Question 1

A takes twice as long as B to finish a job, and C takes three times as long as B. Working together, they finish the job in 12 days. How many days would A take to do the job alone?

Accepted Answer

Introduction / Context: Express all times relative to B. Convert to rates, add to get the joint rate, equate to the given joint time, and back out A’s time. Given Data / Assumptions: A = 2B, C = 3B (in time). Together time = 12 days. Concept / Approach: If B needs t days, then rates are: A = 1/(2t), B = 1/t, C = 1/(3t). Sum to get 11/(6t). Set 12 * 11/(6t) = 1 to solve for t and then 2t. Step-by-Step Solution: Joint rate = 1/(2t) + 1/t + 1/(3t) = 11/(6t)12 * (11/(6t)) = 1 → 22/t = 1 → t = 22A alone time = 2t = 44 days Verification / Alternative check: Compute joint rate with t=22: 1/44 + 1/22 + 1/66 = 1/12 → consistent. Why Other Options Are Wrong: They correspond to incorrect algebra when resolving t or doubling to get A’s time. Common Pitfalls: Adding times rather than rates; mixing up “takes twice the time” with “twice as efficient.” Final Answer: 44

Question 2

Ten men and eight women together complete a work in 5 days. One woman’s 1-day work equals half of one man’s 1-day work. In how many days will four men and six women complete the same work?

Accepted Answer

Introduction / Context: Use the given equivalence (1 woman = 1/2 man) to convert all workers to man-equivalents, then compute time for the smaller team. Given Data / Assumptions: 10 men + 8 women → 5 days. 1 woman = 1/2 man. Concept / Approach: Compute the total work in man-days, then divide by the rate of 4 men + 6 women in man-equivalents. Step-by-Step Solution: Equivalent men for (10, 8) = 10 + 8*(1/2) = 14Job size = 14 * 5 = 70 man-daysEquivalent men for (4, 6) = 4 + 6*(1/2) = 7Time = 70 / 7 = 10 days Verification / Alternative check: Half-efficiency rule applied consistently across both teams yields an exact integer. Why Other Options Are Wrong: They miscompute either the man-equivalent conversion or the final division. Common Pitfalls: Using 2 women = 1 man inconsistently; ensure consistent conversion on both teams. Final Answer: 10 days

Question 3

Five men started a job expected to be done in 15 days. After 5 days, 10 women joined them and together they finished in the next 5 days. If 10 women alone had started the job, in how many days would they have completed it?

Accepted Answer

Introduction / Context: Infer women-to-man efficiency from the mixed phase, then apply it to compute the time for 10 women to complete the full job alone. Given Data / Assumptions: 5 men for 15 days → job size = 75 man-days. After 5 days (25 man-days done), 10 women join and the remaining is done in 5 more days. Concept / Approach: Use remaining work and team rate over the next 5 days to deduce 1 woman in man-equivalents, then compute the women-only time. Step-by-Step Solution: Remaining work after 5 days = 75 - 25 = 50 man-daysIn next 5 days: (5 men + 10 women)*5 days = 50 → 25 man-days + 50 woman-days = 50Thus, 50 woman-days = 25 man-days → 1 woman-day = 1/2 man-dayWomen-only rate (10 women) = 10 * (1/2) = 5 man-days/dayTime = 75 / 5 = 15 days Verification / Alternative check: Back-substitution to mixed phase confirms consistency with the observed completion time. Why Other Options Are Wrong: They reflect incorrect conversion between women and men or wrong total job size. Common Pitfalls: Confusing “man-days” with “men-days” arithmetic; always treat efficiency conversion consistently. Final Answer: 15 days

Question 4

A contract is to be completed in 50 days with 105 men working 8 hours per day. After 25 days, only 2/5 of the work is finished. If each man now works 9 hours per day, how many additional men are required to complete the work on time?

Accepted Answer

Introduction / Context: When hours per day change mid-project, treat work in man-hours. Use completed fraction to infer total work, then determine the remaining workforce at new hours to meet the deadline. Given Data / Assumptions: Initial: 105 men, 8 h/day, 25 days → completed 2/5. Remaining time: 50 - 25 = 25 days at 9 h/day. Concept / Approach: Work W in man-hours. Compute W from the 2/5 completion, then solve M such that M*9*25 equals remaining 3/5 of W. Step-by-Step Solution: Initial man-hours = 105*8*25 = 21000 = 2/5 W → W = 21000*(5/2) = 52500Remaining = 3/5 W = 31500 man-hoursLet required men = M → M*9*25 = 31500 → M = 31500/225 = 140Additional men = 140 - 105 = 35 Verification / Alternative check: Check: 140*9*25 = 31500 → matches remaining workload. Why Other Options Are Wrong: Off-by-one choices (34, 36, 37) test arithmetic; only 35 satisfies the exact equality. Common Pitfalls: Forgetting to convert to man-hours after changing the daily hours. Final Answer: 35

Question 5

Fourteen men and twelve boys finish a work in 4 days, while eight men and sixteen boys finish the same work in 5 days. Compare the 1-day work of 1 man with that of 1 boy (expressed as a ratio of man to boy).

Accepted Answer

Introduction / Context: Two different mixed teams complete the same job in different times. Set up equations in terms of man (m) and boy (b) daily work to solve for the ratio m/b. Given Data / Assumptions: (14m + 12b)*4 = W (8m + 16b)*5 = W Concept / Approach: Equate the two expressions for W, then solve for m in terms of b. Step-by-Step Solution: 56m + 48b = 40m + 80b16m = 32b → m = 2bThus, man : boy (1-day work) = 2 : 1 Verification / Alternative check: Substitute m = 2b into either team equation to confirm both give the same total job W. Why Other Options Are Wrong: Ratios like 1 1/2 or 3 misrepresent the algebraic balance between m and b. Common Pitfalls: Mixing the total-time equality with rate equality; always equate total work W. Final Answer: 2

Question 6

In a factory there are equal numbers of women and children. Women work 6 hours per day and children work 4 hours per day. During festival time, workload increases by 50%. By rule, children cannot work more than 6 hours per day. If women and children…

Accepted Answer

Introduction / Context: When workload increases proportionally, compute new total hours required and distribute under constraints. Since children are capped at 6 hours, women must pick up the remaining hours. Given Data / Assumptions: Equal counts: n women, n children; equal hourly efficiency. Normal: women 6 h/day, children 4 h/day → total = n*(6 + 4) = 10n. Festival: +50% workload → need 15n hours/day. Children capped at 6 h/day → children contribute 6n. Concept / Approach: Compute required women-hours as the difference between total needed and capped children-hours, then compare with normal women-hours to find the extra per woman. Step-by-Step Solution: Required total hours = 15nChildren hours (capped) = 6nWomen required hours = 15n - 6n = 9nNormal women hours = 6n → Extra women hours = 9n - 6n = 3nExtra per woman = (3n)/n = 3 hours Verification / Alternative check: New total = 9n + 6n = 15n which equals 150% of original 10n → consistent. Why Other Options Are Wrong: They miscount either the cap or the proportional increase. Common Pitfalls: Assuming children also bear extra unlimited hours; here the legal cap binds. Final Answer: 3

Question 7

Three women and eighteen children together complete a work in 2 days. If six women alone can complete the same work in 3 days, then how many days will nine children alone take to finish the work?

Accepted Answer

Introduction / Context: Use the women-only completion time to define the total job, then determine a child’s rate by balancing the mixed team against the known total. Finally, scale to nine children. Given Data / Assumptions: 6 women → 3 days → total work W = 18w (where w = woman/day). (3w + 18c) * 2 days = W. Concept / Approach: From (3w + 18c)*2 = 18w, solve for c in terms of w. Then compute time for 9 children = W / (9c). Step-by-Step Solution: (3w + 18c)*2 = 18w → 3w + 18c = 9w18c = 6w → c = w/39 children rate = 9c = 3wTime = W / (3w) = (18w) / (3w) = 6 days Verification / Alternative check: Back-substitute c = w/3 into the mixed team to confirm the total matches W. Why Other Options Are Wrong: They conflict with the derived equivalence c = w/3. Common Pitfalls: Forgetting to multiply by the number of days when forming equations; dropping the factor 2. Final Answer: 6

Question 8

A contractor plans to finish a work in 124 days with 120 men. After 64 days, 2/3 of the work is complete. How many men can be dismissed so that the work still finishes on time?

Accepted Answer

Introduction / Context: If progress is ahead of schedule, the workforce can be reduced while still meeting the deadline. Compute remaining work and divide by remaining days to find the men needed, then subtract from current staff. Given Data / Assumptions: 120 men for 124 days planned. After 64 days, 2/3 of work completed; remaining 1/3 must be done in 124 - 64 = 60 days. Concept / Approach: Treat work in man-days with a constant per-man daily productivity. Infer the total job size W from the completed fraction; compute remaining crew M to finish 1/3 W in 60 days. Step-by-Step Solution: Let k = per-man daily output; 64*120*k = 2/3 W → W = 64*180*k = 11520kRemaining = 1/3 W = 3840kNeed M so that 60*M*k = 3840k → M = 64Men to discharge = 120 - 64 = 56 Verification / Alternative check: Check: 64 men for 60 days produce exactly 3840k which equals remaining work. Why Other Options Are Wrong: Other values do not balance remaining man-days over the remaining 60 days. Common Pitfalls: Assuming proportionality to the original plan rather than the observed progress. Final Answer: 56

Question 9

A contract must be finished in 92 days by 234 men working 16 hours per day. After 66 days, 4/7 of the work is completed. If each man now works 18 hours per day, how many additional men are needed to finish on time?

Accepted Answer

Introduction / Context: Varying daily hours changes throughput, so compute in man-hours. From the completed fraction, determine the remaining man-hours and solve for the headcount at the new hours to meet the deadline. Given Data / Assumptions: Initial setup: 234 men, 16 h/day, 92 days planned. After 66 days → completed 4/7. Remaining days: 92 - 66 = 26; new hours = 18 h/day. Concept / Approach: Let W be total man-hours. Use 66-day production to infer W, then compute men M such that M * 18 * 26 equals remaining 3/7 W. Subtract current 234 to get additional men. Step-by-Step Solution: 234*16*66 = 4/7 W → W = (234*16*66)*(7/4)Remaining = 3/7 W = 3*(234*16*66)/4M = [3*234*16*66] / [4*18*26] = 396Additional men = 396 - 234 = 162 Verification / Alternative check: Dimensional check in man-hours confirms units and cancellation are consistent. Why Other Options Are Wrong: They do not satisfy the precise balance of remaining man-hours within 26 days at 18 h/day. Common Pitfalls: Using man-days instead of man-hours after changing daily hours. Final Answer: 162

Question 10

A, B, and C can finish a work in 16, 32, and 48 days, respectively. They all start together. A works until completion; C leaves 2 days before the work is finished; B leaves 1 day before completion. In how many days is the work completed?

Accepted Answer

Introduction / Context: This problem requires piecewise accounting of contributions because the team composition changes over the last two days. Add work across three intervals with correct rates and durations, then solve for the total time T. Given Data / Assumptions: A = 1/16 per day (works throughout T days). B = 1/32 per day (works up to T - 1 days). C = 1/48 per day (works up to T - 2 days). Concept / Approach: Split the timeline: [0, T-2): A+B+C; [T-2, T-1): A+B; [T-1, T): A only. Sum contributions and set equal to 1 job, then solve for T. Step-by-Step Solution: Work = (A+B+C)*(T-2) + (A+B)*1 + A*1= (1/16 + 1/32 + 1/48)*(T-2) + (1/16 + 1/32) + 1/16Convert to /96: (11/96)*(T-2) + 9/96 + 6/96Total = (11T - 22 + 15)/96 = (11T - 7)/96Set (11T - 7)/96 = 1 → 11T = 103 → T = 103/11 = 9 4/11 days Verification / Alternative check: Approximate check: A alone would take 16 days; team support early on reduces time to a bit under 10 days, consistent with 9.36 days. Why Other Options Are Wrong: Integer guesses (10, 15, 20, 30) ignore the piecewise nature; only 9 4/11 satisfies the exact partitioned-rate equation. Common Pitfalls: Assuming constant team composition or forgetting to split the last two days correctly. Final Answer: 9 4/11 days

Question 11

Rashmi can complete a work in 16 days, Ravina in 12 4/5 (12.8) days, and Gitika in 32 days. They start together; Rashmi leaves after 4 days, and Ravina leaves 3 days before completion. How long does the work take in total?

Accepted Answer

Introduction / Context: Another staged-contribution problem: all three begin, then Rashmi leaves early and, later, Ravina leaves before the end. Sum contributions over the appropriate intervals to solve for total time T. Given Data / Assumptions: Rashmi (R) = 1/16 per day; Ravina (V) = 1/12.8 = 5/64 per day; Gitika (G) = 1/32 per day. All three work for the first 4 days; then only V and G until the last 3 days; finally only G for the last 3 days. Concept / Approach: Partition the timeline into three intervals and add their contributions. Let T be total days. Middle interval length is (T - 7) days. Step-by-Step Solution: Work in first 4 days = 4*(R+V+G) = 4*(1/16 + 5/64 + 1/32)= 4*(0.0625 + 0.078125 + 0.03125) = 0.6875Work in middle (T-7) days = (T-7)*(V+G) = (T-7)*(5/64 + 1/32) = (T-7)*0.109375Work in last 3 days = 3*G = 3*(1/32) = 0.09375Total = 0.6875 + 0.109375*(T-7) + 0.09375 = 10.109375*(T-7) = 0.21875 → T - 7 = 2 → T = 9 days Verification / Alternative check: Rough bounds: With three workers at start then fewer later, 9 days is plausible between the fastest joint and single-worker extremes. Why Other Options Are Wrong: They do not satisfy the partitioned-rate equality. Common Pitfalls: Misinterpreting “3 days before completion” as an isolated day rather than the last 3-day block without Ravina. Final Answer: 9 days

Question 12

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

Accepted Answer

A's 5 days work = 50% B's 5 days work = 33.33% C's 2 days work = 16.66% [100- (50+33.33)] Ratio of contribution of work of A, B and C = 50 : 33 1 3 : 16 2 3 = 3 : 2 : 1 A's total share = Rs. 1500 B's total share = Rs. 1000 C's total share = Rs. 500 A's one day's earning = Rs.300 B's one day's earning = Rs.200 C's one day's earning = Rs.250

Question 13

A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?

Accepted Answer

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10. C leaves 4 days before completion of the work, which means only A and B work during the last 4 days. Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10. Remaining Work = 7/10, which was done by A,B and C in the initial number of days. Number of days required for this initial work = 7 days. Thus, the total numbers of days required = 4 + 7 = 11 days.

Question 14

12 men can complete a work in 8 days. 16 women can complete the same work in 12 days. 8 men and 8 women started working and worked for 6 days. How many more men are to be added to complete the remaining work in 1 day?

Accepted Answer

1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192 Work done in 6 days= 6 8 96 + 8 192 = 6 × 1 8 = 3 4 Remaining work = 1/4 (8 men +8 women)'s 1 day work = 1 8 96 + 8 192 =1/8 Remaining work =1/4 - 1/8 = 1/8 1/96 work is done in 1 day by 1 man Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men

Question 15

P can complete a work in 12 days working 8 hours a day.Q can complete the same work in 8 days working 10 hours a day. If both p and Q work together,working 8 hours a day,in how many days can they complete the work?

Accepted Answer

P can complete the work in (12 x 8) hrs = 96 hrs Q can complete the work in (8 x 10) hrs=80 hrs Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80 (P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480. So both P and Q will finish the work in 480/11 hrs Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11

Question 16

A and B can do a piece of work in 30 days , while B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

Accepted Answer

2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8 =>(A+B+C)'s 1 day's work= 1/16 work done by A,B and C in 10 days=10/16 = 5/8 Remaining work= 3/8 A's 1 day's work= 1 16 - 1 24 = 1 48 Now, 1/48 work is done by A in 1 day. So, 3/8 work wil be done by A in =48 x (3/8) = 18 days

Question 17

A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 20 days and C alone in 60 days ,then B alone could do it in:

Accepted Answer

(A+B)'s 1 day's work=1/20 C's 1 day work=1/60 (A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15 Also A's 1 day's work =(B+C)'s 1 day's work Therefore, we get: 2 x (A's 1 day 's work)=1/15 =>A's 1 day's work=1/30 Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60 So, B alone could do the work in 60 days.

Question 18

A works twice as fast as B.If B can complete a work in 18 days independently,the number of days in which A and B can together finish the work is:

Accepted Answer

Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2 Therefore, A's 1 day's work=1/9 B's 1 day's work=1/18 (A+B)'s 1 day's work= 1/9 + 1/18 = 1/6 so, A and B together can finish the work in 6 days

Question 19

A Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by three-fourth of the scheduled time, s…

Accepted Answer

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days. D * x +(100- D) * 2x= 175x => D= 25 days Now , the work done in 25 days = 25x Total work = 175x Therefore, workdone before increasing the no of workers = 25 x 175 x × 100 % = 14 2 7 %

Question 20

A,B,C together can do a piece of work in 10 days.All the three started workingat it together and after 4 days,A left.Then,B and C together completed the work in 10 more days.In how many days can complete a work alone ?

Accepted Answer

(A+B+C) do 1 work in 10 days. So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5 Remaining work=1-2/5=3/5 (B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50 Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25 A does 1/25 work in in 1 day Therefore 1 work in 25 days.

Time and Work Questions