Question 1

A clock is set correctly at 8:00 a.m. It gains 10 minutes in every 24 hours. What is the true time when this clock shows 1:00 p.m. on the following day?

Accepted Answer

Introduction / Context: This problem tests the concept of clocks that do not keep perfect time. The clock gains time, meaning it runs faster than a correct clock. We are asked to relate the incorrect reading of the faulty clock to the actual or true time. Such questions frequently appear in aptitude exams to check understanding of proportional reasoning and relative rate of gain or loss of time in clocks and watches. Given Data / Assumptions: The clock is correct at 8:00 a.m. on day one. The clock gains 10 minutes in 24 hours. We consider the reading when the clock shows 1:00 p.m. on the following day. We assume the gain of time is uniform over the whole period. Concept / Approach: A correct clock gains zero minutes in 24 hours. Here the faulty clock shows more time than actually passed. In 24 true hours, the faulty clock advances by 24 hours plus 10 minutes. Thus, the rate of the faulty clock relative to true time is: faulty rate factor = (24 hours 10 minutes) / 24 hours Once this factor is known, we can set: shown time elapsed = faulty rate factor * true time elapsed We know the shown time elapsed from 8:00 a.m. to 1:00 p.m. next day and solve for the true time elapsed. Step-by-Step Solution: Step 1: Time shown by the faulty clock from day one 8:00 a.m. to day two 1:00 p.m. From 8:00 a.m. day one to 8:00 a.m. next day = 24 hours. From 8:00 a.m. to 1:00 p.m. on next day = 5 hours. Total shown time elapsed = 24 + 5 = 29 hours. Step 2: Faulty rate factor. In 24 true hours, the clock gains 10 minutes, so it shows 24 hours 10 minutes. Convert to minutes: 24 hours = 1440 minutes, gain is 10 minutes, so total = 1450 minutes. Rate factor = 1450 / 1440. Step 3: Let true time elapsed = T hours. Shown time elapsed = rate factor * T. 29 = (1450 / 1440) * T. T = 29 * 1440 / 1450 hours. Compute T in hours: 29 * 1440 / 1450 = 28.8 hours. Step 4: Convert 28.8 hours. 28 hours + 0.8 hour; 0.8 hour = 0.8 * 60 = 48 minutes. True time elapsed from 8:00 a.m. day one is 28 hours 48 minutes. Step 5: Find the actual clock time. Add 24 hours to return to 8:00 a.m. next day, remaining = 4 hours 48 minutes. 8:00 a.m. next day + 4 hours 48 minutes = 12:48 p.m. So the true time is 48 minutes past 12. Verification / Alternative check: We can check whether 28 hours 48 minutes of true time, multiplied by the rate factor, does give 29 hours shown: 28.8 * 1450 / 1440 = 29 hours This confirms that our computation and reasoning are consistent. Why Other Options Are Wrong: Option b (46 minutes past 12) and option c (45 minutes past 12) underestimate the true elapsed time and would not satisfy the proportional rate equation. Option d (47 minutes past 12) is close but still does not fit the required ratio exactly and is a trap for approximate calculators. Only 48 minutes past 12 satisfies the exact gain rate. Common Pitfalls: Common mistakes include directly adding or subtracting 10 minutes without considering the proportion over 24 hours, or assuming that 29 hours of shown time equals 29 hours of true time. Another frequent error is mishandling unit conversions between minutes and hours, or forgetting that the gain is spread over the whole period, not added just once. Final Answer: The true time when the clock shows 1:00 p.m. on the following day is 48 minutes past 12, that is 12:48 p.m.

Question 2

At what exact time between 4 oclock and 5 oclock will the hour and minute hands of a clock be in opposite directions and form a straight line?

Accepted Answer

Introduction / Context: This question involves relative motion of the hour and minute hands on a clock. We are asked to find the exact time between 4 oclock and 5 oclock when the two hands are in opposite directions, that is when the angle between them is 180 degrees. Such clock problems test a candidate's understanding of angular speed, relative speed, and algebraic setup using simple equations. Given Data / Assumptions: We consider a standard clock face with 12 hours marked equally. The minute hand moves 360 degrees in 60 minutes. The hour hand moves 360 degrees in 12 hours, that is 30 degrees per hour. We are observing the period between 4 oclock and 5 oclock. We want the hands to be in exactly opposite directions, so the angle between them is 180 degrees. Concept / Approach: The minute hand moves faster than the hour hand. We measure angles from 12 oclock in degrees. At any given time: minute hand angle = 6 * m hour hand angle = 30 * h + 0.5 * m Here h is the hour and m is the minutes past that hour. Between 4 and 5, h = 4. For opposite directions, we need the absolute difference between the angles to be 180 degrees. Step-by-Step Solution: Step 1: Set up the angles at time 4:m. Hour hand angle = 30 * 4 + 0.5 * m = 120 + 0.5 m. Minute hand angle = 6 * m. Step 2: Condition for opposite directions. We need |6 m - (120 + 0.5 m)| = 180. Since after 4 oclock the minute hand moves ahead of the hour hand, we can take: 6 m - (120 + 0.5 m) = 180. Step 3: Solve the equation. 6 m - 120 - 0.5 m = 180 5.5 m - 120 = 180 5.5 m = 300 m = 300 / 5.5 = 300 * 2 / 11 = 600 / 11 minutes. m = 54 6/11 minutes. Thus the time is 4 hours 54 6/11 minutes. Verification / Alternative check: We can compute the actual angles to confirm. Minute hand angle at this time: 6 * (600 / 11) = 3600 / 11 degrees. Hour hand angle: 120 + 0.5 * 600 / 11 = 120 + 300 / 11 = (1320 + 300) / 11 = 1620 / 11 degrees. Difference: 3600 / 11 - 1620 / 11 = 1980 / 11 = 180 degrees. So the condition for opposite directions is satisfied exactly. Why Other Options Are Wrong: Option a (54 minutes) is close but ignores the fractional part and gives a slightly different angle. Options b and c propose other fractional minutes but they do not give an exact 180 degree separation when substituted into the formula. Only 54 6/11 minutes produces exactly 180 degrees between the hands. Common Pitfalls: Many learners forget that the hour hand also moves as minutes pass and incorrectly fix the hour hand at 120 degrees for the whole hour. Another common mistake is ignoring the absolute value and choosing the wrong sign for the equation. Rounding off the fractional minute too early also leads to approximate answers which do not match any option exactly. Final Answer: The hands of the clock are in opposite directions at 54 6/11 minutes past 4.

Question 3

A watch that gains uniformly is 5 minutes slow at 8:00 a.m. on Sunday and 5 minutes 48 seconds fast at 8:00 p.m. on the following Sunday. At what time was the watch showing the correct time?

Accepted Answer

Introduction / Context: This question deals with a watch that gains time at a constant uniform rate. We are told how slow it is at one moment and how fast it becomes after a certain number of days and hours. From this information we must find the exact moment when the watch was correct. These problems test understanding of linear change over time and simple proportional reasoning with minutes and hours. Given Data / Assumptions: The watch is 5 minutes slow at 8:00 a.m. on Sunday. The watch is 5 minutes 48 seconds fast at 8:00 p.m. on the following Sunday. The gain or change in error of the watch is uniform over time. We are asked to find when the watch showed the correct time between these two instants. Concept / Approach: The watch error changes linearly from a negative value (slow) to a positive value (fast). The total change in error divided by the time between the two observations gives the rate of change of error per hour. Knowing the rate, we can compute how long it takes for the error to go from minus 5 minutes to zero, which is exactly when the watch is correct. We then add this duration to the starting instant. Step-by-Step Solution: Step 1: Express all errors in minutes. At 8:00 a.m. Sunday: error = -5 minutes (watch is slow). At 8:00 p.m. next Sunday: 5 minutes 48 seconds fast. 48 seconds = 48 / 60 = 0.8 minute. So final error = +5.8 minutes. Total change in error = 5.8 - ( -5 ) = 10.8 minutes. Step 2: Find the time interval between the two observations. From Sunday 8:00 a.m. to next Sunday 8:00 a.m. = 7 days = 7 * 24 = 168 hours. From Sunday 8:00 a.m. to Sunday 8:00 p.m. of the following week is 7 days 12 hours = 7.5 days. Total hours = 7.5 * 24 = 180 hours. Step 3: Rate of change of error per hour. Rate = total change in error / total time. Rate = 10.8 minutes / 180 hours = 0.06 minute per hour. Step 4: Time needed for error to reach zero. At start the error is -5 minutes. We need an increase of 5 minutes. Time needed = 5 minutes / 0.06 minute per hour. Time needed = 5 / 0.06 hours = 83.333... hours. That is 83 hours 20 minutes. Step 5: Add this to the starting instant. From 8:00 a.m. Sunday plus 72 hours = 8:00 a.m. Wednesday (3 days later). Remaining = 11 hours 20 minutes. 8:00 a.m. Wednesday + 11 hours 20 minutes = 7:20 p.m. Wednesday. So the watch is correct at 20 minutes past 7:00 p.m. on Wednesday. Verification / Alternative check: If the watch is correct at 7:20 p.m. Wednesday, then by 8:00 p.m. next Sunday it has gained 5.8 minutes over 180 - 83.333... = 96.666... hours. The gain is: rate * time = 0.06 * 96.666... = 5.8 minutes, which matches the given final error, so our calculation is consistent. Why Other Options Are Wrong: Option a (7:00 p.m. Wednesday) and option c (15 minutes past 7:00 p.m.) correspond to smaller time intervals and would not accumulate exactly 5 minutes of correction from the initial slow reading. Option d (8:00 p.m. Wednesday) overshoots and would make the watch already fast instead of exactly correct at that moment. Common Pitfalls: Typical mistakes include mixing days and hours, or forgetting to convert seconds into fractional minutes. Some learners divide 10.8 by 7.5 instead of by 180 hours, which produces an incorrect rate. Another mistake is treating the change as symmetric at the midpoint in time without checking the exact rate, which only works if the start and end errors are equal in magnitude, which is not the case here. Final Answer: The watch was showing the correct time at 20 minutes past 7:00 p.m. on Wednesday.

Question 4

A clock is set correctly at 5:00 a.m. It loses 16 minutes in every 24 hours. What is the true time when the clock indicates 10:00 p.m. on the fourth day?

Accepted Answer

Introduction / Context: This problem asks us to relate the reading of a slow clock to the actual time. The clock loses time, meaning that in a true 24 hour day it shows less than 24 hours. We are given the starting moment when it is correct and a later moment according to the faulty clock, and we need to deduce the true time at that later reading. These questions check understanding of proportional change and how errors accumulate uniformly over several days. Given Data / Assumptions: The clock is correct at 5:00 a.m. on the starting day. The clock loses 16 minutes in 24 hours, so it runs slow. The clock indicates 10:00 p.m. on the fourth day after it was set right. The loss is uniform (constant rate) over time. Concept / Approach: In 24 true hours, the clock shows 24 hours minus 16 minutes. Thus its speed as a fraction of true time is: rate factor = (1440 - 16) / 1440 = 1424 / 1440. If we let T be the true time elapsed since 5:00 a.m., then: shown time elapsed = rate factor * T. We know the shown time elapsed from 5:00 a.m. on day one to 10:00 p.m. on day four, so we can solve for the true elapsed time and hence the true clock time. Step-by-Step Solution: Step 1: Compute shown time elapsed. From 5:00 a.m. day one to 5:00 a.m. day four = 3 full days = 72 hours. From 5:00 a.m. to 10:00 p.m. on the same day = 17 hours. Total shown elapsed time = 72 + 17 = 89 hours. Step 2: Rate factor of the slow clock. In 24 hours, it shows 24 hours minus 16 minutes = 1440 - 16 = 1424 minutes. Rate factor = 1424 / 1440. Step 3: Relate shown time and true time. Let T be true hours elapsed since 5:00 a.m. shown time = rate factor * T. 89 = (1424 / 1440) * T. T = 89 * 1440 / 1424 hours. Compute T: simplifying gives T = 90 hours. Step 4: Find the true clock time. From 5:00 a.m. day one add 72 hours (3 days) to reach 5:00 a.m. day four. Remaining time = 90 - 72 = 18 hours. 5:00 a.m. day four + 18 hours = 11:00 p.m. day four. So when the slow clock shows 10:00 p.m. on the fourth day, the true time is 11:00 p.m. Verification / Alternative check: Check the loss: in 90 true hours, how much does the clock show? shown = rate factor * 90 = (1424 / 1440) * 90 = 89 hours. Thus in 90 true hours it shows 89 hours, which matches the given 89 hour reading from 5:00 a.m. to 10:00 p.m. on the fourth day. The result is consistent. Why Other Options Are Wrong: Option b (12 p.m.) refers to noon and is far earlier than the computed true time. Option c (1 p.m.) and option d (2 p.m.) also correspond to much smaller true elapsed times and do not match the proportional relationship between shown and true time given by the rate factor. Only 11 p.m. satisfies the calculation. Common Pitfalls: Learners sometimes incorrectly subtract 16 minutes only once instead of considering the effect over multiple days. Others mistake 10:00 p.m. on the fourth day as 4 days exactly instead of computing accurately from the starting time. Confusing 12 p.m. and 12 a.m. is another small but common source of error in clock questions. Final Answer: The true time when the clock indicates 10:00 p.m. on the fourth day is 11:00 p.m.

Question 5

An accurate clock shows 8 oclock in the morning. Through how many degrees will the hour hand rotate by the time the clock shows 2 oclock in the afternoon?

Accepted Answer

Introduction / Context: This is a direct question based on angles formed by the hour hand of a clock. It checks the basic fact that the hour hand moves at a constant angular speed as time passes. Such questions are straightforward but help build the foundation for more complex clock problems involving relative motion of the hour and minute hands. Given Data / Assumptions: The clock is accurate. Initial time: 8:00 a.m. Final time: 2:00 p.m. on the same day. We are concerned only with the hour hand, not the minute hand. Concept / Approach: On a standard clock, the hour hand completes one full circle of 360 degrees in 12 hours. Therefore, the angular movement of the hour hand per hour is: 360 degrees / 12 hours = 30 degrees per hour. Once we know the time interval in hours, we can multiply by 30 degrees to get the total rotation of the hour hand. Step-by-Step Solution: Step 1: Determine the time interval. From 8:00 a.m. to 2:00 p.m. the difference is 6 hours. Step 2: Use the hourly angular movement of the hour hand. The hour hand moves 30 degrees every hour. Step 3: Compute total rotation. Total rotation = 30 degrees per hour * 6 hours. Total rotation = 180 degrees. Thus, the hour hand turns through half a circle between 8:00 a.m. and 2:00 p.m. Verification / Alternative check: We can check by intermediate points. At 9:00 a.m. it moves 30 degrees from 8:00, at 10:00 a.m. 60 degrees, at 11:00 a.m. 90 degrees, at 12:00 noon 120 degrees, at 1:00 p.m. 150 degrees, and at 2:00 p.m. 180 degrees. The stepwise check confirms the calculation. Why Other Options Are Wrong: Option a (360 degrees) would imply a full revolution, which requires 12 hours, not 6 hours. Option c (90 degrees) corresponds to only 3 hours of movement, and option d (60 degrees) corresponds to only 2 hours. Neither matches the 6 hour interval between 8:00 a.m. and 2:00 p.m. Common Pitfalls: A common mistake is to confuse the movement of the minute hand (which completes 360 degrees in 60 minutes) with that of the hour hand. Some students also miscount the number of hours between 8:00 a.m. and 2:00 p.m., forgetting that the time interval is 6 hours, not 4 or 5 hours. Remembering the simple rule of 30 degrees per hour for the hour hand prevents such errors. Final Answer: The hour hand rotates through 180 degrees between 8:00 a.m. and 2:00 p.m.

Question 6

At what time between 5 oclock and 6 oclock are the hour hand and minute hand of a clock exactly 3 minute divisions apart on the dial?

Accepted Answer

Introduction / Context: This problem deals with the relative positions of the hour and minute hands of a clock. The phrase "3 minutes apart" refers to 3 minute divisions on the clock dial. Each minute division corresponds to 6 degrees, so the required angular separation is 18 degrees. The question tests the ability to set up and solve a simple equation using the angular speeds of the two hands. Given Data / Assumptions: We consider the time between 5 oclock and 6 oclock. The hands of the clock are to be 3 minute divisions apart, that is 18 degrees apart. The minute hand moves 6 degrees per minute. The hour hand moves 0.5 degree per minute. Concept / Approach: At any time t minutes after 5 oclock: hour hand angle = 30 * 5 + 0.5 * t = 150 + 0.5 t degrees, minute hand angle = 6 * t degrees. The separation between the two hands is the absolute difference of these angles. We set this difference equal to 18 degrees, solve for t, and ensure that the solution lies between 0 and 60 minutes. Step-by-Step Solution: Step 1: Write the angles. Hour hand angle H = 150 + 0.5 t. Minute hand angle M = 6 t. Step 2: Separation condition. We need |M - H| = 18 degrees. Between 5 and 5:24 the minute hand is behind but catching up, so initially H is ahead. Hence use H - M = 18 for the relevant interval. Step 3: Form the equation. (150 + 0.5 t) - 6 t = 18. 150 + 0.5 t - 6 t = 18. 150 - 5.5 t = 18. 5.5 t = 150 - 18 = 132. t = 132 / 5.5 = 24 minutes. So the hands are 3 minute divisions apart at 24 minutes past 5. Verification / Alternative check: At 5:24: Hour hand angle = 150 + 0.5 * 24 = 150 + 12 = 162 degrees. Minute hand angle = 6 * 24 = 144 degrees. Difference = 162 - 144 = 18 degrees, which is exactly 3 minute divisions (3 * 6 degrees). Therefore, the result is correct. Why Other Options Are Wrong: Option b (12 minutes) and option c (13 minutes) would give smaller angles than 18 degrees between the hands when substituted into the formula, whereas option d (14 minutes) also does not satisfy the 18 degree separation condition. Only at 24 minutes past 5 do we get the exact required separation. Common Pitfalls: Learners sometimes treat "3 minutes apart" as 3 minutes of time instead of 3 minute markings on the dial. Another common mistake is to forget that the hour hand moves as minutes pass, incorrectly fixing it at 150 degrees. Ignoring the absolute value and choosing the wrong sign in the angle difference equation can also lead to a negative separation or a wrong time. Final Answer: The hands of the clock are exactly 3 minute divisions apart at 24 minutes past 5.

Question 7

At what time between 2 oclock and 3 oclock will the hour hand and minute hand of a clock coincide and be exactly together?

Accepted Answer

Introduction / Context: This classic clock question asks when the hands coincide between 2 oclock and 3 oclock. It uses the standard formula involving the relative speed of the hour and minute hands. Such problems are frequently used to test algebraic manipulation and understanding of uniform circular motion on the clock face. Given Data / Assumptions: We are considering a time between 2 oclock and 3 oclock. The minute hand moves 6 degrees per minute. The hour hand moves 0.5 degree per minute. Coincidence means their angles, measured from 12 oclock, are equal. Concept / Approach: Between 2 and 3, let t be the minutes after 2 oclock. Then: hour hand angle = 30 * 2 + 0.5 * t = 60 + 0.5 t, minute hand angle = 6 * t. For coincidence, we set these two angles equal and solve for t. We then convert the answer into mixed fraction minutes to match the options. Step-by-Step Solution: Step 1: Write the angles. Hour hand angle H = 60 + 0.5 t. Minute hand angle M = 6 t. Step 2: Set H = M for coincidence. 60 + 0.5 t = 6 t. Step 3: Solve for t. 6 t - 0.5 t = 60. 5.5 t = 60. t = 60 / 5.5 = 60 * 2 / 11 = 120 / 11 minutes. 120 / 11 = 10 10/11 minutes. So the hands coincide at 10 10/11 minutes past 2. Verification / Alternative check: Substitute t = 120 / 11 into the expressions: Hour hand angle = 60 + 0.5 * 120 / 11 = 60 + 60 / 11 = (660 + 60) / 11 = 720 / 11 degrees. Minute hand angle = 6 * 120 / 11 = 720 / 11 degrees. Both angles are equal, so the hands are indeed together. Why Other Options Are Wrong: Options a, c, and d (9 10/11, 11 10/11, 12 10/11) do not satisfy the equality when substituted in place of t. They give slightly different angles for the hour and minute hands, so the hands will not be exactly together at those times. Common Pitfalls: Learners sometimes use an approximate rule like 5 minute past the hour for all coincidences, which is incorrect except for specific hours. Another mistake is forgetting that the hour hand moves as time passes and treating its angle as fixed at 60 degrees between 2 and 3. Rounding off t too early can also cause mismatch with the accurate mixed fraction answer. Final Answer: The hands of the clock are together at 10 10/11 minutes past 2.

Question 8

Find the angle between the hour hand and the minute hand of a clock when the time shown is exactly 3:25.

Accepted Answer

Introduction / Context: This question asks for the angle between the hour and minute hands at a given time. It is a basic clock angle problem that uses standard formulas for the angular positions of the two hands. These questions are common in aptitude tests as they test accuracy in applying simple arithmetic and geometric reasoning. Given Data / Assumptions: The time is 3:25. The minute hand completes 360 degrees in 60 minutes. The hour hand completes 360 degrees in 12 hours. We need the smaller angle between the two hands. Concept / Approach: The angle of the minute hand from 12 oclock is: minute hand angle = 6 * minutes. The angle of the hour hand from 12 oclock is: hour hand angle = 30 * hours + 0.5 * minutes. We compute both angles and then take the absolute difference. If necessary, we choose the smaller of the two possible angles (theta and 360 minus theta). Step-by-Step Solution: Step 1: Compute the minute hand angle. Minutes = 25. Minute hand angle = 6 * 25 = 150 degrees. Step 2: Compute the hour hand angle. Hours = 3, minutes = 25. Hour hand angle = 30 * 3 + 0.5 * 25. Hour hand angle = 90 + 12.5 = 102.5 degrees. Step 3: Find the absolute difference. Difference = |150 - 102.5| = 47.5 degrees. This is less than 180 degrees, so it is already the smaller angle between the hands. Verification / Alternative check: If we take the reflex angle, it would be: 360 - 47.5 = 312.5 degrees. Since questions about angle between clock hands usually refer to the acute or smaller angle unless the reflex angle is specifically mentioned, 47.5 degrees is the correct interpretation here. Why Other Options Are Wrong: Options b (57.5 degrees), c (45.5 degrees), and d (55.5 degrees) do not match the calculated difference between 102.5 degrees and 150 degrees. They may arise from approximate or incorrect calculations such as ignoring the movement of the hour hand or adding or subtracting an incorrect value. Common Pitfalls: A common error is to assume the hour hand is exactly at 90 degrees at 3:25, which ignores its extra movement during the 25 minutes. Another mistake is to compute the angle as 30 * 3 - 6 * 25, which uses the wrong formula. Being careful with decimal values like 12.5 is also important to avoid rounding errors. Final Answer: The angle between the hour and minute hands at 3:25 is 47.5 degrees.

Question 9

At what angle are the hands of a clock inclined when the time shown is exactly 5:15?

Accepted Answer

Introduction / Context: Here we are asked to find the angle between the hour hand and the minute hand at a specific time, 5:15. This is a standard type of clock angle question and uses known angular speeds of the clock hands. It reinforces understanding of how the hour hand also moves continuously as minutes pass, not just in jumps between hour marks. Given Data / Assumptions: The time is 5:15. Minute hand makes 360 degrees in 60 minutes. Hour hand makes 360 degrees in 12 hours. The required answer is the smaller angle between the hands at that time. Concept / Approach: We use the standard formulas: minute hand angle = 6 * minutes. hour hand angle = 30 * hours + 0.5 * minutes. We then compute their difference and, if needed, use 360 minus that difference to get the smaller angle. Step-by-Step Solution: Step 1: Calculate the minute hand angle. Minutes = 15. Minute hand angle = 6 * 15 = 90 degrees. Step 2: Calculate the hour hand angle. Hours = 5, minutes = 15. Hour hand angle = 30 * 5 + 0.5 * 15. Hour hand angle = 150 + 7.5 = 157.5 degrees. Step 3: Compute the angle between the hands. Difference = |157.5 - 90| = 67.5 degrees. Since 67.5 degrees is less than 180 degrees, it is the smaller angle between the hands. Verification / Alternative check: The reflex angle would be: 360 - 67.5 = 292.5 degrees, which is larger and therefore not the standard answer unless specifically asked for. The computed 67.5 degrees fits the typical interpretation of the angle between clock hands. Why Other Options Are Wrong: Option a (57.5 degrees), option c (77.5 degrees), and option d (87.5 degrees) do not match the exact difference between 157.5 degrees and 90 degrees. These numbers might come from miscalculations such as forgetting the half degree per minute movement of the hour hand or using 30 degrees per minute by mistake. Common Pitfalls: One common mistake is to assume that at 5:15 the hour hand is at exactly 150 degrees, the position at 5:00, and ignore its movement during the additional 15 minutes. Another pitfall is mixing up degrees per hour and degrees per minute, which can lead to completely incorrect angles. Final Answer: The angle between the hands of the clock at 5:15 is 67.5 degrees.

Question 10

A watch gains 5 seconds in every 3 minutes and is set correctly at 7:00 a.m. On the same day, when the watch shows 4:15 p.m., what is the true time?

Accepted Answer

Introduction / Context: In this problem, a watch is running fast and gaining a fixed number of seconds every few minutes. We know the time at which it was set correctly and a later time that it shows. The task is to find the actual correct time at that later reading. This tests understanding of rate of gain, conversion between seconds and minutes, and proportional reasoning. Given Data / Assumptions: The watch gains 5 seconds every 3 minutes. The watch is correct at 7:00 a.m. Later in the afternoon, it shows 4:15 p.m. The gain is uniform over time. Concept / Approach: We first find how fast the watch runs relative to true time. In 3 minutes of true time, the watch gains 5 seconds. So in 3 minutes (180 seconds) of real time, it shows 185 seconds. The speed factor is therefore: rate factor = 185 / 180. If T is the true time elapsed from 7:00 a.m., then: shown elapsed time = rate factor * T. We know the shown elapsed time between 7:00 a.m. and 4:15 p.m., so we can solve for T. Step-by-Step Solution: Step 1: Compute shown time elapsed. From 7:00 a.m. to 4:15 p.m. is 9 hours 15 minutes. Convert to minutes: 9 hours = 540 minutes, plus 15 gives 555 minutes. Step 2: Rate factor of the fast watch. In 3 minutes of true time, watch shows 3 minutes 5 seconds = 3 minutes + 5/60 minute. In seconds: 3 minutes = 180 seconds, plus 5 = 185 seconds. Rate factor = 185 / 180. Step 3: Set up relation between shown time and true time. Let T be true minutes elapsed since 7:00 a.m. Shown minutes = (185 / 180) * T. Shown minutes are 555, so: 555 = (185 / 180) * T. T = 555 * 180 / 185. Simplify: 555 / 185 = 3, so T = 3 * 180 = 540 minutes. Step 4: Convert true elapsed time to hours. 540 minutes = 9 hours. Step 5: Add to starting time. True time = 7:00 a.m. + 9 hours = 4:00 p.m. Thus, when the watch shows 4:15 p.m., the correct time is 4:00 p.m. Verification / Alternative check: From 7:00 a.m. to 4:00 p.m. is 9 hours or 540 minutes of true time. The watch gains 5 seconds every 3 minutes. Number of 3 minute intervals: 540 / 3 = 180 intervals. Total gain = 180 * 5 seconds = 900 seconds = 15 minutes. So the watch will show 9 hours 15 minutes after 7:00 a.m., that is 4:15 p.m., exactly matching the given reading. Why Other Options Are Wrong: Options b, c, and d (5:00 p.m., 6:00 p.m., 7:00 p.m.) would require much larger gains or different rate relationships not supported by the data. Only 4:00 p.m. is consistent with the given gain rate and displayed time. Common Pitfalls: Common errors include treating the gain of 5 seconds as per minute instead of per 3 minutes, or failing to convert seconds properly into minutes. Another mistake is to add 15 minutes directly to the true time instead of reversing the effect of the gain. Careful setup of the proportional relation avoids these problems. Final Answer: The true time when the watch indicates 4:15 p.m. is 4:00 p.m.

Question 11

Find the angle between the minute hand and the hour hand of a clock when the time shown is 8:30.

Accepted Answer

Introduction / Context: This is another angle between hands problem, this time at 8:30. It checks the same core concept as earlier angle questions but at a different time. The key idea is to remember that both hands move and that the hour hand moves half a degree per minute in addition to its 30 degrees per hour component. Given Data / Assumptions: The time is 8:30. Minute hand moves 6 degrees per minute. Hour hand moves 30 degrees per hour plus 0.5 degree per minute. We want the smaller angle between the two hands. Concept / Approach: Use the standard formulas: minute hand angle = 6 * minutes, hour hand angle = 30 * hours + 0.5 * minutes. Compute both angles and take their absolute difference. Choose the smaller angle if the reflex angle is produced as well. Step-by-Step Solution: Step 1: Compute minute hand angle. Minutes = 30. Minute hand angle = 6 * 30 = 180 degrees. Step 2: Compute hour hand angle. Hours = 8, minutes = 30. Hour hand angle = 30 * 8 + 0.5 * 30. Hour hand angle = 240 + 15 = 255 degrees. Step 3: Find the angle between the hands. Difference = |255 - 180| = 75 degrees. Since 75 degrees is less than 180 degrees, this is the required angle. Verification / Alternative check: Reflex angle would be: 360 - 75 = 285 degrees, which is larger and therefore not the standard angle between hands unless specifically requested. The 75 degree result is therefore correct for the usual interpretation. Why Other Options Are Wrong: Option a (80 degrees) and option d (105 degrees) are far from the computed difference. Option c (60 degrees) might appear from an incorrect assumption that the hour hand is at exact 240 degrees, ignoring the half degree per minute term. Only 75 degrees correctly represents the actual angular separation. Common Pitfalls: Students often forget to include the 0.5 degree per minute contribution to the hour hand angle and instead assume it stays at the hour mark for the entire hour. Another issue is mixing up which angle is taken as the difference, but using the absolute difference of the two angles always avoids sign mistakes. Final Answer: The angle between the minute hand and hour hand at 8:30 is 75 degrees.

Question 12

At what time between 7 oclock and 8 oclock will the hands of a clock lie in the same straight line but not coincide, that is, be 180 degrees apart?

Accepted Answer

Introduction / Context: The question asks for the time between 7 and 8 oclock when the clock hands are in one straight line but not together. This means the angle between them must be 180 degrees. It is similar to the earlier opposite direction problem but in a different hour interval. It checks the use of angular positions and solving a linear equation involving the time in minutes after an hour. Given Data / Assumptions: We consider a time between 7:00 and 8:00. The hands must form a straight line but not overlap, so the angle between them is 180 degrees. Minute hand moves 6 degrees per minute. Hour hand moves 0.5 degree per minute and starts from 210 degrees at 7:00. Concept / Approach: Let t be the number of minutes after 7:00. Then: hour hand angle = 30 * 7 + 0.5 * t = 210 + 0.5 t, minute hand angle = 6 * t. We want the angle between the hands to be 180 degrees, so the absolute difference of these two angles must be 180 degrees. We choose the correct sign based on which hand is ahead in that interval. Step-by-Step Solution: Step 1: Set up the separation condition. |6 t - (210 + 0.5 t)| = 180. Step 2: Choose sign. Shortly after 7:00, the hour hand is initially far ahead. The minute hand starts behind and moves faster, so for the first time when they are opposite, the hour hand is still ahead. Thus we use: (210 + 0.5 t) - 6 t = 180. Step 3: Solve the equation. 210 + 0.5 t - 6 t = 180. 210 - 5.5 t = 180. 5.5 t = 210 - 180 = 30. t = 30 / 5.5 = 30 * 2 / 11 = 60 / 11 minutes. 60 / 11 = 5 5/11 minutes. Therefore, the time is 7 hours 5 5/11 minutes. Verification / Alternative check: At t = 60 / 11: Hour hand angle = 210 + 0.5 * 60 / 11 = 210 + 30 / 11 = (2310 + 30) / 11 = 2340 / 11 degrees. Minute hand angle = 6 * 60 / 11 = 360 / 11 degrees. Difference = (2340 / 11) - (360 / 11) = 1980 / 11 = 180 degrees, confirming the hands are opposite. Why Other Options Are Wrong: Option b (6 5/11 minutes) gives a different separation when substituted into the formula. Options c and d (exactly 7:00 or 8:00) correspond to the hands being together or at right angles depending on the time, but not 180 degrees apart in this context. Only 5 5/11 minutes past 7 produces an exact 180 degree separation. Common Pitfalls: A frequent mistake is to forget the 0.5 degree per minute movement of the hour hand and treat it as fixed at the 7 oclock position. Some learners also use the wrong sign in the absolute difference equation, leading to the second solution where t exceeds 60 minutes, which is not within the hour under consideration. Final Answer: The hands of the clock are in a straight line but not together at 5 5/11 minutes past 7.

Question 13

How much time does a watch lose in one day if its hands coincide every 64 minutes instead of at the correct interval?

Accepted Answer

Introduction / Context: This question concerns a faulty watch whose hands coincide more frequently than they should. On a correct clock, the time between two consecutive coincidences of the hour and minute hands is a fixed value. If a watch shows coincidence earlier than that, it is running slow in terms of real time. The goal is to determine how much time the watch loses in one real day. This tests precise use of the formula for coincidence interval and proportional reasoning. Given Data / Assumptions: On a correct clock, the hands coincide at a standard interval. On this faulty watch, the hands coincide every 64 minutes as shown by the watch. The watch is assumed to be uniformly slow. We want the amount of time the watch loses in one real day of 24 hours. Concept / Approach: For a correct clock, the interval between two coincidences is: T_correct = 720 / 11 minutes, which is approximately 65 5/11 minutes. For the faulty watch, let k be the number of real minutes corresponding to 1 minute shown by the watch. Then both hands slow equally, so their relative angular speed in terms of shown minutes is scaled by k, and the interval between coincidences in shown minutes is: T_shown = T_correct / k. We are told T_shown is 64 minutes, so we can solve for k and then find how much real time corresponds to 24 shown hours. Step-by-Step Solution: Step 1: Standard coincidence interval for a correct clock. T_correct = 720 / 11 minutes. Step 2: Relation between real time and shown time for the faulty watch. Let 1 shown minute represent k real minutes. Then for the faulty watch, the relative speed is slower by factor k, so the coincidence interval in shown minutes is: T_shown = T_correct / k. Step 3: Use the given coincidence interval. T_shown = 64 minutes, so: 64 = (720 / 11) / k. k = (720 / 11) / 64 = 720 / (11 * 64). k = 720 / 704 = 90 / 88 = 45 / 44. So 1 shown minute equals 45 / 44 real minutes, which is more than 1, meaning the watch is slow. Step 4: Find how much time is lost in one day. In 24 hours shown by the watch, real time elapsed = 24 * k hours = 24 * 45 / 44 hours. Real time = 1080 / 44 hours = 24 12/22 hours = 24 6/11 hours. Extra real time beyond 24 hours = 6/11 hour. Convert 6/11 hour to minutes: 6/11 * 60 = 360 / 11 = 32 8/11 minutes. Since in 24 hours of true time the watch shows only 24 hours, it has lost 32 8/11 minutes in a day. Verification / Alternative check: The ratio k = 45 / 44 means that in 44 minutes shown, real time is 45 minutes. So the watch loses 1 minute every 45 minutes. In 24 real hours (1440 minutes), loss = 1440 / 45 = 32 minutes exactly plus a fraction which matches 32 8/11 minutes. This alternative way supports the same result. Why Other Options Are Wrong: Options b, c, and d (33 8/11, 34 8/11, 35 8/11) give larger daily losses that do not match the derived ratio from the coincidence interval. These values would correspond to different faulty intervals, not the given 64 minutes between coincidences. Common Pitfalls: A common mistake is to assume that if the hands coincide more often, the watch is fast, whereas here it is actually slow in real time. Another error is to forget that both hands slow at the same rate, so the relative angular speed decreases and the formula must be adjusted via the factor k for shown minutes. Also, mishandling fractions like 720 / 11 often leads to approximate answers instead of an exact one. Final Answer: The watch loses 32 8/11 minutes in one day.

Question 14

What is the reflex angle between the hands of a clock when the time shown is 10:25?

Accepted Answer

Introduction / Context: Here we are asked specifically for the reflex angle between the clock hands at 10:25. A reflex angle is the larger angle formed between the two hands, that is an angle greater than 180 degrees but less than 360 degrees. We first compute the smaller angle between the hands, then subtract that from 360 degrees to get the reflex angle. Given Data / Assumptions: The time is 10:25. Minute hand speed: 6 degrees per minute. Hour hand speed: 30 degrees per hour plus 0.5 degree per minute. Reflex angle means we finally take 360 minus the acute or smaller angle. Concept / Approach: Use: minute hand angle = 6 * minutes, hour hand angle = 30 * hours + 0.5 * minutes. The smaller angle between them is the absolute difference of these two angles. Then the reflex angle is: reflex angle = 360 - smaller angle. Step-by-Step Solution: Step 1: Compute the minute hand angle. Minutes = 25. Minute hand angle = 6 * 25 = 150 degrees. Step 2: Compute the hour hand angle. Hours = 10, minutes = 25. Hour hand angle = 30 * 10 + 0.5 * 25. Hour hand angle = 300 + 12.5 = 312.5 degrees. Step 3: Find the smaller angle between the hands. Difference = |312.5 - 150| = 162.5 degrees. Step 4: Find the reflex angle. Reflex angle = 360 - 162.5 = 197.5 degrees. In mixed form, this is 197 1/2 degrees. So the reflex angle between the hands is 197 1/2 degrees. Verification / Alternative check: We can verify quickly: if the smaller angle is less than 180 degrees and is 162.5 degrees, then the larger angle must be more than 180 degrees and less than 360 degrees, which is exactly 360 - 162.5 = 197.5 degrees. This matches the calculations and the given answer option. Why Other Options Are Wrong: Options b (167 1/2), c (157 1/2), and d (187 1/2) differ from 197 1/2 by 10 or 20 degrees and are not equal to 360 minus the correct smaller angle of 162.5 degrees. They would come from incorrect arithmetic or incorrect initial angle calculations. Common Pitfalls: Sometimes students confuse the smaller and larger angles and stop at 162.5 degrees, forgetting that the question asks for the reflex angle. Another pitfall is to assume the hour hand is fixed at 300 degrees for 10 oclock and ignore its progress during the 25 minutes, which leads to a wrong difference. Final Answer: The reflex angle between the hands of the clock at 10:25 is 197 1/2 degrees.

Question 15

At what time between 8 oclock and 9 oclock will the hands of a clock be in the same straight line but not coincide with each other?

Accepted Answer

Introduction / Context: This problem asks us to find the time between 8 and 9 oclock when the hands lie in the same straight line but do not coincide. That means they are 180 degrees apart. It is similar in structure to other straight line questions, but the starting hour and resulting fraction differ. The method involves setting up the angular positions of both hands and solving for the time in minutes after 8 oclock. Given Data / Assumptions: Time interval is between 8:00 and 9:00. Hands are in the same straight line but not together, so the angle between them is 180 degrees. Minute hand angle = 6 degrees per minute. Hour hand angle = 30 degrees per hour plus 0.5 degree per minute, starting from 240 degrees at 8:00. Concept / Approach: Let t be the number of minutes past 8:00. Then: hour hand angle = 30 * 8 + 0.5 * t = 240 + 0.5 t, minute hand angle = 6 * t. We require the angle between them to be 180 degrees, so: |6 t - (240 + 0.5 t)| = 180. We solve this equation and choose the solution with t between 0 and 60 minutes. Step-by-Step Solution: Step 1: Set up the equation. |6 t - (240 + 0.5 t)| = 180. There are two possibilities, but only one will fall between 0 and 60 minutes. Case 1: 6 t - (240 + 0.5 t) = 180. 5.5 t - 240 = 180. 5.5 t = 420. t = 420 / 5.5 = 840 / 11 = 76 4/11 minutes, which is more than 60 minutes and therefore outside the hour from 8 to 9. Case 2: (240 + 0.5 t) - 6 t = 180. 240 - 5.5 t = 180. 5.5 t = 60. t = 60 / 5.5 = 120 / 11 minutes. 120 / 11 is approximately 10 10/11 minutes, which lies between 0 and 60 minutes. Thus, the time is 120 / 11 minutes past 8. Verification / Alternative check: At t = 120 / 11: Minute hand angle = 6 * 120 / 11 = 720 / 11 degrees. Hour hand angle = 240 + 0.5 * 120 / 11 = 240 + 60 / 11 = (2640 + 60) / 11 = 2700 / 11 degrees. Difference = (2700 / 11) - (720 / 11) = 1980 / 11 = 180 degrees, confirming the hands are in a straight line but opposite each other. Why Other Options Are Wrong: Options a, c, and d (100/11, 90/11, 80/11 minutes) do not yield exactly 180 degrees of separation when substituted into the angle formulas. They correspond to other separations that are either acute or obtuse but not a straight line. Common Pitfalls: A common mistake is to accept the first solution t = 76 4/11 minutes without checking that it exceeds 60 minutes and therefore does not fall in the specified hour. Another is forgetting that the hour hand also moves, leading to equations that use 240 degrees as a fixed value and ignore the 0.5 t term. Final Answer: The hands of the clock are in the same straight line but not together at 120/11 minutes past 8.

Question 16

At what time between 5:30 and 6:00 will the hands of a clock be at right angles to each other?

Accepted Answer

Introduction / Context: This problem asks for a time between 5:30 and 6:00 when the hands of a clock are at right angles, that is, the angle between them is 90 degrees. It is more specific than the usual hour range question because it restricts the time to a half hour interval. It checks whether the candidate can correctly write the angular positions at a general time between two given points and solve for the right angle condition. Given Data / Assumptions: Time interval: from 5:30 to 6:00. Right angle means the angle between hands is exactly 90 degrees. At 5:30, the hour hand is not exactly on 5 but halfway between 5 and 6. The minute hand moves 6 degrees per minute, hour hand 0.5 degree per minute. Concept / Approach: Let t be the number of minutes after 5:30. Then the time is 5:30 + t. We first compute the angle of each hand at 5:30 and then add the further movement over t minutes. We then set the absolute difference of the angles equal to 90 degrees and solve for t in the specified interval. Step-by-Step Solution: Step 1: Angles at 5:30. At 5:00, hour hand angle = 30 * 5 = 150 degrees. In 30 minutes, hour hand moves 0.5 * 30 = 15 degrees. So at 5:30, hour hand angle = 150 + 15 = 165 degrees. Minute hand at 5:30 is at 30 minutes, so angle = 6 * 30 = 180 degrees. Step 2: General angles after t minutes from 5:30. Hour hand angle H = 165 + 0.5 t. Minute hand angle M = 180 + 6 t. Step 3: Right angle condition. We need |M - H| = 90 degrees. Compute M - H: (180 + 6 t) - (165 + 0.5 t) = 15 + 5.5 t. Thus, |15 + 5.5 t| = 90. Since for t between 0 and 30, 15 + 5.5 t is positive, we use: 15 + 5.5 t = 90. Step 4: Solve for t. 5.5 t = 90 - 15 = 75. t = 75 / 5.5 = 150 / 11 minutes. 150 / 11 = 13 7/11 minutes. Step 5: Convert to clock time. Time = 5:30 + 13 7/11 minutes = 5:43 7/11 minutes. So the hands are at right angles at 43 7/11 minutes past 5. Verification / Alternative check: At t = 150 / 11: Hour hand angle = 165 + 0.5 * 150 / 11 = 165 + 75 / 11 = (1815 + 75) / 11 = 1890 / 11 degrees. Minute hand angle = 180 + 6 * 150 / 11 = 180 + 900 / 11 = (1980 + 900) / 11 = 2880 / 11 degrees. Difference = (2880 / 11) - (1890 / 11) = 990 / 11 = 90 degrees, confirming the right angle condition. Why Other Options Are Wrong: Option a (43 5/11 minutes) is very close but does not produce exactly 90 degrees; it is a trap for rounding errors. Options c and d are not within the specified 5:30 to 6:00 interval for right angle formation and lack any calculated basis for this time segment. Common Pitfalls: Some learners treat 5:30 as if the hour hand were still at 150 degrees instead of 165 degrees. Others forget to start the calculation from 5:30 and simply set time as t minutes after 5:00. Failing to restrict t to between 0 and 30 minutes can also lead to an incorrect second solution which may lie outside the allowed interval. Final Answer: The hands of the clock are at right angles at 43 7/11 minutes past 5.

Question 17

At what time between 4 oclock and 5 oclock will the hands of a clock be at a right angle to each other?

Accepted Answer

Introduction / Context: This question asks for a time between 4 and 5 oclock when the hands of a clock are at a right angle, that is, 90 degrees apart. The method is similar to other clock angle problems where we express the angles of both hands as functions of the minutes past the hour and then enforce the condition on their difference. Given Data / Assumptions: Time interval is from 4:00 to 5:00. Right angle means the angle between the hands is exactly 90 degrees. Minute hand moves 6 degrees per minute. Hour hand moves 30 degrees per hour plus 0.5 degree per minute. Concept / Approach: Let t be the number of minutes after 4:00. Then: hour hand angle = 30 * 4 + 0.5 * t = 120 + 0.5 t, minute hand angle = 6 * t. We need: |6 t - (120 + 0.5 t)| = 90. We solve this equation and choose the value of t between 0 and 60 minutes that fits. Step-by-Step Solution: Step 1: Form the equation. |6 t - (120 + 0.5 t)| = 90. This gives two possible equations: Case 1: 6 t - (120 + 0.5 t) = 90. Case 2: (120 + 0.5 t) - 6 t = 90. Step 2: Solve Case 1. 6 t - 120 - 0.5 t = 90. 5.5 t = 210. t = 210 / 5.5 = 420 / 11 minutes, which is about 38 2/11 minutes, within the hour but corresponds to another right angle time. Step 3: Solve Case 2. 120 + 0.5 t - 6 t = 90. 120 - 5.5 t = 90. 5.5 t = 30. t = 30 / 5.5 = 60 / 11 minutes. 60 / 11 = 5 5/11 minutes. Thus, one right angle time between 4 and 5 is 4:05 5/11. The option list refers to this first right angle instant. Verification / Alternative check: At t = 60 / 11: Hour hand angle = 120 + 0.5 * 60 / 11 = 120 + 30 / 11 = (1320 + 30) / 11 = 1350 / 11 degrees. Minute hand angle = 6 * 60 / 11 = 360 / 11 degrees. Difference = (1350 / 11) - (360 / 11) = 990 / 11 = 90 degrees, confirming the right angle condition. Why Other Options Are Wrong: Options a, b, and d (3 5/11, 4 5/11, 6 5/11 minutes past 4) do not yield an exact 90 degree separation when substituted into the formulas. The second right angle within the hour occurs later and does not match any of these choices, while 5 5/11 minutes past 4 matches the first right angle exactly and is given as option c. Common Pitfalls: A typical error is to forget that there are two right angle times in each hour except in special cases, and then mis identify which one matches the options. Another pitfall is solving only one case from the absolute value equation and missing the valid solution, or incorrectly approximating fractions and failing to recognise 5 5/11 minutes as the exact solution. Final Answer: The hands of the clock are at right angles at 5 5/11 minutes past 4.

Question 18

A watch that gains uniformly is 2 minutes slow at noon on Monday and 4 minutes 48 seconds fast at 2:00 p.m. on the following Monday. At what time was the watch exactly correct?

Accepted Answer

Introduction / Context: This question is another example of a uniformly gaining watch where we are given its error at two different times and asked to find when it was correct. The watch transitions from being slow to being fast, and the moment of being correct is when the cumulative gain cancels the initial delay. The reasoning is based on uniform change in error over a known time interval. Given Data / Assumptions: At noon on Monday, the watch is 2 minutes slow, so its error is -2 minutes. At 2:00 p.m. on the following Monday, it is 4 minutes 48 seconds fast, so its error is +4.8 minutes. The gain is uniform over this period. We need to find the exact time between these instances when the watch shows correct time. Concept / Approach: The watch error changes linearly from -2 minutes to +4.8 minutes. The total change in error divided by the total time gives the rate of change of error per hour. The time needed for the error to go from -2 minutes to zero is then found by dividing 2 minutes by this rate. Adding this time to the starting moment gives the required instant when the watch was correct. Step-by-Step Solution: Step 1: Convert errors to minutes. Initial error at Monday noon = -2 minutes. Final error at next Monday 2:00 p.m. = +4 minutes 48 seconds. 48 seconds = 48 / 60 = 0.8 minute. So final error = +4.8 minutes. Total change in error = 4.8 - ( -2 ) = 6.8 minutes. Step 2: Compute time between the two readings. From Monday noon to next Monday noon = 7 days = 7 * 24 = 168 hours. From Monday noon to next Monday 2:00 p.m. = 7 days plus 2 hours = 170 hours. Step 3: Rate of change of error. Rate = total change / total time = 6.8 minutes / 170 hours. Rate = 0.04 minute per hour. Step 4: Time needed for error to go from -2 minutes to zero. We need an increase of 2 minutes. Time needed = 2 minutes / 0.04 minute per hour = 50 hours. Step 5: Add this to the starting time, Monday noon. 50 hours = 2 days plus 2 hours. Monday noon + 2 days = Wednesday noon. Add 2 hours to reach Wednesday 2:00 p.m. Thus the watch is correct at 2:00 p.m. on Wednesday. Verification / Alternative check: From Wednesday 2:00 p.m. (when error is zero) to next Monday 2:00 p.m. is 5 days = 120 hours. At a rate of 0.04 minute per hour, the gain is: 0.04 * 120 = 4.8 minutes. This matches the final error, so the time we computed is consistent with the problem data. Why Other Options Are Wrong: Option a (Tuesday 2:00 p.m.) corresponds to only 26 hours from Monday noon and would not allow the error to grow by the required 2 minutes. Options c and d (Thursday or Friday) correspond to larger time spans and thus give too much gain before the watch becomes correct, which contradicts the sign changes of the error. Common Pitfalls: Some students incorrectly treat the midpoint in time between the two observations as the moment when the watch is correct, but this would only be true if the slow and fast errors had equal magnitude. Another common error is to mix hours and days incorrectly or to forget to convert the 48 seconds into a fraction of a minute. Final Answer: The watch was showing the correct time at 2:00 p.m. on Wednesday.

Question 19

A watch gains 5 seconds in every 3 minutes and is set correctly at 8:00 a.m. What time will it show at 10:00 p.m. on the same day?

Accepted Answer

Introduction / Context: This problem again involves a fast watch, but now we know the true time interval and must find what the watch will indicate at the end of that interval. The watch gains a fixed amount of time every few minutes. We use this rate of gain to find how many extra minutes and seconds the watch will show after running from 8:00 a.m. to 10:00 p.m. on the same day. Given Data / Assumptions: The watch gains 5 seconds in every 3 minutes. It is set correct at 8:00 a.m. We consider the true time 10:00 p.m. on the same day. We assume the gain rate is uniform. Concept / Approach: We first calculate the true time elapsed between 8:00 a.m. and 10:00 p.m. in minutes. We then find how many 3 minute intervals occur in that time and thus how much total gain (in seconds) the watch accumulates. Adding this gain to the true time gives the reading shown by the watch. Step-by-Step Solution: Step 1: Compute the true time interval. From 8:00 a.m. to 10:00 p.m. is from hour 8 to hour 22. So the number of hours elapsed = 22 - 8 = 14 hours. Convert to minutes: 14 hours = 14 * 60 = 840 minutes. Step 2: Determine how many 3 minute intervals occur. Number of intervals = 840 / 3 = 280 intervals. Step 3: Compute total gain in seconds. Each 3 minute interval adds 5 seconds. Total gain = 280 * 5 seconds = 1400 seconds. Step 4: Convert gained seconds to minutes and seconds. 1400 seconds / 60 = 23 minutes and 20 seconds (since 23 * 60 = 1380, remainder 20 seconds). So the watch is 23 minutes 20 seconds fast at 10:00 p.m. true time. Step 5: Find the watch reading. True time: 10:00 p.m. Add the gain of 23 minutes 20 seconds. 10:00 p.m. + 23 minutes 20 seconds = 10:23:20 p.m. Therefore, the watch will show 10:23:20 p.m. Verification / Alternative check: We can express the rate factor as a ratio: in 3 minutes of real time, the watch shows 3 minutes 5 seconds, which is 185 seconds versus 180 seconds of real time. The factor is 185 / 180. Multiply 840 real minutes by 185 / 180 gives 840 * 185 / 180 = 862 * 1? more precisely 840 * 185 / 180 = 840 * (37 / 36) = 863.333... minutes, which is 14 hours 23 minutes 20 seconds. Adding this to 8:00 a.m. indeed reaches 10:23:20 p.m., confirming the earlier calculation. Why Other Options Are Wrong: Options a and b involve times in the morning and do not relate correctly to a gain from 8:00 a.m. to 10:00 p.m. Option c (9:45:15 p.m.) would correspond to a loss rather than a gain. Only option d reflects a time that is ahead of the true time by the correct amount. Common Pitfalls: Learners sometimes forget to convert the total gained seconds into minutes and seconds correctly, or they mis count the number of 3 minute intervals in 840 minutes. Another typical mistake is to apply the gain per hour instead of per 3 minutes without a proper conversion, which leads to inaccurate totals. Final Answer: At 10:00 p.m. true time, the fast watch will show 10:23:20 p.m.

Question 20

A faulty watch has its hour and minute hands coinciding every 64 minutes instead of the usual interval. How many minutes does this watch lose in one full day of 24 hours?

Accepted Answer

Introduction / Context: Clock problems that involve faulty watches often use the fact that, in a correct clock, the hour and minute hands coincide at fixed regular intervals. If the watch is slow or fast, this coincidence interval changes. By comparing the incorrect interval with the correct one, we can find how much time the watch gains or loses in a full day. This question asks you to determine how many minutes a faulty watch loses in twenty four hours when its hands coincide too frequently compared with a perfect clock. Given Data / Assumptions: In a correct clock, the hour and minute hands coincide every 65 5/11 minutes. In the faulty watch, the hands coincide every 64 minutes, as read on that watch. We assume that the fault affects the overall speed of the watch uniformly throughout the day. We must find the total loss of time in one real day of 24 hours, that is, 1440 real minutes. Concept / Approach: In a perfect clock, the relative speed between the minute hand and the hour hand is 5.5 degrees per minute, and they need to cover 360 degrees relative to each other to coincide. This gives the standard interval of 360 / 5.5 = 65 5/11 minutes between coincidences. If a faulty watch shows coincidences every 64 minutes on its own dial, that means that in 64 minutes of indicated time, the amount of real time is actually equal to one true coincidence interval. The ratio between shown time and real time can be used to find the daily loss. Once that rate is known, we simply scale it up to 24 hours. Step-by-Step Solution: Step 1: Correct interval between coincidences for a good clock is 65 5/11 minutes. Write this as an improper fraction: 65 5/11 = 720/11 minutes. Step 2: For the faulty watch, its hands coincide at 64 minutes of indicated time. In that same period, real time must be 720/11 minutes, because the true relative motion is unchanged. Step 3: Therefore, in 720/11 real minutes, the watch shows only 64 minutes. The rate of the faulty watch compared with real time is: rate = shown time / real time = 64 / (720/11) = (64 * 11) / 720 = 704 / 720 = 44 / 45. Step 4: In one real day, there are 1440 minutes. The faulty watch will show: shown time = (44 / 45) * 1440 = 44 * 32 = 1408 minutes. Step 5: The loss per day is real time minus shown time: loss = 1440 - 1408 = 32 minutes. Step 6: Thus the watch loses 32 minutes in one full day. Verification / Alternative check: You can also view the error per coincidence interval. The correct interval is 65 5/11 minutes, while the watch indicates only 64 minutes. So it loses 65 5/11 - 64 = 1 5/11 minutes per coincidence interval. The number of coincidence intervals in a full day for a correct clock is 1440 / (65 5/11) = 1440 / (720/11) = 22. Multiplying 1 5/11 minutes of loss per interval by 22 intervals gives (16/11) * 22 = 32 minutes, exactly the same result as before. Why Other Options Are Wrong: 30 minutes: This is close but underestimates the accumulated loss and does not match the exact ratio calculation. 34 3/11 minutes: This would correspond to a slightly larger error per interval than we actually have. 36 7/9 minutes: This is significantly higher than the computed 32 minutes and does not arise from any consistent use of the coincidence interval. 28 minutes: This underestimates the loss and cannot be obtained from the correct fractions. Common Pitfalls: Students sometimes mix up whether the watch is gaining or losing time by comparing 64 minutes and 65 5/11 minutes in the wrong order. Another common error is to forget that 65 5/11 must be converted to a single fraction, which can lead to arithmetic slips. Some learners try to compare hours instead of using the coincidence interval, which complicates the problem unnecessarily. Carefully setting up the ratio shown time over real time and then scaling to a full day is the safest way to avoid mistakes. Final Answer: The faulty watch loses 32 minutes in one day of 24 hours.

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