Time & Work • Topic 5 of 5

Men-Women-Children

These problems mix workers of different efficiencies — men, women and children — and are solved with the master chained formula M1·D1·H1/W1 = M2·D2·H2/W2, where W is the amount of work. The reliable strategy is to first reduce everyone to a single common unit using the given equivalence (e.g. "2 men = 3 women" means 1 man = 1.5 women, so convert the whole crew into "woman-units"). Once everyone speaks one currency, it collapses to an ordinary man-days problem. A frequent CAT phrasing gives two separate crews finishing the same job in different times and asks about a third mixed crew — set up the equivalence from the first two, then plug into the formula. If the jobs differ in size, the W1 and W2 terms carry that ratio. Keep H1/H2 in only when hours per day are stated; otherwise drop them. The trap is mixing units (counting a man as a woman) — always convert before you add.

✅ Solved examples

1. 2 men = 3 women in efficiency. 8 men finish a job in 12 days. How long for 12 women?
12 women = 8 men (since 2 men = 3 women ⇒ 12 women = 8 men). Same as 8 men ⇒ 12 days.
2. 3 men or 6 women can do a job in 20 days. How many days for 4 men and 8 women together?
3 men = 6 women ⇒ 1 man = 2 women. Total work = 6 women × 20 = 120 woman-days. Crew = 4 men + 8 women = 8 + 8 = 16 women ⇒ 120/16 = 7.5 days.
3. 6 men and 8 boys finish a job in 10 days; 26 men and 48 boys in 2 days. Time for 15 men and 20 boys?
Let man = m, boy = b. 10(6m+8b) = 2(26m+48b) ⇒ 60m+80b = 52m+96b ⇒ 8m = 16b ⇒ m = 2b. Total work = 10(6·2b+8b) = 10×20b = 200b. Crew = 15m+20b = 30b+20b = 50b ⇒ 200b/50b = 4 days.
4. 4 men, 5 women and 6 children can do a job in 5 days. A man, a woman and a child do work in the ratio 3:2:1. How long for 1 man, 1 woman, 1 child together?
In child-units: man = 3, woman = 2, child = 1. Crew 1 rate = 4·3 + 5·2 + 6·1 = 12+10+6 = 28 child-units/day. Total work = 28 × 5 = 140 child-units. New crew = 3+2+1 = 6/day ⇒ 140/6 ≈ 23.33 days.

✏️ Practice — try these, take hints as needed

1. 3 men = 5 women. 9 men finish a job in 20 days. Days for 15 women?
Convert: 15 women = 9 men (3 men = 5 women).
Same workforce as 9 men.
Therefore same time.
20 days
2. 2 men or 3 women do a piece of work in 12 days. Days for 4 men and 3 women together?
2 men = 3 women ⇒ 1 man = 1.5 women; total = 3 women × 12 = 36 woman-days.
4 men + 3 women = 6 + 3 = 9 women.
36 ÷ 9.
4 days
3. 1 man = 2 women = 3 children in efficiency. 3 men, 4 women and 6 children finish a job in 8 days. Time for 1 man, 1 woman, 1 child?
In child-units: man = 3, woman = 1.5, child = 1; crew = 9 + 6 + 6 = 21/day.
Total work = 21 × 8 = 168 child-units.
New crew = 3 + 1.5 + 1 = 5.5/day; 168 ÷ 5.5.
≈ 30.55 days
4. 12 men finish a wall in 8 days; 8 women finish the same wall in 12 days. Days for 12 men and 8 women together?
Total work = LCM(8·12... use rates): men rate 1/8 wall total via man-days = 96 man-days; women = 96 woman-days, so 1 man = 1 woman here.
Combined = 12 + 8 = 20 (man-equivalents).
96 ÷ 20.
4.8 days
5. 5 men and 2 boys do a job in 4 days; 1 man and 1 boy do it in 14 days. Days for 1 man alone?
4(5m+2b) = 14(1m+1b) ⇒ 20m+8b = 14m+14b ⇒ 6m = 6b ⇒ m = b.
Total work = 4(5m+2m) = 28m.
1 man alone: 28m ÷ m.
28 days

📝 Topic test — 8 questions

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