General Aptitude – GATE PI General AptitudePractice Questions & PYQs

This question asks us to identify the most logical conclusion drawn directly from the provided text. The passage states that climate change threatens regular rice production, and Green Super Rice (GSR) is being developed to be tough ($resilient$) in bad weather and still give more grain ($higher$ $yields$).

Let's break down each option:

• Option A says GSR grows only in extreme weather. The passage says GSR is $resilient$ in extreme weather, meaning it can handle it, not that it exclusively grows there. So, this isn't right.
• Option B suggests GSR could be used later to fight the bad impacts of climate change. This perfectly matches the passage's point that GSR is a solution to the threat climate change poses to rice. This is a strong logical inference.
• Option C claims GSR produces less than regular rice. The passage clearly states GSR gives $higher$ $yields$, which contradicts this option. Thus, this is incorrect.
• Option D states regular rice will keep producing well even in extreme weather. The passage explicitly says climate change $poses$ $a$ $threat$ to sustained rice availability, directly opposing this idea. So, this is also incorrect.

is the correct logical inference.

In Injection Blow Molding (IBM) for plastic beverage bottles, we first create a preform (parison) by injection molding. This heated preform is then inflated into the final bottle shape inside a mold. This inflation, or blowing, is done by introducing compressed hot air into the parison. Hot air is preferred because it's easily available, allows for precise control over temperature and pressure, and is safe for plastic inflation. Other options like hot water, hot oil, or alcohol are not suitable due to potential issues like contamination, control difficulties, viscosity, flammability, and safety concerns. Therefore, hot air is used for blowing in injection blow molding.

To find the probability that the sum of the numbers on the two disks is $8$, we first determine the total possible outcomes. Each disk has $8$ equally likely outcomes (numbers $1$ to $8$). Since the two disks are spun independently, the total number of possible outcomes is $8 \times 8 = 64$. Next, we identify the favorable outcomes where the sum of the numbers from the two disks is $8$. These combinations are $(1, 7)$, $(2, 6)$, $(3, 5)$, $(4, 4)$, $(5, 3)$, $(6, 2)$, and $(7, 1)$, totaling $7$ favorable outcomes. Therefore, the probability is calculated as:
$\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} = \frac{7}{64}$

To perfectly fit a sphere of radius $r$ inside a cubical box, the minimum side length of the cube must be equal to the sphere's diameter. The diameter of the sphere is $2 \times r = 2r$ cm. So, the minimum side length of the cube, $s$, must be $2r$ cm.

The volume of a cube is calculated using the formula $V = s^3$.
Substituting $s = 2r$ into the volume formula gives:
$V_{min} = (2r)^3$
$V_{min} = 2^3 \times r^3$
$V_{min} = 8r^3$
Therefore, the minimum volume of the cubical box required is $8r^3$ cubic centimeters.

This problem involves understanding the concept of work rate, specifically how pipes fill or empty a tank. We need to find the tank's total capacity.

First, let's define the rates of each pipe.
Pipe P fills the tank in 10 minutes, so its filling rate is $\frac{1}{10}$ of the tank per minute.
Pipe Q fills the tank in 6 minutes, so its filling rate is $\frac{1}{6}$ of the tank per minute.
Pipe R empties water at a constant rate of 34 litres per minute.

Let $C$ be the capacity of the tank in litres. We can express the filling rates of P and Q in litres per minute:
Rate of Pipe P (filling) = $\frac{C}{10}$ litres per minute.
Rate of Pipe Q (filling) = $\frac{C}{6}$ litres per minute.

When all three pipes operate simultaneously, the net rate of change in the tank's volume is the sum of the filling rates minus the emptying rate:
$\text{Net Rate} = \left( \frac{C}{10} + \frac{C}{6} \right) - 34 \text{ litres/minute}$
We are given that a full tank empties completely in 1 hour (which is 60 minutes) when all pipes are operating. This means the net emptying rate is $\frac{C}{60}$ litres per minute. Since it's emptying, we represent this as a negative net rate:
$\text{Net Rate} = -\frac{C}{60} \text{ litres/minute}$
Now, we equate the two expressions for the Net Rate:
$\left( \frac{C}{10} + \frac{C}{6} \right) - 34 = -\frac{C}{60}$
To solve for $C$, we find a common denominator for the fractions, which is 60:
$\left( \frac{6C}{60} + \frac{10C}{60} \right) - 34 = -\frac{C}{60}$
$\frac{16C}{60} - 34 = -\frac{C}{60}$
Rearranging the terms to isolate $C$:
$\frac{16C}{60} + \frac{C}{60} = 34$
$\frac{17C}{60} = 34$
Solving for $C$:
$17C = 34 \times 60$
$C = \frac{34 \times 60}{17}$
$C = 2 \times 60$
$C = 120$
The capacity of the storage tank is 120 litres.

To form a square with a single round hole near the center without overlapping pieces, we must find a set of shapes that fit together perfectly.
Option 1 and Option 2 show irregular shapes that geometrically cannot form a precise square boundary with a distinct central hole.
Option 3 provides a central piece with a round cutout and four identical edge pieces. When these four edge pieces are arranged around the central piece, they precisely complete a square, with the central cutout forming the required single round hole. This arrangement satisfies all conditions.
Option 4 contains pieces that are not suitable for assembling the desired square with a central circular void.

To determine which expression involving $p+q$ satisfies both inequalities, we'll combine them. We are given:
(i) $3p - q < 4$
(ii) $3q - p < 12$
Adding the two inequalities together:
$(3p - q) + (3q - p) < 4 + 12$
Simplifying the left side by combining like terms ($3p - p$ and $-q + 3q$) and the right side:
$2p + 2q < 16$
Factoring out $2$ from the left side:
$2(p + q) < 16$
Finally, dividing both sides by $2$ to isolate $p+q$:
$p + q < 8$
This result indicates that $p+q$ must be strictly less than $8$ for both original inequalities to hold true. Comparing this with the given options, Option A, $p+q < 8$, is the direct match.

To determine the minimum number of doors to open to travel from point P to point Q, we need to find the shortest path in terms of door crossings. By carefully examining the provided building layout, we identify the most efficient route:

1. From the room containing P, open the door connecting it to the next room. This is the first door.
2. Proceed through this room and open the door that leads towards point Q. This is the second door.
   This path requires opening $2$ doors to reach point Q. Therefore, the minimum number of doors to be opened is $2$.

The order of a differential equation is determined by the order of its highest derivative. In the given differential equation, $\left(\frac{dy}{dx}\right)^2 + 5\frac{dy}{dx} + 4y = 5x^3$ the only derivative present is $\frac{dy}{dx}$. This is a first derivative, meaning its order is 1. Since this is the highest (and only) derivative, the order of the differential equation is 1.

We have 6 people (P, Q, R, S, T, U) sitting around a circular table facing the center. We need to find Q's immediate neighbors.

Let's break down the given statements:

1. "P sits next to S and T": This tells us P is exactly between S and T. So, the sequence will look like S-P-T or T-P-S.
2. "Q sits diametrically opposite to P": In a circular arrangement of 6 people, "diametrically opposite" means there are 2 people between them on either side.
3. "The shortest distance between S and R is equal to the shortest distance between T and U": Here, "distance" refers to the number of people sitting between the two specified individuals along the shorter path around the table. If they are adjacent, the distance is $0$.

Let's start placing them:

• We can fix P at position 1. Due to statement 2, Q must be at position 4 (P(1) -- Person -- Person -- Q(4)).
• Now, using statement 1, S and T must be next to P. This leads to two initial scenarios for placing S and T:
  • Scenario A: S is at position 2, and T is at position 6. Our current partial arrangement is P(1), S(2), Q(4), T(6).
  • Scenario B: T is at position 2, and S is at position 6. Our current partial arrangement is P(1), T(2), Q(4), S(6).
• The remaining people are R and U, and the remaining positions are 3 and 5.

Now, let's test these scenarios with statement 3:

Scenario A (P-S-?-Q-?-T)
Positions: P(1), S(2), Q(4), T(6). Remaining positions for R and U are 3 and 5.

• Possibility A1: R at 3, U at 5. The full arrangement is P(1), S(2), R(3), Q(4), U(5), T(6).

  • Distance(S, R): S(2) and R(3) are adjacent, so 0 people between them. Distance = $0$.
  • Distance(T, U): T(6) and U(5) are adjacent, so 0 people between them. Distance = $0$.
  • Statement 3 (Distance = $0$ for both) is satisfied.
  • In this possibility, Q(4)'s neighbors are R(3) and U(5).

• Possibility A2: U at 3, R at 5. The full arrangement is P(1), S(2), U(3), Q(4), R(5), T(6).

  • Distance(S, R): S(2) to R(5). The people between them are U(3) and Q(4). So, 2 people. Distance = $2$.
  • Distance(T, U): T(6) to U(3). The people between them (clockwise) are P(1), S(2). So, 2 people. Distance = $2$.
  • Statement 3 (Distance = $2$ for both) is satisfied.
  • In this possibility, Q(4)'s neighbors are U(3) and R(5).

Scenario B (P-T-?-Q-?-S)
Positions: P(1), T(2), Q(4), S(6). Remaining positions for R and U are 3 and 5.

• Possibility B1: R at 3, U at 5. The full arrangement is P(1), T(2), R(3), Q(4), U(5), S(6).

  • Distance(S, R): S(6) to R(3). The people between them (clockwise) are P(1), T(2). So, 2 people. Distance = $2$.
  • Distance(T, U): T(2) to U(5). The people between them are R(3), Q(4). So, 2 people. Distance = $2$.
  • Statement 3 (Distance = $2$ for both) is satisfied.
  • In this possibility, Q(4)'s neighbors are R(3) and U(5).

• Possibility B2: U at 3, R at 5. The full arrangement is P(1), T(2), U(3), Q(4), R(5), S(6).

  • Distance(S, R): S(6) and R(5) are adjacent, so 0 people between them. Distance = $0$.
  • Distance(T, U): T(2) and U(3) are adjacent, so 0 people between them. Distance = $0$.
  • Statement 3 (Distance = $0$ for both) is satisfied.
  • In this possibility, Q(4)'s neighbors are U(3) and R(5).

In all valid arrangements (A1, A2, B1, B2), the neighbors of Q are consistently R and U.

This problem tests our ability to manipulate algebraic expressions involving fractions. We are given $p$ and $q$ are positive integers and the equation:
$\frac{p}{q} + \frac{q}{p} = 3$
Our goal is to find the value of:
$\frac{p^2}{q^2} + \frac{q^2}{p^2}$
Let's simplify by substituting $x = \frac{p}{q}$. This means $\frac{1}{x} = \frac{q}{p}$.
The given equation transforms to:
$x + \frac{1}{x} = 3$
The expression we need to find becomes:
$\left(\frac{p}{q}\right)^2 + \left(\frac{q}{p}\right)^2 = x^2 + \left(\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2}$
We can use the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$. Let $a = x$ and $b = \frac{1}{x}$:
$\left(x + \frac{1}{x}\right)^2 = x^2 + \left(\frac{1}{x}\right)^2 + 2(x)\left(\frac{1}{x}\right)$
Simplify the term $2(x)\left(\frac{1}{x}\right)$:
$2(x)\left(\frac{1}{x}\right) = 2 \times \frac{x}{x} = 2 \times 1 = 2$
So, the identity becomes:
$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$
We know that $x + \frac{1}{x} = 3$. Substitute this value into the equation:
$(3)^2 = x^2 + \frac{1}{x^2} + 2$
$9 = x^2 + \frac{1}{x^2} + 2$
To find $x^2 + \frac{1}{x^2}$, subtract 2 from both sides:
$x^2 + \frac{1}{x^2} = 9 - 2$
$x^2 + \frac{1}{x^2} = 7$
Since $x^2 + \frac{1}{x^2} = \frac{p^2}{q^2} + \frac{q^2}{p^2}$, the value we are looking for is 7.

Let's break down each statement to see if it can be concluded from the passage.

Statement (i): Nowadays, computers are present in almost all places.
The passage clearly states, "Computers are ubiquitous," and "They are used to improve efficiency in almost all fields from agriculture to space exploration." "Ubiquitous" means present everywhere, and "almost all fields" supports this widespread presence. So, statement (i) is a valid deduction.

Statement (ii): Computers cannot be used for solving problems in engineering.
The passage mentions computers are used in "almost all fields from agriculture to space exploration." Space exploration heavily involves complex engineering. Therefore, computers can be used in engineering, which contradicts statement (ii). So, statement (ii) is not a valid deduction.

Statement (iii): For humans, there are both positive and negative effects of using computers.
The passage states, "For humans, sitting in front of a computer for long hours can lead to health issues," which is a negative effect. The passage also mentions computers "improve efficiency in almost all fields," implying positive outcomes from their use. Thus, both positive and negative effects are implied. So, statement (iii) is a valid deduction.

Statement (iv): Artificial intelligence can be done without data.
The passage defines AI: "AI enables computers to learn, given enough training data." The phrase "given enough training data" explicitly indicates that data is a requirement for AI to learn. This directly refutes statement (iv). So, statement (iv) is not a valid deduction.

Based on this analysis, only statements (i) and (iii) can be deduced from the passage.

This problem involves finding an initial value before a percentage increase over time, which is a common application of compound growth principles. We need to find the population 2 years ago ($P_0$) given the current population ($P = 11,02,500$), an annual growth rate ($r = 5\%$), and the time period ($n = 2$ years).

The standard formula for compound growth is:
$P = P_0 \left(1 + \frac{r}{100}\right)^n$
To find the population 2 years ago ($P_0$), we rearrange the formula:
$P_0 = \frac{P}{\left(1 + \frac{r}{100}\right)^n}$
Now, let's plug in the values and calculate:
First, convert the percentage rate to a decimal: $\frac{5}{100} = 0.05$.
Next, add 1 to the decimal rate: $1 + 0.05 = 1.05$.
Then, calculate the growth factor for 2 years: $(1.05)^2 = 1.05 \times 1.05 = 1.1025$.
Finally, divide the current population by this growth factor to find $P_0$:
$P_0 = \frac{11,02,500}{1.1025}$
$P_0 = 10,00,000$
The population of the city 2 years ago was $10,00,000$.

This question tests our deductive reasoning skills in analyzing categorical statements.

The statements provided are:

1. All purple are green. (\text{P} \subseteq \text{G}$)
2. All black are green. (\text{B} \subseteq \text{G}$)

These statements mean that the entire set of 'purple' is a subset of 'green', and the entire set of 'black' is also a subset of 'green'. However, these statements do not define the relationship between 'purple' and 'black'.

Now, let's look at the conclusions:
I. Some black are purple. (\text{B} \cap \text{P} \neq \emptyset$) II. No black is purple. (\text{B} \cap \text{P} = \emptyset$)

Conclusion I suggests that there is an overlap between 'black' and 'purple'. Conclusion II suggests that 'black' and 'purple' are entirely separate.
From the given statements, both scenarios (overlap or no overlap between 'black' and 'purple' within 'green') are possible. We cannot definitively say that 'Some black are purple' is true, nor can we definitively say that 'No black is purple' is true based only on the premises.

However, Conclusion I and Conclusion II are contradictory statements. They cannot both be true simultaneously, and they cannot both be false simultaneously. If one is false, the other must be true. Since the premises allow for either possibility (overlap or no overlap), and one of these two contradictory conclusions must hold true, the logically correct option is that "Either conclusion I or II is correct."

Let's break down this geometry problem step-by-step to find the cone's volume!

First, we start with a square sheet of side 1 unit. When we cut it along its main diagonal, we get two identical right-angled triangles. Each of these triangles has two sides of length 1 unit (the original sides of the square) and the third side is the diagonal. Using the Pythagorean theorem, the diagonal length is $\sqrt{1^2 + 1^2} = \sqrt{2}$ units. So, each triangle has sides of length 1, 1, and $\sqrt{2}$ units. The two shorter edges are the sides with length 1 unit.

Next, one of these triangles is revolved about its short edge. When a right-angled triangle is revolved around one of its legs (a short edge), it forms a solid cone. The leg it revolves around becomes the height ($h$) of the cone, and the other leg becomes the radius ($r$) of the cone's base. The hypotenuse becomes the slant height. Since we are revolving about a short edge of length 1 unit, we have:
Height ($h$) = 1 unit
Radius ($r$) = 1 unit
Slant height ($l$) = $\sqrt{2}$ units

Finally, to calculate the volume of this cone, we use the formula: $V = \frac{1}{3}\pi r^2 h$.
Substituting our values for $r$ and $h$:
$V = \frac{1}{3}\pi (1)^2 (1)$
$V = \frac{1}{3}\pi (1)(1)$
$V = \frac{\pi}{3}$ cubic units.

The core concept here is the formula for the Area of an equilateral triangle: $Area = \frac{\sqrt{3}}{4} \times a^2$, where $a$ is the side length.

Calculation:
Let the side of the original large equilateral triangle be $a$. When corners are cut to form a regular hexagon, the side of this regular hexagon becomes $\frac{a}{3}$. The hexagon is formed by removing three smaller equilateral triangles from the corners. Each of these smaller triangles has a side length of $\frac{a}{3}$.

The area of the original equilateral triangle ($A_T$) is $A_T = \frac{\sqrt{3}}{4} \times a^2$.

The regular hexagon can be thought of as consisting of 6 smaller equilateral triangles, each with side length $\frac{a}{3}$.
So, the area of the regular hexagon ($A_H$) is $6 \times (\frac{\sqrt{3}}{4} \times (\frac{a}{3})^2)$.

Now, to find the ratio of the area of the regular convex hexagon to the area of the original equilateral triangle:
$\frac{A_H}{A_T} = \frac{6 \times \frac{\sqrt{3}}{4} \times \left(\frac{a}{3}\right)^2}{\frac{\sqrt{3}}{4} \times a^2}$
$\frac{A_H}{A_T} = \frac{6 \times \frac{\sqrt{3}}{4} \times \frac{a^2}{9}}{\frac{\sqrt{3}}{4} \times a^2}$
Cancel out the common terms $\frac{\sqrt{3}}{4}$ and $a^2$:
$\frac{A_H}{A_T} = \frac{6 \times \frac{1}{9}}{1} = \frac{6}{9} = \frac{2}{3}$
Thus, the ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is $\frac{2}{3}$.

This analogy question tests our understanding of relationships between words. The core relationship here is one of temporal direction.

1. Analyze the given pair:

   • Nostalgia is a feeling focused on the past. We experience nostalgia for something that has already happened.
   • Anticipation is a feeling focused on the future. We anticipate something that is going to happen.
   • So, the relationship is: Past-oriented emotion : Future-oriented emotion.

2. Evaluate the options based on this temporal direction:

   • A. Present : Past $-$ This doesn't match the "past : future" directional flow.
   • B. Future : Past $-$ This is the reverse of our required "past : future" flow.
   • C. Past : Future $-$ This directly mirrors the "past-oriented : future-oriented" relationship established by Nostalgia : Anticipation.
   • D. Future : Present $-$ This doesn't match the "past : future" directional flow.

3. Conclusion: The option that maintains the same logical relation (Past-oriented : Future-oriented) is C. Past, future. Therefore, Nostalgia is to anticipation as Past is to Future.

To find the number of days where one student spent at least $10\%$ more time exercising than the other, we calculate the percentage increase with respect to the smaller value for each day.

We use the formula: Percentage Increase $= \frac{\text{Difference}}{\text{Smaller Value}} \times 100$.

• Sunday: $(\frac{65 - 55}{55}) \times 100 = 18.18\%$
• Monday: $(\frac{70 - 45}{45}) \times 100 = 55.55\%$
• Tuesday: $(\frac{65 - 55}{55}) \times 100 = 18.18\%$
• Wednesday: $(\frac{60 - 50}{50}) \times 100 = 20\%$
• Thursday: $(\frac{60 - 55}{55}) \times 100 = 9.09\%$
• Friday: $(\frac{35 - 20}{20}) \times 100 = 75\%$
• Saturday: $(\frac{60 - 50}{50}) \times 100 = 20\%$

Comparing these percentages to $10\%$, we find that on Sunday, Monday, Tuesday, Wednesday, Friday, and Saturday, the increase is $10\%$ or more. Thus, there are 6 such days.

To identify the grammatically correct sentences, we need to understand the principal forms of the verb 'wake': Base Form: wake; Past Simple: woke; Past Participle: woken.

• Sentence (i) "I woke up from sleep." uses 'woke', which is the correct simple past tense of 'wake', making it grammatically correct.
• Sentence (ii) "I woked up from sleep." uses 'woked', an incorrect past tense form; the correct form is 'woke', so this sentence is grammatically incorrect.
• Sentence (iii) "I was woken up from sleep." correctly uses 'was woken', which is the past tense passive voice structure with the correct past participle 'woken', making it grammatically correct.
• Sentence (iv) "I was wokened up from sleep." uses 'wokened', which is an incorrect past participle form; the correct form is 'woken', so this sentence is grammatically incorrect.

Thus, sentences (i) and (iii) are grammatically correct.

To make the line $P-Q$ a line of symmetry, the figure must be identical on both sides of the line. We need to add squares to one side such that it mirrors the existing squares on the other side. By visually inspecting the given figure and adding squares to complete this mirror image, we find that a total of $6$ squares must be added to achieve symmetry.