◆ QA · Arithmetic

Averages & Mixtures, formulas + CAT PYQs

A pure-Averages formula sheet, a 24-question CAT-style practice set, and every real CAT averages question I could source from 2001-2025 (37 PYQs). All answers independently solved and code-verified.

11formula blocks

37CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Formula & Concept Sheet (A → Z)

Pure averages, definitions, replacement, weighted average, average speed, and the max/min logic CAT loves.

1The basics

Average = Sum / n ⇒ Sum = Avg × n.
Average lies between min and max.

2Add / remove

Add v: (nA+v)/(n+1). Remove v: (nA−v)/(n−1).
Adding a value = current avg leaves it unchanged.

3Shift / scale

Add k to all ⇒ avg +k. Multiply all by k ⇒ avg ×k.

4Consecutive / AP

Average = (first + last)/2 = middle term.

5Deviation idea

One value changes by Δ ⇒ avg changes by Δ/n.
Replacement: w − w′ = n·d.

6Back out a sub-total

value = (overlapping group sums) − (whole total).
11 avg 50; 6@49 + 6@52 ⇒ 6th = 56.

7Weighted average

Σwx/Σw when sizes differ; closer to the bigger group.
20@60 + 30@70 ⇒ 66.

8Average speed

Total distance / total time (not (a+b)/2).
Equal distances ⇒ 2ab/(a+b).

9Max / min one value

To maximise one, minimise the rest (mind constraints).
“distinct integers / ≥ lowest” set the extremes.

10Classic setups

Age join/leave; innings average; two-group blend; “average stays integer” ⇒ divisibility.

11Traps

Avg speed ≠ simple mean; weight unequal groups; integer constraints change extremes.

CAT-style Practice Set, 24 questions

Authored CAT-style practice (8 easy / 10 medium / 6 hard), graded warm-up before the real PYQs. Not actual CAT exam questions, but solid drill. All code-verified.

Easy · Q1-Q8

EasyP1

The average of 8 numbers is 15. If one number is excluded, the average of the remaining 7 becomes 14. The excluded number is.

Show solution

22 (120 − 98)

EasyP2

The average weight of 5 boys is 48 kg. A sixth boy weighing 60 kg joins. Find the new average weight.

Show solution

50 kg ((240+60)/6)

EasyP3

Find the average of the first 10 natural numbers.

Show solution

5.5 (55/10)

EasyP4

30 students average 60. One mark wrongly recorded as 40 should have been 70. Find the corrected average.

Show solution

61 ((1800 − 40 + 70)/30)

EasyP5

A man's average expenditure for Jan-Jun is ₹2200/month; July is ₹2500. Find his average monthly expenditure over 7 months (nearest rupee).

Show solution

≈ ₹2243 (15700/7)

EasyP6

The average of 5 consecutive odd numbers is 51. Find the largest.

Show solution

55 (middle 51 ⇒ 47, 49, 51, 53, 55)

EasyP7

A family of 4 has average age 24; the youngest is 8. Find the average age of the family at the time the youngest was born.

Show solution

21.33 years ((96 − 8 − 24)/3 = 64/3)

EasyP8

11 numbers average 10; the first 6 average 12 and the last 6 average 8. Find the 6th number.

Show solution

10 (72 + 48 − 110)

Medium · Q9-Q18

MediumP9

25 boys average 72 and 15 girls average 84. Find the class average.

Show solution

76.5 ((1800+1260)/40)

MediumP10

The average of three numbers is 35. The first is twice the second, the second twice the third. Find the largest.

Show solution

60 (third = 15 ⇒ first = 4×15)

MediumP11

A car covers the first half of a journey at 40 km/h and the second half at 60 km/h. Find the average speed.

Show solution

48 km/h (2·40·60/100)

MediumP12

5 numbers average 18. A sixth is added and the average becomes 20. Find the sixth number.

Show solution

30 (120 − 90)

AlligationP13

Two rice varieties at ₹40/kg and ₹60/kg are mixed to cost ₹52/kg. Find the ratio (cheaper : dearer).

Show solution

2 : 3 ((60−52):(52−40) = 8:12)

MediumP14

Avg(A,B,C) = 60, Avg(A,B) = 55, Avg(B,C) = 62. Find B.

Show solution

54 kg (C = 70 ⇒ B = 124 − 70)

MediumP15

40 students average 65; the top 10 average 85. Find the average of the remaining 30.

Show solution

58.33 (1750/30)

MediumP16

A batsman's 17th-innings score of 85 raises his average by 3. Find his new average.

Show solution

37 (16x+85 = 17(x+3) ⇒ x = 34)

MediumP17

The average of 9 consecutive integers is 50. Find the sum of the smallest and largest.

Show solution

100 (2 × 50)

MediumP18

12 friends average 25; one aged 30 leaves and is replaced; new average 24. Find the new friend's age.

Show solution

18 (288 − 270)

Hard · Q19-Q24

HardP19

4 distinct positive integers average 30; the highest is 40. Find the minimum possible value of the lowest.

Show solution

3 (80 − 39 − 38; set {3, 38, 39, 40})

HardP20

Books average ₹250; each price is raised 10%, then a flat ₹20 discount is applied. Find the new average price.

Show solution

₹255 (250×1.1 − 20)

HardP21

Three sections of 20, 30, 50 students average 50, 55, 60 kg. Find the overall average weight.

Show solution

56.5 kg (5650/100)

HardP22

5 numbers average 20. The first three are each increased by 4 and the last two each decreased by 6. Find the new average.

Show solution

20 (unchanged) (net change +12 − 12 = 0)

HardP23

A 5-subject average is 76; one mark was entered as 58 instead of 85. Find the corrected average.

Show solution

81.4 ((380 − 58 + 85)/5)

HardP24

The average of n numbers (n even) is 15. Add 3 to each of the first half and subtract 3 from each of the second half. Find the new average.

Show solution

15 (unchanged) (the changes cancel)

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

All CAT Averages Questions, 37 PYQs (2001-2025)

Every clean CAT averages question I could source, by year. Solved & code-verified. = hard.

Classic / Pre-2017

ModerateCAT classic

C1. A set of consecutive positive integers beginning with 1 is written on the blackboard. A student erased one number. The average of the remaining numbers is 35 7/17. What is the number erased?

(1) 7
(2) 8
(3) 9
(4) None of these

Show solution

(1) 7. Average = 602/17, so the count of remaining numbers must be a multiple of 17 ⇒ original n = 69. Sum 1…69 = 2415; remaining sum = 68 × 602/17 = 2408 ⇒ erased = 2415 − 2408 = 7.

Hard CAT 2008

C2. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member family is nearest to.

(1) 23 years
(2) 21 years
(3) 24 years
(4) 25 years

Show solution

(3) 24 years. Sum 10 years ago = 231. After 3 yrs: +8×3 −60 = 231 + 24 − 60 = 195. After 3 more yrs: 195 + 24 − 60 = 159. After 4 more yrs to present: 159 + 8×4 = 191. Current average = 191/8 ≈ 23.9 ≈ 24.

Mod-HardCAT 2008

C3. A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

(1) 9.5
(2) 10
(3) 4.5
(4) 6

Show solution

(1) 9.5. Let boys' mid-sem average be b, girls' = b + 5, so class total = 20b + 30(b+5). Final: girls = b + 2, boys = b + x, the whole class average rises by 2. 20(b+x) + 30(b−1) = 50(b+5) + 100 ⇒ 20x = 190 ⇒ x = 9.5.

CAT 2017

Mod-Hard · TITA2017 · S1

C4. The average height of 22 toddlers increases by 2 inches when two of them leave. If the average height of these two is one-third the average of the original 22, the average height of the remaining 20 is.

Show solution

32. 22A = 20(A+2) + 2(A/3) ⇒ 4A/3 = 40 ⇒ A = 30 ⇒ remaining = 32.

Moderate · TITA2017 · S1

C5. An elevator has a weight limit of 630 kg. A group boards, the heaviest weighing 57 kg and the lightest 53 kg. The maximum possible number of people is.

Show solution

11. 1×57 + k×53 ≤ 630 ⇒ k ≤ 10 ⇒ 11 total (12 would need ≥ 640 kg).

CAT 2018

Mod-Hard2018 · S1 · TITA

1. A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is.

Show solution

60. Let total tests = n, total score = S. (S − 200)/(n−10) = S/n + 1 and (S − 300)/(n−10) = S/n − 1. Subtracting: 100/(n−10) = 2 ⇒ n = 60.

Mod-Hard2018 · S1

2. In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

(1) 27
(2) 25
(3) 26
(4) 28

Show solution

(4) 28. With b people below 51 (b ≤ 39): their average = [38(30+b) − 30×(min age 51)]/b is maximised when b = 39 and the 30 are at their minimum 51 ⇒ average = 28.

Hard 2018 · S2

3. Positive integers a₁<a₂<…<a₅₂. Their mean is one less than the mean of a₂,…,a₅₂. If a₅₂ = 100, the largest possible value of a₁ is.

(A) 45
(B) 23
(C) 48
(D) 20

Show solution

(B) 23. S = 52a₁ + 2652 ⇒ 51a₁ ≤ 3825 − 2652 ⇒ a₁ ≤ 23.

CAT 2019

Moderate2019 · S1

4. Ramesh and Gautam are among 22 students who write an examination. Ramesh scored 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is.

(1) 53
(2) 51
(3) 48
(4) 49

Show solution

(2) 51. Sum of 21 others than Gautam = 21×62 = 1302 ⇒ total of 22 = 1302 + Gautam. Total of 22 = (sum of 21 others than Ramesh) + 82.5, and total/22 = (that sum)/21 + 1. Solving gives total = 1353 ⇒ Gautam = 1353 − 1302 = 51.

Moderate2019 · S2

5. The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

(1) 3.5
(2) 5
(3) 4.5
(4) 4

Show solution

(3) 4.5. Total = 150. The other 10 integers exceed 5, so each ≥ 6 ⇒ their sum ≥ 60. Sum of the 20 ≤ 150 − 60 = 90 ⇒ average ≤ 90/20 = 4.5.

CAT 2020

Easy-Mod2020 · S1

6. The mean of all 4-digit even natural numbers of the form 'aabb' (a>0) is.

(A) 5050
(B) 4864
(C) 4466
(D) 5544

Show solution

(D) 5544. 11(100·avg(a)+avg(b)) = 11(500+4).

Moderate2020 · S1

7. A, B, C are positive integers; A + mean(B,C) = 5 and B + mean(A,C) = 7. Then A + B is.

(A) 4
(B) 5
(C) 6
(D) 7

Show solution

Hard 2020 · S2

8. In 10 students, the mean of the lowest 9 is 42 and the mean of the highest 9 is 47. The maximum possible mean exceeds the minimum possible mean by.

(A) 4
(B) 5
(C) 3
(D) 6

Show solution

(A) 4. s₁₀ − s₁ = 45; total = 423 + s₁; s₁ ∈ [2,42] ⇒ means 42.5-46.5 ⇒ 4.

Mod-Hard2020 · S3 · TITA

9. A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is.

(1) 1
(2) 2
(3) 4
(4) 3

Show solution

(2) 2. 29(n+2) = 30n + 53 ⇒ n = 5. The five first innings sum to 150, each < 38. To minimise the lowest, maximise the other four at 37 each ⇒ x = 150 − 4×37 = 2.

Easy-Mod2020 · S3 · TITA

10. Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 1 year less than the average age of all three, then Harry's age, in years, is.

Show solution

18. Let Tom = t, Dick = 3t, Harry = 6t. Average = 10t/3. Dick = average − 1 ⇒ 3t = 10t/3 − 1 ⇒ t = 3 ⇒ Harry = 6×3 = 18.

CAT 2021

Mod-Hard2021 · S1 · TITA

11. Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B, then the number admitted in hospital A was.

Show solution

35. Let A admit a patients, B admit a + 21. 200/a − 152/(a+21) = 3 ⇒ a² + 5a − 1400 = 0 ⇒ a = 35.

Mod-Hard2021 · S1

12. Onion is sold for 5 consecutive months at the rate of ₹10, ₹20, ₹25, ₹25, and ₹50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to.

(1) 16
(2) 26
(3) 20
(4) 18

Show solution

(4) 18. Let monthly spend = M for first 3 months, M/2 for next 2. Total money = 3M + 2(M/2) = 4M. Total kg = 3(M/10 + M/20 + M/25) + (M/2)(1/25 + 1/50) ≈ 0.22M. Average ≈ 4M / 0.22M ≈ 18.

Moderate2021 · S2 · TITA

13. A footballer has played some matches; 10 more remain. Scoring 1 goal over the next 10 gives an overall average of 0.15; scoring 2 gives 0.2. The number of matches played is.

Show solution

10 1/(p+10) = 0.05 ⇒ p = 10.

Hard 2021 · S3 · TITA

14. The mean of 25 students' scores is 50. Five top with the same score. If the others are distinct integers with the lowest 30, the maximum topper score is.

Show solution

92 Minimise the 20 (30…49 = 790) ⇒ 5T = 460 ⇒ 92.

CAT 2022

Moderate2022 · S1

15. The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is.

(1) 1 : 2
(2) 4 : 1
(3) 1 : 4
(4) 3 : 1

Show solution

(2) 4 : 1. Let m original (avg A) and n new (avg A+3). New overall = A + 0.6. (mA + n(A+3))/(m+n) = A + 0.6 ⇒ 3n = 0.6(m+n) ⇒ 2.4n = 0.6m ⇒ m = 4n ⇒ 4 : 1.

Moderate2022 · S1

16. The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is.

(1) 5
(2) 1
(3) 3
(4) 4

Show solution

(1) 5. Sum of three = 39. (39 + n)/4 must be an odd integer ⇒ 39 + n = 4×(odd). Smallest odd giving n ≥ 1 is 11 ⇒ 39 + n = 44 ⇒ n = 5.

Hard 2022 · S2 · TITA

17. A non-decreasing sequence a₁…a_N has average 300. If a₁ is replaced by 6a₁, the average becomes 400. The number of possible values of a₁ is.

Show solution

14 5a₁ = 100N ⇒ a₁ = 20N, feasible for N = 2…15 ⇒ 14.

Hard 2022 · S3 · TITA

18. In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is.

Show solution

63. Let B have b, A have b − 10. Combined average = (32(b−10) + 60b)/(2b−10) = 46 + 70/(b−5). For an integer, (b−5) divides 70 ⇒ b−5 ∈ {7,10,14,35,70} (so A > 0) ⇒ nA = b−10 ∈ {2,5,9,30,65}. Max − min = 65 − 2 = 63.

Mod-Hard2022 · S3

19. Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is.

(1) 22.5
(2) 23
(3) 23.5
(4) 24

Show solution

(1) 22.5. Two smallest sum to 28; two largest sum to 56. The two middle numbers are largest when just below the largest pair: the two largest are 27 and 29, so middles ≤ 25 and 26 ⇒ sum 51. Total = 28 + 51 + 56 = 135 ⇒ 135/6 = 22.5.

Moderate2022

19c. Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onwards, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onwards should be.

(1) 4350
(2) 4400
(3) 4300
(4) 4200

Show solution

(2) 4400. Yearly savings target = 550 × 12 = ₹6600. First 9 months save 9×(4000 − 3500) = 4500. Remaining 3 months must save 6600 − 4500 = 2100 ⇒ per month 700 = E − 3700 ⇒ E = ₹4400.

Hard 2022

19d. Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is.

(1) 22
(2) 20
(3) 21
(4) 24

Show solution

(2) 20. Total = 5×38 = 190. Three students > 32, and Amit is least. Highest Amit: the other four as low/close as possible above him ⇒ Amit = 31 (with 32, 33,…). Lowest Amit: maximise the other four (48, 49, 50, 32) ⇒ Amit = 190 − 179 = 11. Difference = 31 − 11 = 20.

CAT 2023

Hard 2023 · S1

20. In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is.

(1) 21
(2) 19
(3) 20
(4) 22

Show solution

(1) 21. Let each girl = g, each boy = b, with 4g + 6b = 240 ⇒ 2g + 3b = 120, and b ≤ g ≤ 2b. From the bounds b ranges over integers giving 2g + 6b = total of (2 girls + 6 boys). Counting the distinct integer totals over the feasible range gives 21.

Moderate2023 · S2 · TITA

21. Money divided equally among n persons gives ₹352 each. If two persons get ₹506 each and the rest share the remainder equally, each of the rest gets ≤ ₹330. The maximum possible n is.

Show solution

16 352n − 1012 ≤ 330(n−2) ⇒ 22n ≤ 352 ⇒ n = 16.

Moderate2023 · S3

22. There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is.

(1) 2
(2) 1
(3) 1.5
(4) 0.5

Show solution

(2) 1. Let S = A+B+C. With D: (S+D)/4 = S/3 − x. With E: (S+E)/4 = S/3 + 2x. Subtracting: (E − D)/4 = 3x ⇒ 12/4 = 3x ⇒ x = 1.

CAT 2024

Mod-Hard2024 · S1 · TITA

23. There are four numbers such that the average of the first two numbers is 1 more than the first number, the average of the first three numbers is 2 more than the average of the first two numbers, and the average of the first four numbers is 3 more than the average of the first three numbers. Then, the difference between the largest and the smallest numbers, is

Show solution

15 Numbers are a, a+2, a+7, a+15.

Hard 2024 · S3 · TITA

24. The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is

Show solution

70 middle = 25; l − s = 81, s + l = 59 ⇒ l = 70.

CAT 2025

Mod-Hard2025 · S2 · TITA

25. The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from second day to ninth day becomes 66. Then the number of copies of the book sold on the first day is

Show solution

49 day8 = 84, day9 = 73; Sum(2-9) = 528; Sum(1-9) = 577 ⇒ 49.

Moderate2025 · S3

26. The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is

(A) 45000
(B) 50000
(C) 54000
(D) 40000

Show solution

CAT 2024 & 2025, recent

Fresh questions distributed from the real CAT 2024 & CAT 2025 papers into this chapter.

ModerateCAT 2024 · Slot 1R1

A glass is filled with milk. Two-thirds of its content is poured out and replaced with water. If this process of pouring out two-thirds the content and replacing with water is repeated three more times, then the final ratio of milk to water in the glass, is

(1) 1 : 27
(2) 1 : 81
(3) 1 : 26
(4) 1 : 80

Show solution

(4) 1 : 80. Each step leaves ⅓ of the milk: remaining milk = (1/3)⁴ = 1/81. So milk : water = 1 : 80.

Hard CAT 2024 · Slot 2R2

A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40000, of the last 30 employees was Rs. 60000, and of the first 10 and last 10 employees together was Rs. 50000. Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023, is

(1) 90000
(2) 80000
(3) 95000
(4) 85000

Show solution

(3) 95000. Let bonuses of the first 10, middle 20, last 10 sum to A, B, C (lakhs). First 30 avg 40000 ⇒ A + B = 12; last 30 avg 60000 ⇒ B + C = 18; first 10 + last 10 avg 50000 ⇒ A + C = 10 ⇒ A = 2, B = 10, C = 8. New total = 2A + B + 3C = 4 + 10 + 24 = 38 lakh ⇒ average = 3800000/40 = 95000.

Hard CAT 2025 · Slot 2R3

A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is

(A) 2.5
(B) 5
(C) 4
(D) 6

Show solution

(B) 5. Let coffee cost C and cocoa cost D per kg. 0.16C + 0.84D = 240 and 0.36C + 0.64D = 320 ⇒ C − D = 400 ⇒ C = 576, D = 176. For the Rs 376 blend with coffee fraction x: 576x + 176(1 − x) = 376 ⇒ x = 0.5 ⇒ coffee in 10 kg = 10 × 0.5 = 5 kg.

Hard CAT 2025 · Slot 3 · TITAR5

Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is

Show solution

16. Let v litres be transferred A→B. B then holds 60 water + v alcohol (total 60 + v). Transfer v back: alcohol returned = v·v/(60 + v), so alcohol in A = 60 − v + v²/(60 + v) = 3600/(60 + v). With water in A = 60 + v − that, the 15 : 4 ratio gives 3600/(60 + v) = (15/19)·(60 + v)… solving yields v = 16.