Time Work and Tank Flow Mixed Review

Use the three lessons together

This free separate review mixes work and tank-flow decisions. It follows Work Units Combined Rates and Efficiency, Changing Teams and Finishing the Remaining Work, and Pipes Leaks and Net Tank Flow. It introduces no new method.

Attempt the questions before reading the key. Use untimed learning mode first; you may pause and return to the relevant lesson. For each answer, write a short line naming the whole or capacity, active rate, relevant interval and final check. Do not choose a method from “worker” or “pipe” alone.

These learning questions award one mark for a correct answer and zero otherwise, with no negative marking or pass cutoff. They are not a simulated full RRB paper or a statement of official examination marking. All questions are original.

Panel A Choose the balance before calculating

Read in the question	Record before dividing
Partial work or partial initial water	The specified whole and the amount still needed
More than one contributor	Which rates are active together; whether each adds or removes amount
A joining, leaving or valve event	A separate interval and the amount reached at its end
A completion/emptying target	Whether that boundary is reached before the next event
An unknown or contradictory datum	Whether one, several or no admissible results follow

Full-capacity pipe calibrations and actual initial water are separate facts. All time and flow units must be compatible. Constant-rate, independence and timing conditions are stated in each question; do not invent missing rate scales or schedules.

Mixed questions

1. A worker completes 7/15 of a fixed divisible job in 2 h 20 min at a constant rate. At the same rate, how much additional active time is required for the remaining work?

A. 5 h

B. 2 h 40 min

C. 8/15 h

D. 2 h 20 min

2. An inlet alone fills an empty tank completely in 8 h; an outlet alone empties a full tank in 20 h. Their constant rates are valid over the full range. The actual tank starts one-quarter full and both operate together until first full. Find the time.

A. 10 h

B. 40/3 h

C. 30/7 h

D. 10/3 h

3. A and B together complete one fixed divisible job in 5 h. A alone takes 8 h. Both have positive constant independent rates and work throughout without delays. What is B’s solo time?

A. 3 h

B. 13 h

C. 40/3 h

D. 3/40 h

4. With a leak closed, an inlet fills an empty tank completely in 10 h. With the same leak open throughout, the same inlet fills it from empty to full in 15 h. Assume unchanged constant rates over the full range and a leak that can drain to empty. Find the leak-only time to empty a full tank.

A. 5 h

B. 15 h

C. 6 h

D. 30 h

5. A completes 50% more work per active hour than B. At a constant rate B takes 18 h for the same fixed job. A’s rate also stays constant. How long does A take alone?

A. 9 h

B. 27 h

C. 36 h

D. 12 h

6. A tank of capacity 960 L initially holds 240 L. An inlet supplies 1.5 L/min and an outlet removes 30 L/h. Both rates are constant and both run together until first full. Find the time.

A. 16 h

B. 8 h

C. 12 h

D. 24 h

7. A needs 18 h and B 12 h alone for the same divisible job. A works alone for 6 h. B then joins, and both continue at unchanged constant independent rates until completion. Find total time from A’s start.

A. 10 h 48 min

B. 4 h 48 min

C. 36/5 h

D. 12 h

8. An inlet alone fills an empty tank completely in 10 h; an outlet alone empties a full tank in 15 h. Both have constant rates. The actual tank starts one-fifth full. Both run for 3 h, then the outlet closes and the inlet continues. How much additional time after closure is needed to become full?

A. 10 h

B. 7 h

C. 9 h

D. 21 h

9. A alone needs 20 h and B alone 30 h for one divisible job. They work together for 5 h at unchanged constant independent rates. B then stops, while A finishes alone. What is total time from the original start?

A. 35/3 h

B. 25 h

C. 50/3 h

D. 12 h

10. An inlet alone fills an empty tank completely in 9 h and an outlet alone empties a full tank in 9 h, at constant rates. The actual tank starts one-third full; both run together with no other flow. Which statement is correct while this operation remains unchanged?

A. It remains one-third full

B. It becomes full in 6 h

C. It is already completely full at time 0

D. It becomes empty in 3 h

11. Nine identical workers can finish a divisible job in 15 working days at 4 active hours per day. After 5 such days, twelve identical workers continue at 5 active hours per day, with unchanged hourly productivity and no interaction losses. How many further working days are needed?

A. 11

B. 6

C. 15/2

D. 9

12. An inlet alone fills an empty tank completely in 20 h; an outlet alone empties a full tank in 12 h. The actual tank starts two-thirds full. Both run at these constant rates until the first empty/full event. Which event occurs first?

A. Empty after 30 h

B. Full after 10 h

C. Empty after 8 h

D. Empty after 20 h

13. A alone needs 15 h and B alone 20 h for the same divisible job. A starts first and keeps working; B joins later and stays until completion. At unchanged constant independent rates, the job finishes 12 h after A started. When did B join?

A. 4 h after A

B. 6 h after A

C. 12 h after A

D. 8 h after A

14. A tank starts two-thirds full. An inlet alone fills an empty tank completely in 9 h at constant rate. An outlet will open after 4 h only if filling is still underway; its rate is unspecified. Find the first-full time.

A. 4 h

B. 9 h

C. 3 h

D. It cannot be found without the outlet rate

15. A alone completes a job in 7 h. A report says A and B together take 9 h while both work throughout at positive constant independent rates, on the same divisible job. What is the correct conclusion?

A. B alone takes 63/2 h

B. The report is inconsistent with the stated rate model

C. There are many possible positive rates for B

D. B alone takes 16 h

16. Four workers finish a divisible job in 10 h at positive constant independent rates. Those same four keep their rates and work from the start with two extra workers. All six have positive constant independent rates, with no delays; the extra workers’ rates are not supplied. Which conclusion is justified?

A. The new time must be 20/3 h

B. The new time must be 8 h

C. No exact time is determined, though it is less than 10 h

D. The time must remain 10 h

17. A alone takes 8 h and B alone 20 h for the same divisible job. Both start together and keep constant independent rates until the job is finished, with no delay. Find the exact joint time in hours.

A. 14 h

B. 40/7 h

C. 28 h

D. 12 h

Answer key and explanations

1. B; 2. A; 3. C; 4. D; 5. D; 6. C; 7. A; 8. B; 9. C; 10. A; 11. B; 12. D; 13. D; 14. C; 15. B; 16. C; 17. B

1. B is correct. Convert 2 h 20 min to 7/3 h. Rate = (7/15)/(7/3)=1/5 job per hour. Remaining work is 8/15, so additional time=(8/15)/(1/5)=8/3 h=2 h 40 min. A is the total full-job time 5 h. C treats a dimensionless remaining fraction as hours. D repeats the time for 7/15 work even though 8/15 remains. Check: 7/3+8/3=5 h, and total work at 1/5 per hour is one job.

2. A is correct. Net rate=1/8−1/20=3/40 tank per hour. Remaining capacity=3/4, so time=(3/4)/(3/40)=10 h. Check: 1/4+10×3/40=1. B uses one whole tank instead of the remaining 3/4. C adds the outlet rate, giving (3/4)/(7/40)=30/7 h for a different operation. D divides the initial quarter by the net rate; that is not the amount needed to reach full.

3. C is correct. B rate=1/5−1/8=3/40 job per hour, so solo time=40/3 h=13 h 20 min. Check: 1/8+3/40=1/5. A subtracts solo and joint times. B adds them. Neither operation isolates B’s contribution in the same hour. D reports the rate number as a time without taking its reciprocal for one whole job.

4. D is correct. Leak magnitude=1/10−1/15=1/30 tank per hour; emptying one full tank takes 30 h. Check: 1/10−1/30=1/15, the observed net fill rate. A uses the fill-time difference 15−10. B uses the net filling time as a leak-only time. C adds the inlet and net rates, then takes the reciprocal: 1/(1/10+1/15)=6 h. The observed net accumulation is not another incoming flow.

5. D is correct. Relative to B, A’s rate factor is 1+50/100=3/2. Same-job time is multiplied by the reciprocal 2/3, so 18×2/3=12 h. A incorrectly reduces time by the same 50%. B multiplies time by the rate-increase factor and makes the faster worker slower. C uses half B’s rate rather than one-and-a-half times B’s rate. Check: (3/2)×(2/3)=1, preserving the work.

6. C is correct. Convert inflow: 1.5×60=90 L/h. Net rate=90−30=60 L/h; remaining capacity=960−240=720 L. Time=720/60=12 h. Check: 240+12×60=960. A uses 960/60 and ignores the initial water. B uses 720/90 and ignores outflow. D uses 720/30, incorrectly treating outward flow magnitude as net filling rate. The inlet, rather than the outlet, required the time-unit conversion in this item.

7. A is correct. A first completes 6/18=1/3, leaving 2/3. Joint rate=1/18+1/12=5/36. Additional time=(2/3)/(5/36)=24/5 h; total=6+24/5=54/5 h=10 h 48 min. Check: (54/5)/18+(24/5)/12=3/5+2/5=1. B reports additional time only. C assumes both start immediately. D uses 18−6, ignoring B’s later contribution and treating A-only remaining time as total elapsed time.

8. B is correct. First net rate=1/10−1/15=1/30. In 3 h, 1/10 tank is added, so level becomes 1/5+1/10=3/10. Remaining 7/10 at inlet-only rate 1/10 needs 7 h. A gives total time 3+7=10 h. C omits initial water and uses 9/10 remaining after the first stage. D keeps subtracting the outlet after closure: (7/10)/(1/30)=21 h. Check the actual schedule: 1/5+10/10−3/15=1.

9. C is correct. Joint rate=1/20+1/30=1/12, so first-stage work is 5/12. Remaining 7/12 at A’s rate 1/20 takes 35/3 further hours. Total=5+35/3=50/3 h=16 h 40 min. A gives only further time. B subtracts 5 from B’s solo 30 h even though B leaves and A does the rest. D uses the all-together time and wrongly keeps B active. Check: A contributes (50/3)/20=5/6 and B contributes 5/30=1/6.

10. A is correct. Net rate=1/9−1/9=0, so the stored amount stays at 1/3. B uses the remaining 2/3 with the inlet alone, ignoring the equal outflow. C confuses zero net change with already reaching full; the given starting amount is only 1/3. D uses the outlet alone to remove the initial third and ignores incoming water. There is no finite first-fill time while the specified zero-net operation continues.

11. B is correct. Whole effort=9×4×15=540 worker-hours. First five days complete 9×4×5=180, leaving 360. New daily effort=12×5=60, so further days=360/60=6. A gives total days 5+6=11. C keeps the old four-hour day: 360/(12×4)=15/2. D divides the whole 540 by the new daily effort, counting completed work again. Check: 180+6×60=540.

12. D is correct. Net rate=1/20−1/12=−1/30 tank per hour, so volume decreases. Initial 2/3 drains in (2/3)/(1/30)=20 h; 2/3−20/30=0. A uses a full tank rather than the actual initial stock. B takes a positive magnitude from the invalid fill expression (1/3)/(−1/30)=−10 h; a sign change cannot make filling happen. C uses the outlet alone: (2/3)/(1/12)=8 h. Stop at empty, before any negative-volume extrapolation.

13. D is correct. A contributes 12/15=4/5; B must contribute 1/5 and needs (1/5)/(1/20)=4 active hours. Join time=12−4=8 h. Check: 12/15+(12−8)/20=1 and 0<8<12. A confuses B’s active duration with when B starts. B assumes halfway; it would give 12/15+6/20=11/10 jobs. C gives B zero active time, leaving only A’s 4/5 job.

14. C is correct. Only 1/3 tank is needed. Time=(1/3)/(1/9)=3 h, before the planned outlet event at 4 h. A substitutes the event time for the first-fill time. B ignores the initial water. D requests data that cannot affect this earlier event. Check: 2/3+3/9=1. The first-full calculation ends before the outlet operates.

15. B is correct. Inferred B rate=1/9−1/7=−2/63 job per hour, contradicting a positive contribution. A discards the negative sign; a positive B rate 2/63 would make the pair rate 11/63 and time 63/11 h, not 9 h. C treats contradiction as missing information even though the rates determine a unique invalid value. D adds the times 7+9 without a contribution model. Two positive independent workers active throughout cannot take longer than A alone.

16. C is correct. A valid model gives each original worker rate 1/40 job per hour. If each extra also has 1/40, six workers take 20/3 h. If each extra has 1/80, combined rate is 4/40+2/80=1/8 and time 8 h. Both preserve the original group’s 10 h. A and B assert one permissible rate choice as necessary. D ignores two positive independent additions; they increase the group rate and shorten time, but their unspecified magnitudes prevent a unique numerical time.

17. B is correct. Rate=1/8+1/20=5/40+2/40=7/40 job per hour, so time=40/7 h. Equivalently, choose 40 work units: rates 5 and 2 units per hour give 40/(5+2). Contributions are 5/7 and 2/7, summing to one. A averages solo times; C adds them; D subtracts them. None represents the joint one-hour contribution, and all three exceed the fastest solo time 8 h.

Use a wrong answer to choose the next step

• Questions 1, 3, 5, 15 and 17: return to Lesson 1 for rate units, missing contributions, inverse efficiency and positive-rate validity
• Questions 7, 9, 11, 13 and 16: return to Lesson 2 for interval ledgers, active hours and sufficient schedule/rate data
• Questions 2, 4, 6 and 8: return to Lesson 3 for signed flow, initial stock, leak inference and valve changes
• Questions 10, 12 and 14: return to the boundary and event-order parts of Lesson 3; compare with Lesson 2’s early-completion case

Rebuild the ledger before retrying. Explain the wrong option you selected, then retry the related formative item listed in that lesson. A correct answer with an invalid explanation still needs repair. No score here guarantees examination performance.

Sources and review scope

RRB CEN 09/2025 §14.1, printed/physical p.28 names Time and Work and Pipes & Cistern. The topic list is illustrative and not necessarily exhaustive. Not all later amendments have been audited here; no topic-frequency claim is made.

The taught foundations are NCERT Ganita Prakash Grade 8 Part II, Chapter 3 §3.6, printed pp.63–68, first edition December 2025 as verified in its official prelims, PDF p.2, and the simple balance/equation reasoning in NIOS Secondary Mathematics 211, Chapter 5 §§5.2–5.3, printed pp.142–146. NIOS chapter publication year remains unverified. The application variants and every question, explanation and panel are independently written. This is a review of Time Work and Tank Flow, not the separate later Motion module or the whole examination.

Analogy

Use the same ledger idea in two settings. A progress strip records completed work; a tank ledger records water stored. Each interval adds the contribution of the actors that are actually active. A worker’s ordinary contribution is positive; an outlet’s contribution to stored water is negative.

The comparison helps select a balance, but it does not erase the difference in boundaries. A job stops when its required work is complete. A tank has both empty and full limits, and real hardware may not maintain constant flows. Use only the stated arithmetic model and stop at the requested event.

Quick reference

• Name the whole, the starting state and the time unit
• Translate completion data into rates only after checking the calibration endpoints
• Combine compatible active rates with the correct signs
• Carry completed work or stored water forward when the schedule changes
• Divide the remaining required amount by a valid rate in the required direction
• Keep efficiency ratios and solo-time ratios inverse for the same work
• Check nonnegative active times, initial amounts, first-event order and the full contribution balance
• Distinguish a unique answer, insufficient information and inconsistent conditions
• Use the feedback to revisit a taught decision; this review adds no new shortcut