Question 1

6 men + 8 boys finish a job in 10 days; 26 men + 48 boys finish the same job in 2 days. In how many days will 15 men + 20 boys finish it?

Accepted Answer

Problem restatement Given two group-productivity scenarios, deduce the individual efficiency of a man and a boy, then compute the time for a third team. Given data (6m + 8b) × 10 = 1 job → 6m + 8b = 1/10. (26m + 48b) × 2 = 1 job → 26m + 48b = 1/2. Concept/Approach Let m, b be daily job-fractions of one man and one boy. Solve the linear system for m and b; then compute (15m + 20b) and invert to get time. Step-by-step calculation 6m + 8b = 1/10 … (1) 26m + 48b = 1/2 … (2) From (1): m = (1/10 − 8b)/6 Substitute in (2): 26(1/10 − 8b)/6 + 48b = 1/2 (13/3)(1/10 − 8b) + 48b = 1/2 13/30 − (104/3)b + 48b = 1/2 13/30 + (40/3)b = 1/2 → (40/3)b = 1/2 − 13/30 = 1/15 b = (1/15) × (3/40) = 1/200 Then from (1): 6m = 1/10 − 8(1/200) = 1/10 − 1/25 = 3/50 → m = 1/100 For 15 men + 20 boys: rate = 15m + 20b = 15/100 + 20/200 = 0.15 + 0.10 = 0.25 = 1/4 job/day Time = 1 ÷ (1/4) = 4 days Verification Plug m = 1/100, b = 1/200 back into (2): 26/100 + 48/200 = 0.26 + 0.24 = 0.50 = 1/2 (correct). Common pitfalls Forgetting to convert group-time statements into per-day equations first. Arithmetic errors when handling fractions; keep exact fractions to avoid rounding mistakes. Final Answer 4 days

Question 2

Sakshi can finish a job in 20 days. Tanya is 25% more efficient than Sakshi. In how many days can Tanya finish the same job?

Accepted Answer

Problem restatement Given Tanya is 25% more efficient (faster) than Sakshi, compute Tanya's time from Sakshi's time. Given data Sakshi's time = 20 days → rate S = 1/20 job/day. Tanya's efficiency = 1.25 × S → rate T = 1.25 × (1/20) = 1/16 job/day. Concept/Approach Time is inverse of rate. If efficiency increases by 25%, time multiplies by 1/1.25. Step-by-step calculation Tanya's time = 20 ÷ 1.25 = 16 days Verification Direct inverse check: rate 1/16 indeed corresponds to 16 days. Common pitfalls Adding 25% to time (incorrect). The 25% applies to rate, not time. Final Answer 16 days

Question 3

A finishes a work in 18 days, B in 15 days. B works for 10 days and leaves. In how many days will A alone finish the remaining work?

Accepted Answer

Problem restatement With different solo times for A and B, B contributes for 10 days first. Compute the leftover work for A and the time needed. Given data A's time = 18 days → A's rate = 1/18 job/day. B's time = 15 days → B's rate = 1/15 job/day. B works 10 days. Concept/Approach Work done = rate × time. Remaining = 1 − (B's contribution). Step-by-step calculation Work by B in 10 days = 10 × (1/15) = 2/3 Remaining work = 1 − 2/3 = 1/3 Time for A = (1/3) ÷ (1/18) = 6 days Verification Equivalent unit method: LCM 90 units. B does 6 units/day × 10 = 60; remaining 30 units; A does 5/day → 6 days. Common pitfalls Subtracting days instead of subtracting work. Final Answer 6 days

Question 4

A and B together finish a job in 30 days. A works alone for 16 days; B then finishes the remainder alone in 44 days. In how many days can B alone finish the whole job?

Accepted Answer

Problem restatement We know the joint time for (A + B), a head start by A, and the finishing time by B alone. Find B's solo time. Given data a + b = 1/30 (job/day). A works 16 days alone: work = 16a. B finishes the rest in 44 days: 44b = 1 − 16a. Concept/Approach Express a = 1/30 − b and substitute into the remainder equation to solve for b. Step-by-step calculation 44b = 1 − 16(1/30 − b) = 1 − 16/30 + 16b = 1 − 8/15 + 16b = 7/15 + 16b 44b − 16b = 7/15 → 28b = 7/15 → b = 1/60 Therefore, B's time = 1 ÷ b = 60 days Verification Then a = 1/30 − 1/60 = 1/60; A's 16 days produce 16/60 = 4/15 of the job; remainder 11/15 finished by B in 44 days → 44 × (1/60) = 11/15 (consistent). Common pitfalls Using 30 − 16 − 44 style arithmetic on days (invalid); must use rates and work. Final Answer 60 days

Question 5

A, B, C can complete a work in 24, 6, and 12 days respectively. Working together, how long will they take to complete the work?

Accepted Answer

Problem restatement Combine individual rates of A, B, and C to get a joint completion time. Given data A's time = 24 days → rate = 1/24. B's time = 6 days → rate = 1/6. C's time = 12 days → rate = 1/12. Concept/Approach Independent workers' rates add: joint rate = 1/24 + 1/6 + 1/12. Step-by-step calculation Joint rate = 1/24 + 1/6 + 1/12 = 1/24 + 4/24 + 2/24 = 7/24 job/day Joint time = 1 ÷ (7/24) = 24/7 days 24/7 days ≈ 3.4286 days ≈ 3 days 10 hours 17 minutes Verification In 24/7 days, total work = (7/24) × (24/7) = 1 (exact). Common pitfalls Taking average of times (24, 6, 12) rather than adding rates. Final Answer 24/7 days (≈ 3 days 10 hours)

Question 6

A can finish a job in 24 days, B in 9 days, and C in 12 days. B and C work together for 3 days; A completes the remainder. How many days does A take for the remainder?

Accepted Answer

Problem restatement Two faster workers (B, C) start and then leave; compute remaining work for A and the time A needs. Given data A's rate = 1/24 job/day. B's rate = 1/9 job/day. C's rate = 1/12 job/day. B and C work together for 3 days. Concept/Approach First compute the fraction completed by (B + C) in 3 days, subtract from 1 to get the remainder, then divide by A's rate. Step-by-step calculation (B + C) rate = 1/9 + 1/12 = 7/36 Work done in 3 days = 3 × (7/36) = 7/12 Remainder = 1 − 7/12 = 5/12 Time for A = (5/12) ÷ (1/24) = (5/12) × 24 = 10 days Verification Unit-check: 10 days at 1/24 per day completes 10/24 = 5/12, equal to the remainder. Common pitfalls Multiplying instead of dividing by A's rate when converting remainder to time. Final Answer 10 days

Question 7

A alone can do a job in 4 hours; (B + C) together in 3 hours; (A + C) together in 2 hours. How long will B alone take?

Accepted Answer

Problem restatement Given three composite times involving A, B, and C, find B's solo time. Given data a = 1/4 job/hour. b + c = 1/3 job/hour. a + c = 1/2 job/hour. Concept/Approach Isolate c from (a + c), then subtract from (b + c) to get b. Step-by-step calculation c = (a + c) − a = 1/2 − 1/4 = 1/4 b = (b + c) − c = 1/3 − 1/4 = (4 − 3)/12 = 1/12 job/hour B's time = 1 ÷ (1/12) = 12 hours Verification Check consistency: a = 1/4, c = 1/4 → (a + c) = 1/2 (given). Then (b + c) = 1/12 + 1/4 = 1/3 (given). Common pitfalls Adding or subtracting the times rather than the rates. Final Answer 12 hours

Question 8

A completes 80% of a work in 20 days. A and B together finish the remaining work in 3 days. How long would B alone take to do the whole work?

Accepted Answer

Problem restatement From A's partial-completion time and the joint finishing time with B, find B's solo time for the entire job. Given data A completes 80% (0.8) in 20 days → A's rate = 0.8/20 = 0.04 = 1/25 job/day. Remaining work = 20% = 0.2. (A + B) finish 0.2 in 3 days → (A + B) rate = 0.2/3 = 1/15 job/day. Concept/Approach Compute B's rate by subtracting A's rate from the combined rate, then invert to get B's time for 1 job. Step-by-step calculation A's rate = 1/25 (A + B) rate = 1/15 B's rate = 1/15 − 1/25 = (5 − 3)/75 = 2/75 job/day B's time = 1 ÷ (2/75) = 75/2 = 37.5 days Verification/Alternative Check the finish segment: In 3 days, A contributes 3 × 1/25 = 0.12; B contributes 3 × (2/75) = 0.08; total = 0.20 (the remaining 20%). Common pitfalls Assuming A's 20-day time is for the full job; it is for 80% only. Subtracting times instead of subtracting rates to get B's speed. Final Answer 37.5 days

Question 9

A is 30% more efficient than B. If A alone can complete a job in 23 days, how long will A and B together take to finish it?

Accepted Answer

Problem restatement A is 30% more efficient than B. A alone takes 23 days. Find the joint time of A and B working together. Given data A's time alone = 23 days ⇒ A's rate = 1÷23 job/day. A is 30% more efficient than B ⇒ rA = 1.3 × rB. Concept/Approach Convert the efficiency statement into rates. From rA = 1.3 rB, obtain rB. Then add rates to get the combined time. Step-by-step calculation rA = 1÷23r = r÷1.3 = (1÷23)÷1.3 = 1÷(23 × 1.3)Combined rate = r + r = 1÷23 + 1÷(23 × 1.3) = (1 + 1÷1.3) ÷ 231 + 1÷1.3 = 1 + 10÷13 = 23÷13So combined rate = (23÷13) ÷ 23 = 1÷13 job/dayTime together = 1 ÷ (1÷13) = 13 days Verification/Alternative Let B's rate be b. Then A's rate = 1.3b. If A = 1÷23, then 1.3b = 1÷23 ⇒ b = 1÷29.9. Sum = 1÷23 + 1÷29.9 ≈ 0.076923… ⇒ time ≈ 13 days. Common pitfalls Interpreting “30% more efficient” as time reduction instead of rate increase. The percentage applies to rate, not days. Final Answer 13 days

Question 10

A and B can complete a work together in 30 days. They work together for 20 days; then B leaves, and A finishes the remaining work in the next 20 days. In how many days can A alone finish the work?

Accepted Answer

Problem restatement The pair completes a project in 30 days. After 20 days of working together, B leaves; A alone completes the rest in 20 days. Find A's solo time. Given data (A + B) time = 30 days ⇒ (A + B) rate = 1÷30 per day. First phase: 20 days together. Second phase: 20 days by A alone. Concept/Approach Compute the fraction completed in phase 1; the remainder is done by A in phase 2, which gives A's rate. Step-by-step calculation Work done in first 20 days = 20 × (1÷30) = 2÷3Remaining work = 1 − 2÷3 = 1÷3A completes 1÷3 in 20 days ⇒ A's rate = (1÷3) ÷ 20 = 1÷60 per dayTherefore, A's time alone = 60 days Verification/Alternative Check units: rate in job/day times days gives fraction of job, consistent in both phases. Common pitfalls Forgetting that only A works in the last 20 days; mixing average of 30 and 20 without using rates is incorrect. Final Answer 60 days

Question 11

A man, on tour travel first 160 km, at 64 km/hr. and the next 160km at 80km/hr. the average speed for the first 320 km. of the tour is?

Accepted Answer

Average speed = 2t1 x t2/( t1 + t2) = (2x64x80)/(64+80)km = (2 x 64 x 80 ) / 144 km/hr. =71.11km/hr.

Question 12

By walking at 3/4 of his usual speed, a man reaches his office 20 minute later than usual. this usual time is?

Accepted Answer

At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time = (4/3 of usual time ) – (usual time ) = 20min. ⇒ 4x/3 -x = 20 ⇒ x/3 =20 ⇒ x=60 min

Question 13

A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is

Accepted Answer

Suppose the man covers first distance in x hrs. and second distance in y hrs. Then , 4x + 5y = 35 and 5x +4y= 37 Solving these equation , we get X=5 and y=3 Total time taken = (5+3) hrs= 8hrs

Question 14

6 men or 10 women can reap a field in 15 days, then the number of days that 12 men and 5 women will take to reap the same field is :

Accepted Answer

Work done by 6 men = work done by 10 women. Work done by 1 man = work done by 10 / 6 = 5/3 women ∴ 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women ∴ W1 x D 1 = W2 x D2 W = women, D=days 10 X 15 = 25 x D2 D2 = 6

Question 15

6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?

Accepted Answer

6C + 2M = 6days 36C + 12M = 1 days Again 1M = 2C ∴ 36+12X2 = 1 day 60 children can do the work in 1 day Now, 5 men = 10 children ∴ 10 children can do the work in 6 days.

Question 16

2 men or 5 women or 7 boys can finish a work in 469 days, then the number of days taken by 7 men, 5 women and 2 boys to finish the work is :

Accepted Answer

2 men = 7 boys ⇒ 1 man = 7/2 days 5 women = 7 boys ⇒ 1 women = 7/5 boys 7 men + 5 women + 2 boys = 7 x 7/2 + 5 x 7/5 + 2 = 67/2 boys Now, B1 x D1 = B2 x D2 7 X 469 = 67/2 x D2 D2 = 98 days

Question 17

314 weavers weaves 6594 shawls in 1/6 hours. What is the number of shawls weaved per hour by an average weaver?

Accepted Answer

In 1 hour 314 weavers weave = 6594 x 6 shawls In 1 hour 1 weaver weaves = ( 6594 x 6 ) / 314 shawls = 126 shawls

Question 18

The Ratio of efficiencies of A, B and C is 6 : 5 : 4 and if A, B and C received total Rs. 27000 for the whole work. What is the share of C, if the money is distributed in the ratio of amount of work done, individually?

Accepted Answer

Ratio of efficiencies of A,B and C = 6 : 5 : 4 ∴ Share of C = [4/(6+5+4)] x 27000 = 7200

Question 19

If 2 men or 3 women or 4 boys can do a piece of work in 52 days. Then the same piece of work will be done by 1 man, 1 woman and 1 boy in?

Accepted Answer

Work done by 2 men = 3 woman = 4 boys ∴ 1 man = 2 boys 1 woman = 4/3 boys ∴ boys X days = 4 X 52 (boys - days) Again 1 man + 1 woman + 1 boy = 2 + 4/3 + 1 = 13/3 boys B1 x D1 = B2 x D2 B = boys, D = days 4 X 52 = 13/3 x D2 D2 = 48 days

Question 20

A contractor undertook a work to complete in 60 days. But just after 20 days he observed that only 1/5th of the project work had been completed. To complete the work in time (i.e,In rest days) minimum how many workers he had to increase, if there we…

Accepted Answer

Work done = 1/5 remaining work = 4/5 ∴ 4 (20 x 75) = 40 x N N = 150 Therefore 75 men should be increased.

Time and Work Questions