Question 1

A clock gains 15 minutes every day. It is set correctly at exactly 12:00 noon. When the actual time is 4:00 a.m. the next day, what time will this fast clock show?

Accepted Answer

Introduction / Context: This question tests your understanding of clocks that are not perfectly accurate, specifically a clock that gains time uniformly each day. You must relate the actual elapsed time to the time shown on a faulty clock that gains 15 minutes in 24 hours. Such problems are common in aptitude tests under the topic of clocks and time gain or loss, and they build your skills in proportional reasoning and time calculations. Given Data / Assumptions: • The clock gains 15 minutes in every 24 hours (one full day). • The clock is set correctly at 12:00 noon (actual time = clock time at that moment). • We are interested in the clock reading when the actual time is 4:00 a.m. the next day. • From noon of one day to 4:00 a.m. next day is 16 hours of actual elapsed time. • The gain is uniform over time. Concept / Approach: If a clock gains time uniformly, then its rate is slightly faster than the correct clock. The idea is to find how much time the faulty clock shows for a given actual time. The gain of 15 minutes in 24 hours means that when 24 hours of real time pass, the faulty clock shows 24 hours 15 minutes. Using a proportional relationship, we can find what the faulty clock will show after 16 hours of actual time. Step-by-Step Solution: Step 1: Gain per day = 15 minutes in 24 hours. Step 2: The rate factor of the faulty clock = (24 hours + 15 minutes) / 24 hours. Step 3: Convert 15 minutes to hours: 15 minutes = 15 / 60 = 0.25 hours. Step 4: So, in 24 hours of real time, the clock shows 24.25 hours. Step 5: Rate factor = 24.25 / 24 = 97/96 (since 24.25 = 24 + 0.25 = 24 + 1/4 = 97/4, and 24 = 96/4). Step 6: Actual elapsed time between noon and 4:00 a.m. next day = 16 hours. Step 7: Time shown by the faulty clock for 16 hours actual = 16 * (97/96) hours. Step 8: 16 * (97/96) = (16 * 97) / 96 = 1552 / 96 = 16 + 10/60 hours = 16 hours 10 minutes. Step 9: 16 hours 10 minutes after 12:00 noon corresponds to 4:10 a.m. on the next day. Verification / Alternative check: Think of the gain per hour. The gain is 15 minutes in 24 hours, so gain per hour = 15 / 24 minutes = 0.625 minutes per hour. For 16 hours of real time, total gain = 0.625 * 16 = 10 minutes. Therefore, when the actual time is 4:00 a.m., the faulty clock will be 10 minutes fast and will show 4:10 a.m. This provides a quick alternative check. Why Other Options Are Wrong: • 4:20 a.m.: This would imply a gain of 20 minutes in 16 hours, which is larger than the calculated 10 minutes. • 4:30 a.m.: This suggests a massive gain of 30 minutes in 16 hours, inconsistent with a gain of only 15 minutes per day. • 4:02 a.m.: This indicates a gain of only 2 minutes in 16 hours, which is far smaller than the proportional gain from 15 minutes per 24 hours. Common Pitfalls: A common mistake is to add 15 minutes directly to the next day's time, forgetting that the gain of 15 minutes occurs over a full 24-hour period, not over 16 hours. Another error is miscounting the elapsed time between noon and 4:00 a.m. the next day. Remember that from 12:00 noon to midnight is 12 hours, plus 4 more hours to 4:00 a.m., giving 16 hours in total. Using the correct proportional relation avoids these mistakes. Final Answer: When the actual time is 4:00 a.m. the next day, the fast clock will show 4:10 a.m..

Question 2

Between 4 o'clock and 5 o'clock, at what exact time (in minutes past 4) will the hour and minute hands of a clock first be at a right angle of 90 degrees to each other?

Accepted Answer

Introduction / Context: This problem is a classic example of relative speed and angular movement of clock hands. You are asked to determine the first time between 4 o'clock and 5 o'clock when the hour hand and the minute hand are at right angles (90 degrees). Such questions typically require you to use the standard formula for the angle between the hands and solve a simple linear equation in minutes past a given hour. Given Data / Assumptions: • We are working between 4:00 and 5:00 on a clock. • Let M be the number of minutes past 4 o'clock. • The hour hand moves 0.5 degrees per minute. • The minute hand moves 6 degrees per minute. • The angle between the hands must be 90 degrees for a right angle. Concept / Approach: The standard expression for the angle between the hour hand and the minute hand at a given time is based on their positions from 12 o'clock. At H hours and M minutes, the hour hand is at 30H + 0.5M degrees, and the minute hand is at 6M degrees. The angle between them is the absolute difference of these two positions. For a right angle, we set this difference equal to 90 degrees and solve for M. Step-by-Step Solution: Step 1: Here H = 4, and M is the unknown minutes past 4 o'clock. Step 2: Hour hand angle = 30 * 4 + 0.5 * M = 120 + 0.5M degrees. Step 3: Minute hand angle = 6M degrees. Step 4: Angle between hands = |(120 + 0.5M) - 6M| = |120 - 5.5M| degrees. Step 5: For the first time they are at right angles after 4:00, we set 120 - 5.5M = 90 (because initially the hour hand is ahead). Step 6: 120 - 5.5M = 90 implies 5.5M = 30. Step 7: M = 30 / 5.5 = (30 * 2) / 11 = 60/11 minutes. Step 8: So the first time they are at right angles between 4 and 5 o'clock is 60/11 minutes past 4, which is approximately 5 minutes and 27 seconds after 4:00. Verification / Alternative check: As an approximate check, 60/11 is about 5.4545 minutes. At roughly 4:05, the hour hand has moved slightly from its 4 o'clock position, and the minute hand has moved a bit beyond the 1-minute mark region. The calculated angle using M ≈ 5.45 minutes will still be very close to 90 degrees. Plugging M = 60/11 back into the expression |120 - 5.5M| confirms the angle is exactly 90 degrees, verifying our solution. Why Other Options Are Wrong: • 58/11 minutes: This value does not satisfy the equation |120 - 5.5M| = 90 and leads to a slightly different angle. • 422/11 minutes and 420/11 minutes: These are far beyond 60 minutes and thus cannot represent a time between 4:00 and 5:00. They are clearly invalid in the given interval. Common Pitfalls: Learners sometimes forget that the hour hand moves during the hour and treat it as fixed at 120 degrees (4 * 30), leading to incorrect equations. Another frequent error is solving for both 120 - 5.5M = 90 and 5.5M - 120 = 90 without selecting the solution that lies in the correct interval or represents the first occurrence. Always ensure that the computed M is between 0 and 60 and corresponds to the required first time in that hour. Final Answer: The two hands of the clock are at right angles for the first time between 4 and 5 o'clock at 60/11 minutes past 4 o'clock.

Question 3

A clock is started at exactly 12:00 noon. By 5:10 p.m. on the same day, through how many degrees has the hour hand turned from its initial position at 12 o'clock?

Accepted Answer

Introduction / Context: This question examines your understanding of how far the hour hand of a clock moves over a given time interval. Unlike the minute hand, which makes a full revolution every 60 minutes, the hour hand moves more slowly and covers 360 degrees in 12 hours. Knowing the rate of movement of the hour hand in degrees per minute allows you to convert elapsed time into angular displacement quickly and accurately. Given Data / Assumptions: • The clock starts at 12:00 noon, with the hour hand at 0 degrees reference position. • We are interested in the position of the hour hand at 5:10 p.m. on the same day. • The hour hand completes 360 degrees in 12 hours. • Therefore, it moves 30 degrees per hour, or 0.5 degrees per minute. Concept / Approach: The idea is straightforward: find the total elapsed time in minutes from 12:00 noon to 5:10 p.m. and then multiply by the hour hand's rate of 0.5 degrees per minute. The product gives the total angular displacement of the hour hand from its starting position at 12:00. This avoids confusion with the minute hand and keeps the calculation simple and accurate. Step-by-Step Solution: Step 1: From 12:00 noon to 5:00 p.m. is 5 hours. Step 2: From 5:00 p.m. to 5:10 p.m. is an additional 10 minutes. Step 3: Total elapsed time = 5 hours 10 minutes = 5 * 60 + 10 = 310 minutes. Step 4: The hour hand moves 0.5 degrees per minute. Step 5: Total angle turned = 310 * 0.5 degrees. Step 6: 310 * 0.5 = 155 degrees. Step 7: Hence, by 5:10 p.m., the hour hand has rotated 155 degrees from its original 12 o'clock position. Verification / Alternative check: We can cross-check by breaking the movement into hours and minutes. In 5 hours, the hour hand moves 5 * 30 = 150 degrees. In the extra 10 minutes, it moves 10 * 0.5 = 5 degrees. Total = 150 + 5 = 155 degrees. This matches our earlier computation, confirming that 155 degrees is correct. Why Other Options Are Wrong: • 145 degrees: This would correspond to less elapsed time or a wrong rate per minute. • 165 degrees: This overestimates the movement and would require 330 minutes (5 hours 30 minutes) at 0.5 degrees per minute, not the given 310 minutes. • 175 degrees: This is even farther from the correct value and does not correspond to the actual elapsed time from 12:00 noon to 5:10 p.m. Common Pitfalls: Students sometimes mistakenly use the minute hand's rate (6 degrees per minute) instead of the hour hand's rate, leading to very large and incorrect angles. Another error is miscounting the total minutes between the given times, especially when converting hours to minutes. Carefully converting the time interval into minutes and using the correct rate of 0.5 degrees per minute ensures a correct answer. Final Answer: By 5:10 p.m., the hour hand of the clock has turned through 155 degrees from its starting position at 12:00 noon.

Question 4

In a period of 24 hours (one full day), how many times are the hour hand and the minute hand of a clock in a straight line but pointing in exactly opposite directions (forming an angle of 180 degrees)?

Accepted Answer

Introduction / Context: This question focuses on how often the hour and minute hands of a clock are directly opposite each other in a day. When the hands are in a straight line but opposite in direction, they form a 180 degree angle. Understanding how frequently this condition occurs requires knowledge of the relative speeds of the hands and the standard results for their various alignments. This concept forms an important part of clock-based aptitude questions. Given Data / Assumptions: • We consider a 12-hour analog clock with hour and minute hands. • The minute hand completes 1 full revolution (360 degrees) in 60 minutes. • The hour hand completes 1 full revolution (360 degrees) in 12 hours (720 minutes). • We are interested in how many times per day (24 hours) the hands are in a straight line but opposite in direction, i.e., separated by 180 degrees. Concept / Approach: The minute hand gains on the hour hand because it moves faster. The relative speed of the minute hand compared to the hour hand determines how frequently certain alignments occur. It is known that in 12 hours, the hands coincide 11 times and are opposite each other 11 times. Since a full day has 24 hours, which is two such 12-hour cycles, we simply double the number of occurrences for the 12-hour period to obtain the result for 24 hours. Step-by-Step Solution: Step 1: In 12 hours, the minute hand completes 12 full revolutions, while the hour hand completes 1 revolution. Step 2: The difference in the number of revolutions between the minute and hour hands in 12 hours is 11. Step 3: Between consecutive times when the hands are opposite (180 degrees apart), the relative position changes by one-half revolution. Step 4: Over each 12-hour period, the hands are opposite each other exactly 11 times (a standard clock result). Step 5: A full day is 24 hours, which consists of two 12-hour cycles. Step 6: Therefore, in 24 hours, the number of times the hands are opposite = 2 * 11 = 22 times. Verification / Alternative check: We can reason based on symmetry: the behavior of the clock from 12:00 to 12:00 occurs identically in each 12-hour block (for example, from 12:00 midnight to 12:00 noon and then from 12:00 noon to 12:00 midnight). Since we know there are 11 times the hands are opposite in each 12-hour interval, adding the two intervals gives 22. This symmetry-based reasoning confirms the calculation without needing to list every occurrence. Why Other Options Are Wrong: • 11 times: This is the count for 12 hours, not for the full 24-hour day. • 33 times: This would imply an additional half-cycle of occurrences, which does not fit the regular pattern of the clock's motion. • 44 times: This would incorrectly double the already doubled result, and there is no mathematical basis for such a number in a 24-hour period. Common Pitfalls: A typical error is to confuse the number of times the hands coincide with the number of times they are opposite, or to misremember the standard results. Another common mistake is to think that in 12 hours they are opposite 12 times, which is incorrect. Remembering that the number of times the minute hand gains a full revolution on the hour hand in 12 hours is 11 helps in deriving the right counts for both coincidences and opposite alignments. Final Answer: In one full day of 24 hours, the hands of a clock are in a straight line but opposite in direction 22 times.

Question 5

A watch that gains uniformly is 2 minutes slow at 12:00 noon on Tuesday and 4 minutes 48 seconds fast at 2:00 p.m. on the following Tuesday. At what exact time did this watch show the correct time?

Accepted Answer

Introduction / Context: In this problem, we deal with a watch that gains time at a uniform rate. We are given two readings: initially the watch is slow, and later it is fast. The task is to determine when the watch must have been exactly correct. Such questions involve understanding uniform gain (or loss) in time and applying concepts of linear variation over a given interval, which are very common in aptitude tests dealing with faulty clocks and watches. Given Data / Assumptions: • At 12:00 noon on Tuesday, the watch is 2 minutes slow (that is, it shows 2 minutes less than the correct time). • At 2:00 p.m. on the following Tuesday, the same watch is 4 minutes 48 seconds fast. • One week plus 2 hours elapses between these two instants. • The watch gains time uniformly, meaning its error changes at a constant rate. • We must find the exact time between these two instants when the error is zero (the watch is correct). Concept / Approach: Since the gain is uniform, the time error changes linearly from -2 minutes to +4 minutes 48 seconds over a known time span. First, convert all errors to minutes and find the total change in error. Then compute the rate of change of error per hour. Next, determine how long it takes to move from -2 minutes to 0 minutes at that constant rate. Adding this duration to the initial time gives the moment when the watch showed the correct time. Step-by-Step Solution: Step 1: Convert the final error to minutes. 4 minutes 48 seconds = 4 + 48/60 = 4.8 minutes fast. Step 2: Initial error at noon on Tuesday = -2 minutes (slow). Step 3: Final error at 2:00 p.m. on the following Tuesday = +4.8 minutes (fast). Step 4: Total change in error = 4.8 - (−2) = 6.8 minutes. Step 5: Time between noon Tuesday and 2:00 p.m. the next Tuesday = 7 days + 2 hours. Step 6: In hours, this is 7 * 24 + 2 = 168 + 2 = 170 hours. Step 7: Rate of change of error per hour = 6.8 minutes / 170 hours = 0.04 minutes per hour. Step 8: To go from −2 minutes to 0 error, the change needed is +2 minutes. Step 9: Time required to correct this −2 minute error = 2 / 0.04 = 50 hours after noon on Tuesday. Step 10: 48 hours after Tuesday noon is Thursday noon, and 2 more hours gives Thursday 2:00 p.m. Verification / Alternative check: We can verify by checking the fraction of total time. The total error change is 6.8 minutes over 170 hours. The fraction needed to move from −2 minutes to 0 is 2 / 6.8 = 1 / 3.4. Multiplying this fraction by the total time: 170 * (1 / 3.4) = 50 hours. This matches our earlier calculation, confirming that the watch is correct exactly 50 hours after noon on Tuesday, which is 2:00 p.m. on Thursday. Why Other Options Are Wrong: • 12:00 noon on Wednesday: This is only 24 hours after the start, which is too early to correct the full −2 minute error at the given rate. • 3:00 p.m. on Thursday: This is 51 hours after noon Tuesday and would mean the watch has gained slightly more than needed to reach zero error, making it fast instead of correct at that time. • 2:00 p.m. on Wednesday: This is 26 hours after the start, still not enough time to eliminate the entire −2 minute error at the calculated rate of 0.04 minutes per hour. Common Pitfalls: Students often miscalculate the total time interval between the two observations, forgetting that “the following Tuesday” is a full week later. Others may forget to convert seconds into minutes correctly or may accidentally treat the gain as per day rather than per hour in the calculation. Carefully converting all quantities into consistent units (hours and minutes) and recognizing the linear nature of the error change helps avoid these mistakes. Final Answer: The watch showed the correct time at 2:00 p.m. on Thursday.

Question 6

An analog clock shows exactly 5 o'clock. If you look at this clock in a vertical mirror, what time will the mirror image appear to show?

Accepted Answer

Introduction / Context: Mirror-image clock questions test your spatial reasoning and understanding of how an analog clock looks when reflected in a vertical mirror. When a clock showing a given time is placed in front of a mirror, the reflected image shows a different time that can be computed using a simple rule. Here, the clock shows exactly 5 o'clock, and you must determine what time appears in the mirror image. Given Data / Assumptions: • The clock is an analog 12-hour clock with hour and minute hands. • The actual time on the clock is exactly 5:00 (5 o'clock sharp). • The mirror is vertical, so we see a left-right reversed image. • At 5:00, the minute hand points to 12, and the hour hand points to 5. Concept / Approach: The standard approach for mirror-image clock problems is to use the idea that a full hour scale from 12:00 to 12:00 is symmetric around 6 o'clock. For times not exactly at 6:00, the mirror-image time can generally be found by subtracting the given time from 11:60 (or 12:00 with an adjustment). The basic rule is: Mirror time = 11 hours 60 minutes − actual time (when expressed in hours and minutes). This gives the reflected time as seen in the mirror. Step-by-Step Solution: Step 1: Express the actual time 5:00 in hours and minutes. Step 2: Actual time = 5 hours 00 minutes. Step 3: Use the mirror-time formula: Mirror time = 11 hours 60 minutes − actual time. Step 4: Compute: 11:60 − 5:00 = (11 − 5) hours and (60 − 0) minutes = 6 hours 60 minutes. Step 5: 6 hours 60 minutes is equivalent to 7 hours 00 minutes. Step 6: Therefore, the mirror image of a clock showing exactly 5:00 appears to show 7:00. Verification / Alternative check: You can also reason visually: at 5:00, the hour hand points to the 5 on the right side of the clock. In a mirror, positions on the right become positions on the left. The 5 on the right side corresponds to the 7 on the left side (symmetrically placed around the vertical line passing through 12 and 6). The minute hand at 12 remains at the top in the mirror image. So the reflected time shows the hour hand at 7 and the minute hand at 12, which is 7:00. Why Other Options Are Wrong: • 12:00: At 12:00, both hands are at the top. This is not the mirror image of 5:00, where the hour hand is not at 12. • 5:00: This would be the original time, not the mirrored time. A mirror changes the positions of the numbers on the dial left to right. • 6:00: At 6:00, both hands are vertically aligned (hour at 6, minute at 12), which does not match the mirror reflection of 5:00. It is a symmetric time but not the correct reflection of 5:00. Common Pitfalls: Many students assume the time remains the same in a mirror or confuse the rule by subtracting from 12:00 instead of 11:60, which can lead to errors in cases involving exact hours. Others misread the mirrored positions of numbers on the dial. Remember that the mirror reverses left and right, and using the formula Mirror time = 11:60 − actual time is a reliable way to obtain the correct mirrored time. Final Answer: The mirror image of a clock that shows exactly 5 o'clock will appear to show 7:00.

Question 7

Kanchan bought a clock at 25% discount on its marked price and later sold it at a 75% gain on the price she paid. What was her profit percentage calculated on the original marked price of the clock?

Accepted Answer

Introduction / Context: This question belongs to the topic of profit, loss, and discount in arithmetic. Kanchan buys a clock at a discount on the marked price (label price) and then sells it at a profit based on her purchase price. The key is to compute her final selling price relative to the original marked price and then express the gain as a percentage of the marked price, not of her cost price. Such questions are very common in competitive exams and require clear understanding of successive percentage changes. Given Data / Assumptions: • Let the marked price (MP) of the clock be taken as Rs. 100 for convenience. • Discount given on the marked price = 25%. • Kanchan buys the clock after discount and later sells it at a 75% gain on her cost price. • We must find profit percentage with respect to the original marked price. Concept / Approach: To solve such problems easily, we assume a convenient marked price, typically Rs. 100. After applying the discount, we get Kanchan's cost price. Then, using the given percentage gain, we compute her selling price. Finally, we compare this selling price with the original marked price to obtain the required profit percentage on the marked price. Using a convenient assumed value simplifies percentage calculations without affecting the final percentage result. Step-by-Step Solution: Step 1: Assume the marked price (MP) = Rs. 100. Step 2: A discount of 25% on the marked price means Kanchan pays 75% of the MP. Step 3: Cost price (CP) for Kanchan = 75% of 100 = Rs. 75. Step 4: She sells the clock at a 75% gain on her cost price. Step 5: Selling price (SP) = CP * (1 + 75/100) = 75 * 1.75. Step 6: 75 * 1.75 = 75 * (7/4) = (75 * 7) / 4 = 525 / 4 = Rs. 131.25. Step 7: Profit calculated on the marked price = SP − MP = 131.25 − 100 = Rs. 31.25. Step 8: Profit percentage on the marked price = (31.25 / 100) * 100% = 31.25%. Verification / Alternative check: We can also compute directly as a combined effect: effective value of SP as a percentage of MP = (remaining after discount) * (multiplier for profit) = 75% of MP * 175% (1.75) of CP. So SP as % of MP = 0.75 * 1.75 * 100% = 1.3125 * 100% = 131.25% of MP. Thus, profit on MP = 131.25% − 100% = 31.25%, which confirms the earlier result. Why Other Options Are Wrong: • 50%: This would imply a selling price of 150% of MP, inconsistent with the successive changes (25% discount then 75% gain). • 56.25%: This might arise from misapplying percentage gain directly to marked price or confusing compounding effects. • 60%: This is a large overestimate and does not match the exact calculation from cost price to selling price relative to the original MP. Common Pitfalls: A frequent mistake is to calculate profit percentage on the cost price and then present that as the profit on the marked price. Another error is to add or subtract percentages directly (for example, thinking 75% gain minus 25% discount equals 50% profit on MP). Successive percentages must be applied multiplicatively, not simply added or subtracted. Using an assumed marked price of Rs. 100 helps keep the logic and arithmetic clear. Final Answer: Kanchan's profit percentage calculated on the original marked price of the clock is 31.25%.

Question 8

Two people, A and B, are 260 km apart at 9:00 a.m. A immediately starts travelling towards B with a speed of 25 km/h. At 11:00 a.m., B starts travelling towards A with a speed of 10 km/h. At what time in the afternoon (in p.m.) will they meet each o…

Accepted Answer

Introduction / Context: This question belongs to the topic of relative speed and motion in one dimension. Two people start from different points and move towards each other at different times and speeds. You need to find the time at which they meet. Such problems require you to account for unequal start times, calculate distances covered by each person, and use the concept of relative speed once both are moving simultaneously. Given Data / Assumptions: • Initial distance between A and B at 9:00 a.m. = 260 km. • A starts moving towards B at 9:00 a.m. with a speed of 25 km/h. • B starts moving towards A at 11:00 a.m. with a speed of 10 km/h. • Both travel along the same straight line towards each other. • We need to find the meeting time in the afternoon (p.m.). Concept / Approach: First, compute how far A travels before B even starts, that is, from 9:00 a.m. to 11:00 a.m. Then, once B starts at 11:00 a.m., both are moving towards each other, so we can consider their relative speed which is the sum of their individual speeds. Using the reduced distance between them at 11:00 a.m. and their combined speed, we can calculate how long it takes for them to meet and then convert that into clock time. Step-by-Step Solution: Step 1: Time from 9:00 a.m. to 11:00 a.m. = 2 hours (only A is moving during this time). Step 2: Distance covered by A in 2 hours = speed * time = 25 * 2 = 50 km. Step 3: Remaining distance between A and B at 11:00 a.m. = 260 − 50 = 210 km. Step 4: From 11:00 a.m., both A and B start moving towards each other. Step 5: Relative speed when moving towards each other = 25 km/h + 10 km/h = 35 km/h. Step 6: Time taken to meet after 11:00 a.m. = remaining distance / relative speed = 210 / 35 hours. Step 7: 210 / 35 = 6 hours. Step 8: Therefore, they meet 6 hours after 11:00 a.m., which is at 5:00 p.m. Verification / Alternative check: We can verify by checking the distances each travels by 5:00 p.m. A travels from 9:00 a.m. to 5:00 p.m., which is 8 hours. Distance by A = 25 * 8 = 200 km. B travels from 11:00 a.m. to 5:00 p.m., which is 6 hours. Distance by B = 10 * 6 = 60 km. Total distance covered together = 200 + 60 = 260 km, which matches the original distance between them. Thus, they indeed meet at 5:00 p.m. Why Other Options Are Wrong: • 6:00 p.m.: If they met at 6:00 p.m., A would have travelled 9 hours (225 km) and B 7 hours (70 km), totaling 295 km, which exceeds the initial 260 km. • 6:30 p.m.: This gives even more total distance, making it even more impossible for them to meet at that time if they started 260 km apart. • 7:00 p.m.: The total distance covered by then would be much larger than 260 km, so meeting earlier must have already occurred at 5:00 p.m. Common Pitfalls: A common mistake is to assume both start at the same time and directly use relative speed for the entire distance, ignoring the delay in B's start. Another error is miscounting the elapsed time or miscomputing the remaining distance after A's initial solo movement. Carefully separating the motion into two phases (before and after B starts) and then applying relative speed in the second phase avoids these issues. Final Answer: A and B will meet each other at 5:00 p.m..

Question 9

In a standard analog clock, when the hour hand and the minute hand are at right angles to each other (forming a 90 degree angle), how many minute spaces apart are the two hands on the dial?

Accepted Answer

Introduction / Context: This problem tests a basic but important concept about the division of a clock face into minute spaces and how these correspond to angles. When the hour and minute hands are at right angles to each other, they form a 90 degree angle. You are asked to express this separation as a number of minute spaces on the dial. Understanding how degrees relate to minute spaces simplifies many clock problems involving angles. Given Data / Assumptions: • A clock face is a full circle of 360 degrees. • The dial is divided into 60 equal minute spaces. • Each minute space corresponds to an equal angle at the center of the clock. • A right angle is 90 degrees. Concept / Approach: Since the entire 360 degree circle is divided into 60 minute spaces, each minute space represents 360 / 60 degrees. Knowing the degree measure of a right angle, we simply divide that angle by the degrees per minute space to find the number of minute spaces that correspond to a 90 degree separation between the hands. Step-by-Step Solution: Step 1: Total angle in a full circle of the clock = 360 degrees. Step 2: Total number of minute spaces on the dial = 60. Step 3: Angle per minute space = 360 / 60 = 6 degrees. Step 4: For a right angle, we need an angular separation of 90 degrees between the hands. Step 5: Number of minute spaces for a right angle = 90 / 6. Step 6: 90 / 6 = 15 minute spaces. Step 7: Therefore, when the hands are at right angles, they are 15 minute spaces apart on the clock dial. Verification / Alternative check: You can also reason that half a circle (180 degrees) corresponds to 30 minute spaces (as each space is 6 degrees), so a right angle, which is half of 180 degrees, would correspond to half of 30 minute spaces. Half of 30 is 15, confirming that 15 minute spaces represent a right angle between the hands. Why Other Options Are Wrong: • 13 minute spaces: 13 * 6 = 78 degrees, which is less than 90 degrees, so the hands are not at a right angle. • 14 minute spaces: 14 * 6 = 84 degrees, still an acute angle and short of 90 degrees. • 16 minute spaces: 16 * 6 = 96 degrees, which is slightly greater than 90 degrees, so the hands are more than a right angle apart. Common Pitfalls: A typical mistake is to mix up hour divisions (12 large divisions) with minute spaces (60 divisions), leading to incorrect angle calculations. Some students also forget that each minute space is 6 degrees, not 5 or any other number. Remembering the simple relation that 360 degrees divided by 60 spaces equals 6 degrees per space helps quickly handle such problems. Final Answer: When the two hands of a clock are at right angles, they are separated by 15 minute spaces on the dial.

Question 10

What is the circular (radian) measure of the angle between the hour hand and the minute hand of a clock at exactly 3:00 p.m.?

Accepted Answer

Introduction / Context: This question involves the relationship between degree measure and radian measure on a clock face. At exactly 3:00 p.m., the hour hand and minute hand form a specific angle, and you are asked to express that angle in circular measure (radians). Understanding the conversion between degrees and radians is fundamental for higher mathematics, trigonometry, and aptitude problems involving angular motion. Given Data / Assumptions: • At 3:00 p.m., the minute hand is at 12 and the hour hand is at 3 on a 12-hour clock. • The clock face is a full circle, 360 degrees, corresponding to 2π radians. • The numbers 12, 3, 6, and 9 on the clock divide the circle into four equal quadrants. • We must find the angle between the hands in radians. Concept / Approach: At 3:00 p.m., the minute hand points to 12 and the hour hand points to 3. The angle between them is clearly one-quarter of the full circle. One-quarter of 360 degrees is 90 degrees. We then convert this 90 degree angle into radians using the fact that 180 degrees corresponds to π radians. Thus, 90 degrees corresponds to π/2 radians. This direct reasoning makes the problem simple and quick to solve. Step-by-Step Solution: Step 1: At 3:00 p.m., the minute hand is at 12, and the hour hand is at 3. Step 2: The positions 12, 3, 6, and 9 divide the clock into four equal parts. Step 3: Each part represents a right angle, which is 360 / 4 = 90 degrees. Step 4: Therefore, the angle between the hands at 3:00 p.m. is 90 degrees. Step 5: Use the conversion: 180 degrees = π radians. Step 6: Hence, 90 degrees = (90 / 180) * π = π/2 radians. Step 7: Thus, the circular measure of the angle between the hands at 3:00 p.m. is π/2 radians. Verification / Alternative check: We can also reason that each hour marking on a 12-hour clock corresponds to 30 degrees (since 360 / 12 = 30). From 12 to 3 is 3 hour marks, so the angle between them is 3 * 30 = 90 degrees. Converting 90 degrees to radians again gives π/2 radians. This independently confirms the same result. Why Other Options Are Wrong: • π/4 radians: This corresponds to 45 degrees, which is half of the right angle and would occur at different times, not at exactly 3:00 p.m. • π/3 radians: This is 60 degrees, smaller than the angle between 12 and 3 on the clock. • 5π/12 radians: This equals 75 degrees, which also does not correspond to the standard quarter-circle separation at 3:00 p.m. Common Pitfalls: Students sometimes confuse degree and radian measures or apply the conversion factor incorrectly. Another error is misidentifying the angle between the hands at 3:00 p.m., sometimes thinking it is 60 degrees instead of 90 degrees. Remember that each hour mark is 30 degrees apart, and three such marks give 90 degrees. Using 180 degrees = π radians as the key conversion helps avoid mistakes. Final Answer: The circular measure of the angle between the hour hand and the minute hand at exactly 3:00 p.m. is π/2 radians.

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