SSC CGL Preparation: Concepts and Practice

Learning outcome

Convert efficiency ratios into time ratios, combine linked ratios, and calculate mixed-team capacity.

Concepts and assumptions

Assume constant individual productivity, comparable work, and contributions that add without interference. Unless hours are explicitly changed, every worker has the same daily working hours. Workers within each stated category are equally efficient.

A daily efficiency ratio of m:n means outputs of mk and nk for some common positive scale k. For a fixed job, completion times are inversely proportional to these outputs, so the time ratio is n:m.

When two ratios share a worker, first make that worker’s ratio value identical before combining them. Unmatched middle terms cannot be joined directly.

Worker equivalence follows from equal output. If p trained workers match q assistants over the same time, then:

p × trained-worker rate = q × assistant rate.

Therefore one trained worker equals q/p assistants in productive capacity.

Choose a reference worker, convert the entire team into equivalent reference workers, and multiply by days to measure total work. If daily hours change, use equivalent worker-hours instead:

Equivalent workers × hours per day × days.

This works because each reference worker-hour contributes the same amount.

Worked examples

Example 1 — Reversing an efficiency ratio. A:B efficiency is 3:5. B finishes a job in 18 days. Find A’s time.

B’s rate is 1/18 job daily. A’s rate = (3/5) × (1/18) = 1/30. Therefore A takes 30 days. Check: time ratio 30:18 = 5:3, the reverse of 3:5.

Example 2 — A mixed team. Four trained workers match six assistants. Eight trained workers complete a job in 15 days. How long will six trained workers and six assistants need?

Six assistants equal four trained workers. The new team therefore equals 6 + 4 = 10 trained workers. The job requires 8 × 15 = 120 trained-worker-days. New time = 120/10 = 12 days. This uses equivalent productive capacity, not the new team’s actual headcount of 12.

Example 3 — Linked ratios. A:B efficiency is 2:3 and B:C is 6:5. Together they finish in 8 days. Find each individual completion time.

Multiply A:B by 2 to obtain 4:6. Hence A:B:C = 4:6:5. Choose outputs of 4, 6 and 5 units daily. Joint output = 15 units daily; total job = 15 × 8 = 120 units. A needs 120/4 = 30 days; B needs 120/6 = 20 days; C needs 120/5 = 24 days. Their daily fractions add to 1/30 + 1/20 + 1/24 = 1/8.

Common mistakes

Keeping time and efficiency ratios in the same order; combining unmatched ratios; treating different worker categories as equally productive; ignoring changed daily hours.

Practice questions

1. A:B efficiency is 4:7. A needs 35 days for a job. Find B’s time.
2. Three trained workers match five assistants. Nine trained workers finish in 10 days. Find the time for six trained workers and ten assistants.
3. A:B efficiency is 3:4 and B:C is 2:3. Together they finish in 12 days. Find C’s time alone.
4. Eight equally efficient workers finish a job in 15 days at 6 hours daily. How many such workers are needed for 10 days at 8 hours daily?

Worked solutions

1. Time ratio A:B = 7:4. Since 7 parts represent 35 days, one part is 5 days. B needs 4 × 5 = 20 days.
2. Ten assistants equal six trained workers. New capacity = 6 + 6 = 12 trained workers. Required work = 9 × 10 = 90 trained-worker-days. Time = 90/12 = 7 1/2 days.
3. Scale B:C to 4:6, giving A:B:C = 3:4:6. Daily total = 13 units; job size = 13 × 12 = 156 units. C needs 156/6 = 26 days.
4. Work requires 8 × 6 × 15 = 720 worker-hours. Each new worker supplies 10 × 8 = 80 hours. Required workers = 720/80 = 9 workers.