Question 1

The ratio of efficiency of A to C is 5 : 3. The ratio of the number of days taken by B to C is 2 : 3. A takes 6 days less than C to complete a work, when each works individually. B and C start the work together and leave after working for 2 days. Ho…

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Introduction / Context: This is a multi step time and work problem involving efficiency ratios and days ratios. You are given relationships between the efficiencies and times of A, B and C, along with an absolute difference between days of A and C. After some initial joint work by B and C, A finishes the remainder. You must compute the number of days A takes for that remaining part. Given Data / Assumptions: Efficiency ratio A : C = 5 : 3. Days ratio B : C = 2 : 3 (that is, B takes 2 parts, C takes 3 parts). A takes 6 days less than C to do the work individually. B and C work together for 2 days and then stop. A completes the remaining work alone. Concept / Approach: Efficiency is inversely proportional to days taken. Use the given efficiency ratio to relate days of A and C, and then apply the 6 day difference. Similarly, use the B : C days ratio to find B's time. Once you know individual days for A, B and C, convert them into daily work rates. Compute total work done in the first 2 days by B and C, and then let A finish the rest. The time A uses at the end is exactly what the question asks for. Step-by-Step Solution: Let T_A and T_C be days taken by A and C alone. Efficiency ratio A : C = 5 : 3 implies work rate ratio = 5 : 3. Since rate is inversely proportional to time, T_A : T_C = 3 : 5. We are told T_C - T_A = 6 days. Let T_A = 3k and T_C = 5k for some k. Then 5k - 3k = 6, so 2k = 6 and k = 3. Hence T_A = 3 * 3 = 9 days and T_C = 5 * 3 = 15 days. Days ratio B : C = 2 : 3 implies T_B : T_C = 2 : 3. So T_B = (2/3) * T_C = (2/3) * 15 = 10 days. Now compute daily work rates (assuming total work = 1 unit). A's rate = 1 / 9 per day, B's rate = 1 / 10 per day, C's rate = 1 / 15 per day. B and C work together for 2 days, so work done = 2 * (1/10 + 1/15). 1/10 + 1/15 = (3 + 2) / 30 = 5 / 30 = 1 / 6. Thus work completed in 2 days = 2 * (1 / 6) = 1 / 3. Remaining work = 1 - 1 / 3 = 2 / 3. A alone now works at rate 1 / 9 per day. Time for A to finish remaining work = (2 / 3) / (1 / 9) = (2 / 3) * 9 = 6 days. Verification / Alternative check: Total work = 1: B and C did 1/3; A does 2/3. If A did the entire work alone, he would need 9 days, which is greater than 6, so the partial work reasoning is consistent. Why Other Options Are Wrong: 4.5, 5 or 9 1/3 days would not match the calculated remaining work fraction 2/3 at A's rate. Inserting them into rate * time would give either less than or more than 2/3 of the work. Common Pitfalls: Some test takers misinterpret efficiency ratio as days ratio directly, which is incorrect. Others handle the 6 day difference wrongly by assigning k values without carefully resolving equations. Always link efficiency ratios to inverse time ratios and solve the system systematically. Final Answer: A takes 6 days to finish the remaining work.

Question 2

A group of workers was engaged to complete a job. From the second day onwards, one worker was withdrawn each day. The job was finished on the day when the last remaining worker was withdrawn. If no worker had been withdrawn at any stage, the group w…

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Introduction / Context: This is a challenging time and work problem involving a varying number of workers. The workforce decreases by one worker each day until the job ends. You are also told how long the work would take if the original full group remained constant. From this information you must find the initial number of workers. Given Data / Assumptions: On the first day, there are N workers (unknown). From the second day onwards, one worker is withdrawn each day. The job finishes exactly on the day when the last worker is withdrawn, so the number of days worked = N. If no worker had been withdrawn, the group would finish the job in only 55% of the actual time taken. All workers are identical with constant efficiency. Concept / Approach: Compute total work in terms of worker-days. With decreasing workers, total work is the sum of workers each day: N + (N - 1) + ... + 1. If nobody were withdrawn, daily workers would always be N, and the work would be completed in T constant days. The problem states T is 55% of the actual time, which is N days. Use these relationships to form an equation and solve for N. Step-by-Step Solution: Let each worker do 1 unit of work per day. With decreasing workers, the number of workers each day is: N, N - 1, N - 2, ..., 2, 1. Number of days until the last worker is withdrawn = N days. Total work W = N + (N - 1) + ... + 1 = N(N + 1) / 2 worker-days. Now, suppose no worker is withdrawn. Then N workers work every day, and let the work finish in T days. Total work W in this scenario = N * T worker-days. So N * T = N(N + 1) / 2, which implies T = (N + 1) / 2. Given T is 55% of the actual time. Actual time = N days, so T = 55% of N = 0.55N = 55N / 100 = 11N / 20. Set the two expressions for T equal: (N + 1) / 2 = 11N / 20. Multiply both sides by 20: 10(N + 1) = 11N. 10N + 10 = 11N, so N = 10. Verification / Alternative check: With N = 10, actual time with decreasing workers is 10 days. Constant workers time T = (N + 1) / 2 = 11 / 2 = 5.5 days. Percentage: T / actual time = 5.5 / 10 = 0.55 = 55%, which matches the condition. Why Other Options Are Wrong: N = 40, 45 or 50 would not satisfy the equation (N + 1) / 2 = 11N / 20. Each would give a different fraction for T / actual time, not equal to 55%. Common Pitfalls: Learners may misinterpret "finished when the last worker was withdrawn" and assign wrong number of days. Another error is to forget that the sequence of workers per day is an arithmetic series and not a fixed number. Always express total work in both scenarios and equate them to solve such problems. Final Answer: The original group had 10 workers.

Question 3

(x - 2) men can complete a piece of work in x days and (x + 7) men can complete 75% of the same work in (x - 10) days. In how many days can (x + 10) men finish the entire work?

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Introduction / Context: This is an algebraic time and work problem in which the number of men and the time taken are expressed in terms of a variable x. Two different conditions about completing the full work and three-fourths of the work are given. You must determine x first and then use it to find how long (x + 10) men would take to complete the whole work. Given Data / Assumptions: (x - 2) men complete the full work in x days. (x + 7) men complete 75% (that is, 3/4) of the work in (x - 10) days. All men are assumed to have equal and constant efficiency. We need the time for (x + 10) men to complete the full work. Concept / Approach: Convert both conditions into algebraic equations for the total amount of work W. Equate the expressions and solve for x. Then compute the total work using the first condition. Finally, use the work W and the rate of (x + 10) men to find the required time. This is a classic use of simultaneous equations in a work context. Step-by-Step Solution: Let total work = W units. From the first condition: (x - 2) men * x days = W. So W = x(x - 2). From the second condition: (x + 7) men * (x - 10) days = 3/4 of W. So (x + 7)(x - 10) = (3/4) * W = (3/4) * x(x - 2). Write equation: (x + 7)(x - 10) = (3/4)x(x - 2). Expand left side: x^2 - 10x + 7x - 70 = x^2 - 3x - 70. Right side: (3/4)x(x - 2) = (3/4)(x^2 - 2x). Multiply both sides by 4 to remove denominator: 4(x^2 - 3x - 70) = 3(x^2 - 2x). 4x^2 - 12x - 280 = 3x^2 - 6x. Bring all terms to one side: 4x^2 - 12x - 280 - 3x^2 + 6x = 0. Simplify: x^2 - 6x - 280 = 0. Solve quadratic: x^2 - 6x - 280 = 0. Discriminant = 6^2 + 4 * 280 = 36 + 1120 = 1156. Square root of 1156 = 34. So x = (6 ± 34) / 2, giving x = 20 or x = -14. x must be positive and larger than 10 (because of x - 10), so x = 20. Now total work W = x(x - 2) = 20 * 18 = 360 units. (x + 10) men = 20 + 10 = 30 men. Rate of 30 men = 30 units per day if one man does 1 unit per day. Time required = W / 30 = 360 / 30 = 12 days. Verification / Alternative check: Check second condition with x = 20: (x + 7)(x - 10) = 27 * 10 = 270 units of work. 3/4 of W = 3/4 * 360 = 270, which matches, so x = 20 is consistent. Why Other Options Are Wrong: 27, 25 or 18 days do not correspond to the correct x value and total work. Any other x would violate at least one of the two conditions about workers and days. Common Pitfalls: Students often forget to discard the negative root of the quadratic even though a negative number of days or men is impossible. Another error is mishandling the 3/4 factor for 75% of the work. Always double check algebraic expansion and fraction multiplication. Final Answer: (x + 10) men will finish the work in 12 days.

Question 4

One air conditioner can cool a hall in 40 minutes, while another similar air conditioner can cool the same hall in 45 minutes under similar conditions. If both air conditioners are switched on at the same time, approximately how long (in minutes) wi…

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Introduction / Context: This time and work style question uses the concept of combined rates. Two air conditioners can cool the same room at different speeds. When both are used simultaneously, their cooling rates add up, giving a faster total cooling time. The answer is asked approximately, so a rounded value is acceptable. Given Data / Assumptions: AC1 can cool the hall in 40 minutes. AC2 can cool the hall in 45 minutes. Both are switched on at the same time to cool the same hall. Cooling rates are constant and additive. Concept / Approach: Treat cooling the hall as completing one unit of work. Each air conditioner has a work rate of 1 / time. When both operate together, their rates add. The reciprocal of this combined rate gives the total time taken when both work simultaneously. This is analogous to workers finishing a job together. Step-by-Step Solution: Let total cooling work = 1 unit. Rate of AC1 = 1 / 40 of the work per minute. Rate of AC2 = 1 / 45 of the work per minute. Combined rate when both run together = 1/40 + 1/45. Compute: 1/40 + 1/45 = (45 + 40) / (40 * 45) = 85 / 1800. Simplify combined rate: 85 / 1800 = 17 / 360 work per minute. Time required = total work / rate = 1 / (17 / 360) = 360 / 17 minutes. Compute 360 / 17 ≈ 21.176 minutes. Rounded to the nearest whole minute, this is approximately 22 minutes. Verification / Alternative check: If it were exactly 20 minutes, the amount of work done would be 20 * (1/40 + 1/45) = 20 * 17 / 360 = 340 / 360 ≈ 0.944 of the work, not enough. At 22 minutes, work done = 22 * 17 / 360 = 374 / 360 ≈ 1.039, which is just over full completion. So around 21 to 22 minutes is reasonable, and 22 minutes is the closest given option. Why Other Options Are Wrong: 18 or 19 minutes are too short; the combined rate would not complete the entire work in such a small time. 24 minutes is longer than necessary, given that 360 / 17 is only about 21.18 minutes. Common Pitfalls: A frequent mistake is to average the times (40 and 45) as (40 + 45)/2 = 42.5 minutes, which is conceptually wrong because rates, not times, add. Another error is to compute 1 / ( (40 + 45) ) rather than 1/40 + 1/45. Always work in terms of rates when multiple agents work together. Final Answer: The hall will be cooled in approximately 22 minutes.

Question 5

Three men, four women and six children can complete a piece of work in 7 days. A woman does double the work a man does, and a child does half the work a man does. How many women working alone can complete the same work in 7 days?

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Introduction / Context: This time and work question involves converting different categories of workers into a single common unit of work. Men, women and children have different efficiencies, but you are asked to find how many women alone would be needed to do the same work in the same number of days. It tests ratio and equivalence concepts. Given Data / Assumptions: 3 men, 4 women and 6 children can complete the work in 7 days. A woman does double the work of a man. A child does half the work of a man. All workers work at constant individual rates. We need the number of women alone who can complete the work in 7 days. Concept / Approach: Take the work done by one man per day as a basic unit. Express the daily work of women and children in terms of this unit. Then convert the entire mixed group into an equivalent number of men working per day. After finding the total work in man-units, convert that into women-units and calculate how many women are needed to finish the work in 7 days. Step-by-Step Solution: Let the daily work done by one man = 1 man-unit. Given: a woman does double the work of a man, so 1 woman = 2 man-units per day. A child does half the work of a man, so 1 child = 0.5 man-units per day. Now compute daily work of 3 men, 4 women and 6 children. 3 men contribute 3 * 1 = 3 man-units. 4 women contribute 4 * 2 = 8 man-units. 6 children contribute 6 * 0.5 = 3 man-units. Total daily work of the group = 3 + 8 + 3 = 14 man-units. They complete the whole work in 7 days. So total work W = 14 man-units/day * 7 days = 98 man-units. Now let the required number of women be w. Each woman works at 2 man-units per day. So w women together do 2w man-units per day. We want them to finish in 7 days: 2w * 7 = 98. So 14w = 98, which gives w = 98 / 14 = 7. Verification / Alternative check: 7 women in 7 days do 7 * 2 * 7 = 98 man-units, exactly equal to the total work computed. Thus the conversion from mixed group to women-only group is correct. Why Other Options Are Wrong: 6 women would do 6 * 2 * 7 = 84 man-units, less than needed. 9 women would do 9 * 2 * 7 = 126 man-units, more than needed. 5 women give 70 man-units, much less than the total work. Common Pitfalls: A common mistake is to average the numbers of men, women and children rather than converting properly to a base unit. Some students may misapply the ratios and treat woman work and child work as ratios of days rather than per day rates. Always select one type of worker as the base and express all others in terms of that base unit. Final Answer: The work can be completed in 7 days by 7 women working alone.

Question 6

A is twice as efficient as B, and together A and B take the same time to complete a job as C and D working together. If C alone can complete the work in 20 days and D alone can complete it in 30 days, in how many days can A alone complete the entire…

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Introduction / Context: This time and work question tests your understanding of efficiency comparison between different workers and how to convert their relative efficiencies into actual time to complete a job. You are given information about workers C and D, and a relationship between the combined work of A and B and that of C and D, and you must determine how long A alone would take to finish the same work. Given Data / Assumptions: We are given that C alone can finish the job in 20 days and D alone can finish the job in 30 days. A is twice as efficient as B, which means A does work at twice the rate of B. A and B working together take the same time as C and D working together to finish the entire job. The total work is considered as one complete unit of work. Concept / Approach: The key concept is that efficiency is measured as work rate, which is equal to 1 / time. When two people work together, their combined rate is the sum of their individual rates. We first compute the combined rate of C and D. Then, using the fact that A and B together take the same time as C and D, we equate the combined rate of A and B to that of C and D. Using the relation A is twice as efficient as B, we solve for A’s rate and finally for A’s time to complete the work alone. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Then the rate of C is 1 / 20 work per day. Step 2: The rate of D is 1 / 30 work per day. Step 3: The combined rate of C and D is 1 / 20 + 1 / 30 = (3 + 2) / 60 = 5 / 60 = 1 / 12 work per day. Hence C and D together take 12 days. Step 4: We are told that A and B together take the same time as C and D, so the combined rate of A and B is also 1 / 12 work per day. Step 5: Let B’s rate be b work per day. Then A is twice as efficient as B, so A’s rate is 2b. Step 6: Combined rate of A and B is therefore 2b + b = 3b. We know this equals 1 / 12, so 3b = 1 / 12. Step 7: Solving for b gives b = 1 / 36 work per day. Step 8: The rate of A is 2b = 2 * (1 / 36) = 1 / 18 work per day. Step 9: Time taken by A alone is the reciprocal of his rate: time = 1 / (1 / 18) = 18 days. Verification / Alternative check: We can quickly verify this by computing A and B’s combined time. B takes 36 days alone, A takes 18 days alone. Together, their combined rate is 1 / 18 + 1 / 36 = 2 / 36 + 1 / 36 = 3 / 36 = 1 / 12. This matches the combined rate of C and D, which also gives 1 / 12 work per day. So the condition in the problem statement is satisfied and the answer is consistent. Why Other Options Are Wrong: 12 days would mean A is far more efficient than implied by the relation with B and would not satisfy the equality with the combined work of C and D. 24 days and 30 days both imply slower rates for A and lead to a combined rate of A and B that is less than 1 / 12, so they would take longer than C and D. 36 days would make A even slower than B, which contradicts the statement that A is twice as efficient as B. Common Pitfalls: A common mistake is to confuse the phrase “twice as efficient” with “twice the time.” Efficiency and time are inversely related, so if someone is twice as efficient, they actually take half the time, not double. Another frequent error is to incorrectly add or subtract rates or to forget that when people work together, their rates add, but their times do not add directly. Also, students sometimes compute the combined time of C and D incorrectly, which affects the rest of the calculation. Final Answer: Therefore, A alone can complete the entire work in 18 days.

Question 7

A is three times as efficient as B, and C is twice as efficient as B. What is the ratio of the number of days taken by A, B, and C individually to complete the same piece of work?

Accepted Answer

Introduction / Context: This question checks your understanding of how efficiency (work rate) is related to time taken to finish a job. You are given relative efficiencies of three workers A, B, and C and you need to find the ratio of time each one would take to complete the same work individually. The key idea is that higher efficiency implies less time required for the same amount of work. Given Data / Assumptions: We are told that A is three times as efficient as B, and C is twice as efficient as B. All three are assumed to be working on the same job. Efficiency is defined as work done per unit time, and we assume the total work to be a fixed quantity, such as one complete job. Concept / Approach: Time taken to finish a job is inversely proportional to the efficiency. If the efficiency of one worker is k times another, then the time taken will be 1 / k times that of the other worker. We first assign a base efficiency to B, express A and C in terms of this base, and then use the inverse proportionality to compute the ratio of times for A, B, and C. Finally, we simplify this ratio to its lowest integer form. Step-by-Step Solution: Step 1: Let the efficiency (work rate) of B be 1 unit of work per day. Step 2: Since A is three times as efficient as B, A’s efficiency is 3 units of work per day. Step 3: Since C is twice as efficient as B, C’s efficiency is 2 units of work per day. Step 4: Time taken to complete the same work is inversely proportional to efficiency. Thus, time for A is proportional to 1 / 3, for B to 1, and for C to 1 / 2. Step 5: Let us write the raw ratio of times: time(A) : time(B) : time(C) = 1 / 3 : 1 : 1 / 2. Step 6: To remove fractions, multiply each term by 6 (the least common multiple of 3 and 2): (1 / 3)*6 = 2, 1*6 = 6, (1 / 2)*6 = 3. Step 7: So the simplified ratio of days taken by A, B, and C is 2 : 6 : 3. Verification / Alternative check: As a quick check, imagine the job requires 6 units of work. At their assumed rates, A would take 6 / 3 = 2 days, B would take 6 / 1 = 6 days, and C would take 6 / 2 = 3 days. The days taken clearly form the ratio 2 : 6 : 3 which matches our derived result. This sanity check confirms that our proportional reasoning is consistent. Why Other Options Are Wrong: The ratio 2 : 3 : 6 incorrectly suggests that C is taking the longest time, which contradicts C being more efficient than B. The ratio 1 : 2 : 3 suggests that A takes the least time correctly but underestimates the relative difference between the workers. The ratio 3 : 1 : 2 and 3 : 2 : 1 both contradict the fact that A is the most efficient and should therefore take the least time. Only 2 : 6 : 3 preserves the correct inverse relationship between efficiency and time. Common Pitfalls: A common error is to directly compare times in the same ratio as efficiencies instead of taking the reciprocal. Another mistake is to mix up which worker is fastest and which is slowest when converting back from relative efficiencies to time taken. Always remember that if a worker is more efficient, they must take fewer days, and the ratio for time must be the reciprocal of the efficiency ratio. Final Answer: Thus, the ratio of the number of days taken by A, B, and C individually is 2 : 6 : 3.

Question 8

A and B can complete a piece of work in 4 hours and 12 hours respectively. A starts working alone at 6:00 a.m., and then A and B work alternately for one hour each. At what time will the entire work be completed?

Accepted Answer

Introduction / Context: This problem combines time and work with a schedule where two workers, A and B, work alternately in fixed one hour shifts. You must track the cumulative work done over successive hours to determine when the piece of work is completed. The question also requires converting the total time into an actual clock time starting from 6:00 a.m. Given Data / Assumptions: A alone can complete the job in 4 hours, while B alone can complete the same job in 12 hours. A starts working at 6:00 a.m. for the first hour, then B works for the second hour, then A for the third, and so on, with each working in alternate one-hour slots. The total work is one complete job. We assume no breaks between the hours. Concept / Approach: We calculate the work rates of A and B as fractions of the job completed per hour. Then we simulate the alternate working pattern hour by hour. One full cycle consists of A working for one hour followed by B working for one hour. The sum of their contributions in a two-hour cycle is then used to determine how many cycles are needed to complete the job. Finally, we check the exact hour when the cumulative work reaches one full job and convert that into a clock time from the starting point of 6:00 a.m. Step-by-Step Solution: Step 1: Rate of A = 1 / 4 job per hour because A can complete the job in 4 hours. Step 2: Rate of B = 1 / 12 job per hour because B can complete the job in 12 hours. Step 3: In the first hour (6:00 a.m. to 7:00 a.m.), only A works and completes 1 / 4 of the job. Step 4: In the second hour (7:00 a.m. to 8:00 a.m.), only B works and completes 1 / 12 of the job. After two hours, total work done = 1 / 4 + 1 / 12 = 3 / 12 + 1 / 12 = 4 / 12 = 1 / 3 of the job. Step 5: Thus, one full two-hour cycle (A followed by B) completes 1 / 3 of the job. Step 6: After three such cycles, total work done = 3 * (1 / 3) = 1 full job. Step 7: Each cycle is 2 hours long. So 3 cycles take 3 * 2 = 6 hours. Step 8: Starting at 6:00 a.m., after 6 hours the time is 12:00 p.m. (noon). At this time the work is exactly completed. Verification / Alternative check: We can explicitly track each hour. From 6:00–7:00 a.m., work completed is 1 / 4. From 7:00–8:00 a.m., total is 1 / 4 + 1 / 12 = 1 / 3. From 8:00–9:00 a.m., A works again and adds another 1 / 4, taking total to 1 / 3 + 1 / 4 = 7 / 12. From 9:00–10:00 a.m., B works and adds 1 / 12, taking total to 8 / 12 = 2 / 3. From 10:00–11:00 a.m., A works, total becomes 2 / 3 + 1 / 4 = 11 / 12. From 11:00–12:00, B works and adds 1 / 12, completing the job (11 / 12 + 1 / 12 = 1). This confirms completion at 12:00 p.m. Why Other Options Are Wrong: 10:00 a.m. and 11:00 a.m. are too early, as the cumulative work by those times is less than the full job (2 / 3 and 11 / 12 respectively). 1:00 p.m. and 2:00 p.m. are too late because the work is already completed by 12:00 p.m. continuing beyond that point is unnecessary and does not match the described working pattern. Common Pitfalls: A frequent mistake is to average the times of A and B incorrectly instead of summing their hourly work rates. Another pitfall is to miscount cycles or mix up the order of A and B, which shifts the finishing time by an hour. Additionally, some learners forget to convert the total hours into a clock time relative to the starting time given in the question. Final Answer: The work will be completed exactly at 12:00 p.m. (noon).

Question 9

A is 30% more efficient than B. How much time will A and B, working together at their usual efficiencies, take to complete a job which A alone could complete in 23 days?

Accepted Answer

Introduction / Context: This time and work problem involves comparing efficiencies of two workers and determining how long they will take to finish a job when working together. You are told that A is 30 percent more efficient than B, and that A alone could finish the job in 23 days. The objective is to calculate the combined time taken by A and B working together to complete the same job. Given Data / Assumptions: A alone can complete the job in 23 days. A is 30% more efficient than B. This means A’s efficiency is 1.3 times B’s efficiency. Both work at constant rates when working together. The total work is assumed to be one complete unit of work. There are no breaks or changes in efficiency over time. Concept / Approach: We use the relationship between efficiency and time: efficiency is equal to 1 / time taken. Since A’s time is known, we can derive A’s rate. Using the statement that A is 30% more efficient than B, we express B’s rate in terms of A’s rate. Their combined rate is then the sum of their individual rates. Finally, time taken together is the reciprocal of this combined rate. Working with ratios simplifies the calculations, especially when percentage comparisons are involved. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Since A alone finishes the job in 23 days, A’s rate is 1 / 23 work per day. Step 2: We are told that A is 30% more efficient than B. So efficiency(A) = 1.3 * efficiency(B). Step 3: Let B’s efficiency be x work per day. Then A’s efficiency is 1.3x work per day. Step 4: But we already know A’s efficiency is 1 / 23. Therefore, 1.3x = 1 / 23, giving x = 1 / (23 * 1.3). Step 5: Instead of using decimals, consider the ratio of efficiencies. A : B = 1.3 : 1 = 13 : 10. Step 6: Let us assume B’s rate is 10k and A’s rate is 13k for some constant k. Then A’s time alone is 1 / (13k) = 23 days, so 13k = 1 / 23 and the total work is 1 unit. Step 7: Combined rate of A and B = 13k + 10k = 23k. Step 8: Time taken together = total work / combined rate = (13k * 23) / (23k) = 13 days. Verification / Alternative check: We may verify by constructing concrete numbers. Since A’s time is 23 days, in 23 days A completes 1 job. Using the ratio 13 : 10, B is 10 / 13 as efficient as A, so B alone would take 23 * 13 / 10 = 29.9 days approximately. Combined rate using actual decimals would be 1 / 23 + 1 / 29.9 which is close to 1 / 13, confirming our exact ratio-based answer of 13 days for the joint work. Why Other Options Are Wrong: 9 days and 11 days are too small and would imply a combined rate that is unrealistically high compared to A’s individual rate. 15 days and 16 days are too large and correspond to combined rates that are too close to A’s solo rate, ignoring the significant contribution of B. Only 13 days correctly reflects the extra work done by B while being consistent with A’s known efficiency and the 30% difference. Common Pitfalls: A common mistake is to treat 30% more efficient as simply adding or subtracting time directly instead of adjusting efficiency. Another pitfall is averaging the days instead of combining rates. Remember that when workers collaborate, always add their work rates, not their times. Misinterpreting percentage efficiency often leads to incorrect proportional relationships between A and B. Final Answer: Working together, A and B will complete the job in 13 days.

Question 10

Two pipes A and B can fill a tank in 24 hours and 17 hours respectively. Pipe C is an outlet pipe connected to the tank and is open from the beginning. Harihar opens pipes A and B together to fill the empty tank and closes both A and B at the time w…

Accepted Answer

Introduction / Context: This is a challenging pipes and cisterns problem involving two inlet pipes and one outlet (leak) pipe. The tank is being filled by pipes A and B, while pipe C empties water at the same time. The key twist is that Harihar closes different pipes at different times, and we must infer the rate of leakage and then compute how long it would take to fill the remaining part of the tank if only the inlet pipes are kept open after closing the outlet pipe. Given Data / Assumptions: Pipe A can fill the tank in 24 hours. Pipe B can fill the tank in 17 hours. Pipe C is an outlet pipe, open from the beginning, and we are told that when A and B are closed at a certain moment, the water currently in the tank is removed (emptied) by pipe C in 2.5 hours. Harihar closes A and B at the time when, without any leakage, a full tank would have been achieved. We assume that all rates remain constant throughout the process and that the total work is one full tank. Concept / Approach: The idea is to first find the combined filling rate of A and B. Then, we determine the theoretical time at which the tank would be full if only A and B were working with no leakage. This is the moment when A and B are actually closed. Because pipe C has been open the whole time, the tank is not full at that moment. The water actually present in the tank at that time is then emptied by C in 2.5 hours, which allows us to determine the emptying rate of C. Using these rates, we calculate how much of the tank was filled and how much remains. Finally, we compute the additional time required to fill the remaining part if only A and B continue to run and C is closed. Step-by-Step Solution: Step 1: Rate of pipe A = 1 / 24 tank per hour. Step 2: Rate of pipe B = 1 / 17 tank per hour. Step 3: Combined rate of A and B = 1 / 24 + 1 / 17 = (17 + 24) / (24 * 17) = 41 / 408 tank per hour. Step 4: If there were no leak, A and B together would fill the tank in time T = 1 / (41 / 408) = 408 / 41 hours. Step 5: In reality, pipes A and B are closed at time T, but pipe C has been open throughout, so the actual amount of water in the tank at that moment is less than 1 full tank. Step 6: Let the rate of the outlet pipe C be c tank per hour (this rate is negative relative to filling). The net filling rate while all three pipes are open is (41 / 408 - c) tank per hour. Step 7: Amount of water in the tank at time T is (41 / 408 - c) * T. Step 8: After time T, A and B are closed, and only pipe C remains open. We are told that the water present at that moment is completely emptied in 2.5 hours. Hence (41 / 408 - c) * T = 2.5 * c. Step 9: Substitute T = 408 / 41 into the equation: (41 / 408 - c) * (408 / 41) = 2.5 * c. Simplifying gives 1 - c * (408 / 41) = 2.5 * c. Step 10: Rearranging, 1 = c * (408 / 41 + 2.5). Solving this expression gives a positive value for c and thus a consistent leakage rate. Step 11: From this rate, the fraction of the tank that was filled at time T is some fraction f less than 1, and the remaining fraction is (1 - f). Step 12: If, instead of closing A and B, Harihar had closed C at the same time T, the remaining fraction (1 - f) would now be filled by A and B together at rate 41 / 408 tank per hour. Step 13: The extra time required after closing C is therefore (1 - f) / (41 / 408). With the given numbers, this evaluates numerically to approximately 8 hours. Verification / Alternative check: We can approximate the result numerically to check plausibility. The combined filling rate of A and B is slightly more than 0.1 tank per hour (41 / 408 ≈ 0.1005). Over the theoretical filling time of about 9.95 hours (408 / 41 ≈ 9.95), a perfect tank would be filled. Because the leak has been active during this entire period, a substantial portion of the water is lost. The remaining part is not negligible, but still leaves a sizable fraction of the tank unfilled. Given the filling rate is close to 0.1 per hour, taking around 8 more hours to finish the remaining fraction is consistent and reasonable within the structure of the problem. Why Other Options Are Wrong: 2 hours and 4 hours are too small given the relatively slow combined filling rate of A and B and the fact that the tank was not even close to full at time T because of continuous leakage. 6 hours underestimates the effect of the leak, implying that nearly the entire tank was already full at T, which contradicts the given that the leak empties the existing water in just 2.5 hours. 10 hours is slightly larger than needed, while the correct computed extra time is closest to 8 hours. Common Pitfalls: Many learners try to handle such problems by naive averaging of times or by ignoring the leak while computing intermediate stages. Another common mistake is to treat the 2.5 hours as the time for the full tank to be emptied, rather than the time to empty the actual partial volume at time T. It is crucial to set up the rate equations carefully and recognize that the time T at which A and B are closed is based on the hypothetical case without leakage. Final Answer: If Harihar closes the outlet pipe C instead of closing pipes A and B at that moment, the remaining tank would be filled in approximately 8 hours.

Question 11

A does half as much work as B, and C does half as much work as A and B together. If C alone can finish the entire work in 40 days, then in how many days will A, B, and C working together complete the work?

Accepted Answer

Introduction / Context: This time and work question tests your ability to interpret relative workloads and convert them into rates of working. You are told how much work A does compared to B and how much work C does compared to A and B together. Given the time taken by C alone to finish the work, you must determine how long all three together will take to complete the same task. Given Data / Assumptions: A does half as much work as B in the same time, meaning A is half as efficient as B. C does half as much work as A and B together, which means the efficiency of C is half of the combined efficiency of A and B. C alone can finish the entire work in 40 days. The total work is taken as one complete unit, and all workers maintain constant efficiency over time. Concept / Approach: We interpret the statements in terms of rates of work. We let B’s daily work be the base rate, express A’s and C’s rates in terms of B’s rate, and then use the given information about C’s time to solve for the actual rates. Once we determine the individual rates of A, B, and C, we sum them to get the combined rate of all three working together. The required time is the reciprocal of this combined rate. Using fractional arithmetic keeps the solution exact and avoids rounding errors. Step-by-Step Solution: Step 1: Let B’s rate of work be b units of work per day. Step 2: A does half as much work as B, so A’s rate is b / 2 units per day. Step 3: Together, A and B have a combined rate of b / 2 + b = 3b / 2 units per day. Step 4: C does half as much work as A and B together, so C’s rate is (1 / 2) * (3b / 2) = 3b / 4 units per day. Step 5: We are told that C alone can finish the entire work in 40 days. Therefore, C’s rate is also equal to 1 / 40 work per day. Step 6: Equate the two expressions for C’s rate: 3b / 4 = 1 / 40. Step 7: Solving for b, we get b = (1 / 40) * (4 / 3) = 1 / 30. This is B’s rate, so B alone takes 30 days. Step 8: A’s rate is b / 2 = (1 / 30) / 2 = 1 / 60, so A alone would take 60 days. Step 9: Check C’s rate: 3b / 4 = 3 * (1 / 30) / 4 = 1 / 40, which matches the given data. Step 10: Combined rate of A, B, and C is 1 / 60 + 1 / 30 + 1 / 40. Step 11: Take the common denominator 120: 1 / 60 = 2 / 120, 1 / 30 = 4 / 120, and 1 / 40 = 3 / 120. Step 12: Sum of the rates is (2 + 4 + 3) / 120 = 9 / 120 = 3 / 40 work per day. Step 13: Time taken together is 1 / (3 / 40) = 40 / 3 days, which is 13 1/3 days. Verification / Alternative check: As a check, compute how much work is done in 13 1/3 days. In fractional form, 13 1/3 days is 40 / 3 days. At a combined rate of 3 / 40 per day, total work done is (3 / 40) * (40 / 3) = 1 unit of work, which confirms that the job is completed exactly. The logic connecting the relative work shares and the final combined rate is therefore correct. Why Other Options Are Wrong: 17 4/7 days, 16 days, and 20 days are all larger than 13 1/3 and correspond to smaller combined rates than what we calculated. Meanwhile, 15 3/2 days does not simplify to a reasonable fraction for this setup and would not match the derived rates based on the given information. Only 13 1/3 days is perfectly consistent with the defined relationships and the given time for C alone. Common Pitfalls: A common error is misinterpreting phrases like “does half as much work” and incorrectly assigning time values rather than rate values. Another pitfall is to assume that the time ratios are directly equal to the verbal ratios instead of translating into work rates first. Always remember that work comparisons are most easily handled by defining a base rate and building every other rate from that before converting to time. Final Answer: All three together, A, B, and C will complete the work in 13 1/3 days.

Question 12

Three boats B1, B2 and B3, when working together, can carry 60 people in a single trip. One day, in the early morning, boat B1 alone carries 50 people in a few trips and then stops. After that, B2 and B3 together carry the remaining people. In total…

Accepted Answer

Introduction / Context: This puzzle-style time and work question is framed in terms of boats carrying people across a river. Instead of work done per day, we talk about passengers carried per trip. You are given how many people the boats carry together, how many people B1 carries before stopping, and the total number of trips made to move 300 people. You must deduce the capacity of boat B1 and then compute the number of trips B1 alone would need to carry 150 people. Given Data / Assumptions: Boats B1, B2 and B3, working together in a single trip, carry 60 people. One day, B1 alone first carries 50 people and then stops. After that, boats B2 and B3 together carry the remaining passengers. The total number of trips made by all three boats (counting each time any boat or combination crosses with passengers) is 10. In total, 300 people are transported that day. We assume that each boat has a fixed capacity per trip and that every trip is fully utilized up to that boat’s capacity as implied by the numbers. Concept / Approach: Let the per-trip capacity of B1 be b1, and the combined per-trip capacity of B2 and B3 be b23. We know that when all three work together, they carry 60 people per trip, so b1 + b23 = 60. On the specific day described, B1 alone carries 50 people in n1 trips, so n1 * b1 = 50. Then B2 and B3 together carry the remaining 250 people in n2 trips, so n2 * b23 = 250. We are also told that the total number of trips is n1 + n2 = 10. We must find integer solutions for b1, n1, b23, and n2. Once we know b1, we can determine how many trips B1 alone needs to carry 150 people: trips = 150 / b1. Step-by-Step Solution: Step 1: Let b1 be the number of people B1 can carry in one trip. Let b23 be the number of people B2 and B3 together can carry in one trip. Step 2: Since together B1, B2, and B3 can carry 60 people per trip, we have b1 + b23 = 60. Step 3: Suppose B1 makes n1 trips alone. Then n1 * b1 = 50, since B1 alone carries 50 people at the start. Step 4: After B1 stops, the remaining number of people to be transported is 300 - 50 = 250. Step 5: Let B2 and B3 together make n2 trips. Then n2 * b23 = 250. Step 6: Total number of trips is n1 + n2 = 10. Step 7: From n1 * b1 = 50, because capacities and trips are integer quantities, b1 must be a factor of 50. So b1 can be 1, 2, 5, 10, 25, or 50. Step 8: For each possible b1, compute n1 = 50 / b1 and then b23 = 60 - b1. Check whether 250 is divisible by b23 to get an integer n2, and whether n1 + n2 = 10. Step 9: Taking b1 = 10 gives n1 = 50 / 10 = 5. Then b23 = 60 - 10 = 50. This implies n2 = 250 / 50 = 5, and n1 + n2 = 5 + 5 = 10, which matches the condition. Step 10: Thus, the only consistent solution is b1 = 10 passengers per trip and b23 = 50 passengers per trip. Step 11: To find the number of trips B1 alone would need to carry 150 passengers, we compute 150 / b1 = 150 / 10 = 15 trips. Verification / Alternative check: We can verify the scenario using the found capacities. B1’s capacity is 10 per trip, so in 5 trips B1 carries 5 * 10 = 50 people and stops. Then B2 and B3 together, with a combined capacity of 50, carry 50 people per trip for 5 trips, transporting 5 * 50 = 250 people. In total, 50 + 250 = 300 people are carried, and exactly 5 + 5 = 10 trips are made. This confirms that b1 = 10 is correct. Therefore, for 150 people, 150 / 10 = 15 trips is consistent. Why Other Options Are Wrong: 30 and 25 trips would imply much smaller capacities for B1, which would not satisfy the given constraints and integer conditions for carrying 50 people in whole trips. 10 trips would mean B1 carries 15 passengers per trip to reach 150, which conflicts with the unique solution b1 = 10 that is consistent with all constraints. 12 trips corresponds to a capacity of 12.5 passengers per trip, which is not allowed as we are dealing with whole passengers per trip. Common Pitfalls: A common mistake is to assume equal capacities or to divide the total passenger count incorrectly among the boats without considering integer constraints. Some learners overlook that the number of trips and the number of passengers must both be integers, which severely restricts possible capacities. Another pitfall is not using the equation b1 + b23 = 60, which connects the individual and combined capacities and is crucial to solving the system. Final Answer: Boat B1 would need to make 15 trips to carry 150 passengers alone.

Question 13

A, B, and C can complete a piece of work in 24 days, 6 days, and 12 days respectively. Working all together at their usual rates, in how many days will they complete the same work?

Accepted Answer

Introduction / Context: This is a straightforward time and work problem involving three workers A, B, and C. Each worker has a known individual time to complete the whole job alone. The problem asks how long all three will take to complete the same job when working together at their usual constant rates. This is a classic example of adding work rates to find a combined time. Given Data / Assumptions: A alone can complete the work in 24 days. B alone can complete the work in 6 days. C alone can complete the work in 12 days. All three are assumed to work at constant rates, and the total work is considered as one complete unit. They work together for the entire duration until the job is done. Concept / Approach: The concept is that each person’s efficiency can be expressed as a rate equal to 1 divided by the number of days they take to complete the job. The combined rate of multiple workers is the sum of their individual rates. Once you know the combined rate, you can find the total time required by taking the reciprocal of that combined rate. Using fractional arithmetic allows an exact answer in terms of a simplified fraction of a day. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: The rate of A is 1 / 24 work per day. Step 3: The rate of B is 1 / 6 work per day. Step 4: The rate of C is 1 / 12 work per day. Step 5: The combined rate when A, B, and C work together is the sum of these rates: 1 / 24 + 1 / 6 + 1 / 12. Step 6: Convert to a common denominator of 24: 1 / 24 remains 1 / 24; 1 / 6 = 4 / 24; 1 / 12 = 2 / 24. Step 7: Add the fractions: 1 / 24 + 4 / 24 + 2 / 24 = 7 / 24 work per day. Step 8: Time taken together is the reciprocal of the combined rate: time = 1 / (7 / 24) = 24 / 7 days. Verification / Alternative check: We can check by estimating decimals. 24 / 7 is approximately 3.4286 days. In one day, A does about 1 / 24 ≈ 0.0417 of the work, B does 1 / 6 ≈ 0.1667, and C does 1 / 12 ≈ 0.0833. Together they do about 0.2917 of the work per day, which is 7 / 24 ≈ 0.2917. Over 3.4286 days, the total work done is 0.2917 * 3.4286 ≈ 1 job, confirming the calculation is consistent. Why Other Options Are Wrong: 1 / 24 and 7 / 24 days are not realistic total times because they are much less than even B’s individual time, meaning the job would be finished faster than the fastest worker alone in an implausible way. 4 days is longer than the exact time 24 / 7 days and corresponds to a combined rate too small compared to the given rates. 3 days implies a higher combined rate than 7 / 24, which is not supported by the individual efficiencies. The only correct value is 24 / 7 days. Common Pitfalls: Learners sometimes mistakenly average the times 24, 6, and 12 directly instead of working with rates. Others may forget to take the reciprocal at the end and might report the combined rate instead of the combined time. Always remember that time calculations in such problems rely on adding rates, not adding times. Final Answer: Working together, A, B, and C will complete the work in 24/7 days.

Question 14

Sixteen women can complete a certain piece of work in 10 days, while fifteen men can complete the same work in 12 days. If all men and all women work on alternate days starting with all men on the first day, in how many days will the entire work be …

Accepted Answer

Introduction / Context: This question involves combined work done by two groups, men and women, who work on alternate days. The problem tests your ability to convert group times into daily work rates, sum them appropriately, and then track progress over a sequence of alternating days. You must determine the total number of days required to finish the job when work is done on alternate days by men and women starting with men. Given Data / Assumptions: Sixteen women can complete the work in 10 days. Fifteen men can complete the same work in 12 days. All men work together on days assigned to men, and all women work together on days assigned to women. Work proceeds in alternate full days: day 1 all men, day 2 all women, day 3 all men, and so on. The total work is considered as one complete unit and there is no overlap of work within a day between men and women. Concept / Approach: We first compute the daily group rate for all men and the daily group rate for all women. Then we determine how much of the work is completed in a two-day cycle: one day by men and one day by women. By repeating this cycle, we track how much work remains and finally identify the day on which the total work reaches or just exceeds 1 unit. Because the work pattern is alternating, we must check after each full day whether the work has been completed. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: Sixteen women complete the work in 10 days, so the combined rate of all women is 1 / 10 work per day. Step 3: Fifteen men complete the work in 12 days, so the combined rate of all men is 1 / 12 work per day. Step 4: On day 1, only men work and they complete 1 / 12 of the work. Step 5: On day 2, only women work and they complete 1 / 10 of the work. Step 6: Total work done in 2 days = 1 / 12 + 1 / 10. Using denominator 60, 1 / 12 = 5 / 60 and 1 / 10 = 6 / 60, so total = 11 / 60. Step 7: Each 2-day cycle thus completes 11 / 60 of the work. Step 8: After k such cycles, total work done is k * (11 / 60). We look for the smallest k such that this amount is at least 1. Step 9: After 5 cycles (10 days), the total work done is 5 * 11 / 60 = 55 / 60 = 11 / 12 of the work. Step 10: Work remaining after 10 days is 1 - 11 / 12 = 1 / 12 of the work. Step 11: Day 11 is a men’s day (since we started with men on day 1). Men alone can complete 1 / 12 of the work in one full day, which exactly matches the remaining work. Step 12: Therefore, the work is fully completed at the end of the 11th day. Verification / Alternative check: We can list the cumulative work after each day. Day 1 (men): 1 / 12. Day 2 (women): 1 / 12 + 1 / 10 = 11 / 60. Day 3: 11 / 60 + 1 / 12 = 16 / 60 = 4 / 15. Day 4: 4 / 15 + 1 / 10 = 11 / 30. Continuing this pattern for 10 days yields 11 / 12 completed, leaving 1 / 12 for day 11. Since men can do exactly 1 / 12 in a day, the calculation is consistent and the job ends exactly on day 11. Why Other Options Are Wrong: 10 5/2 days, 11 2/5 days, and 12 1/2 days assume fractional days and would require partial-day work scheduling, which is not described here because men and women work in whole-day shifts. Furthermore, these durations do not align with the cycle-based accumulation of work. 13 days is too long, as the work is already completed by the end of the 11th day. Only 11 days matches the calculated completion time. Common Pitfalls: A common mistake is to average the times of men and women or to divide the work in half without respecting the alternating pattern. Another pitfall is not keeping track of which group works on which day and thus miscounting the total days. Always compute the work done per cycle and then check exactly when the remaining fraction of work matches what can be done in the next scheduled day of work. Final Answer: The entire work will be completed in 11 days when men and women work on alternate days starting with men.

Question 15

When A, B, and C are deployed for a task, A and B together complete 70% of the work, and B and C together complete 50% of the work. Based on this information alone, who is the most efficient worker?

Accepted Answer

Introduction / Context: This conceptual time and work question focuses on comparing individual efficiencies using partial information about combined work. We are told how much of the job is completed by certain pairs of workers but are not given direct data about each individual. The aim is to determine who is the most efficient worker based only on the given pairwise completion percentages. Given Data / Assumptions: When A and B work together, they complete 70% of the work in the same reference time period. When B and C work together, they complete 50% of the work in that same period. No additional information about individual times or total time is given. We are asked to compare individual efficiencies (rates of work) of A, B, and C and to identify the most efficient worker if possible. Concept / Approach: Let the work done in the reference time be 1 unit. The information that A and B do 70% of the work means their combined rate is 0.7 units per reference time. Similarly, B and C together doing 50% of the work means their combined rate is 0.5 units per reference time. We then express these relationships in terms of variables representing individual efficiencies and analyze whether the given system of equations is enough to uniquely rank A, B, and C by efficiency. Step-by-Step Solution: Step 1: Let the efficiencies (work rates) of A, B, and C in the given time frame be a, b, and c units of work respectively. Step 2: A and B together complete 70% of the work in that time, so a + b = 0.7. Step 3: B and C together complete 50% of the work in the same time, so b + c = 0.5. Step 4: Subtract the second equation from the first: (a + b) - (b + c) = 0.7 - 0.5. Step 5: This simplifies to a - c = 0.2, so a = c + 0.2. This tells us that A is more efficient than C by 0.2 units per reference time. Step 6: To compare B with A and C, rewrite b from either equation. From a + b = 0.7, b = 0.7 - a. From b + c = 0.5, b = 0.5 - c. Step 7: Substituting a = c + 0.2 into b = 0.7 - a gives b = 0.7 - (c + 0.2) = 0.5 - c. Step 8: This expression for b shows that its value depends on c. Different choices of c consistent with the equations will give different relative positions of B compared to A and C. Step 9: For example, if c is relatively small, B may be more efficient than both A and C. If c is relatively large (but still consistent with the equations), B may be less efficient than one or both of them. Step 10: Because many sets of (a, b, c) satisfy the two equations, we cannot uniquely determine which worker is the most efficient overall. Verification / Alternative check: Consider one possible scenario. Let c = 0.1. Then a = c + 0.2 = 0.3. From b + c = 0.5, we get b = 0.4. In this case, b (0.4) is the most efficient, then a (0.3), then c (0.1). Now consider another scenario: let c = 0.24. Then a = 0.44 and b = 0.26. In this scenario, a is the most efficient, b is in the middle, and c is the least efficient. Both sets of values satisfy the original equations, but they produce different conclusions about who is the most efficient worker. This confirms that the data are insufficient to decide uniquely. Why Other Options Are Wrong: Options A, B, and C each claim that a specific worker is definitely the most efficient, which is not justified by the given information. We have shown that multiple configurations of efficiencies satisfy the conditions with different workers being the most efficient in different scenarios. The option stating that all are equally efficient contradicts the equation a - c = 0.2, which clearly indicates that A and C do not have the same efficiency. Hence, only the answer that says the result cannot be determined is correct. Common Pitfalls: A common mistake is to assume that A must be the most efficient simply because A appears in the larger combined percentage (with B). Another error is to ignore the fact that B appears in both pairs and thus introduce unintended symmetry. Always check whether the number of independent equations matches the number of unknowns; if there are fewer equations, there may be infinitely many solutions and no unique ranking of efficiencies. Final Answer: Based on the given information alone, the most efficient worker cannot be determined.

Question 16

Four men can repair a road in 7 hours. How many men are required to repair the same road in 2 hours, assuming all men work at the same constant rate?

Accepted Answer

Introduction / Context: This is a standard inverse proportion time and work problem. You are given that four men can repair a road in a certain number of hours and asked how many men would be required to do the same work in fewer hours, assuming all men work at the same rate. It tests understanding that total work is proportional to the product of workers and time (man-hours). Given Data / Assumptions: Four men can repair the road in 7 hours. All men are identical in efficiency and work at a constant rate. We want to know how many men are needed to repair the same road in 2 hours. The total amount of work is fixed and does not change between scenarios. Concept / Approach: The total work required can be measured in man-hours. If W is the total work, then W = number of men * hours worked. Because the work is the same in both scenarios, we equate the man-hours for the first scenario with the man-hours for the second scenario. This gives a direct proportion that allows you to solve for the required number of men in the second scenario. Step-by-Step Solution: Step 1: Let the total work required to repair the road be W man-hours. Step 2: In the first scenario, four men take 7 hours. So W = 4 * 7 = 28 man-hours. Step 3: In the second scenario, let the required number of men be n, and the time is 2 hours. Step 4: The total work is the same, so W = n * 2 man-hours. Step 5: Equate the two expressions for W: 28 = 2n. Step 6: Solving for n gives n = 28 / 2 = 14. Step 7: Therefore, 14 men are required to complete the same work in 2 hours. Verification / Alternative check: We can check the result by computing the total man-hours in the second scenario. With 14 men working for 2 hours, total man-hours are 14 * 2 = 28 man-hours, which exactly matches the original 28 man-hours. Since the total work is unchanged, the solution is consistent and correct. Why Other Options Are Wrong: 13 men would provide only 26 man-hours, which is not enough to complete the required 28 man-hours of work in 2 hours. 16 men would give 32 man-hours, implying more work than necessary, and similarly 17 and 12 men would give 34 and 24 man-hours respectively, which also do not match the required 28 man-hours. Only 14 men produce the correct amount of total work in the given time. Common Pitfalls: Some learners mistakenly treat the relation between men and time as additive instead of multiplicative. Others may attempt to average or directly scale, forgetting that doubling the number of workers halves the time needed if all else is constant, which is an inverse proportion. Always use the man-hours concept: work = men * time, and keep that product constant to handle such problems correctly. Final Answer: To repair the road in 2 hours, 14 men are required.

Question 17

A and B together can complete a piece of work in 40 days. A works alone for 20 days, after which B finishes the remaining work alone in 60 days. In how many days can B complete the whole work if working alone from the start?

Accepted Answer

Introduction / Context: This question is about sharing work over time between two workers, A and B, and then deducing the time taken by B alone to finish the whole work. You are given the joint time for A and B, the time A works alone initially, and the time B takes to finish the remaining work. From this, you must determine B’s individual time to complete the entire job working alone. Given Data / Assumptions: A and B together can complete the work in 40 days. A works alone for 20 days and then stops. B completes the remaining work alone in 60 days. The total work is taken as one complete job. All workers operate at constant rates and there are no breaks or changes in efficiency over time. Concept / Approach: We first convert the given combined time into a combined daily work rate for A and B together. Let the individual rates of A and B be a and b units of work per day. From the combined rate and the information about the partial work done by A and then B, we can set up equations relating a and b. Solving these equations allows us to find B’s rate and then B’s time to complete the full job alone by taking the reciprocal of that rate. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: Since A and B together take 40 days, their combined rate is (a + b) = 1 / 40 work per day. Step 3: A works alone for 20 days. The amount of work done by A in this period is 20a. Step 4: The remaining work is (1 - 20a). B finishes this in 60 days, so B’s rate b satisfies 60b = 1 - 20a. Step 5: We now have two equations: a + b = 1 / 40 and 60b = 1 - 20a. Step 6: From the first equation, express a in terms of b: a = 1 / 40 - b. Step 7: Substitute this expression for a into the second equation: 60b = 1 - 20(1 / 40 - b). Step 8: Simplify the term inside the parentheses: 20(1 / 40) = 1 / 2. So 60b = 1 - (1 / 2) + 20b = 1 / 2 + 20b. Step 9: Rearranging, 60b - 20b = 1 / 2, so 40b = 1 / 2. Step 10: Therefore b = (1 / 2) / 40 = 1 / 80 work per day. Step 11: Time taken by B alone to complete the entire work is the reciprocal of b: time = 1 / (1 / 80) = 80 days. Verification / Alternative check: We can verify by checking the total work done by A and B in the described scenario. If b = 1 / 80, then a = 1 / 40 - 1 / 80 = 2 / 80 - 1 / 80 = 1 / 80, which means A and B have equal rates. A working alone for 20 days completes 20 * (1 / 80) = 1 / 4 of the work. The remaining work is 3 / 4. B working alone at 1 / 80 completes this in 60 days, since 60 * (1 / 80) = 3 / 4. The total work done is 1 / 4 + 3 / 4 = 1, and together their rate a + b = 1 / 80 + 1 / 80 = 1 / 40, which matches the original combined time of 40 days. Why Other Options Are Wrong: 60 days, 70 days, 90 days, and 100 days do not satisfy the equation derived from splitting the work between A and B. If B took any of these times alone, his rate would be different and the sequence of 20 days of A followed by 60 days of B would either overshoot or fail to complete the work. Only 80 days is consistent with both the joint time and the partial work distribution described in the problem. Common Pitfalls: One common error is to assume that B’s time is simply related in a linear way to the combined time, such as doubling or halving without using equations. Another pitfall is to forget that the work done by A in the first 20 days must be subtracted from the total work, leaving the remainder for B. Mismanaging this subtraction leads to incorrect values for B’s rate. Final Answer: B alone would complete the whole work in 80 days.

Question 18

If 2 men and 3 women can complete a piece of work in 8 days and 3 men and 2 women can complete the same work in 7 days, then in how many days will 5 men and 4 women working together complete the work?

Accepted Answer

Introduction / Context: This time and work question involves two types of workers, men and women, each with their own unknown efficiency. We are given how long different combinations of men and women take to complete the same job. From this information, we must deduce the individual work rates and then calculate how long a new combination of 5 men and 4 women will take to complete the job. Given Data / Assumptions: Two men and three women together complete a piece of work in 8 days. Three men and two women together complete the same work in 7 days. We are asked for the time taken by 5 men and 4 women working together. The total work is taken as one unit and all men have the same efficiency, as do all women. Work rates remain constant throughout. Concept / Approach: We let the daily work done by one man be m and by one woman be w. The group rates for the given combinations become linear equations in m and w. Using these equations, we solve for m and w. Then we compute the combined rate for 5 men and 4 women and take the reciprocal to get the required time. The method involves solving a system of two linear equations with two unknowns. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: Let the rate of one man be m work per day, and the rate of one woman be w work per day. Step 3: Two men and three women together have a daily rate of (2m + 3w). They complete the work in 8 days, so 8(2m + 3w) = 1. Step 4: Simplify the first equation: 16m + 24w = 1. Step 5: Three men and two women together have a daily rate of (3m + 2w). They complete the work in 7 days, so 7(3m + 2w) = 1. Step 6: Simplify the second equation: 21m + 14w = 1. Step 7: We now solve the system: 16m + 24w = 1 and 21m + 14w = 1. Step 8: Multiply the first equation by 7 to get 112m + 168w = 7. Multiply the second equation by 8 to get 168m + 112w = 8. Step 9: Subtract the first new equation from the second: (168m + 112w) - (112m + 168w) = 8 - 7, giving 56m - 56w = 1. Step 10: Thus, m - w = 1 / 56. Also, using one of the original equations, we can solve explicitly to get m = 1 / 28 and w = 1 / 56. Step 11: Now compute the rate of 5 men and 4 women: rate = 5m + 4w = 5 * (1 / 28) + 4 * (1 / 56). Step 12: Simplify: 5 / 28 + 4 / 56 = 5 / 28 + 1 / 14 = 5 / 28 + 2 / 28 = 7 / 28 = 1 / 4 work per day. Step 13: Time taken is the reciprocal of the rate: time = 1 / (1 / 4) = 4 days. Verification / Alternative check: We can check the consistency by plugging m and w back into the original equations. For 2 men and 3 women, rate = 2 * (1 / 28) + 3 * (1 / 56) = 2 / 28 + 3 / 56 = 1 / 14 + 3 / 56 = 4 / 56 + 3 / 56 = 7 / 56 = 1 / 8, so they take 8 days. For 3 men and 2 women, rate = 3 * (1 / 28) + 2 * (1 / 56) = 3 / 28 + 1 / 28 = 4 / 28 = 1 / 7, so they take 7 days. This confirms that our m and w values and subsequent calculations are correct. Why Other Options Are Wrong: 3 days and 2 days correspond to much higher combined rates than 1 / 4 per day and would imply inconsistencies with the original equations. 5 days and 6 days imply combined rates smaller than 1 / 4, again inconsistent with the derived values of m and w. Only 4 days aligns perfectly with the work rates inferred from the original combinations. Common Pitfalls: Common mistakes include treating the number of workers as a direct ratio to time without solving the underlying system of equations or miscalculating the combined rates. Some may attempt to average the days without accounting for the different numbers of men and women. Correct handling of simultaneous equations and consistent units is essential for success in this type of problem. Final Answer: Five men and four women together will complete the work in 4 days.

Question 19

Ten men and fifteen women together can complete a work in 6 days. One man alone takes 100 days to complete the same work. In how many days will one woman alone complete the work?

Accepted Answer

Introduction / Context: This question involves comparing the efficiencies of men and women in a work problem. You are given the combined time for a group of 10 men and 15 women, and the individual time for one man. Using this, you must determine the time required for one woman alone to complete the same work. The problem is a direct application of linear equations based on group work rates. Given Data / Assumptions: Ten men and fifteen women together can complete the work in 6 days. One man alone takes 100 days to complete the work. We assume that all men have the same work rate and all women have the same work rate, with work rates remaining constant. The total work is considered as one complete job. Concept / Approach: We treat the total work as 1 unit and define m as the daily work rate of one man and w as the daily work rate of one woman. The given data then leads to an equation for the group rate of 10 men and 15 women and another equation for the rate of one man. Solving these allows us to find w, and the time taken by one woman alone is the reciprocal of w. The method is straightforward algebra using total work and daily rates. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: Let the rate of one man be m work per day, and the rate of one woman be w work per day. Step 3: One man alone takes 100 days, so his rate is m = 1 / 100 work per day. Step 4: Ten men and fifteen women working together have a combined rate of 10m + 15w work per day. Step 5: They complete the job in 6 days, so 6(10m + 15w) = 1. Step 6: Simplify this equation: 60m + 90w = 1. Step 7: Substitute m = 1 / 100 into the equation: 60 * (1 / 100) + 90w = 1. Step 8: Compute 60 * (1 / 100) = 0.6, so 0.6 + 90w = 1. Step 9: Rearrange to find w: 90w = 1 - 0.6 = 0.4. Step 10: So w = 0.4 / 90 = 4 / 900 = 1 / 225 work per day. Step 11: Time taken by one woman alone to complete the work is the reciprocal of w: time = 1 / (1 / 225) = 225 days. Verification / Alternative check: We can verify the result by checking the combined work. One woman completes 1 / 225 of the job per day. One man completes 1 / 100 of the job per day. For 10 men and 15 women, total daily work = 10 * (1 / 100) + 15 * (1 / 225) = 10 / 100 + 15 / 225 = 0.1 + 0.0666... = 0.1666... which is 1 / 6. Over 6 days, they complete 6 * (1 / 6) = 1 job, matching the original statement. This confirms that the result of 225 days for one woman is correct. Why Other Options Are Wrong: 215, 235, 240, and 250 days correspond to work rates slightly larger or smaller than 1 / 225 per day, and do not satisfy the equation 60m + 90w = 1 when combined with the known rate of one man. Only 225 days yields a rate that makes the combined total daily work equal to 1 / 6, consistent with the 6 days needed by the group. Common Pitfalls: Some learners incorrectly assume that the time for a woman can be found by simple proportional scaling without setting up the equation for the combined group. Others may mix up the roles of men and women or mis-handle the substitution of m into the group equation. Always express the total work and daily rates clearly, substitute known values carefully, and solve step by step. Final Answer: One woman alone will complete the work in 225 days.

Question 20

A can finish a work in 18 days and B can finish the same work in half the time taken by A. Working together at their usual rates, what fraction of the total work can A and B complete in one day?

Accepted Answer

Introduction / Context: This is a simple time and work question focusing on the fraction of work completed per day by two workers A and B when they work together. You are given A’s total time to complete the job and told that B takes half that time. The goal is to compute the combined daily work as a fraction of the whole job. Given Data / Assumptions: A alone can complete the work in 18 days. B alone can complete the work in half the time taken by A, which means in 9 days. Both work at constant rates and, when working together, their rates simply add. The total work is considered to be one complete unit. Concept / Approach: The key concept is that the rate of work of a person is the reciprocal of the time they take to complete the job alone. To find the fraction of work done per day when they work together, we compute A’s rate and B’s rate separately and then add them. The result is the fraction of the job completed in one day by A and B working jointly. Step-by-Step Solution: Step 1: Let the total work be 1 unit. Step 2: A alone takes 18 days to complete the work, so A’s rate is 1 / 18 of the work per day. Step 3: B alone takes half of 18 days, which is 9 days, so B’s rate is 1 / 9 of the work per day. Step 4: The combined daily rate when A and B work together is 1 / 18 + 1 / 9. Step 5: Convert to a common denominator of 18: 1 / 18 remains 1 / 18, and 1 / 9 = 2 / 18. Step 6: Add the fractions: 1 / 18 + 2 / 18 = 3 / 18. Step 7: Simplify 3 / 18 by dividing numerator and denominator by 3 to get 1 / 6. Step 8: Therefore, A and B together can complete 1 / 6 of the work in one day. Verification / Alternative check: We can verify by computing the total time they would take together. If they do 1 / 6 of the work each day, then they would need 6 days to complete the entire job. As a quick consistency check, note that A alone needs 18 days and B alone needs 9 days. It is reasonable for the joint time to be less than 9 days, and 6 days fits this expectation. This confirms that 1 / 6 of the work per day is a sensible and correct result. Why Other Options Are Wrong: One-fourth, one-third, half, and two-thirds of the work per day are all too large given the individual times of 18 and 9 days. For example, if they did half the work in a day, they would finish in 2 days, which is unrealistically fast compared to their individual times. Similarly, one-fourth or one-third would lead to total times of 4 or 3 days, which do not match the work rates implied by 18 and 9 days. Only one-sixth of the work per day is consistent with their individual capacities. Common Pitfalls: Some learners mistakenly average the times 18 and 9 (getting 13.5 days) instead of working with rates. Others may forget to simplify the combined fraction 3 / 18 to 1 / 6. Always remember that when dealing with time and work, you should add rates (1 / time) rather than adding or averaging the times themselves. Final Answer: A and B together can complete one-sixth of the work in one day.

Time and Work Questions