We need to select a team of five members (four students and one teacher) from a given group of five students (A, Q, B, S, C) and three teachers (D, E, F), adhering to specific conditions. We will evaluate each option against these conditions.
Let's list the given conditions:
Now, let's examine each option:
All conditions are satisfied by Option D.
Correct Option: D) ABSCE
The question asks about the most common shape of Harappan seals. To answer this, we need to recall the typical characteristics of Harappan seals as described in historical and archaeological studies.
Correct Option: C) The most common Harappan seals were square in shape. This is a well-established fact in the study of the Indus Valley Civilization.
The question asks for the full form of the acronym ENIAC, which was the first electronic digital programmable computer device. This is a factual recall question related to the history of computing.
B) Electronic Numerical Integrator and Computer
ENIAC stands for Electronic Numerical Integrator and Computer. It was built at the University of Pennsylvania's Moore School of Electrical Engineering between 1943 and 1945. It was the first electronic general-purpose digital computer capable of being reprogrammed to solve a full range of computing problems.
The question asks to identify the designer of the 'first' small computer for home use. This requires knowledge of early personal computing history and the pioneers involved in its development.
D) John Blankenbaker is widely credited with designing the Kenbak-1, which is considered by many to be the world's first personal computer. It was first sold in 1971, predating many other well-known early personal computers and was marketed as a small computer for home or educational use, even though it lacked a microprocessor and was programmed using switches.
The question asks to identify the battle associated with Henry Lawrence. This requires knowledge of key figures and events during significant periods of Indian history, particularly the Indian Rebellion of 1857.
Correct Option: D) Chinhat. Henry Lawrence led the British forces at the Battle of Chinhat on June 30, 1857, where they suffered a significant defeat against the Indian rebels. This battle preceded the famous Siege of Lucknow, where Lawrence was fatally wounded.
To find the area of the circle, we first need to determine its radius. We are given the difference between the circumference and the radius. We will use the formulas for circumference and area of a circle to solve the problem.
B) 154 cm² — The calculated area of the circle is 154 cm², which matches this option.
The question asks to identify the glands responsible for controlling blood pressure. This requires knowledge of the endocrine system and the specific hormones produced by various glands that impact cardiovascular function.
Correct Option: C) Adrenal glands. The adrenal glands produce hormones like adrenaline, noradrenaline, and aldosterone, all of which play significant roles in regulating blood pressure by affecting heart rate, vasoconstriction, and blood volume.
If tanα = \( \frac{5}{3} \) and tanβ = 4, then α + β is:
The problem asks us to find the value of \(\alpha + \beta\) given the values of \(\tan\alpha\) and \(\tan\beta\). We can use the tangent addition formula, which states that \[\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\]. After calculating \(\tan(\alpha + \beta)\), we can find \(\alpha + \beta\) by taking the inverse tangent.
We are given:
The formula for \(\tan(\alpha + \beta)\) is:
\[ \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} \]Substitute \(\tan\alpha = \frac{5}{3}\) and \(\tan\beta = 4\) into the formula:
\[ \tan(\alpha + \beta) = \frac{\frac{5}{3} + 4}{1 - \left(\frac{5}{3}\right)(4)} \]The numerator is \(\frac{5}{3} + 4\). To add these, find a common denominator:
\[ \frac{5}{3} + 4 = \frac{5}{3} + \frac{4 \times 3}{3} = \frac{5}{3} + \frac{12}{3} = \frac{5 + 12}{3} = \frac{17}{3} \]The denominator is \(1 - \left(\frac{5}{3}\right)(4)\). First, multiply the terms:
\[ \left(\frac{5}{3}\right)(4) = \frac{20}{3} \]Now, subtract this from 1:
\[ 1 - \frac{20}{3} = \frac{3}{3} - \frac{20}{3} = \frac{3 - 20}{3} = -\frac{17}{3} \]Now, substitute the simplified numerator and denominator back into the formula:
\[ \tan(\alpha + \beta) = \frac{\frac{17}{3}}{-\frac{17}{3}} \]This simplifies to:
\[ \tan(\alpha + \beta) = -1 \]We need to find the angle whose tangent is \(-1\). We know that \(\tan(45^\circ) = 1\). Since the tangent is negative, the angle must be in the second or fourth quadrant. Given the options, we are looking for an angle between \(90^\circ\) and \(180^\circ\) (second quadrant) or between \(270^\circ\) and \(360^\circ\) (fourth quadrant).
In the second quadrant, the angle is \(180^\circ - 45^\circ = 135^\circ\).
So, \(\alpha + \beta = 135^\circ\).
Correct Option: A) \(135^\circ\)
The problem presents an analogy involving letter-clusters separated by a slash. We need to identify the relationship between the letters before the slash and the letters after the slash in the given pairs, and then apply the same logic to find the missing letter-cluster.
M \(\rightarrow\) N
This shows a progression of +1 in the alphabetical order (M is the 13th letter, N is the 14th letter).
FH \(\rightarrow\) FI
The first letter 'F' remains the same.
The second letter 'H' \(\rightarrow\) 'I' shows a progression of +1 in the alphabetical order (H is the 8th letter, I is the 9th letter).
The pattern is:
A) A/FO is the correct option because it follows the established pattern: the first letter before the slash increments by one (A \(\rightarrow\) B), the first letter after the slash remains the same (F \(\rightarrow\) F), and the second letter after the slash increments by one (O \(\rightarrow\) P).
The problem asks us to identify the pair of numbers that follows a different pattern compared to the other three pairs. We will analyze the relationship between the two numbers in each pair, specifically looking at their difference.
Correct Option: D) 47 − 34. The difference between the numbers in this pair is 13, which is different from the difference of 17 found in the other three pairs.