The problem presents a series of letter groups, and we need to find the next group in the sequence. We will analyze the pattern of each letter's position in the English alphabet across the given groups.
S (19) \(\rightarrow\) Q (17) \(\rightarrow\) O (15) \(\rightarrow\) M (13)
The pattern for the first letter is a decrease of 2 positions each time: \(19 - 2 = 17\), \(17 - 2 = 15\), \(15 - 2 = 13\). Following this pattern, the next first letter should be \(13 - 2 = 11\), which corresponds to the letter K.
U (21) \(\rightarrow\) S (19) \(\rightarrow\) Q (17) \(\rightarrow\) O (15)
The pattern for the second letter is also a decrease of 2 positions each time: \(21 - 2 = 19\), \(19 - 2 = 17\), \(17 - 2 = 15\). Following this pattern, the next second letter should be \(15 - 2 = 13\), which corresponds to the letter M.
X (24) \(\rightarrow\) V (22) \(\rightarrow\) T (20) \(\rightarrow\) R (18)
The pattern for the third letter is also a decrease of 2 positions each time: \(24 - 2 = 22\), \(22 - 2 = 20\), \(20 - 2 = 18\). Following this pattern, the next third letter should be \(18 - 2 = 16\), which corresponds to the letter P.
Based on the analysis, the next group in the series should be KMP.
C) KMP — This option matches our derived sequence by following the consistent pattern of decreasing by 2 for each corresponding letter in the alphabetical order.
The question asks to identify the group of islands based on their geographical coordinates (latitude and longitude). This requires knowledge of the location of major island groups of India.
C) Lakshadweep — The Lakshadweep Islands are located in the Arabian Sea and fall within the specified geographical coordinates of 8° – 12° 13′ North latitude and 71° – 74° East longitude.
To find the percentage gain or loss, we need to determine the Cost Price (CP), Marked Price (MP), and Selling Price (SP). We'll assume a convenient Cost Price, calculate the Marked Price based on the given percentage above CP, then calculate the Selling Price after applying the discount on the Marked Price. Finally, we compare the Selling Price with the Cost Price to find the gain or loss percentage.
Let the Cost Price (CP) of the goods be \(₹100\).
The salesman marks his goods 64% above the Cost Price.
\[ \text{MP} = \text{CP} + 64\% \text{ of CP} \]\[ \text{MP} = 100 + \left(\frac{64}{100} \times 100\right) \]\[ \text{MP} = 100 + 64 = ₹164 \]A discount of 25% is given on the Marked Price.
\[ \text{Discount Amount} = 25\% \text{ of MP} \]\[ \text{Discount Amount} = \frac{25}{100} \times 164 \]\[ \text{Discount Amount} = \frac{1}{4} \times 164 = ₹41 \]\[ \text{SP} = \text{MP} - \text{Discount Amount} \]\[ \text{SP} = 164 - 41 = ₹123 \]Compare the Selling Price (SP) with the Cost Price (CP).
Since \(\text{SP} (₹123) > \text{CP} (₹100)\), there is a gain.
\[ \text{Gain} = \text{SP} - \text{CP} \]\[ \text{Gain} = 123 - 100 = ₹23 \]The percentage gain is calculated on the Cost Price.
\[ \text{Percentage Gain} = \left(\frac{\text{Gain}}{\text{CP}}\right) \times 100\% \]\[ \text{Percentage Gain} = \left(\frac{23}{100}\right) \times 100\% \]\[ \text{Percentage Gain} = 23\% \]Correct Option: B) 23% gain
What is the median of the following distribution?
| Class | 110-160 | 160-210 | 210-260 | 260-310 | 310-360 |
| Frequency | 16 | 15 | 31 | 24 | 38 |
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 \((N/2)^{\text{th}}\) observation. Finally, we use the median formula for grouped data.
First, let's list the given classes and their frequencies, and then calculate the cumulative frequencies (cf).
| Class Interval | Frequency (f) | Cumulative Frequency (cf) |
|---|---|---|
| 110-160 | 16 | 16 |
| 160-210 | 15 | 16 + 15 = 31 |
| 210-260 | 31 | 31 + 31 = 62 |
| 260-310 | 24 | 62 + 24 = 86 |
| 310-360 | 38 | 86 + 38 = 124 |
The total number of observations \(N\) is the sum of all frequencies, which is the last value in the cumulative frequency column.
\[ N = 124 \]The median position is given by \(N/2\).
\[ \frac{N}{2} = \frac{124}{2} = 62 \]The median class is the class interval whose cumulative frequency is just greater than or equal to \(N/2\). In this case, \(N/2 = 62\). Looking at the cumulative frequency column, the class 210-260 has a cumulative frequency of 62. Therefore, the median class is 210-260.
The formula for the median of grouped data is:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - \text{cf}}{f} \right) \times h \]From our median class 210-260:
Substitute these values into the formula:
\[ \text{Median} = 210 + \left( \frac{62 - 31}{31} \right) \times 50 \]\[ \text{Median} = 210 + \left( \frac{31}{31} \right) \times 50 \]\[ \text{Median} = 210 + (1) \times 50 \]\[ \text{Median} = 210 + 50 \]\[ \text{Median} = 260 \]D) 260
This problem involves tracking Rishi's movements using a coordinate system or by drawing a diagram. We need to find his final position relative to his starting point (Point A) and then determine the shortest distance and direction required to return to Point A.
Therefore, Rishi needs to drive 4 km to the west to reach Point A again.
D) 4 km to the west
To find the median of a set of observations, we first need to arrange the observations in ascending order. Then, we identify the middle value. If the number of observations is odd, the median is the single middle value. If the number of observations is even, the median is the average of the two middle values.
B) 38. As calculated, after arranging the observations in ascending order, the 6th term, which is 38, is the median.
The question asks for the date when the Lokpal and Lokayukta Act came into force in India. This requires knowledge of the specific enactment date of this significant anti-corruption legislation.
Correct Option: B) The Lokpal and Lokayukta Act, 2013, received presidential assent on 1 January 2014, but it officially came into force on 16 January 2014, as notified in the Gazette of India. This date marks the commencement of the Act's provisions.
The Goods and Services Tax (GST) was introduced in India to simplify the indirect tax structure by subsuming multiple central and state taxes. To answer this question, one needs to identify which of the listed taxes were replaced by GST.
Correct Option: C) 1, 2 and 3
The question asks to identify the author of the Tamil novel "Andhi Malai" who writes under the pen name Ambai and is a well-known feminist writer. This requires knowledge of prominent Indian authors, particularly those from Tamil literature, and their pen names.
CS Lakshmi is a renowned Indian feminist writer who writes primarily in Tamil. She uses the pen name Ambai. Her works often explore themes of women's experiences, identity, and societal constraints. "Andhi Malai" (Evening Ragam) is one of her notable works.
D) CS Lakshmi — CS Lakshmi is indeed the well-known feminist writer who uses the pen name Ambai and is the author of the Tamil novel "Andhi Malai".
The problem describes a right-angled triangle with its sides given in terms of a variable \(x\). We need to find the area of this triangle. The key to solving this is to use the Pythagorean theorem to find the value of \(x\), and then calculate the lengths of the perpendicular sides (base and height) to determine the area.
Identify the sides of the right triangle:
The sides are given as \((x - 18)\), \((x - 25)\), and \(x\). In a right-angled triangle, the hypotenuse is the longest side. Comparing the given expressions, \(x\) must be the hypotenuse because \((x - 18)\) and \((x - 25)\) are both smaller than \(x\). Also, for the side lengths to be positive, we must have \(x - 25 > 0\), which implies \(x > 25\).
Apply the Pythagorean theorem:
For a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (legs).
\[ (\text{Hypotenuse})^2 = (\text{Side}_1)^2 + (\text{Side}_2)^2 \]In this case:
\[ x^2 = (x - 18)^2 + (x - 25)^2 \]
Expand and simplify the equation:
Using the formula \((a - b)^2 = a^2 - 2ab + b^2\):
\[ x^2 = (x^2 - 2 \cdot x \cdot 18 + 18^2) + (x^2 - 2 \cdot x \cdot 25 + 25^2) \]\[ x^2 = (x^2 - 36x + 324) + (x^2 - 50x + 625) \]\[ x^2 = 2x^2 - 86x + 949 \]
Rearrange into a quadratic equation:
Move all terms to one side to form a standard quadratic equation \(ax^2 + bx + c = 0\):
\[ 0 = 2x^2 - x^2 - 86x + 949 \]\[ x^2 - 86x + 949 = 0 \]
Solve the quadratic equation for \(x\):
We can solve this by factoring. We need two numbers that multiply to \(949\) and add up to \(-86\). Let's test factors of \(949\). Since \(949\) ends in \(9\), its factors might end in \(1, 9\) or \(3, 3\) or \(7, 7\). Let's try dividing by small prime numbers. \(949\) is not divisible by \(2, 3, 5\). Let's try \(7\): \(949 / 7 \approx 135\). Let's try \(13\): \(949 / 13 = 73\).
So, the factors are \(13\) and \(73\). Since their sum should be \(-86\), the numbers are \(-13\) and \(-73\).
\[ (x - 13)(x - 73) = 0 \]This gives two possible values for \(x\): \(x = 13\) or \(x = 73\).
Check the validity of \(x\) values:
Recall that for the side lengths to be positive, \(x - 25 > 0\), which means \(x > 25\).
If \(x = 13\), then \(x - 25 = 13 - 25 = -12\), which is not a valid side length. So, \(x = 13\) is rejected.
If \(x = 73\), then:
Side 1: \(x - 18 = 73 - 18 = 55\) cm
Side 2: \(x - 25 = 73 - 25 = 48\) cm
Hypotenuse: \(x = 73\) cm
All side lengths are positive, so \(x = 73\) is the correct value.
Calculate the area of the triangle:
The area of a right-angled triangle is given by \(\frac{1}{2} \times \text{base} \times \text{height}\). The two perpendicular sides are \(55\) cm and \(48\) cm.
\[ \text{Area} = \frac{1}{2} \times 55 \times 48 \]\[ \text{Area} = 55 \times \frac{48}{2} \]\[ \text{Area} = 55 \times 24 \]To calculate \(55 \times 24\):
\(55 \times 20 = 1100\)
\(55 \times 4 = 220\)
\(1100 + 220 = 1320\)
So, the area is \(1320\) cm\(^2\).
Correct Option: D) 1320