Past Questions — 2025

The relationship in the original pair is 'dwelling : inhabitant'—an aerie is the nest or home of an eagle. Similarly, a house is the dwelling or home of a person. Option 'a' (capital : government) represents a seat of power rather than a physical dwelling. Option 'b' (buffalo : lion) represents a prey-predator relationship. Option 'c' (unit : apartment) represents a part-to-whole relationship.

Step 1: Analyze the relationship between the first pair of words. 'Alleviate' means to make a problem or pain less severe, while 'Aggravate' means to make a problem or injury worse. Therefore, Alleviate and Aggravate are antonyms (opposites). Step 2: Apply the same relationship to the second pair. The word 'Elastic' refers to a material or object that is flexible and can return to its original shape after being stretched or compressed. Step 3: Evaluate the options to find the antonym of 'Elastic': - (a) Flexible: This is a synonym of elastic. - (b) Rigid: This means stiff, unyielding, or unable to bend. This is the direct opposite of elastic. - (c) Strong: This refers to power or durability, not necessarily the ability to bend or stretch. - (d) Malleable: This refers to the ability to be hammered or pressed permanently out of shape, which is a form of flexibility, not the opposite. Conclusion: Since the first pair are antonyms, the second pair must also be antonyms. The opposite of Elastic is Rigid.

1. Find the side of the square park: Since the area of the square park is 169 m², its side length is √169 = 13 m. 2. Find the side of the outer square (park + path): The path has a uniform width of 2 m all around, so we add 2 m to both sides of the park's length. Outer side = 13 + 2 + 2 = 17 m. 3. Find the total area including the path: Outer Area = 17² = 289 m². 4. Find the area of the path: Area of the path = Outer Area - Area of the park = 289 - 169 = 120 m². Other options (a, c, and d) are incorrect because they do not equal the calculated path area of 120 m².

To find the number of letters between the two specified positions: 1. The 20th letter from the left in the English alphabet is T (position 20). 2. The 18th letter from the right corresponds to the 9th letter from the left (since 27 - 18 = 9), which is I (position 9). 3. The letters between the 9th letter (I) and the 20th letter (T) are those from position 10 to 19 (J, K, L, M, N, O, P, Q, R, S). 4. The count of these letters is 20 - 9 - 1 = 10. Therefore, option d (10) is correct. Options a (6), b (12), and c (5) are incorrect because they do not represent the correct count of letters between I and T.

Step-by-step solution: 1. **Find the percentage of failures**: Since 78% of the students were successful, the percentage of students who failed is 100% - 78% = 22%. 2. **Determine the total number of students**: We are given that the total number of failures is 176. Therefore, 22% of the total students (T) = 176. Calculation: 0.22 * T = 176 T = 176 / 0.22 T = 800 students. 3. **Calculate the number of students who got first class**: The problem states that 34% of the total students got first class. Calculation: 34% of 800 = 0.34 * 800 0.34 * 800 = 272 students. Verification of options: - Option (a) 272: Matches our calculation. - Option (b) 112: Incorrect. - Option (c) 210: Incorrect. - Option (d) 254: Incorrect.

To find the number of ways the letters of the word 'APPLE' can be arranged, we use the formula for permutations of a multiset: n! / (n1! * n2! * ... * nk!), where 'n' is the total number of items and 'n1, n2...' are the frequencies of the repeating items. 1. Count the total number of letters in 'APPLE': There are 5 letters (A, P, P, L, E). So, n = 5. 2. Identify repeating letters: The letter 'P' appears 2 times. All other letters (A, L, E) appear only once. 3. Apply the formula: Total arrangements = 5! / 2! 5! (5 factorial) = 5 × 4 × 3 × 2 × 1 = 120 2! (2 factorial) = 2 × 1 = 2 4. Calculate the final result: 120 / 2 = 60. Therefore, there are 60 distinct ways to arrange the letters. Option (b) is correct. Option (d) 120 would be correct if all letters were unique, but since there are two 'P's, we must divide by 2! to avoid double-counting identical arrangements.

Step-by-step reasoning: 1. Identify the standard English alphabet sequence: A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22), W(23), X(24), Y(25), Z(26). 2. Understand the switching rule: The letters are swapped in pairs (1st with 2nd, 3rd with 4th, etc.). This means for any even position 'n', the letter at that position in the new series is the letter that was at position 'n-1' in the original series. 3. Apply the rule to the 20th letter: Since 20 is an even number, the 20th letter in the new series will be the letter that was originally in the 19th position. 4. Identify the 19th letter of the standard English alphabet: The 19th letter is 'S'. 5. Verification: - 1st (A) ↔ 2nd (B) - 3rd (C) ↔ 4th (D) - ... - 19th (S) ↔ 20th (T) In the new series, the 19th position is occupied by T and the 20th position is occupied by S. Therefore, the 20th letter from the left is S.

Let's break down the relationships step-by-step: 1. **Lura and Liani**: Lura is the only son of Liani. (Liani is the mother, Lura is the son). 2. **Liani and Lawma**: Liani is the only daughter of Lawma. (Lawma is the grandparent of Lura). 3. **Sawmi and Liani**: Sawmi is the daughter-in-law of Liani. Since Lura is Liani's only son, Sawmi must be married to Lura. 4. **Sawmi and Lala**: Sawmi is the mother of Lala. Since Sawmi is married to Lura, Lala is the child of Lura and Sawmi. 5. **Lala and Liani**: Liani is the mother of Lura, and Lura is the father of Lala. Therefore, Lala is the grandchild of Liani. Evaluating the options: - (a) Son: Incorrect. Lura is the son of Liani. - (b) Grandson: Correct. Lala is the child of Liani's son (Lura). While the gender of Lala is not explicitly stated, in the context of standard competitive exam blood relation questions, 'Grandson' is the only viable familial link provided among the options that fits the generational gap. - (c) Husband: Incorrect. Lala is the child of Liani's daughter-in-law. - (d) None of these: Incorrect, as 'Grandson' fits the relationship.

Let's map the movements on a coordinate plane where 'M' (the capital) is the origin (0, 0). 1. **M to A**: 'A' is 4 km North-West of 'M'. In a coordinate system, North-West is a 45-degree angle. X-coordinate of A = -4 * cos(45°) = -4 * (1/√2) = -2√2 Y-coordinate of A = 4 * sin(45°) = 4 * (1/√2) = 2√2 2. **A to B**: 'B' is 4 km South-West of 'A'. South-West is a 45-degree angle downward. X-coordinate of B = X_A - 4 * cos(45°) = -2√2 - 2√2 = -4√2 Y-coordinate of B = Y_A - 4 * sin(45°) = 2√2 - 2√2 = 0 3. **B to E**: 'E' is 4 km North-West of 'B'. X-coordinate of E = X_B - 4 * cos(45°) = -4√2 - 2√2 = -6√2 Y-coordinate of E = Y_B + 4 * sin(45°) = 0 + 2√2 = 2√2 4. **E to C**: 'C' is 2 km South-West of 'E'. X-coordinate of C = X_E - 2 * cos(45°) = -6√2 - (2 * 1/√2) = -6√2 - √2 = -7√2 Y-coordinate of C = Y_E - 2 * sin(45°) = 2√2 - (2 * 1/√2) = 2√2 - √2 = √2 Now, let's determine the position of 'C' (-7√2, √2) relative to 'M' (0, 0): - The X-coordinate is negative (-7√2), meaning it is to the West. - The Y-coordinate is positive (√2), meaning it is slightly to the North. However, looking at the movement pattern: - M -> A (NW) - A -> B (SW) - B -> E (NW) - E -> C (SW) Total displacement in X (Westward) = 4cos45 + 4cos45 + 4cos45 + 2cos45 = 14 * (1/√2) = 7√2 km West. Total displacement in Y (North/South) = 4sin45 - 4sin45 + 4sin45 - 2sin45 = 2 * (1/√2) = √2 km North. Since the westward displacement (7√2 ≈ 9.9) is significantly larger than the northward displacement (√2 ≈ 1.4), the point C is located almost directly West of M. In standard competitive exam direction sense questions of this type, when a point is significantly shifted in one cardinal direction with a minor offset, or when following a zig-zag pattern that returns toward the axis, we evaluate the dominant direction. Wait, let's re-verify the coordinates: C is at (-7√2, √2). This is technically North-West. However, let's check the options. If we look at the net movement: North movements: 4km (NW) + 4km (NW) = 8km total diagonal North components. South movements: 4km (SW) + 2km (SW) = 6km total diagonal South components. Net North = 2km diagonal North component. Net West = 4+4+4+2 = 14km diagonal West components. Since the net result is West and slightly North, the direction is North-West. Let's re-read: - M to A: NW - A to B: SW (B is now due West of M because the North component of MA is cancelled by the South component of AB) - B to E: NW - E to C: SW From B (which is West of M), we go NW then SW. B -> E (NW) adds North and West. E -> C (SW) adds South and West. Since the distance B->E (4km) is greater than E->C (2km), the North component is not fully cancelled. Net position of C relative to B: West and slightly North. Since B is already West of M, C is West and slightly North of M. Comparing options: North, North-West, West, South-West. Technically, C is North-West of M. Let's double check the math: Net Y = 4sin45 - 4sin45 + 4sin45 - 2sin45 = 2sin45 = √2 (Positive = North) Net X = -4cos45 - 4cos45 - 4cos45 - 2cos45 = -14cos45 = -7√2 (Negative = West) Direction is North-West. Correction: Based on the coordinates (-7√2, √2), the direction is North-West. Wait, let's re-calculate: M(0,0) A(-2√2, 2√2) B(-4√2, 0) E(-6√2, 2√2) C(-7√2, √2) C is in the second quadrant (X negative, Y positive), which is North-West. Correct Option: b. North-West

Let Liana's present age be $L$ and his son's present age be $S$. 1. **Ten years ago:** Liana was four times as old as his son. $$L - 10 = 4(S - 10)$$ $$L - 4S = -30 \quad \text{--- (Equation 1)}$$ 2. **Ten years hence:** Liana will be twice as old as his son. $$L + 10 = 2(S + 10)$$ $$L - 2S = 10 \quad \text{--- (Equation 2)}$$ Subtracting Equation 1 from Equation 2: $$(L - 2S) - (L - 4S) = 10 - (-30)$$ $$2S = 40 \implies S = 20$$ Substitute $S = 20$ into Equation 2: $$L - 2(20) = 10 \implies L = 50$$ Thus, the present ages of Liana and his son are 50 and 20 respectively, which corresponds to option **c**. * **Why other options are incorrect:** * **Option a (55 & 25):** 10 years ago, their ages would be 45 and 15. 45 is not 4 times 15. * **Option b (50 & 25):** 10 years ago, their ages would be 40 and 15. 40 is not 4 times 15. * **Option d (55 & 20):** 10 years ago, their ages would be 45 and 10. 45 is not 4 times 10.

Step-by-step solution: 1. Let the Selling Price (SP) be denoted as 'S'. 2. According to the question, the loss is 1/3 of the selling price. Therefore, Loss = (1/3)S. 3. We know the formula for Cost Price (CP) when there is a loss: CP = Selling Price + Loss. 4. Substituting the values: CP = S + (1/3)S = (4/3)S. 5. The formula for Loss Percentage is: (Loss / Cost Price) × 100. 6. Substituting the values into the formula: Loss% = [(1/3)S / (4/3)S] × 100. 7. The 'S' and the denominators '3' cancel out: Loss% = (1/4) × 100. 8. Loss% = 25%. Comparison with options: - Option a (15%), b (20%), and c (30%) are incorrect because they do not satisfy the mathematical relationship between the loss and the cost price derived from the given condition.

Step-by-step solution: 1. Let the Cost Price (CP) of one article be 'x'. 2. Therefore, the Cost Price of 18 articles = 18x. 3. According to the problem, the Selling Price (SP) of 15 articles is equal to the Cost Price of 18 articles. So, SP of 15 articles = 18x. 4. To find the profit or loss, we need to compare the CP and SP of the same number of articles. Let's find the SP of one article: SP of 1 article = 18x / 15 = 1.2x. 5. Now we compare CP and SP for one article: CP = x SP = 1.2x Since SP > CP, it is a profit. 6. Profit = SP - CP = 1.2x - x = 0.2x. 7. Profit Percentage = (Profit / CP) * 100 Profit % = (0.2x / x) * 100 = 0.2 * 100 = 20%. Alternative Shortcut Method: Profit % = [(Difference in quantity) / (Quantity Sold)] * 100 Profit % = [(18 - 15) / 15] * 100 Profit % = (3 / 15) * 100 = (1 / 5) * 100 = 20%.

Step-by-step solution: 1. **Understand the Problem**: We need to find the side of the largest possible square slab that can perfectly pave a rectangular floor. This means the side of the square must be a common divisor of both the length and the breadth of the room. To find the 'largest' such square, we must calculate the Highest Common Factor (HCF) of the dimensions. 2. **Convert Units to a Single Standard**: To avoid calculation errors, convert all dimensions to centimeters (cm). - Length = 2m 50cm = (2 × 100) + 50 = 250 cm - Breadth = 1m 50cm = (1 × 100) + 50 = 150 cm 3. **Find the HCF of 250 and 150**: - Prime factorization of 250: 2 × 5 × 5 × 5 (or 2 × 5³) - Prime factorization of 150: 2 × 3 × 5 × 5 (or 2 × 3 × 5²) - The common factors are 2 and 5² (5 × 5). - HCF = 2 × 5 × 5 = 50 cm. 4. **Verification**: - 250 ÷ 50 = 5 (fits exactly 5 times along the length) - 150 ÷ 50 = 3 (fits exactly 3 times along the breadth) Conclusion: The side of the largest square slab is 50 cm. - Option (a) 25 cm is a common factor, but not the highest. - Option (b) 30 cm does not divide 250 evenly. - Option (d) 72 cm does not divide either dimension evenly.

Step-by-step solution: 1. **Identify the given values**: - Diameter of the spherical ball (D) = 21 cm. - Radius of the spherical ball (r) = D / 2 = 21 / 2 = 10.5 cm. - Side of each cube (a) = 1 cm. 2. **Calculate the volume of the spherical ball**: The formula for the volume of a sphere is V = (4/3)πr³. Using π ≈ 22/7: V_sphere = (4/3) * (22/7) * (21/2) * (21/2) * (21/2) V_sphere = (4/3) * (22/7) * (9261 / 8) V_sphere = (4 * 22 * 9261) / (3 * 7 * 8) V_sphere = 813408 / 168 V_sphere = 4841.25 (approx) or more precisely: V_sphere = (4/3) * (22/7) * (10.5)³ = 4851 cm³. Calculation check: (4/3) * (22/7) * (21/2)^3 = (4/3) * (22/7) * (9261/8) = (22 * 9261) / (3 * 7 * 2) = 203742 / 42 = 4851 cm³. 3. **Calculate the volume of one cube**: The formula for the volume of a cube is V = a³. V_cube = 1³ = 1 cm³. 4. **Find the number of cubes**: Number of cubes = Total Volume of Sphere / Volume of one Cube Number of cubes = 4851 / 1 = 4851. Therefore, the number of cubes formed is 4851, which corresponds to option (d).

In Microsoft Excel, referencing a cell from a different worksheet requires a specific syntax: 'SheetName!CellAddress'. 1. **Analysis of Option (a):** `=B1+Sheet2!A1` correctly uses the addition operator (+) and the exclamation mark (!) to separate the sheet name (Sheet2) from the cell reference (A1). This is the standard syntax for cross-sheet calculations. 2. **Analysis of Option (b):** `=B1+A1(Sheet2)` is syntactically incorrect. Excel does not use parentheses to denote sheet names; it would interpret this as an attempt to call a function named 'A1', which does not exist. 3. **Analysis of Option (c):** `=B1&Sheet2!A1` uses the ampersand (&) operator. In Excel, the ampersand is used for concatenation (joining text together), not for mathematical addition. This would result in the two values being displayed side-by-side as text (e.g., if B1=10 and Sheet2!A1=5, the result would be '105'). 4. **Analysis of Option (d):** `=SUM(B1,Sheet2A1)` is incorrect because it lacks the exclamation mark (!) between the sheet name and the cell reference. Excel would treat 'Sheet2A1' as an invalid named range or a typo, resulting in a #NAME? error.

When copying a chart from Microsoft Excel and pasting it into another MS Office application (like Word or PowerPoint), the 'Paste Link' option creates a dynamic connection between the source file (Excel) and the destination document. If the data in the original Excel spreadsheet is modified, the linked chart in the destination document will automatically update to reflect those changes. Here is why the other options are incorrect: - a. Paste as Picture: This converts the chart into a static image. It cannot be edited or updated if the source data changes. - b. Embed Object: Embedding places a copy of the data into the destination file. While you can edit the embedded data using Excel tools within the document, it is no longer connected to the original external Excel file; changes to the original file will not update the embedded object. - d. Paste Values: This option is typically used for cells rather than charts; it strips away all formatting and formulas, leaving only the raw text or numbers, which are static.

In Microsoft Word, the shortcut key Ctrl + H is specifically designed to open the 'Find and Replace' dialog box, allowing the user to search for a specific word or phrase and replace it with another throughout the document. Analysis of other options: - Option (a) Copy: The shortcut key for Copy is Ctrl + C. - Option (b) Find: The shortcut key for Find is Ctrl + F. - Option (d) Highlight: There is no standard single-combination Ctrl shortcut for highlighting text; this is typically done via the ribbon menu or mouse selection.

A web cookie (HTTP cookie) is a small piece of data that a server sends to a user's web browser. The browser stores this data and sends it back to the same server with every subsequent request. This allows websites to 'remember' stateful information, such as items in a shopping cart, user login sessions, or personalized site preferences. Analysis of options: - Option (a) is correct because it accurately describes the flow: Website (Server) -> Web Browser (Client storage). - Option (b) is incorrect because cookies are primarily stored on the client-side (browser), not the server-side (though server-side sessions exist, the 'cookie' itself resides with the user). - Option (c) and (d) are incorrect as they describe server-to-server communication or DNS root server operations, which have nothing to do with the mechanism of web cookies.

Computer processing speed refers to the clock speed of the Central Processing Unit (CPU), which is the number of cycles a CPU can execute per second. This is measured in Hertz (Hz). Modern processors operate at billions of cycles per second, which is expressed as Gigahertz (GHz). Why other options are incorrect: - a. Bits per second (bps): This is a measure of data transfer rate or network bandwidth, not processing speed. - b. Dots Per Inch (DPI): This is a measure of spatial printing or image scanner resolution. - d. Megabytes per sec (MB/s): This is a measure of data throughput or storage read/write speed (e.g., for a hard drive or SSD), not the CPU's processing speed.

In the Windows Operating System, the Recycle Bin acts as a temporary storage area for files deleted from the local internal storage. 1. **Internal Storage (Options a, b, and c):** Files deleted from the Desktop, the internal hard disk (C: drive, etc.), and the Documents folder are moved to the Recycle Bin. This allows users to restore them if they were deleted by mistake. 2. **Removable Storage (Option d):** When a file is deleted from a removable drive, such as a USB flash drive, an SD card, or a network drive, Windows does not move the file to the Recycle Bin. Instead, the file is permanently deleted from the file system immediately. This is because removable drives are designed for portability and may not have a dedicated hidden system folder (like $Recycle.Bin) configured to manage deleted files across different computers. Therefore, files deleted from removable drives bypass the Recycle Bin entirely.

In the context of GUI (Graphical User Interface) and User Experience (UX) design, 'affordance' is a term coined by James J. Gibson and later popularized by Don Norman. It refers to the properties of an object that indicate to the user how it can be interacted with. For example, a button that looks 'raised' or 3D suggests that it can be clicked (pushed), and a slider suggests that it can be dragged. Analysis of other options: - Option (a) refers to 'Flexibility' or 'Redundancy' in design. - Option (b) refers to 'Feedback', which is a separate principle ensuring the user knows the result of an action. - Option (c) refers to 'Consistency', which ensures that similar elements behave the same way across the system. Therefore, option (d) is the only correct definition of affordance.

The given formula is =IF(AND(A1>50,B1<100), “Pass”,”Fail”). 1. The AND function is a logical operator that returns TRUE only if all its arguments are true. In this case, the two conditions are (A1 > 50) and (B1 < 100). 2. The IF function follows the syntax: =IF(logical_test, value_if_true, value_if_false). 3. Therefore, the formula will return 'Pass' (the value_if_true) only if the AND condition is satisfied. 4. For the AND condition to be true, A1 must be strictly greater than 50 AND B1 must be strictly less than 100. Evaluating the options: - Option (a) is the opposite of the required conditions. - Option (b) matches the logical requirements of the AND function exactly. - Option (c) is incorrect because if A1=100 and B1=100, the condition B1<100 becomes false, resulting in 'Fail'.

In Microsoft Excel, the 'Protect Sheet' feature is used to prevent unauthorized changes to the contents of a worksheet. By default, all cells in an Excel sheet are marked as 'Locked'. However, this locking mechanism only takes effect once the 'Protect Sheet' command is activated. To allow changes in specific cells while restricting others, a user must first select the cells they want to remain editable, go to Format Cells > Protection, and uncheck the 'Locked' box. After this, applying 'Protect Sheet' (found under the Review tab) will lock all cells except those that were explicitly unlocked. Why other options are incorrect: - a. Freeze Sheet: This is not a standard Excel term. 'Freeze Panes' exists, but it relates to visibility, not editing restrictions. - c. Data Validation: This feature restricts the *type* of data or values that can be entered into a cell (e.g., only numbers between 1 and 10), but it does not prevent a user from editing the cell entirely. - d. Lock Panes: This is a misnomer for 'Freeze Panes', which keeps specific rows or columns visible while scrolling through the rest of the sheet; it has no impact on editing permissions.

In Microsoft Word, the keyboard shortcut 'Ctrl + Shift + N' is specifically designed to apply the 'Normal' style to the selected text or the current paragraph. This resets the formatting of the text to the default style defined in the document's style gallery. Analysis of other options: - Option (a) 'New page': The shortcut for a Page Break is 'Ctrl + Enter'. - Option (b) 'Numbered list': There is no single default global shortcut for numbered lists, though they can be managed via the Ribbon or custom shortcuts. - Option (d) 'New document': The shortcut to create a new blank document is 'Ctrl + N'.

In MS Word, these two features serve different structural purposes: 1. **Page Break**: This simply tells Word to end the current page and start a new one. The formatting (margins, orientation, headers/footers) remains consistent across the break. It is used when you want to start a new chapter or a new page without changing the layout. 2. **Section Break (Next Page)**: This not only starts a new page but also creates a 'Section'. Sections allow you to apply different formatting to different parts of the same document. For example, you can have Page 1 in Portrait orientation and Page 2 in Landscape orientation, or have different headers and footers for different chapters, only by using Section Breaks. **Analysis of other options:** - **Option a** is incorrect because they have fundamentally different impacts on document structure. - **Option c** is incorrect because neither a Page Break nor a Section Break deletes content; they only manage the placement of content. - **Option d** is incorrect because Page Breaks have nothing to do with headers specifically, and Section Breaks are used for much more than just footnotes (e.g., columns, margins, and page numbering).