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 and B can do the same work in 15 days. They work together for 5 days, and the remaining work is completed by C alone in 2 more days. If the total wages for the whole work are Rs 3000, what are the daily wages of A…

Accepted Answer

Introduction / Context: This time and work problem focuses on sharing wages in proportion to the work done by each worker. A, B and C complete different portions of the total work, and the total wage amount must be divided fairly among them. You are then asked to find the effective daily wage for each person. Given Data / Assumptions: A can finish the entire work in 10 days. B can finish the entire work in 15 days. A and B work together for 5 days. C completes the remaining work alone in 2 days. Total wages for the complete work = Rs 3000. Wages are divided in proportion to work done by each person. Concept / Approach: We first convert total work into abstract units using a convenient least common multiple of days. Each person then gets a work rate per day. From this, we can compute how many units A, B and C complete. Their wage shares are proportional to these completed units. Finally, divide each person's share by the number of days they worked to find individual daily wages. Step-by-Step Solution: Assume total work W has 30 units (LCM of 10 and 15). A's rate = 30 / 10 = 3 units per day. B's rate = 30 / 15 = 2 units per day. Let C's rate be c units per day. In 5 days, work done by A and B together = 5 * (3 + 2) = 25 units + 5c? Actually C is not working yet, so they do 5 * (3 + 2) = 25 units. Remaining work = 30 - 25 = 5 units. C completes 5 units in 2 days, so C's rate c = 5 / 2 = 2.5 units per day. Now total work by A = 3 * 5 = 15 units. Total work by B = 2 * 5 = 10 units. Total work by C = 2.5 * 2 = 5 units. Work ratio A : B : C = 15 : 10 : 5 = 3 : 2 : 1. Total wage Rs 3000, so one share = 3000 / 6 = Rs 500. A's wage = 3 * 500 = Rs 1500. B's wage = 2 * 500 = Rs 1000. C's wage = 1 * 500 = Rs 500. Daily wages: A worked 5 days, so 1500 / 5 = Rs 300 per day. B worked 5 days, so 1000 / 5 = Rs 200 per day. C worked 2 days, so 500 / 2 = Rs 250 per day. Verification / Alternative check: The sum of daily wages times days = 300 * 5 + 200 * 5 + 250 * 2 = 1500 + 1000 + 500 = Rs 3000, which matches the total wage. Why Other Options Are Wrong: 200, 250, 300 would not respect the correct work ratio; the highest wage should go to A, not C. 200, 300, 400 and similar patterns give total wages or sharing ratios different from 3 : 2 : 1. Therefore only 300, 200, 250 matches both proportional work and total amount. Common Pitfalls: Some learners mistakenly distribute wages according to time worked alone, ignoring different efficiencies. Another error is to forget that C works only for 2 days, but A and B work for 5 days, so daily wages are not the same as total wages. Always base wage distribution on work done (units completed), not merely on days present. Final Answer: The daily wages are Rs 300, Rs 200 and Rs 250 for A, B and C respectively.

Question 13

A, B and C can complete a piece of work in 24 days, 30 days and 40 days respectively when working alone. They begin the work together, but C leaves 4 days before the completion of the work. In how many days was the entire work completed?

Accepted Answer

Introduction / Context: This time and work problem involves three workers with different efficiencies. All start together, but one worker (C) leaves before the work is over. The question checks if you can translate this situation into a simple equation using daily work rates and total work units. Given Data / Assumptions: A alone finishes in 24 days. B alone finishes in 30 days. C alone finishes in 40 days. All three start together. C leaves 4 days before completion. We assume constant daily work rates. Concept / Approach: Use the idea of work as units. Choose a convenient total work value as the least common multiple of 24, 30 and 40 so that each person's daily work rate is an integer. Compute total work contributed by each person up to completion, using the fact that C works D - 4 days if D is total days taken. Step-by-Step Solution: Let total work = LCM of 24, 30 and 40 = 120 units. A's rate = 120 / 24 = 5 units per day. B's rate = 120 / 30 = 4 units per day. C's rate = 120 / 40 = 3 units per day. Let the total time taken be D days. A works D days and does 5D units. B works D days and does 4D units. C leaves 4 days early, so works D - 4 days and does 3(D - 4) units. Total work equation: 5D + 4D + 3(D - 4) = 120. Simplify: 9D + 3D - 12 = 120, so 12D - 12 = 120. Therefore 12D = 132 and D = 132 / 12 = 11 days. Verification / Alternative check: In 11 days A does 5 * 11 = 55 units. B does 4 * 11 = 44 units. C works 7 days and does 3 * 7 = 21 units. Total = 55 + 44 + 21 = 120 units, which matches the total work. Why Other Options Are Wrong: 12 days would yield larger total work than 120 units using the same rates. 13 or 14 days would increase work even more, so the workers would overshoot the required job by a large margin. Thus only 11 days satisfies the exact work requirement. Common Pitfalls: Students sometimes forget that C works fewer days, mistakenly using D instead of D - 4 for C. Another error is computing LCM or daily work rates incorrectly, which carries as a mistake through the equation. Always write a clear expression for total work and check the arithmetic carefully. Final Answer: The work is completed in 11 days.

Question 14

Twelve men can complete a work in 8 days and sixteen women can complete the same work in 12 days. Eight men and eight women start the work together and work for 6 days. How many more men must be added so that the remaining work is finished in 1 more…

Accepted Answer

Introduction / Context: This time and work question involves converting men and women into a common unit of work efficiency. After part of the work is done by a mixed group, you must determine how many additional men are needed to finish the remaining work in a given time. It tests proportional reasoning and equation setup skills. Given Data / Assumptions: 12 men complete the work in 8 days. 16 women complete the same work in 12 days. 8 men and 8 women work together for 6 days. The remaining work must be completed in 1 more day by 8 men, 8 women and some extra men. Work rate is constant for each man or woman. Concept / Approach: First express total work in terms of man-days. Then convert women into equivalent men using their given performance. Find how much of the work is done in the first 6 days. Compute the remaining fraction of work, and then determine how many men are needed in the last day to finish it. Step-by-Step Solution: Let total work = W units. 12 men in 8 days: 12 * 8 = 96 man-days = W, so 1 man-day = W / 96. 16 women in 12 days: 16 * 12 = 192 woman-days = W, so 1 woman-day = W / 192. Compare: 1 man-day = W / 96 = 2 * (W / 192) = 2 woman-days. So 1 man is equivalent to 2 women in work rate. Now, 8 men and 8 women per day = 8 men + 8 women. Convert women: 8 women = 4 men-equivalent. So daily rate in men-equivalent = 8 + 4 = 12 men. Work per day = 12 * (W / 96) = W / 8. In 6 days they complete 6 * W / 8 = 3W / 4. Remaining work = W - 3W / 4 = W / 4. Let x be the extra men added for the final day. For the last day, total men-equivalent = (8 + x) men + 8 women. Convert women: 8 women = 4 men-equivalent, so total = 8 + x + 4 = 12 + x men-equivalent. Work in one day by these workers = (12 + x) * W / 96. Set equal to remaining work: (12 + x) * W / 96 = W / 4. Cancel W and solve: (12 + x) / 96 = 1 / 4, so 12 + x = 24. Therefore x = 12 extra men. Verification / Alternative check: With 12 + x = 24 men-equivalent, daily work is 24 * W / 96 = W / 4, which is exactly the remaining work. Why Other Options Are Wrong: 8 or 16 extra men would underwork or overwork the remaining fraction and not match exactly W / 4 in one day. 24 extra men would be far too many, doing more work than needed in the final day. Common Pitfalls: A common mistake is to average the days for men and women directly instead of converting them into a single equivalent unit. Some students forget that women are less efficient based on the numbers, leading to wrong equivalence. Always express all workers in one common efficiency unit, such as man-equivalents, before computing. Final Answer: The number of additional men required is 12.

Question 15

P can complete a certain piece of work in 12 days working 8 hours per day. Q can complete the same work in 8 days working 10 hours per day. If P and Q work together, each working 8 hours per day, in how many days can they finish the work?

Accepted Answer

Introduction / Context: This time and work problem mixes the number of days with different daily working hours. It checks if you can convert total work into a standard measure of worker-hours and then recompute the duration when both workers contribute simultaneously for a fixed number of hours per day. Given Data / Assumptions: P completes the work in 12 days at 8 hours per day. Q completes the work in 8 days at 10 hours per day. When working together, both work 8 hours per day. Work rate is constant for each person per hour. Concept / Approach: The key idea is to express the entire work in terms of worker-hours. Then, compute how many hours P and Q each need individually. Next, sum their hourly rates to get a combined hourly rate. Finally, divide total hours of work by the combined hourly rate, then convert into days using 8 working hours per day. Step-by-Step Solution: Total work in worker-hours using P: 12 days * 8 hours/day = 96 hours of P. So P's hourly rate = 1 / 96 of the work per hour. Total work using Q: 8 days * 10 hours/day = 80 hours of Q. So Q's hourly rate = 1 / 80 of the work per hour. Combined hourly rate when both work together = 1/96 + 1/80. Compute: 1/96 + 1/80 = (80 + 96) / (96 * 80) = 176 / 7680. Simplify: 176 / 7680 = 11 / 480. Therefore, together P and Q complete 11 / 480 of the work per hour. Total hours needed together = 1 / (11 / 480) = 480 / 11 hours. They work 8 hours per day, so number of days = (480 / 11) / 8 = 480 / (88) = 60 / 11 days. Verification / Alternative check: 60 / 11 is approximately 5.45 days. This lies between P's 12 days and Q's 8 days, which is reasonable for both working together. If only P worked 8 hours a day, it would take 12 days; with help from faster Q, total time must reduce, which it does. Why Other Options Are Wrong: 61/11, 71/11 or 72/11 days are larger and do not correspond to the true combined hourly rate calculation. Such values either do not match the computed worker-hours or contradict the logical expectation that teamwork finishes earlier than either individual time. Common Pitfalls: A common mistake is to treat "12 days" and "8 days" directly as work rates without adjusting for different hours per day. Another error is to average their days or hours instead of summing their hourly rates. Always convert to a consistent base (worker-hours) and then compute combined efficiencies. Final Answer: P and Q together can complete the work in 60/11 days.

Question 16

A and B together can complete a piece of work in 30 days, B and C together can complete it in 24 days, and C and A together can complete it in 20 days. All three work together for 10 days and then B and C leave. How many additional days will A alone…

Accepted Answer

Introduction / Context: This question is a classic example of using pairwise work rates to determine individual efficiencies. Three workers A, B and C have known times in pairs, and you must find how long A alone will need to complete the remaining work after a joint working period. Given Data / Assumptions: A + B together finish the work in 30 days. B + C together finish in 24 days. C + A together finish in 20 days. All three work together for 10 days. After 10 days, only A continues working. Each worker's daily work rate is constant. Concept / Approach: Let a, b and c be daily work rates of A, B and C. Use the three given pairwise equations to solve for a, b and c. Then find the combined rate a + b + c, calculate how much work is finished in 10 days, and finally compute how long A alone, at rate a, takes to complete the remainder. Step-by-Step Solution: Let total work = 1 unit for convenience. Then a + b = 1 / 30. b + c = 1 / 24. c + a = 1 / 20. Add all three: 2(a + b + c) = 1/30 + 1/24 + 1/20. Compute the right side: LCM of 30, 24, 20 is 120. 1/30 = 4/120, 1/24 = 5/120, 1/20 = 6/120, sum = 15/120 = 1/8. So 2(a + b + c) = 1/8, hence a + b + c = 1/16. Now find A's rate: a = (a + b + c) - (b + c) = 1/16 - 1/24. 1/16 - 1/24 = (3 - 2) / 48 = 1 / 48. Thus A's rate = 1/48 of work per day. In 10 days together, work done = 10 * (1/16) = 10/16 = 5/8. Remaining work = 1 - 5/8 = 3/8. Time for A alone to finish remaining = (3/8) / (1/48) = 3/8 * 48 = 18 days. Verification / Alternative check: Check that individual rates are consistent: b = (a + b) - a = 1/30 - 1/48 and c = (b + c) - b; all remain positive. The remaining work fraction is less than half, so a time less than A's full 48 days is expected; 18 days is reasonable. Why Other Options Are Wrong: 24, 30 or 36 days would imply A is much slower than computed and would contradict the earlier pairwise information. They also do not satisfy the proportion of remaining work 3/8 with rate 1/48. Common Pitfalls: Many aspirants forget to divide by 2 when adding the three pair equations. Some also treat total work as 100 or 120 without correctly normalizing the rates, which is fine only if done consistently. Always handle pair equations systematically and double check fraction arithmetic. Final Answer: A will take 18 days more to finish the work alone.

Question 17

A can do a certain work in the same time as B and C together can do it. A and B together can complete the work in 20 days, and C alone can do the work in 60 days. In how many days can B alone complete the work?

Accepted Answer

Introduction / Context: This time and work question uses a relationship between the time taken by A alone and the time taken by B and C together. You must translate that relationship into equations, use the given combined time for A and B, and the individual time for C, and then solve for B's time alone. Given Data / Assumptions: A alone takes the same time as B and C working together. A and B together finish the work in 20 days. C alone finishes the work in 60 days. Work rate of each person is constant. Concept / Approach: Let T_A, T_B and T_C denote the days taken by A, B and C alone. Using work rate form, 1 / T_A is A's daily work. The condition that A alone takes as long as B and C together implies that 1 / T_A = 1 / T_B + 1 / T_C. Also, 1 / T_A + 1 / T_B = 1 / 20 and T_C = 60. Solve these equations to get T_B. Step-by-Step Solution: Let T_A, T_B be days for A and B alone, and T_C = 60 days. Condition 1: A alone takes same time as B and C together. So 1 / T_A = 1 / T_B + 1 / 60. Condition 2: A and B together complete in 20 days. So 1 / T_A + 1 / T_B = 1 / 20. From condition 1, 1 / T_A = 1 / T_B + 1 / 60. Substitute into condition 2: (1 / T_B + 1 / 60) + 1 / T_B = 1 / 20. So 2 / T_B + 1 / 60 = 1 / 20. 1 / 20 - 1 / 60 = (3 - 1) / 60 = 2 / 60 = 1 / 30. Hence 2 / T_B = 1 / 30, so T_B = 60 days. Verification / Alternative check: Now find T_A: 1 / T_A = 1 / T_B + 1 / 60 = 1 / 60 + 1 / 60 = 1 / 30, so T_A = 30 days. Check A and B together: their daily work = 1/30 + 1/60 = 1/20, so they finish in 20 days, which matches given data. Why Other Options Are Wrong: 20 or 40 days for B would give different combined rates and would not satisfy both given conditions. 50 days does not satisfy the precise equations either; substituting would give inconsistent times. Common Pitfalls: One common error is misreading the phrase "same time in which B and C together can do it" and writing 1 / T_A = 1 / T_B + 1 / T_C incorrectly. Some students also manipulate fractions incorrectly when subtracting 1/60 from 1/20. Careful equation setup and fraction arithmetic are essential in such problems. Final Answer: B alone can complete the work in 60 days.

Question 18

A works twice as fast as B. If B can complete a piece of work in 18 days working alone, then in how many days can A and B together finish the same work?

Accepted Answer

Introduction / Context: This is a straightforward time and work question in which one worker is twice as efficient as another. You are asked to determine the combined time taken when both work together, given the slower worker's time alone. Given Data / Assumptions: B alone can finish the work in 18 days. A works twice as fast as B. Both A and B work together on the same piece of work. Each works at a constant daily work rate. Concept / Approach: Since A is twice as fast as B, A's daily work rate is double that of B. Convert B's 18-day completion time into a daily rate and multiply by 2 to find A's rate. Then add the two rates to get the joint rate and invert it to find the total time needed when both work together. Step-by-Step Solution: Let total work = 1 unit. B's time alone = 18 days, so B's rate = 1 / 18 of the work per day. A works twice as fast, so A's rate = 2 * (1 / 18) = 1 / 9 of the work per day. Combined rate of A and B = 1 / 9 + 1 / 18. Compute: 1/9 + 1/18 = (2 + 1) / 18 = 3 / 18 = 1 / 6. Therefore, A and B together complete 1/6 of the work each day. Time taken together to finish the entire work = 1 / (1 / 6) = 6 days. Verification / Alternative check: In 6 days, A does 6 * (1/9) = 6/9 = 2/3 of the work. B does 6 * (1/18) = 6/18 = 1/3 of the work. Total = 2/3 + 1/3 = 1 full work, so the answer is consistent. Why Other Options Are Wrong: 4 days would require a combined rate of 1/4 which is larger than 1/6 and does not match A + B's computed rate. 8 or 10 days would imply too slow a combined rate, again contradicting the calculated sum of individual rates. Common Pitfalls: Some learners mistakenly assume A's time is half of B's and then average the two times, which is not correct for work rates. Another error is adding days directly instead of adding rates. Always convert days into daily work fractions and then combine the rates. Final Answer: A and B together can finish the work in 6 days.

Question 19

A contractor employed a certain number of workers to finish a road within a scheduled time. After some time, he realised that at the current rate the work would be delayed by three-fourths of the scheduled time. He immediately doubled the number of …

Accepted Answer

Introduction / Context: This is a conceptual time-and-work puzzle about planning and adjusting workforce. The contractor initially underestimates the number of workers needed and later doubles the workforce to avoid a large delay. You have to determine what fraction of the total work had been finished before the adjustment, expressed as a percentage. Given Data / Assumptions: Scheduled completion time for the road = S (unknown) days. Initial number of workers = N (unknown). At some checkpoint, the contractor realises the work would be delayed by three-fourths of S (that is, total time would become S + 3S/4 = 7S/4) if he keeps N workers. He then doubles the workforce to 2N and finishes exactly in S days from the start. We assume constant work rate per worker and no other interruptions. Concept / Approach: The total quantity of work is fixed. We express it in terms of the original workforce N and the time that would have been needed without any change (7S/4). We then equate that with the actual scenario where the contractor works with N workers for some time and 2N workers for the remaining time but completes in S days. That gives us the time of switching and the fraction of work done by that moment. Step-by-Step Solution: Let total work = W units. Original workforce = N workers, each with unit rate per worker. If no change is made, total time needed at rate N is 7S/4 days. So W = N * (7S / 4). In reality, the contractor uses N workers for t days and then 2N workers for S - t days, finishing in S days. Actual work done: W = N * t + 2N * (S - t) = N t + 2N S - 2N t = 2N S - N t. Equate both expressions for W: N * (7S / 4) = 2N S - N t. Divide both sides by N: 7S / 4 = 2S - t. Solve for t: t = 2S - 7S / 4 = (8S / 4 - 7S / 4) = S / 4. So the workforce was doubled after t = S / 4 days from the start. Work done up to that time with N workers = N * (S / 4). Total work W = N * (7S / 4). Fraction of work completed = [N * (S / 4)] / [N * (7S / 4)] = 1 / 7. Convert to percentage: 1 / 7 = 14 2/7 % approximately. Verification / Alternative check: Remaining work fraction after S / 4 days = 6/7 of W. Rate after doubling = 2N, so time needed for remaining work = (6/7 W) / (2N). Since W = N * 7S / 4, substitute: (6/7 * N * 7S / 4) / (2N) = (6S / 4) / 2 = 3S / 4. Total time = S / 4 + 3S / 4 = S, so it matches the scheduled time. Why Other Options Are Wrong: 10% is less than 1/7 and does not satisfy the time algebra. 20% equals 1/5, again inconsistent with the equation derived from actual and projected times. "Can not be determined" is incorrect because the system of equations yields a unique fraction independent of actual S and N values. Common Pitfalls: Some students misinterpret "delayed by three-fourths of the scheduled time" as a fraction of remaining time, not of S. Others assume an explicit numerical S value without realising that it cancels out in the ratio. The key is to treat S and N as symbols and recognise that only the fraction of work completed matters. Final Answer: The work completed before increasing the number of workers was 1/7 of the total work, that is 14 2/7 %.

Question 20

A, B and C together can do a piece of work in 10 days. All three start the work together, but A leaves after 4 days. B and C together then complete the remaining work in 10 more days. In how many days can A alone complete the whole work?

Accepted Answer

Introduction / Context: This problem uses combined work rates initially and then a reduced team to finish the remaining work. The information allows you to determine the shared rate of B and C and then extract A's individual rate. Finally, you compute the time A would take working alone. Given Data / Assumptions: A, B and C together complete the work in 10 days. All three start together. A leaves after 4 days. B and C together finish the work in 10 more days. Work rate is constant for each worker. Concept / Approach: Let the total work be one unit. From the overall time for A, B and C, you get their combined daily work rate. Calculate how much work is done in the first 4 days. The remaining work is then handled by B and C alone over 10 days, which gives you the daily work rate of B + C. Subtracting this from the total combined rate gives A's individual rate and thus A's solo completion time. Step-by-Step Solution: Assume total work = 1 unit. A + B + C together finish in 10 days, so (a + b + c) = 1 / 10 per day. In the first 4 days, work done by A, B and C together = 4 * (1 / 10) = 4 / 10 = 2 / 5. Remaining work = 1 - 2 / 5 = 3 / 5. B and C together complete this 3 / 5 in 10 days. Thus (b + c) = (3 / 5) / 10 = 3 / 50 per day. Now (a + b + c) = 1 / 10 = 5 / 50. So A's rate a = (a + b + c) - (b + c) = 5 / 50 - 3 / 50 = 2 / 50 = 1 / 25 per day. Therefore A alone takes 25 days to finish the work. Verification / Alternative check: In 4 days, A does 4 * (1 / 25) = 4 / 25 of the work. In the same 4 days, B + C do 4 * (3 / 50) = 12 / 50 = 6 / 25. Total after 4 days = 4 / 25 + 6 / 25 = 10 / 25 = 2 / 5 of the work, which matches the earlier calculation. Then B and C do 10 * (3 / 50) = 30 / 50 = 3 / 5, completing the whole work. Why Other Options Are Wrong: 24, 23 or 21 days would imply different daily rates for A, which would not match the observed combined and B + C work rates. Substituting these values back would either leave unfinished work or produce more than the total work. Common Pitfalls: Learners sometimes incorrectly treat "10 more days" as total days, not extra days after the first 4. Another mistake is miscomputing the remaining work fraction after the first phase. Always carefully separate phases of work and recompute remaining work step by step. Final Answer: A alone can finish the work in 25 days.

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