Which is the smallest fraction among the given fractions?
\( \frac{4}{9} \), \( \frac{5}{4} \), \( \frac{3}{8} \), \( \frac{6}{7} \)
\( \frac{6}{7} \)
\( \frac{5}{4} \)
\( \frac{4}{9} \)
\( \frac{3}{8} \)
To find the smallest fraction among the given options, we can convert each fraction into its decimal equivalent. Comparing decimal values is straightforward: the smallest decimal corresponds to the smallest fraction.
The decimal values are approximately \( 0.444 \), \( 1.25 \), \( 0.375 \), and \( 0.857 \).
Comparing these values, \( 0.375 \) is the smallest.
The decimal \( 0.375 \) corresponds to the fraction \( \frac{3}{8} \).
Correct Option: D) \( \frac{3}{8} \)
The decimal equivalent of \( \frac{3}{8} \) is \( 0.375 \), which is the smallest among all the calculated decimal values.
If x = cos-1(cos 4) and y = sin-1(sin 3) then which of the following conditions is true?
tan(x−y)=tan \( \frac{1}{2} \)
x+2y=2
tan(x+y)=−tan7
x+y+1=0
The problem involves finding the values of \(x\) and \(y\) using the inverse trigonometric functions \(\cos^{-1}(\cos \theta)\) and \(\sin^{-1}(\sin \theta)\). The key is to remember the principal value branches for these functions. For \(\cos^{-1} A\), the principal value lies in \([0, \pi]\), and for \(\sin^{-1} A\), it lies in \[[-\frac{\pi}{2}, \frac{\pi}{2}]\]. We will then substitute these values into the given options to find the true condition.
We are given \(x = \cos^{-1}(\cos 4)\).
The principal value branch of \(\cos^{-1} A\) is \([0, \pi]\). We know that \(\pi \approx 3.14159\) and \(2\pi \approx 6.28318\).
Since \(4\) radians is not in \([0, \pi]\), we need to find an angle \(\theta\) in \([0, \pi]\) such that \(\cos \theta = \cos 4\).
The cosine function is positive in the first and fourth quadrants. \(4\) radians is in the third quadrant (since \(\pi < 4 < \frac{3\pi}{2}\), i.e., \(3.14 < 4 < 4.71\)). In the third quadrant, \(\cos 4\) is negative.
We can use the identity \(\cos(2\pi - \theta) = \cos \theta\). Let \(\theta' = 2\pi - 4\).
Then \(2\pi - 4 \approx 6.28318 - 4 = 2.28318\).
This value \(2\pi - 4\) is in the first quadrant (since \(0 < 2\pi - 4 < \pi\), i.e., \(0 < 2.28318 < 3.14159\)).
However, \(\cos 4\) is negative. We need an angle in \([0, \pi]\) whose cosine is equal to \(\cos 4\). Since \(4\) is in the third quadrant, \(\cos 4\) is negative. The angle in \([0, \pi]\) that has the same cosine value as \(4\) is \(2\pi - 4\) if \(4\) were in the fourth quadrant. However, \(4\) is in the third quadrant. The angle in \([0, \pi]\) that has the same cosine value as \(\cos 4\) is \(2\pi - 4\). Let's check this: \(\cos(2\pi - 4) = \cos 4\). And \(0 < 2\pi - 4 < \pi\) is true (as \(2.28 \in [0, 3.14]\)).
So, \(x = 2\pi - 4\).
We are given \(y = \sin^{-1}(\sin 3)\).
The principal value branch of \(\sin^{-1} A\) is \[[-\frac{\pi}{2}, \frac{\pi}{2}]\]. We know that \(\frac{\pi}{2} \approx 1.5708\) and \(\pi \approx 3.14159\).
Since \(3\) radians is not in \[[-\frac{\pi}{2}, \frac{\pi}{2}]\] (as \(3 > 1.5708\)), we need to find an angle \(\theta\) in \[[-\frac{\pi}{2}, \frac{\pi}{2}]\] such that \(\sin \theta = \sin 3\).
\(3\) radians is in the second quadrant (since \(\frac{\pi}{2} < 3 < \pi\), i.e., \(1.5708 < 3 < 3.14159\)).
In the second quadrant, \(\sin 3\) is positive.
We can use the identity \(\sin(\pi - \theta) = \sin \theta\). Let \(\theta' = \pi - 3\).
Then \(\pi - 3 \approx 3.14159 - 3 = 0.14159\).
This value \(\pi - 3\) is in the first quadrant (since \(0 < \pi - 3 < \frac{\pi}{2}\), i.e., \(0 < 0.14159 < 1.5708\)).
So, \(y = \pi - 3\).
C) \(\tan(x+y) = -\tan 7\)
If x2 + y2 = 156 and x-y = 12, then xy =?
This problem can be solved by using the algebraic identity for \((a-b)^2\), which relates \((a-b)^2\), \(a^2+b^2\), and \(ab\). We are given the values of \(x^2+y^2\) and \(x-y\), and we need to find \(xy\).
The relevant algebraic identity is:
\[ (x-y)^2 = x^2 + y^2 - 2xy \]We are given \(x^2 + y^2 = 156\) and \(x-y = 12\). Substitute these values into the identity:
\[ (12)^2 = 156 - 2xy \]Subtract 156 from both sides:
\[ 144 - 156 = -2xy \] \[ -12 = -2xy \]Divide both sides by -2:
\[ xy = \frac{-12}{-2} \] \[ xy = 6 \]Correct Option: D) 6
This problem involves determining directions based on initial orientation and subsequent turns. We need to visualize the directions and the effect of each turn.
Let's visualize the turns:
Correct Option: B) North-West
The question asks to identify the political party that came to power immediately after the national emergency in India. This requires knowledge of Indian political history, specifically the period following the Emergency (1975-1977).
D) Janata Party — The Janata Party was formed as a coalition of opposition parties and won the general elections held immediately after the lifting of the National Emergency in 1977, forming the government at the Centre.
The question asks to identify which animal gives birth to very poorly developed young ones. This relates to the reproductive strategies and developmental stages of different mammals. We need to recall the characteristics of marsupials versus placental mammals.
Correct Option: C) Kangaroo
The question asks to identify the location of Dachigam National Park. This is a factual question related to Indian geography and national parks, which are common topics in competitive exams. Recalling the locations of major national parks is key to answering this type of question.
Correct Option: A) Dachigam National Park is indeed located in Jammu and Kashmir. It is situated about 22 kilometers from Srinagar, the capital city of Jammu and Kashmir. The park was initially established to protect the Hangul (Kashmir Stag), which is a critically endangered species.
What is the mean of the following data?
| Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 8 | 12 | 16 | 11 | 13 |
To find the mean of grouped data, we use the formula for the assumed mean method or the direct method. For this problem, we will use the direct method, which involves calculating the midpoint of each class interval, multiplying it by its corresponding frequency, summing these products, and then dividing by the total frequency.
C) 36.5 is the correct mean calculated from the given grouped data.
The question asks to identify the Indian archer who was the subject of the 2017 documentary 'Ladies First'. This requires knowledge of prominent Indian sports personalities and related media.
A) Deepika Kumari — 'Ladies First' is indeed a 2017 documentary that tells the inspiring story of Deepika Kumari, an Indian archer who overcame significant challenges to become a world champion.
To find the time required, we will use the simple interest formula: \[\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\]. We are given the Principal (P), Simple Interest (SI), and Rate (R), and we need to solve for Time (T).
Identify the given values:
Recall the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
Substitute the given values into the formula:
\[ 375 = \frac{1800 \times 5 \times \text{T}}{100} \]
Simplify the equation:
\[ 375 = 18 \times 5 \times \text{T} \]
\[ 375 = 90 \times \text{T} \]
Solve for T:
\[ \text{T} = \frac{375}{90} \]
\[ \text{T} = \frac{125}{30} \]
\[ \text{T} = \frac{25}{6} \text{ years} \]
Convert the fractional years into years and months:
\[ \text{T} = 4 \frac{1}{6} \text{ years} \]
This means 4 full years and \( \frac{1}{6} \) of a year.
To convert \( \frac{1}{6} \) of a year into months, multiply by 12 (since there are 12 months in a year):
\[ \text{Months} = \frac{1}{6} \times 12 = 2 \text{ months} \]
So, the time is 4 years and 2 months.
D) 4 years, 2 months is the correct answer, as calculated.