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Aptitude and Logical Reasoning Interview Questions & Answers
43 questions with detailed answers — for freshers and experienced candidates.
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Q1. What comes next in the series: 2, 5, 10, 17, 26, ?
This is a number series question. The pattern involves adding consecutive odd numbers to the previous term. Starting from 2, we add 3 to get 5, then add 5 to get 10, then add 7 to get 17, and add 9 to get 26. The next odd number in sequence after 9 is 11. Therefore, the next term in the series will be 26 + 11 = 37. This question assesses basic pattern recognition and arithmetic skills, which are fundamental for problem-solving and understanding sequences in data. Identifying such patterns quickly is a key aptitude.
Q2. If a car travels at 60 km/h for the first hour and 80 km/h for the next two hours, what is its average speed for the entire journey?
To calculate the average speed, we need the total distance traveled and the total time taken. For the first hour, distance = 60 km/h * 1 hour = 60 km. For the next two hours, distance = 80 km/h * 2 hours = 160 km. The total distance is 60 km + 160 km = 220 km. The total time taken is 1 hour + 2 hours = 3 hours. Average speed = Total Distance / Total Time = 220 km / 3 hours = 73.33 km/h (approximately). This problem tests understanding of speed, distance, and time relationships, and the concept of weighted average.
Q3. A shopkeeper sells an item for $120, making a profit of 20%. What was the cost price of the item?
Let the cost price (CP) be 'x'. The selling price (SP) is $120, and the profit is 20% of the cost price. So, SP = CP + (Profit Percentage * CP / 100). This can be written as $120 = x + (20/100)x = x + 0.2x = 1.2x. To find x, we divide the selling price by 1.2. So, x = $120 / 1.2 = $100. The cost price of the item was $100. This question evaluates basic percentage and profit/loss calculations, essential for business logic.
Q4. In a certain code, 'APPLE' is written as 'BQQMF'. How would 'ORANGE' be written in that code?
Let's analyze the pattern for 'APPLE' to 'BQQMF'. Each letter in 'APPLE' is shifted one position forward in the alphabet to get the corresponding letter in 'BQQMF'. A becomes B, P becomes Q, P becomes Q, L becomes M, E becomes F. Applying this same rule to 'ORANGE': O becomes P, R becomes S, A becomes B, N becomes O, G becomes H, E becomes F. So, 'ORANGE' would be written as 'PSBOHF'. This tests basic logical pattern recognition and character manipulation.
Q5. If 6 men can complete a piece of work in 10 days, how many days will 15 men take to complete the same work?
This is an inverse proportion problem. The amount of work is constant. If more men work, it will take fewer days. Let 'M' be the number of men and 'D' be the number of days. The total 'man-days' required for the work is constant. So, M1 * D1 = M2 * D2. Given M1 = 6 men, D1 = 10 days. We need to find D2 when M2 = 15 men. So, 6 * 10 = 15 * D2. 60 = 15 * D2. D2 = 60 / 15 = 4 days. Therefore, 15 men will take 4 days to complete the work. This problem tests understanding of inverse proportionality.
Q6. Find the missing number: 4, 9, 16, 25, ?, 49.
This series consists of perfect squares. The numbers are 2^2, 3^2, 4^2, 5^2. Following this pattern, the next number should be 6^2. And the number after that is 7^2 (which is 49), confirming the pattern. So, 6^2 = 36. The missing number is 36. This question assesses basic number pattern recognition, specifically identifying squares of integers, which is a common type of numerical reasoning problem.
Q7. A train 120 meters long passes a pole in 6 seconds. What is the speed of the train in km/h?
When a train passes a pole, the distance covered is equal to the length of the train. So, distance = 120 meters. Time taken = 6 seconds. Speed = Distance / Time = 120 meters / 6 seconds = 20 meters/second. To convert meters/second to km/h, we multiply by (18/5). So, speed = 20 * (18/5) = 4 * 18 = 72 km/h. This problem requires unit conversion and understanding of basic speed, distance, time calculations, which are common in real-world applications.
Q8. Three friends, A, B, and C, divide a sum of money in the ratio 5:7:8. If B gets $210, what is the total sum of money?
The ratio of shares for A:B:C is 5:7:8. Let the common multiplier for the shares be 'x'. So, A's share is 5x, B's share is 7x, and C's share is 8x. We are given that B gets $210, so 7x = 210. From this, we can find x = 210 / 7 = 30. The total sum of money is the sum of all shares: 5x + 7x + 8x = 20x. Substituting x = 30, the total sum is 20 * 30 = $600. This question tests understanding of ratios and proportional distribution.
Q9. What is the smallest number that is divisible by 12, 15, and 20?
This question asks for the Least Common Multiple (LCM) of 12, 15, and 20. To find the LCM, we can use prime factorization. Prime factors of 12 = 2^2 * 3. Prime factors of 15 = 3 * 5. Prime factors of 20 = 2^2 * 5. To find the LCM, we take the highest power of all prime factors present: 2^2 * 3^1 * 5^1 = 4 * 3 * 5 = 60. So, the smallest number divisible by 12, 15, and 20 is 60. This problem assesses fundamental number theory concepts, important for tasks like scheduling or resource allocation.
Q10. If you are the 10th person in a queue from the front and the 15th person from the back, how many people are in the queue?
This is a simple positional reasoning problem. If you are the 10th from the front, there are 9 people ahead of you. If you are the 15th from the back, there are 14 people behind you. To find the total number of people, we sum the people ahead of you, the people behind you, and yourself. Total people = (People ahead) + (People behind) + 1 (for yourself) = 9 + 14 + 1 = 24 people. Alternatively, you can sum the front position and back position and subtract 1 (since you are counted twice): 10 + 15 - 1 = 24. This tests basic logical counting.
Q11. If the day before yesterday was Thursday, what day will be the day after tomorrow?
This is a temporal reasoning question. If the day before yesterday was Thursday, then yesterday was Friday. This means today is Saturday. The day after today will be Sunday. Therefore, the day after tomorrow will be Monday. This tests sequential thinking and understanding of time relationships. It's a straightforward logical deduction problem often used to check basic comprehension and focus.
Q12. Three friends, X, Y, and Z, are sitting in a row. X is to the left of Y. Z is to the right of Y. Who is in the middle?
Let's place the friends based on the given information. 'X is to the left of Y' means the order starts with X then Y (X Y). 'Z is to the right of Y' means Y is followed by Z (Y Z). Combining these two, we get the order X Y Z. Therefore, Y is in the middle. This is a basic linear arrangement problem testing logical placement.
Q13. Find the next term in the series: A, C, F, J, O, ?
This is an alphabetical series. Let's look at the difference in position of each letter in the alphabet: A (1), C (3), F (6), J (10), O (15). The differences between consecutive terms are: C-A = 3-1 = 2. F-C = 6-3 = 3. J-F = 10-6 = 4. O-J = 15-10 = 5. The differences are increasing by 1 each time (2, 3, 4, 5). The next difference should be 6. So, the next letter will be the 15 + 6 = 21st letter of the alphabet. The 21st letter is U. This tests pattern recognition in alphabetical sequences.
Q14. A man walks 5 km East, then turns South and walks 12 km. What is his shortest distance from the starting point?
This problem involves direction and distance, forming a right-angled triangle. The man walks 5 km East (one leg of the triangle) and then 12 km South (the other leg). The shortest distance from the starting point is the hypotenuse of this right-angled triangle. Using the Pythagorean theorem (a^2 + b^2 = c^2): Distance^2 = 5^2 + 12^2 = 25 + 144 = 169. Distance = sqrt(169) = 13 km. The shortest distance from his starting point is 13 km. This tests spatial reasoning and application of the Pythagorean theorem.
Q15. A father is twice as old as his son. 20 years ago, the father was four times as old as his son. What are their current ages?
Let the son's current age be S and the father's current age be F. From the first statement, F = 2S. 20 years ago, the son's age was S - 20, and the father's age was F - 20. From the second statement, F - 20 = 4(S - 20). Now substitute F = 2S into the second equation: 2S - 20 = 4S - 80. Rearranging the terms: 80 - 20 = 4S - 2S. 60 = 2S. So, S = 30. The son's current age is 30 years. The father's current age is F = 2S = 2 * 30 = 60 years. This problem tests algebraic formulation and solution of age-related word problems.
Q16. If 'water' is called 'food', 'food' is called 'tree', 'tree' is called 'sky', 'sky' is called 'wall', where do birds fly?
This is a coding-decoding type of verbal reasoning question. Birds normally fly in the 'sky'. According to the given code, 'sky' is called 'wall'. Therefore, in this coded language, birds fly in the 'wall'. The trick is to identify the actual action's location and then apply the one-step transformation. This tests careful reading and direct application of a given rule, without overthinking or applying real-world logic beyond the code.
Q17. If it takes 5 machines 5 minutes to make 5 widgets, how long would it take 100 machines to make 100 widgets?
This is a classic logical puzzle. The key is to understand the rate of work per machine. If 5 machines make 5 widgets in 5 minutes, it implies that each machine makes 1 widget in 5 minutes. The number of machines and widgets scale proportionally. Therefore, if one machine takes 5 minutes to make one widget, then 100 machines, working independently and in parallel, would also take 5 minutes to make 100 widgets (each machine making one widget). The time required per widget per machine remains constant. This tests proportional reasoning and avoiding common pitfalls.
Q18. In a race, A is 10 meters ahead of B, and B is 5 meters ahead of C. If A, B, and C run at constant speeds, and A finishes the race, what is the distance between B and C when A finishes?
The key here is that they run at constant speeds. The relative distance between B and C remains constant throughout the race, provided their speeds are the same. The problem states that B is 5 meters ahead of C. This relative distance will be maintained even when A finishes the race, assuming B and C continue running at their constant speeds. Therefore, when A finishes, the distance between B and C will still be 5 meters. This tests understanding of relative speeds and distances in a simplified scenario.
Intermediate Level
Q1. A sum of money doubles itself in 5 years at simple interest. In how many years will it become four times itself?
Let the principal amount be P. If it doubles in 5 years, the interest earned (I) is P. Using the simple interest formula I = (P * R * T) / 100, we have P = (P * R * 5) / 100. This simplifies to 1 = (R * 5) / 100, so R = 20% per annum. Now, we want the amount to become four times itself, meaning the final amount is 4P, and the interest earned is 3P. So, 3P = (P * 20 * T) / 100. This simplifies to 3 = (20 * T) / 100, or 3 = T / 5. Thus, T = 15 years. This problem tests understanding of simple interest concepts and proportional reasoning.
Q2. A bag contains 5 red balls and 3 blue balls. If two balls are drawn at random without replacement, what is the probability that both are red?
To find the probability that both balls drawn are red without replacement, we calculate the probability of each event sequentially. Initially, there are 8 balls in total (5 red + 3 blue). The probability of drawing the first red ball is 5/8. After drawing one red ball, there are now 4 red balls left and a total of 7 balls. The probability of drawing a second red ball is 4/7. The probability of both events occurring is the product of their individual probabilities: P(both red) = (5/8) * (4/7) = 20/56 = 5/14. This problem assesses understanding of conditional probability and combinations.
Q3. If 'A+B' means 'A is the father of B', 'A-B' means 'A is the wife of B', 'A*B' means 'A is the brother of B', and 'A/B' means 'A is the daughter of B', how is P related to T in the expression 'P+Q-R*S/T'?
Let's break down the expression: P+Q means P is the father of Q. Q-R means Q is the wife of R. Therefore, R is Q's husband, and P is R's father-in-law. R*S means R is the brother of S. So, S is R's sibling. S/T means S is the daughter of T. Since S is R's sibling and R's father is P, and S is T's daughter, this implies T is S's mother, and thus R's mother. Therefore, P is T's husband. So, P is the husband of T. This question tests logical deduction based on defined relationships.
Q4. You have a 3-liter jug and a 5-liter jug. How can you measure exactly 4 liters of water?
This is a classic water jug puzzle. Here's a sequence of steps: 1. Fill the 5-liter jug completely. (Jugs: 3L=0, 5L=5) 2. Pour water from the 5-liter jug into the 3-liter jug until the 3-liter jug is full. (Jugs: 3L=3, 5L=2) 3. Empty the 3-liter jug. (Jugs: 3L=0, 5L=2) 4. Pour the 2 liters from the 5-liter jug into the 3-liter jug. (Jugs: 3L=2, 5L=0) 5. Fill the 5-liter jug completely. (Jugs: 3L=2, 5L=5) 6. Pour water from the 5-liter jug into the 3-liter jug until it's full. Since the 3-liter jug already has 2 liters, you will pour 1 liter from the 5-liter jug. (Jugs: 3L=3, 5L=4) You now have exactly 4 liters in the 5-liter jug. This tests problem-solving through sequential logic.
Q5. A and B can do a piece of work in 10 days, B and C in 12 days, and C and A in 15 days. In how many days can C alone do it?
Let the work done by A, B, C in one day be a, b, c respectively. (a+b) = 1/10, (b+c) = 1/12, (c+a) = 1/15. Adding these equations: 2(a+b+c) = 1/10 + 1/12 + 1/15 = (6+5+4)/60 = 15/60 = 1/4. So, (a+b+c) = 1/8 (A, B, C together complete 1/8 of the work in one day). To find C's work rate, subtract (a+b) from (a+b+c): c = (a+b+c) - (a+b) = 1/8 - 1/10 = (5-4)/40 = 1/40. Therefore, C alone can complete the work in 40 days. This tests system of equations and work-rate problems.
Q6. A clock shows 4:30. If the minute hand points East, in what direction will the hour hand point?
At 4:30, the minute hand points exactly at 6. If the minute hand (which is at 6) points East, then 6 on the clock face corresponds to East. Standard compass directions: East is opposite West, North is opposite South. If 6 is East, then 12 is West. This means 3 is North and 9 is South. At 4:30, the hour hand is exactly halfway between 4 and 5. Since 3 is North and 6 is East, the hour hand (between 4 and 5) will point in the North-East direction. This tests spatial reasoning and understanding of clock positions.
Q7. What is the probability of getting a sum of 7 when rolling two fair dice?
When rolling two fair dice, there are 6 possible outcomes for each die, resulting in a total of 6 * 6 = 36 possible combinations. To get a sum of 7, the possible pairs are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). There are 6 such combinations. The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. So, the probability of getting a sum of 7 is 6/36 = 1/6. This problem evaluates fundamental probability concepts.
Q8. A container has 100 liters of milk. 10 liters of milk are removed and replaced with water. This process is repeated. How much milk is left after the second operation?
Initial milk = 100L. After 1st operation: Milk removed = 10L. Milk remaining = 90L. Water added = 10L. Total volume = 100L (90L milk, 10L water). After 2nd operation: 10L of mixture is removed. The proportion of milk in the mixture is 90/100 = 0.9. So, milk removed in 2nd operation = 0.9 * 10L = 9L. Milk remaining = 90L - 9L = 81L. 10L of water is added. The final amount of milk is 81L. This problem evaluates understanding of mixture problems and iterative proportional changes.
Q9. What is the angle between the hour hand and the minute hand of a clock at 3:30?
At 3:30, the minute hand points exactly at 6. The hour hand is exactly halfway between 3 and 4. A full circle is 360 degrees, and there are 12 hours, so each hour mark represents 360/12 = 30 degrees. At 3:30, the minute hand is at 6 (6 * 30 = 180 degrees from 12). The hour hand is at 3.5 (midway between 3 and 4). So, its angle is 3.5 * 30 = 105 degrees from 12. The angle between them is the absolute difference: |180 - 105| = 75 degrees. This tests spatial reasoning and clock arithmetic.
Q10. If all 'P' are 'Q', and some 'Q' are 'R', then which of the following is definitely true: (a) All R are Q, (b) Some P are R, (c) Some Q are P, (d) No P is R.
This is a syllogism problem. From 'All P are Q', we can infer that 'Some Q are P' is definitely true. This is because if every P is a Q, then the set of P is a subset of the set of Q, meaning there must be some Qs that are also Ps. The other options are not necessarily true: (a) 'All R are Q' is not true; some Q are R doesn't mean all R are Q. (b) 'Some P are R' cannot be concluded without more information. (d) 'No P is R' also cannot be concluded. Therefore, (c) 'Some Q are P' is the only definite conclusion. This tests logical deduction and understanding of set relationships.
Q11. How many times do the hands of a clock coincide in 12 hours?
The hands of a clock coincide 11 times in 12 hours. They coincide at approximately 12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55. They don't coincide between 11 and 12, as the 12th coincidence is exactly at 12:00 (which counts as the start of the next 12-hour cycle or the end of the previous one). If considering a 12-hour period starting just after 12:00 and ending at 12:00, there are 11 coincidences. This tests understanding of relative speeds of clock hands and cyclical patterns.
Q12. Four people, P, Q, R, S, are to be seated in a row. In how many ways can they be seated if P and Q must sit together?
If P and Q must sit together, treat them as a single unit. So, we now have 3 'units' to arrange: (PQ), R, S. These 3 units can be arranged in 3! (3 factorial) ways. 3! = 3 * 2 * 1 = 6 ways. Additionally, within the (PQ) unit, P and Q can swap positions (PQ or QP). This can be done in 2! (2 factorial) ways. 2! = 2 * 1 = 2 ways. So, the total number of ways to seat them is the product of the arrangements of the units and the arrangements within the unit: 6 * 2 = 12 ways. This problem involves permutations with constraints.
Q13. A container has red and blue marbles in the ratio 2:3. If 10 red marbles are added, the ratio becomes 3:2. How many blue marbles are there?
Let the initial number of red marbles be 2x and blue marbles be 3x. After adding 10 red marbles, the number of red marbles becomes 2x + 10. The number of blue marbles remains 3x. The new ratio is (2x + 10) : 3x = 3:2. So, (2x + 10) / (3x) = 3/2. Cross-multiplying gives 2(2x + 10) = 3(3x). This simplifies to 4x + 20 = 9x. Subtracting 4x from both sides gives 20 = 5x. So, x = 4. The number of blue marbles is 3x = 3 * 4 = 12. This tests algebraic problem-solving with ratios.
Q14. In a group of 15 people, 7 can speak French, 8 can speak German, and 3 can speak both. How many people can speak neither French nor German?
This is a set theory problem, often solved using Venn diagrams or the Principle of Inclusion-Exclusion. Total people = 15. Let F be French speakers and G be German speakers. |F| = 7, |G| = 8, |F ∩ G| = 3 (both). The number of people who can speak at least one language is |F ∪ G| = |F| + |G| - |F ∩ G| = 7 + 8 - 3 = 15 - 3 = 12. The number of people who can speak neither language is Total People - |F ∪ G| = 15 - 12 = 3. So, 3 people speak neither French nor German. This tests basic set theory and logical counting.
Q15. A train passes a platform 180 meters long in 15 seconds and passes a man standing on the platform in 9 seconds. What is the length of the train?
Let the length of the train be L meters and its speed be S m/s. When the train passes a man, the distance covered is its own length (L). So, S = L / 9. When the train passes a platform, the distance covered is the sum of the train's length and the platform's length (L + 180). So, S = (L + 180) / 15. Now we equate the two expressions for S: L/9 = (L + 180) / 15. Cross-multiplying: 15L = 9(L + 180). 15L = 9L + 1620. 6L = 1620. L = 1620 / 6 = 270 meters. The length of the train is 270 meters. This tests speed, distance, time problems with relative lengths.
Q16. What is the probability of drawing a king or a heart from a standard deck of 52 cards?
In a standard deck of 52 cards, there are 4 kings and 13 hearts. However, the King of Hearts is counted in both categories. To find the probability of drawing a king OR a heart, we use the formula P(A or B) = P(A) + P(B) - P(A and B). P(King) = 4/52. P(Heart) = 13/52. P(King and Heart) = P(King of Hearts) = 1/52. So, P(King or Heart) = (4/52) + (13/52) - (1/52) = (4 + 13 - 1) / 52 = 16/52. This simplifies to 4/13. This problem assesses understanding of probability with overlapping events.
Q17. If a rectangle's length is increased by 20% and its width is decreased by 10%, what is the percentage change in its area?
Let the original length be L and the original width be W. The original area is A = L * W. New length = L + 0.20L = 1.2L. New width = W - 0.10W = 0.9W. New area = (1.2L) * (0.9W) = 1.08LW. The percentage change in area = ((New Area - Original Area) / Original Area) * 100. Percentage change = ((1.08LW - LW) / LW) * 100 = (0.08LW / LW) * 100 = 0.08 * 100 = 8%. Since the result is positive, it's an 8% increase in area. This tests percentage calculations and area changes.
Advanced Level
Q1. A certain sum of money invested at compound interest doubles in 4 years. In how many years will it become 8 times itself?
For compound interest, if a sum doubles in 'n' years, it will become 2^k times itself in k*n years. Here, the sum doubles (2^1 times) in 4 years. We want it to become 8 times itself. Since 8 = 2^3, we have k=3. Therefore, it will become 8 times itself in 3 * 4 = 12 years. This is a classic compound interest pattern problem, demonstrating exponential growth. Understanding this relationship is crucial for financial calculations and growth models.
Q2. You have a balance scale and 9 identical-looking balls. One ball is slightly heavier than the others. How many weighings are needed to find the heavier ball?
This is a classic weighing puzzle. With a balance scale and 9 balls, you can find the heavier ball in a minimum of 2 weighings. Here's how: 1. Divide the 9 balls into three groups of 3 (A, B, C). 2. Weigh group A against group B. a. If A is heavier, the heavier ball is in group A. b. If B is heavier, the heavier ball is in group B. c. If they balance, the heavier ball is in group C. 3. Take the identified group of 3 balls (let's say it's A). Weigh one ball from A against another ball from A. a. If one is heavier, that's your ball. b. If they balance, the third ball from group A is the heavier one. This problem tests logical deduction and optimization.
Q3. A, B, C, D, E are sitting in a circle. D is not next to C. E is to the right of A. B is to the left of E. C is to the left of B. Who is sitting to the right of D?
Let's place them step-by-step: 1. 'E is to the right of A': A E (clockwise) 2. 'B is to the left of E': A B E (clockwise) 3. 'C is to the left of B': A C B E (clockwise) Now we have A, C, B, E in sequence. The only person left is D. 4. 'D is not next to C'. If we place D, the sequence would be A C B E D. In this arrangement, D is not next to C. So the order is A, C, B, E, D (clockwise). To the right of D is A. This tests circular arrangement puzzles and careful constraint checking.
Q4. What is the remainder when 2^100 is divided by 5?
This is a number theory problem involving modular arithmetic. We need to find 2^100 mod 5. Let's look at the pattern of powers of 2 modulo 5: 2^1 = 2 (mod 5), 2^2 = 4 (mod 5), 2^3 = 8 = 3 (mod 5), 2^4 = 16 = 1 (mod 5). The pattern of remainders is 2, 4, 3, 1, and it repeats every 4 powers. To find 2^100 mod 5, we need to find the remainder of 100 when divided by 4. 100 / 4 = 25 with a remainder of 0. A remainder of 0 means it's the last element in the cycle, which is 1. So, 2^100 mod 5 = 1. This tests modular arithmetic and pattern recognition.
Q5. You have a 10x10 grid. How many squares are there in total (including 1x1, 2x2, etc.)?
This is a combinatorial reasoning problem. For an NxN grid, the total number of squares is the sum of the squares of integers from 1 to N. For a 10x10 grid: Number of 1x1 squares = 10 * 10 = 100. Number of 2x2 squares = 9 * 9 = 81. Number of 3x3 squares = 8 * 8 = 64. ... Number of 10x10 squares = 1 * 1 = 1. Total squares = 1^2 + 2^2 + 3^2 + ... + 10^2. This sum is given by the formula N(N+1)(2N+1)/6. For N=10, Total squares = 10(10+1)(2*10+1)/6 = 10 * 11 * 21 / 6 = 5 * 11 * 7 = 385. This tests pattern recognition and summation formulas.
Q6. A cube is painted on all its faces and then cut into 64 smaller identical cubes. How many of the smaller cubes have exactly two faces painted?
If a cube is cut into 64 smaller identical cubes, it means it's a 4x4x4 cube (since 4*4*4 = 64). Cubes with exactly two faces painted are those located along the edges but not at the corners. Each edge of the original cube will have (n-2) such smaller cubes, where 'n' is the number of divisions along one side (here, n=4). So, (4-2) = 2 cubes per edge have exactly two faces painted. A cube has 12 edges. Therefore, the total number of cubes with exactly two faces painted is 12 * 2 = 24 cubes. This tests spatial reasoning and cube cutting concepts.
Q7. You have an unlimited supply of 5-cent and 7-cent stamps. What is the largest amount of postage you cannot make?
This is a variation of the Frobenius Coin Problem (or coin problem). For two relatively prime positive integers a and b, the largest number that cannot be expressed in the form ax + by for non-negative integers x, y is (a*b) - a - b. Here, a=5 and b=7. The largest amount of postage not formable is (5 * 7) - 5 - 7 = 35 - 5 - 7 = 23 cents. This means any postage value greater than 23 cents can be formed. This problem tests number theory and problem-solving through mathematical concepts. It's often solved by trying values systematically or knowing the formula.
Q8. There are three boxes. One contains only apples, one contains only oranges, and one contains both apples and oranges. All three boxes are incorrectly labeled. You can only pick one fruit from one box. How can you correctly label all the boxes?
This is a classic logic puzzle. Since all boxes are incorrectly labeled, the box labeled 'Apples & Oranges' must contain either only apples or only oranges. Pick one fruit from the box labeled 'Apples & Oranges'. 1. If you pick an apple: This box must contain only apples (because it's mislabeled, it can't be 'Apples & Oranges', and if it contained only oranges, you'd pick an orange). So, label this box 'Apples'. 2. Now consider the box originally labeled 'Oranges'. Since it can't be 'Oranges' (it's mislabeled) and the 'Apples & Oranges' box is now 'Apples', this box must be 'Apples & Oranges'. 3. The remaining box, originally labeled 'Apples', must then be 'Oranges'. If you pick an orange, the logic is symmetrical. This tests deductive reasoning and constraint satisfaction.
Prepared by iCampusLink. 43 Aptitude and Logical Reasoning interview questions.