| 7 | F |
| 6 | D |
| 5 | |
| 4 | E |
| 3 | K |
| 2 | |
| 1 |
"Only four boxes are kept between D and G."
For Case 1 (D at position 2): If D is at position 2, then G must be at position 7 (2 + 4 + 1 = 7) or position 1 (2 - 4 - 1 = -3, not possible). But position 7 is already occupied by F. So, Case 1 is not possible.
For Case 2 (D at position 6): If D is at position 6, then G must be at position 1 (6 - 4 - 1 = 1). (6 + 4 + 1 = 11, not possible).
| Position | Box |
|---|---|
| 7 | F |
| 6 | D |
| 5 | |
| 4 | E |
| 3 | K |
| 2 | |
| 1 | G |
"J is kept at some place above I." The remaining boxes are I and J, and the remaining positions are 5 and 2. For J to be above I, J must be at position 5 and I at position 2.
| Position | Box |
|---|---|
| 7 | F |
| 6 | D |
| 5 | J |
| 4 | E |
| 3 | K |
| 2 | I |
| 1 | G |
Now, we need to find the box kept second below J. J is at position 5. The box immediately below J is at position 4 (E). The box second below J is at position 3 (K).
C) K is the box kept second below J.
We need to find the discount percentage (X) given the cost price markup and the final profit percentage. We can assume a cost price (CP) to simplify calculations and then use the given percentages to find the marked price (MP) and selling price (SP). Finally, we can calculate the discount percentage.
Assume Cost Price (CP):
Let the Cost Price (\(\text{CP}\)) of the goods be \(\text{Rs. } 100\).
Calculate Marked Price (MP):
The grocer marked his goods 40% above their cost price.
\[ \text{MP} = \text{CP} + 40\% \text{ of CP} \]\[ \text{MP} = 100 + \frac{40}{100} \times 100 \]\[ \text{MP} = 100 + 40 = \text{Rs. } 140 \]Calculate Selling Price (SP) using Profit:
He gained 33% profit.
\[ \text{Profit} = 33\% \text{ of CP} \]\[ \text{Profit} = \frac{33}{100} \times 100 = \text{Rs. } 33 \]\[ \text{SP} = \text{CP} + \text{Profit} \]\[ \text{SP} = 100 + 33 = \text{Rs. } 133 \]Calculate Discount Amount:
The discount is the difference between the Marked Price and the Selling Price.
\[ \text{Discount} = \text{MP} - \text{SP} \]\[ \text{Discount} = 140 - 133 = \text{Rs. } 7 \]Calculate Discount Percentage (X):
The discount percentage is calculated on the Marked Price.
\[ \text{X} = \frac{\text{Discount}}{\text{MP}} \times 100\% \]\[ \text{X} = \frac{7}{140} \times 100\% \]\[ \text{X} = \frac{1}{20} \times 100\% \]\[ \text{X} = 5\% \]C) 5
Hemant and Irfan can complete a certain piece of work in 6 and 10 days, respectively. They started to work together, and after 2 days, Irfan left. In how many days will Hemant complete the remaining work?
\( \frac{9}{5} \)
\( \frac{17}{5} \)
\( \frac{11}{5} \)
\( \frac{14}{5} \)
This is a work and time problem. We need to calculate the individual work rates, then the combined work done, and finally the remaining work and the time taken by Hemant to complete it.
Hemant can complete the work in 6 days. So, Hemant's 1-day work is:
\[ \text{Hemant's 1-day work} = \frac{1}{6} \text{ of the total work} \]Irfan can complete the work in 10 days. So, Irfan's 1-day work is:
\[ \text{Irfan's 1-day work} = \frac{1}{10} \text{ of the total work} \]When they work together, their combined 1-day work is:
\[ \text{Combined 1-day work} = \frac{1}{6} + \frac{1}{10} = \frac{5+3}{30} = \frac{8}{30} = \frac{4}{15} \text{ of the total work} \]They worked together for 2 days. So, the work done in 2 days is:
\[ \text{Work done in 2 days} = 2 \times \frac{4}{15} = \frac{8}{15} \text{ of the total work} \]The total work is 1. The remaining work after Irfan left is:
\[ \text{Remaining work} = 1 - \frac{8}{15} = \frac{15-8}{15} = \frac{7}{15} \text{ of the total work} \]Hemant's 1-day work is \(\frac{1}{6}\). To find the number of days Hemant will take to complete the remaining work, divide the remaining work by Hemant's 1-day work:
\[ \text{Days for Hemant} = \frac{\text{Remaining work}}{\text{Hemant's 1-day work}} = \frac{\frac{7}{15}}{\frac{1}{6}} = \frac{7}{15} \times 6 = \frac{42}{15} \]Simplify the fraction:
\[ \frac{42}{15} = \frac{14}{5} \text{ days} \]Correct Option: D) \( \frac{14}{5} \)
Find the median from the following data.
| Age in years | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 |
| No. of patients | 37 | 11 | 26 | 10 | 39 | 25 | 18 |
To find the median from grouped data, we first need to construct a cumulative frequency table. Then, we identify the median class using the formula for the median position. Finally, we apply the median formula for grouped data.
The given data is:
| Age in years (Class Interval) | No. of patients (Frequency, \(f\)) | Cumulative Frequency (\(cf\)) |
|---|---|---|
| 15-25 | 37 | 37 |
| 25-35 | 11 | \(37+11=48\) |
| 35-45 | 26 | \(48+26=74\) |
| 45-55 | 10 | \(74+10=84\) |
| 55-65 | 39 | \(84+39=123\) |
| 65-75 | 25 | \(123+25=148\) |
| 75-85 | 18 | \(148+18=166\) |
The total number of patients is the sum of all frequencies, which is \(N = 166\).
The median position is given by \(\frac{N}{2}\).
\[ \frac{N}{2} = \frac{166}{2} = 83 \]The median class is the class interval where the cumulative frequency is just greater than or equal to the median position (83). From the table, the cumulative frequency 84 corresponds to the class interval 45-55. So, the median class is 45-55.
The formula for the median is:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]Substitute these values into the formula:
\[ \text{Median} = 45 + \left( \frac{83 - 74}{10} \right) \times 10 \]\[ \text{Median} = 45 + \left( \frac{9}{10} \right) \times 10 \]\[ \text{Median} = 45 + 9 \]\[ \text{Median} = 54 \]D) 54 years is the calculated median.
The question asks to identify the tribal group associated with the Hornbill Festival in Nagaland. This requires knowledge of Indian festivals and the tribal communities that celebrate them, particularly in the Northeastern states.
Correct Option: C) Naga — The Hornbill Festival is a significant annual cultural event celebrated in Nagaland, India. It is often referred to as the "Festival of Festivals" and is celebrated by all the Naga tribes to showcase their rich cultural heritage, traditions, and customs. The festival aims to revive and protect the unique culture of the Naga people.