A ratio compares quantities measured in the same unit. It has no unit after simplification, but the original quantities still do. A proportion states that two ratios are equal. These ideas appear in sharing, scale, recipes, direct variation, inverse variation and unitary-method questions.
1. Read the order and convert units
If A:B = 3:5, A is represented by three equal parts and B by five. Reversing the order gives B:A = 5:3. Before forming a ratio, convert to a common unit: 2 metres : 50 centimetres = 200:50 = 4:1. Writing 2:50 ignores the unit difference and gives the wrong comparison.
Ratios can be multiplied or divided by a common non-zero factor without changing the relationship. If two numbers are 4x and 7x, their total is 11x and their difference is 3x. A given sum or difference can determine x. A ratio alone cannot determine the actual numbers; it gives only their relative sizes.
Worked example 1. Two amounts are in the ratio 4:7 and their sum is ₹1,650. There are 11 parts, so one part is ₹150. The amounts are ₹600 and ₹1,050. Their sum returns ₹1,650, giving a quick check.
2. Combining two ratios
To join A:B and B:C, make the B parts equal. If A:B = 2:3 and B:C = 4:5, scale the first ratio by 4 and the second by 3: A:B = 8:12 and B:C = 12:15. Therefore A:B:C = 8:12:15. Do not concatenate 2:3:5; the two instances of B represent different numbers of parts until adjusted.
For quantities that change over time, write their original forms before adding or removing values. If A:B = 3:4, put A=3x and B=4x. After adding the same 5 to each, the new ratio is (3x+5):(4x+5); it is generally not still 3:4. Solve the resulting linear equation using cross multiplication.
Worked example 2. A:B = 3:5. After adding 12 to each, the ratio becomes 5:7. Write (3x+12)/(5x+12)=5/7. Cross multiply: 21x+84=25x+60, so x=6. The original numbers are 18 and 30; after the addition they are 30 and 42, or 5:7.
3. Proportion and the cross-product check
The statement a:b = c:d means a/b=c/d, provided b and d are non-zero. Cross multiplication gives ad=bc. This is a test and a solving tool. If a missing value x appears, solve the equation and then substitute it back. The arithmetic mean and geometric mean are different ideas; a ratio question asking for “mean proportional” refers to a geometric mean.
Worked example 3. Find x if 8:12 = x:21. The equation 8/12=x/21 gives 12x=168 and x=14. Check: 8/12 and 14/21 both simplify to 2/3.
4. Direct and inverse variation
When a quantity is directly proportional to another, their ratio stays constant: y=kx. Doubling x doubles y. For a fixed job done by equally efficient workers, time varies inversely with worker count: workers × days is constant. Doubling workers halves the time only when work, individual efficiency and daily hours are unchanged.
Many exam questions combine direct and inverse changes. Production may rise with workers and hours but fall with more days only when a target output is fixed. Write a base equation from the actual physical relationship rather than memorising an arrow diagram. Units help: workers × days × hours per day × output per worker-hour gives total output.
Worked example 4. Eight workers make 240 parts in six days at five hours per day. Their output rate is 240/(8×6×5)=1 part per worker-hour. Twelve workers working four hours per day need 240/(12×4)=5 days for the same 240 parts.
5. The unitary method
The unitary method finds the value of one unit before scaling to the required number. It is especially useful when the relationship is linear and the “one unit” has a clear meaning. If 7 notebooks cost ₹210, one notebook costs ₹30, and 11 cost ₹330. This assumes identical notebooks at a fixed unit price; a bulk discount would change the model.
In work problems, one worker's rate per day can be a fraction, such as 1/12 of a job. In speed problems, one hour's distance is the speed. Unit labels protect the reasoning: ₹/notebook multiplied by notebooks produces rupees; parts/(worker-hour) multiplied by worker-hours produces parts.
Recall before practice
1. Convert 1.5 km : 300 m to a unit-free ratio. 2. Combine A:B=3:4 and B:C=6:7. 3. Explain why workers and days have an inverse relationship only for a fixed job and fixed individual output. 4. State the physical unit behind a unitary-method calculation.
Chapter 5 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. The third proportional to 9 and 21 is:
A. 189 B. 49 C. 33 D. 50
2. The mean proportional between 80 and 320 is:
A. 160 B. 25600 C. 120 D. 200
3. If k, 15, 45 and 135 are in continued proportion, then the value of k is:
A. 15 B. 6 C. 10 D. 5
4. A sum of ₹2,730 is divided among Sunita, Geeta and Neha in the ratio 3 : 4 : 6. What is the share of Geeta?
A. ₹1,260 B. ₹630 C. ₹840 D. ₹910
5. A sum of ₹5,270 is divided among Asha, Anita and Sunita in the ratio 2 : 6 : 9. What is the difference between the largest and the smallest share?
A. ₹2,790 B. ₹620 C. ₹1,240 D. ₹2,170
6. The wages of 13 labourers for 15 days are ₹87,750. What will be the wages of 24 labourers for 14 days at the same rate?
A. ₹1,51,200 B. ₹1,62,000 C. ₹81,900 D. ₹50,926
7. 8 men can complete a piece of work in 25 days. In how many days will 40 men complete a piece of work of the same size?
A. 7 days B. 3 days C. 125 days D. 5 days
8. If 14 kg of wheat cost ₹588, what is the cost of 9 kg of wheat?
A. ₹168 B. ₹378 C. ₹420 D. ₹915
9. 10 labourers can complete a piece of work in 7 days. How many labourers are needed to do the same work in 5 days?
A. 4 B. 14 C. 12 D. 7
10. In a school there are 575 teachers and 100 non-teaching staff. 4/5 of the teachers and 1/5 of the non-teaching staff are women. How many men are there in all?
A. 135 B. 480 C. 540 D. 195
11. A sum of ₹37,400 is divided among Rohit, Kavita and Sanjay in the ratio 3 : 6 : 2. What is the difference between the largest and the smallest share?
A. ₹6,800 B. ₹13,600 C. ₹20,400 D. ₹10,200
12. If A : B = 5 : 2 and B : C = 10 : 3, then A : B : C is:
A. 5 : 2 : 3 B. 25 : 10 : 3 C. 5 : 10 : 3 D. 15 : 6 : 20
13. The compounded ratio of 2 : 7 and 3 : 7 is:
A. 5 : 14 B. 2 : 3 C. 6 : 49 D. 49 : 6
14. A sum of ₹20,250 is divided among Sunita, Ravi and Priya in the ratio 2 : 8 : 5. What is the share of Ravi?
A. ₹10,125 B. ₹6,750 C. ₹2,700 D. ₹10,800
15. 22 men can pack 990 boxes in 9 hours. If 4 men more join, how many boxes will be packed in 6 hours?
A. 780 B. 660 C. 1257 D. 1170
16. 5 labourers can make 25 toys in 5 hours. If 2 labourers more join, how many toys will be made in 10 hours?
A. 70 B. 9 C. 50 D. 35
17. If 5/8 of A is equal to 8/11 of B, then A : B is:
A. 64 : 55 B. 55 : 64 C. 5 : 11 D. 5 : 8
18. If 8 erasers cost as much as 2 boxes, 9 boxes cost as much as 3 notebooks, and one eraser costs ₹23, what is the cost of one notebook?
A. ₹552 B. ₹92 C. ₹276 D. ₹69
19. A bike covers 270 km on 15 litres of petrol. How far will it travel on 10 litres of petrol?
A. 405 km B. 198 km C. 90 km D. 180 km
20. If 10A = 3B = 4C, then A : B : C is:
A. 10 : 3 : 4 B. 6 : 20 : 15 C. 4 : 3 : 10 D. 15 : 20 : 6
21. Find the mean proportional between 1.21 and 0.81.
A. 0.99 B. 0.9801 C. 1.01 D. 9.9
22. Find the mean proportional between 0.1 and 0.9.
A. 0.3 B. 0.09 C. 0.5 D. 3
23. If A : B = 7 : 9, B : C = 7 : 9 and A + B + C = 27,020, then the value of C is:
A. 6,860 B. 11,480 C. 11,340 D. 8,820
24. A car covers 286 km on 13 litres of petrol. How far will it travel on 20 litres of petrol?
A. 186 km B. 594 km C. 440 km D. 462 km
25. If 3 bags cost as much as 1 pens, 2 pens cost as much as 1 erasers, and one eraser costs ₹444, what is the cost of one bag?
A. ₹37 B. ₹148 C. ₹74 D. ₹2,664
26. ₹4,500 is distributed among 4 boys, 5 girls and 4 women so that the shares of a boy, a girl and a woman are in the ratio 6 : 7 : 4. How much does each boy get?
A. ₹240 B. ₹420 C. ₹1,440 D. ₹360
27. Find the mean proportional between 0.1 and 1.6.
A. 4 B. 0.85 C. 0.4 D. 0.16
28. A stock of food is enough for 300 soldiers in a garrison for 60 days. If 200 more soldiers join, for how many days will the same food last?
A. 20 days B. 60 days C. 36 days D. 100 days
29. If 1/8 of A is equal to 6/11 of B, then A : B is:
A. 48 : 11 B. 3 : 44 C. 1 : 6 D. 11 : 48
30. If 2x = 7y, then (x + y) : (x − y) is equal to:
A. 7 : 2 B. 5 : 9 C. 2 : 7 D. 9 : 5
31. The incomes of Arjun and Rahul are in the ratio 3 : 8 and their expenditures are in the ratio 1 : 5. If each of them saves ₹84,000, what is the income of Arjun?
A. ₹60,000 B. ₹1,26,000 C. ₹1,44,000 D. ₹3,84,000
32. If 27, 18, 12 and k are in continued proportion, then the value of k is:
A. 8 B. 216 C. 24 D. 6
33. 6 workers can type 72 pages in 12 hours. If 4 workers more join, how many pages will be typed in 6 hours?
A. 36 B. 86 C. 120 D. 60
34. ₹2,250 is distributed among 2 boys, 3 girls and 3 women so that the shares of a boy, a girl and a woman are in the ratio 3 : 7 : 6. How much does each girl get?
A. ₹350 B. ₹300 C. ₹1,050 D. ₹150
35. 18 men can complete a piece of work in 12 days. How many more men must be employed to finish the same work in 6 days?
A. 18 B. 6 C. 9 D. 36
36. Find the mean proportional between 0.49 and 0.09.
A. 0.21 B. 0.0441 C. 2.1 D. 0.29
37. Find the mean proportional between 0.25 and 0.16.
A. 2 B. 0.04 C. 0.205 D. 0.2
38. In a school there are 75 teachers and 30 non-teaching staff. 1/3 of the teachers and 4/5 of the non-teaching staff are women. What is the ratio of the total number of men to the total number of women?
A. 8 : 7 B. 7 : 8 C. 2 : 1 D. 1 : 4
39. 22 men can type 330 pages in 5 hours. If 8 men leave, how many pages will be typed in 10 hours?
A. 660 B. 420 C. 210 D. 259
40. The population of four towns and the ratio of men to women in each is given as: P: 3,438 (5 : 4); Q: 815 (3 : 2); R: 1,512 (3 : 4); S: 2,144 (3 : 5). What is the total number of men in towns P and Q?
A. 4,253 B. 2,449 C. 2,399 D. 1,854
41. The mean proportional between 3√3 and 27√3 is:
A. 9√9 B. 27 C. 81√3 D. 9√3
42. The mean proportional between √7 and 4√7 is:
A. 14 B. 2√49 C. 2√7 D. 4√7
43. 5 workers working 9 hours a day can do a piece of work in 5 days. In how many days will 3 workers working 5 hours a day do a similar piece of work that is three times as large?
A. 25 days B. 15 days C. 14 days D. 45 days
44. If (7a + 5b)/(7a − 5b) = 47/2, then a : b is:
A. 9 : 7 B. 49 : 45 C. 7 : 9 D. 47 : 2
45. The fourth proportional to 2√3, 4√3 and 8√7 is:
A. 32√7 B. 16√3 C. 16√7 D. 4√7
46. A sum is divided among Geeta, Mohan and Ravi in the ratio 4 : 5 : 2. If Geeta gets ₹160 more than Ravi, what is the total sum?
A. ₹720 B. ₹440 C. ₹293.33 D. ₹880
47. 10 men can weave 450 metres of cloth in 9 hours. In how many hours will 16 men weave 640 metres of cloth?
A. 6 hours B. 13 hours C. 20 hours D. 8 hours
48. A man gives 1/4 of his money to his elder son, 15% to his younger son and the rest to his three daughters in the ratio 1 : 1 : 2. If one son gets ₹6,100 more than the other, what is the largest share received by a daughter?
A. ₹18,300 B. ₹6,100 C. ₹9,150 D. ₹12,200
49. A garrison of 180 people has provisions for 47 days. After 34 days, 15 more people join them. For how many more days will the remaining provisions last?
A. 9 days B. 12 days C. 13 days D. 43 days
50. The population of four towns and the ratio of men to women in each is given as: P: 207 (2 : 1); Q: 1,290 (2 : 3); R: 1,176 (4 : 3); S: 297 (2 : 1). What is the total number of men in towns Q, R and S?
A. 1,377 B. 1,436 C. 2,763 D. 1,386
Chapter 5 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. B. If c is the third proportional to a and b, then a : b = b : c, so c = b²/a = 21²/9 = 441/9 = 49. (Continuing as an AP, 33, is the trap — it is a ratio, not a difference.)
APAR26-05-03 | Third proportional | Easy
2. A. If x is the mean proportional between a and b, then a : x = x : b, so x² = ab. x = √(80 × 320) = √25600 = 160. (The arithmetic mean (80 + 320)/2 = 200 is not the mean proportional.)
APAR26-05-06 | Mean proportional | Easy
3. D. In continued proportion each term divided by the previous is the same constant: 45/15 = 135/45 = 3. So k = 15 ÷ 3 = 5. (Continuing by a constant difference instead of a constant ratio is the usual error.)
APAR26-05-05 | Continued proportion: find a term | Easy
4. C. Sum of ratio terms = 3 + 4 + 6 = 13; one part = ₹2,730/13 = ₹210. Geeta's share = 4 × ₹210 = ₹840.
APAR26-05-07 | Dividing a sum in a ratio | Easy
5. D. Sum of ratio terms = 2 + 6 + 9 = 17; one part = ₹5,270/17 = ₹310. Largest share = 9 parts, smallest = 2 parts; difference = 7 × ₹310 = ₹2,170.
APAR26-05-04 | Dividing a sum in a ratio | Easy
6. A. Wages ∝ (men × days). Wage per man per day = ₹87,750/(13 × 15) = ₹450. Required wages = ₹450 × 24 × 14 = ₹1,51,200. (Changing only the men gives ₹1,62,000; only the days gives ₹81,900.)
APAR26-05-36 | Wages ∝ men × days | Easy
7. D. men × days is constant for the same work: 8 × 25 = 40 × d ⇒ d = 200/40 = 5 days. (More men ⇒ fewer days; the direct-proportion value 125 goes the wrong way.)
APAR26-05-38 | Inverse proportion — days for a different crew | Easy
8. B. Cost is directly proportional to quantity. Unitary method: cost of 1 kg = ₹588/14 = ₹42; cost of 9 = 9 × ₹42 = ₹378. (Or 588 × 9/14 = 378.)
APAR26-05-39 | Direct proportion — cost | Easy
9. B. Fewer days need more labourers (inverse proportion): labourers × days = constant. 10 × 7 = x × 5 ⇒ x = 70/5 = 14 labourers. (Setting up a direct proportion, 10 × 5/7 ≈ 7.1, is the usual error.)
APAR26-05-37 | Inverse proportion — workers needed | Easy
10. D. Women among teachers = 4/5 × 575 = 460; among non-teaching staff = 1/5 × 100 = 20. Men = (575 − 460) + (100 − 20) = 115 + 80 = 195. (Do not apply one fraction to the combined total 675.)
APAR26-05-10 | Caselet: fractions of sub-groups | Easy
11. B. Sum of ratio terms = 3 + 6 + 2 = 11; one part = ₹37,400/11 = ₹3,400. Largest share = 6 parts, smallest = 2 parts; difference = 4 × ₹3,400 = ₹13,600.
APAR26-05-08 | Dividing a sum in a ratio | Easy
12. B. B appears in both ratios as 2 and 10; make it common (LCM = 10). A : B = 5 : 2 = 50 : 20 and B : C = 10 : 3 = 20 : 6. So A : B : C = 50 : 20 : 6 = 25 : 10 : 3. (Writing 5 : 2 : 3 directly ignores that B has two different values.)
APAR26-05-01 | Combining two ratios | Easy
13. C. The compounded ratio of a : b and c : d is ac : bd. Here (2 × 3) : (7 × 7) = 6 : 49 = 6 : 49. (Adding the terms, 5 : 14, is wrong.)
APAR26-05-09 | Compounded ratio | Easy
14. D. Sum of ratio terms = 2 + 8 + 5 = 15; one part = ₹20,250/15 = ₹1,350. Ravi's share = 8 × ₹1,350 = ₹10,800.
APAR26-05-02 | Dividing a sum in a ratio | Easy
15. A. Output per man-hour = 990/(22 × 9) = 5. Now 26 men work 6 hours = 156 man-hours ⇒ boxes: 5 × 156 = 780. (Chain rule: 990 × 26/22 × 6/9 = 780; changing only one factor gives 660 or 1170.)
APAR26-05-47 | Chain rule — persons, hours, output (find output) | Medium
16. A. Output per man-hour = 25/(5 × 5) = 1. Now 7 labourers work 10 hours = 70 man-hours ⇒ toys: 1 × 70 = 70. (Chain rule: 25 × 7/5 × 10/5 = 70; changing only one factor gives 50 or 35.)
APAR26-05-45 | Chain rule — persons, hours, output (find output) | Medium
17. A. (5/8)A = (8/11)B ⇒ A/B = (8/11) ÷ (5/8) = (8 × 8)/(11 × 5) = 64/55, so A : B = 64 : 55. (Multiplying the fractions straight across, 5 : 11, is wrong.)
APAR26-05-27 | Ratio from fractional parts | Medium
18. C. Convert each equivalence to a per-item ratio: 1 box = 8/2 eraser; 1 notebook = 9/3 box. Multiplying along the chain, 1 notebook = 72/6 eraser = 72/6 × ₹23 = ₹276. (Inverting the ratio gives ₹1.92.)
APAR26-05-14 | Chain of cost equivalences | Medium
19. D. Distance ∝ petrol (more petrol, more distance). Per litre: 270/15 = 18 km. For 10 litres: 18 × 10 = 180 km. (Direct proportion: 270/15 = x/10.)
APAR26-05-42 | Direct proportion — fuel and distance | Medium
20. B. Let 10A = 3B = 4C = k. Then A : B : C = k/10 : k/3 : k/4 = 1/10 : 1/3 : 1/4. Multiplying by LCM(10, 3, 4) = 60: A : B : C = 6 : 20 : 15 = 6 : 20 : 15. (Writing 10 : 3 : 4 directly is the usual trap — the ratio is of the reciprocals.)
APAR26-05-19 | Equal products to a ratio | Medium
21. A. Mean proportional = √(ab) = √(1.21 × 0.81) = √0.9801 = 0.99. Check: 1.21 : 0.99 = 0.99 : 0.81. (Do not take the average of the two numbers.)
APAR26-05-21 | Mean proportional of decimals | Medium
22. A. Mean proportional = √(ab) = √(0.1 × 0.9) = √0.09 = 0.3. Check: 0.1 : 0.3 = 0.3 : 0.9. (Do not take the average of the two numbers.)
APAR26-05-13 | Mean proportional of decimals | Medium
23. C. A : B : C = 49 : 63 : 81 = 49 : 63 : 81. Sum of parts = 193, so one part = 27,020/193 = 140. C = 81 × 140 = 11,340. (Using 7 : 9 : 9 as the ratio gives 9727.2, which is wrong.)
APAR26-05-20 | Chained ratios with a sum | Medium
24. C. Distance ∝ petrol (more petrol, more distance). Per litre: 286/13 = 22 km. For 20 litres: 22 × 20 = 440 km. (Direct proportion: 286/13 = x/20.)
APAR26-05-43 | Direct proportion — fuel and distance | Medium
25. C. Convert each equivalence to a per-item ratio: 1 bag = 1/3 pen; 1 pen = 1/2 eraser. Multiplying along the chain, 1 bag = 1/6 eraser = 1/6 × ₹444 = ₹74. (Inverting the ratio gives ₹2,664.)
APAR26-05-23 | Chain of cost equivalences | Medium
26. D. Total ratio units = 4×6 + 5×7 + 4×4 = 75. One unit = ₹4,500/75 = ₹60. Each boy gets 6 units = ₹360. (Dividing by 6 + 7 + 4 ignores the number of persons in each group.)
APAR26-05-28 | Per-head shares in groups | Medium
27. C. Mean proportional = √(ab) = √(0.1 × 1.6) = √0.16 = 0.4. Check: 0.1 : 0.4 = 0.4 : 1.6. (Do not take the average of the two numbers.)
APAR26-05-26 | Mean proportional of decimals | Medium
28. C. Total food = 300 × 60 = 18000 person-days. New strength = 300 + 200 = 500. Days = 18000/500 = 36 days. (More people ⇒ fewer days; a direct proportion would give 100, which is wrong.)
APAR26-05-40 | Provisions — people join or leave | Medium
29. A. (1/8)A = (6/11)B ⇒ A/B = (6/11) ÷ (1/8) = (6 × 8)/(11 × 1) = 48/11, so A : B = 48 : 11. (Multiplying the fractions straight across, 3 : 44, is wrong.)
APAR26-05-15 | Ratio from fractional parts | Medium
30. D. 2x = 7y ⇒ x : y = 7 : 2. Put x = 7k, y = 2k: (x + y) : (x − y) = (7 + 2)k : (7 − 2)k = 9 : 5 = 9 : 5. (Taking x : y = 2 : 7 makes the difference negative — a sign that the ratio was inverted.)
APAR26-05-22 | Sum-to-difference ratio from a linear equation | Medium
31. C. Let incomes be 3x, 8x and expenditures 1y, 5y. Savings: 3x − 1y = 8x − 5y = ₹84,000. Subtracting: 5x = 4y ⇒ y = 5x/4. Substituting in 3x − 1y = 84,000 gives x = 48,000, y = 60,000. Income of Arjun = 3 × 48,000 = ₹1,44,000.
APAR26-05-18 | Income, expenditure and savings | Medium
32. A. In continued proportion each term divided by the previous is the same constant: 18/27 = 12/18 = 2/3. So k = 12 × 2/3 = 8. (Continuing by a constant difference instead of a constant ratio is the usual error.)
APAR26-05-17 | Continued proportion: find a term | Medium
33. D. Output per man-hour = 72/(6 × 12) = 1. Now 10 workers work 6 hours = 60 man-hours ⇒ pages: 1 × 60 = 60. (Chain rule: 72 × 10/6 × 6/12 = 60; changing only one factor gives 36 or 120.)
APAR26-05-41 | Chain rule — persons, hours, output (find output) | Medium
34. A. Total ratio units = 2×3 + 3×7 + 3×6 = 45. One unit = ₹2,250/45 = ₹50. Each girl gets 7 units = ₹350. (Dividing by 3 + 7 + 6 ignores the number of persons in each group.)
APAR26-05-16 | Per-head shares in groups | Medium
35. A. Fewer days need more men (inverse proportion): men × days = constant. 18 × 12 = x × 6 ⇒ x = 216/6 = 36 men. Extra men = 36 − 18 = 18. (Setting up a direct proportion, 18 × 6/12 ≈ 9, is the usual error.)
APAR26-05-46 | Inverse proportion — extra workers | Medium
36. A. Mean proportional = √(ab) = √(0.49 × 0.09) = √0.0441 = 0.21. Check: 0.49 : 0.21 = 0.21 : 0.09. (Do not take the average of the two numbers.)
APAR26-05-24 | Mean proportional of decimals | Medium
37. D. Mean proportional = √(ab) = √(0.25 × 0.16) = √0.04 = 0.2. Check: 0.25 : 0.2 = 0.2 : 0.16. (Do not take the average of the two numbers.)
APAR26-05-25 | Mean proportional of decimals | Medium
38. A. Women = 1/3 × 75 + 4/5 × 30 = 25 + 24 = 49. Men = (75 − 25) + (30 − 24) = 50 + 6 = 56. Ratio men : women = 56 : 49 = 8 : 7.
APAR26-05-12 | Caselet: fractions of sub-groups | Medium
39. B. Output per man-hour = 330/(22 × 5) = 3. Now 14 men work 10 hours = 140 man-hours ⇒ pages: 3 × 140 = 420. (Chain rule: 330 × 14/22 × 10/5 = 420; changing only one factor gives 660 or 210.)
APAR26-05-44 | Chain rule — persons, hours, output (find output) | Medium
40. C. Split each total in the given ratio (men share = total × men-part/(men + women)). P: 3,438 × 5/9 = 1,910; Q: 815 × 3/5 = 489. Total men = 2,399.
APAR26-05-11 | Caselet: totals split by ratio | Medium
41. D. Mean proportional = √(3√3 × 27√3) = √(81 × 3) = √81 × √3 = 9√3 = 9√3. (The product of the surds is 243, not 81√3.)
APAR26-05-32 | Mean proportional of surds | Difficult
42. C. Mean proportional = √(√7 × 4√7) = √(4 × 7) = √4 × √7 = 2√7 = 2√7. (The product of the surds is 28, not 4√7.)
APAR26-05-33 | Mean proportional of surds | Difficult
43. D. M1 D1 H1/W1 = M2 D2 H2/W2 ⇒ 5 × 5 × 9/1 = 3 × D2 × 5/3 ⇒ D2 = 5 × 5 × 9 × 3/(3 × 5) = 675/15 = 45 days. (Inverting the hours ratio gives 14; dropping the work multiple gives 15.)
APAR26-05-49 | Men, days, hours/day and job size | Difficult
44. C. By componendo and dividendo: (7a + 5b + 7a − 5b)/(7a + 5b − 7a + 5b) = (47 + 2)/(47 − 2) ⇒ 7a/5b = 49/45 ⇒ a/b = (49 × 5)/(45 × 7) = 245/315 = 7/9. So a : b = 7 : 9.
APAR26-05-30 | Componendo and dividendo | Difficult
45. C. If d is the fourth proportional to a, b, c then a : b = c : d, so d = bc/a = (4√3 × 8√7)/(2√3) = (4×8/2)√7 = 16√7. The √3 cancels; only √7 remains.
APAR26-05-29 | Fourth proportional (surds) | Difficult
46. D. Geeta − Ravi = (4 − 2) parts = 2 parts = ₹160, so one part = ₹80. Total = (4 + 5 + 2) parts = 11 × ₹80 = ₹880. (Dividing ₹160 by 4 instead of 2 is the usual slip.)
APAR26-05-34 | Total from a difference of shares | Difficult
47. D. Hours ∝ output (direct) and ∝ 1/men (inverse). H2 = 9 × (640/450) × (10/16) = 8 hours. Check: each man does 5 metres of cloth per hour, so 16 men do 80 per hour and need 640/80 = 8 h. (Inverting the men ratio gives 20.)
APAR26-05-48 | Chain rule — persons, hours, output (find hours) | Difficult
48. A. Let the total be T. Sons get T × 1/4 and T × 15/100; their difference = T × |1/4 − 15/100| = ₹6,100 ⇒ T = ₹61,000. Sons: ₹15,250 and ₹9,150; remainder for daughters = ₹61,000 − ₹15,250 − ₹9,150 = ₹36,600. Split 1 : 1 : 2 (one part = ₹9,150): ₹9,150, ₹9,150, ₹18,300. Largest daughter's share = ₹18,300.
APAR26-05-31 | Multi-stage division of property | Difficult
49. B. After 34 days the food left = 180 × (47 − 34) = 2340 person-days. New strength = 180 + 15 = 195. Remaining days = 2340/195 = 12 days. (Using the full 47 days instead of the remaining 13 gives 43.4 — the standard trap.)
APAR26-05-50 | Provisions — reinforcement after some days | Difficult
50. D. Split each total in the given ratio (men share = total × men-part/(men + women)). Q: 1,290 × 2/5 = 516; R: 1,176 × 4/7 = 672; S: 297 × 2/3 = 198. Total men = 1,386.
APAR26-05-35 | Caselet: totals split by ratio | Difficult