Time And Work Questions for SSC GD

Core Concept: Work Rate

Work is a job to be completed. Rate is how much work one person/machine does per unit time. If a person completes a job in 10 days, their rate is 1/10 of the job per day. This fraction (1/time) is the KEY to all time-and-work problems. Always think in terms of 'fraction of work per day', not absolute numbers.

• Work Rate = 1 / Time taken (if working alone)
• If A completes work in 10 days, A's rate = 1/10 per day
• If B completes work in 15 days, B's rate = 1/15 per day
• Combined rate = Sum of individual rates
• Time taken together = 1 / Combined rate

Two-Person Work Problems

Most SSC GD questions involve two workers with different rates. The trick is to set up the rate equation and solve. Common scenario: A and B work together for some days, then one leaves, the other finishes. Break the problem into phases and use Work = Rate × Time for each phase.

• Always convert 'days to complete' into daily rate (fraction)
• If A and B work together for x days, work done = (1/T_A + 1/T_B) × x
• If A works alone for y days after B leaves, work done = (1/T_A) × y
• Total work = 1 (always, unless stated otherwise)
• Set up equation: Phase1_work + Phase2_work + ... = 1

Efficiency and Work Ratio

Efficiency is another word for work rate. If A is twice as efficient as B, A does twice the work in the same time, or same work in half the time. Efficiency ratio directly translates to rate ratio. This is useful when the problem gives relative efficiencies instead of absolute times.

• Efficiency ratio = Inverse of time ratio
• If A takes 10 days and B takes 20 days, efficiency ratio A:B = 20:10 = 2:1
• If A is 3 times more efficient than B, A takes 1/3 the time B takes
• When work is divided by efficiency, each person gets work proportional to their rate
• Payment or work share is usually proportional to efficiency

Pipe and Cistern (Water Tank) Problems

Pipes filling/emptying tanks follow the same rate logic as work problems. Inlet pipes add water (positive rate), outlet pipes remove water (negative rate). Net rate = Inlets - Outlets. Time to fill = 1 / Net rate. This is a common SSC GD variant.

• Inlet pipe rate = 1 / Time to fill tank alone (positive)
• Outlet pipe rate = 1 / Time to empty tank alone (negative)
• Net rate = Sum of inlet rates - Sum of outlet rates
• If net rate is positive, tank fills; if negative, tank empties
• Time to fill/empty = 1 / |Net rate|

Work Done by Multiple Workers Over Different Periods

Advanced scenario: A works for x days, B works for y days, C works for z days (possibly overlapping or sequential). Calculate work done in each period separately, sum them, and set equal to 1 (or given total work). This requires careful tracking of who works when.

• Break problem into time phases (when worker composition changes)
• For each phase, calculate: Work = (Sum of rates in that phase) × Duration
• Sum all phases' work and equate to total work (usually 1)
• Solve for the unknown (usually time or rate)

• Confusing time with rate. If A takes 10 days, rate is 1/10 per day, NOT 10. Aspirants often use 10 directly and get wrong answers.
• Forgetting to convert to common denominator when adding fractions. 1/12 + 1/18 requires LCM(12,18)=36, not just adding numerators.
• In pipe problems, forgetting to assign negative sign to outlet pipes. Outlet rate should be subtracted, not added.
• Setting up the equation wrong when work is divided into phases. Always track which workers are active in which phase.
• Assuming all work = 1. Some problems state work in units (e.g., 100 units). Read carefully and adjust the equation accordingly.
• Misinterpreting 'A is twice as efficient as B'. This means A's rate is 2x B's rate, so A takes half the time, NOT double.

• RATE = 1/TIME. This is the golden rule. Every time-and-work problem starts here. Memorize this fraction inversion.
• COMBINED = SUM OF RATES. When workers team up, add their individual rates. Simple addition of fractions.
• WORK = RATE × TIME. The universal formula. Use it in every phase of a multi-phase problem.
• INLET POSITIVE, OUTLET NEGATIVE. In pipe problems, inlets add (positive), outlets subtract (negative). Think: filling vs draining.
• EFFICIENCY RATIO = INVERSE TIME RATIO. If A:B time is 2:3, then A:B efficiency is 3:2. Flip the ratio.

• SSC GD typically asks 2-3 time-and-work questions per paper. Expect straightforward two-person problems (A and B type) and one pipe-cistern variant. Rarely asks complex three-person scenarios.
• Time limit is tight (15 seconds per question in Tier-1). Pre-calculate common LCMs (12, 15, 18, 20, 24) to save time. Avoid long decimal calculations; stick to fractions.
• Recent papers show preference for 'work done in x days' followed by 'remaining work done by one person' structure. Practice this pattern repeatedly.
• Pipe problems in SSC GD are usually straightforward (2-3 pipes max). Outlet pipes are common; watch for 'leak' or 'drain' keywords.
• If a problem mentions 'efficiency' or 'work ratio', convert to time equivalents immediately. Don't try to solve in efficiency units; it confuses.
• Always verify your answer: plug back into the original problem. If A takes 20 days and B takes 30 days, together should be less than 20 days (usually 12 days). Sanity check saves marks.