An average condenses a total into an equal share. It does not say every observation has that value. Age problems add a second structure: each person's age advances by the same number of years. Keeping the total and the timeline visible makes both topics easier.
1. Average as total divided by count
Arithmetic mean = sum of observations / number of observations. Therefore total = average × count. If 12 scores average 18, their total is 216. A new score of 30 raises the total to 246 over 13 scores, so the new average is 246/13, not (18+30)/2. The latter would give equal weight to one new score and the whole earlier group.
When one observation is corrected, adjust the total by the correction difference. If 56 was wrongly recorded as 65, subtract 9 from the earlier total. If one member leaves, remove that member's value and reduce the count. Changing only the numerator or only the denominator produces a false average.
Worked example 1. The average of 20 values is 35. One value recorded as 48 should have been 84. The corrected total is 20×35−48+84=736. The corrected average is 736/20=36.8.
2. Combined and weighted averages
Two groups with counts n₁,n₂ and averages a₁,a₂ have a combined average of (n₁a₁+n₂a₂)/(n₁+n₂). This is a weighted average. The combined value lies between the two group averages when counts are positive, and it lies closer to the average of the larger group.
An alligation-style shortcut can recover a count ratio from two group averages and a combined mean. If the group means are 40 and 60 and the combined mean is 45, the lower group must be three times as large as the higher group: (60−45):(45−40)=15:5=3:1. Always check by taking three values at 40 and one at 60, whose average is 45.
Worked example 2. Twenty students average 62 marks, while thirty average 72. Their combined total is 1,240+2,160=3,400 over 50 students. The combined average is 68, not the unweighted mean 67.
3. Replacement, insertion and removal
If a group of n people has average A, its total is nA. When one member of value x is replaced by one of value y, the new average is A+(y−x)/n. This avoids recalculating every observation. A rise of 2 in the average of 15 people means the group's total rose by 30.
Worked example 3. A team of 11 has average age 24. A 19-year-old leaves and a new member joins; the average becomes 25. The old total was 264, and the new total is 275. The new member is 275−(264−19)=30 years old.
4. Ages are linear timelines
Let a person's present age be x. Five years ago it was x−5; five years later it will be x+5. Two people's age difference stays constant, because the same number of years is added to both. Their age ratio usually changes, often moving closer to 1:1 as both grow older.
Translate a sentence into an equation before computing. “A is twice as old as B was five years ago” means A's present age is 2(B−5), unless the sentence also places A in the past. The placement of “was” matters. Draw a table with rows for persons and columns for past, present and future if the wording is dense.
Worked example 4. A father is three times his son's present age. In ten years he will be twice the son's age. Let the son be x; the father is 3x. Then 3x+10=2(x+10), so x=10 and the father is 30. After ten years they are 20 and 40, matching the condition.
5. Average age of a group over time
If the same n people remain together for k years, each person's age rises by k, so their average also rises by k. A birth, death, arrival or departure changes the group; the shortcut no longer applies without adjusting total and count. If a group average rises by less or more than elapsed years, a membership change or data error must be present.
Worked example 5. Six years ago, the average age of five friends was 18. If the membership is unchanged, the current average is 24 and the current total is 120. There is no need to know any individual age.
6. Estimation and error checks
The mean lies between the minimum and maximum observations. A combined average lies between the component averages. An age cannot be negative in an ordinary age word problem. These bounds identify arithmetic mistakes quickly. In ratio-of-ages problems, substitute the solved ages into every stated time condition, not only the last equation.
Recall before practice
1. Explain why the average of two group averages requires weights. 2. State how much a 3-mark rise in average changes the total of 12 scores. 3. Describe what stays constant for two people's ages over time. 4. Show how to reconstruct a missing person after a replacement.
Chapter 7 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. The ages of Deepak and Vikas are in the ratio 3 : 5 and Vikas is 18 years older than Deepak. What will be the age of Vikas after 5 years?
A. 50 years B. 32 years C. 95 years D. 45 years
2. The present ages of Anita and Pooja are in the ratio 3 : 4. If after 2 years the ratio of their ages will be 10 : 13, what is Pooja's present age?
A. 6 years B. 24 years C. 18 years D. 26 years
3. Find the average of the first 97 even numbers.
A. 196 B. 98 C. 97 D. 99
4. The ages of Vikas, Sanjay and Suresh are in the ratio 2 : 3 : 5 and the sum of their ages is 50 years. What will be the age of Vikas after 4 years?
A. 29 years B. 14 years C. 19 years D. 10 years
5. The present ages of Rekha and Suresh are in the ratio 7 : 8. After 10 years, Rekha will be 45 years old. What is the present age of Suresh?
A. 30 years B. 50 years C. 40 years D. 35 years
6. The average of 8 observations is 77. If the average of 7 of them is 83, find the value of the remaining observation.
A. 83 B. 616 C. 77 D. 35
7. The average age of 5 members is recorded. When a member aged 55 years is replaced by a new one, the average age rises by 2 years. Find the age of the new member.
A. 65 years B. 45 years C. 57 years D. 10 years
8. The sum of the present ages of Vikas and Pooja is 49 years. 2 years ago, Vikas was twice as old as Pooja. What is the present age of Vikas?
A. 30 years B. 17 years C. 32 years D. 34 years
9. The average of 27 numbers is 34. Find the sum of these numbers.
A. 918 B. 891 C. 945 D. 34
10. Find the average of the first 63 odd numbers.
A. 126 B. 63 C. 62 D. 64
11. The average age of 25 workers is 44 years. When the new worker's age is included, the average falls by 1. Find the new worker's age.
A. 18 years B. 69 years C. 19 years D. 43 years
12. The average age of 5 girls is recorded. When a girl aged 46 years is replaced by a new one, the average age rises by 2 years. Find the age of the new girl.
A. 10 years B. 56 years C. 36 years D. 48 years
13. The average of 15 numbers is 430. Find the sum of these numbers.
A. 6450 B. 430 C. 6435 D. 6465
14. The average age of 9 family members is 29 years. When the newborn's age is included, the average rises by 3. Find the newborn's age.
A. 2 years B. 56 years C. 32 years D. 59 years
15. The sum of the present ages of Anita and Geeta is 51 years. 3 years ago, Anita was twice as old as Geeta. What is the present age of Anita?
A. 36 years B. 30 years C. 34 years D. 33 years
16. 6 years ago, a mother was five times as old as her daughter. 7 years from now, the mother will be three times as old as the daughter. What is the present age of the daughter?
A. 26 years B. 13 years C. 19 years D. 21 years
17. A father is six times as old as his son. After 9 years, the father will be three times as old as the son. What is the sum of their present ages?
A. 42 years B. 60 years C. 105 years D. 33 years
18. 3 years ago, a mother was three times as old as her daughter. 12 years from now, the mother will be twice as old as the daughter. What is the present age of the mother?
A. 54 years B. 45 years C. 60 years D. 48 years
19. The average age of 4 players is recorded. When a player aged 60 years is replaced by a new one, the average age rises by 5 years. Find the age of the new player.
A. 65 years B. 40 years C. 20 years D. 80 years
20. The average age of 5 players is recorded. When a player aged 36 years is replaced by a new one, the average age falls by 1 years. Find the age of the new player.
A. 35 years B. 41 years C. 31 years D. 5 years
21. The average age of 12 students is 60 years. When the teacher's age is included, the average falls by 1. Find the teacher's age.
A. 72 years B. 48 years C. 47 years D. 59 years
22. The average age of 21 workers is 43 years. When the new worker's age is included, the average rises by 2. Find the new worker's age.
A. 85 years B. 45 years C. 87 years D. 1 years
23. The average age of 11 girls is recorded. When a girl aged 42 years is replaced by a new one, the average age falls by 1 years. Find the age of the new girl.
A. 11 years B. 41 years C. 31 years D. 53 years
24. A batsman scores 154 runs in his 11th innings and thus raises his average by 7. Find his new batting average.
A. 91 B. 77 C. 84 D. 14
25. The average of 30 numbers is 73. If each number is multiplied by 5, find the new average.
A. 335 B. 395 C. 365 D. 73
26. The average of 8 observations is 21. If the average of 7 of them is 22, find the value of the remaining observation.
A. 22 B. 21 C. 168 D. 14
27. The average of P, Q and R is 100; the average of Q, R and S is 51; and P + S = 199. Find P.
A. 148 B. 100 C. 173 D. 26
28. The average age of 8 members is recorded. When a member aged 17 years is replaced by a new one, the average age rises by 1 years. Find the age of the new member.
A. 18 years B. 8 years C. 25 years D. 9 years
29. The average age of 11 workers is 37 years. When the new worker's age is included, the average rises by 3. Find the new worker's age.
A. 40 years B. 4 years C. 70 years D. 73 years
30. The average of 30 numbers is 21. If 10 is added to each number, find the new average.
A. 31 B. 21 C. 61 D. 1
31. The average of 23 numbers is 53. If each number is multiplied by 4, find the new average.
A. 53 B. 212 C. 189 D. 235
32. The present ages of Mohan and Meena are in the ratio 4 : 5. If the difference between their ages is 5 years, what will be the ratio of their ages after 8 years?
A. 12 : 13 B. 33 : 28 C. 28 : 33 D. 4 : 5
33. 3 years ago, a mother was seven times as old as her daughter. 15 years from now, the mother will be three times as old as the daughter. What is the present age of the daughter?
A. 27 years B. 12 years C. 9 years D. 14 years
34. The sum of the present ages of Pooja and Priya is 78 years. 3 years ago, Pooja was five times as old as Priya. What is the present age of Priya?
A. 18 years B. 12 years C. 15 years D. 13 years
35. The sum of the present ages of Neha and Meena is 27 years. 3 years ago, Neha was twice as old as Meena. What is the present age of Meena?
A. 10 years B. 9 years C. 7 years D. 13 years
36. A father is five times as old as his son. After 4 years, the father will be four times as old as the son. What is the sum of their present ages?
A. 80 years B. 72 years C. 68 years D. 96 years
37. The average of P, Q and R is 66; the average of Q, R and S is 55; and P + S = 149. Find P.
A. 66 B. 94 C. 58 D. 91
38. The present ages of Geeta and Vikas are in the ratio 2 : 3. If the difference between their ages is 7 years, what will be the ratio of their ages after 6 years?
A. 27 : 20 B. 8 : 9 C. 20 : 27 D. 2 : 3
39. The sum of the present ages of Meena and Anita is 58 years. 7 years ago, Meena was three times as old as Anita. What is the present age of Meena?
A. 33 years B. 40 years C. 47 years D. 18 years
40. A batsman scores 95 runs in his 19th innings and thus raises his average by 1. Find his previous batting average.
A. 75 B. 76 C. 77 D. 5
41. The average salary (in thousand rupees) of 77 staff is 23; men average 26 and women average 19. How many women are there?
A. 34 B. 44 C. 33 D. 54
42. A batsman scores 199 runs in his 20th innings and thus raises his average by 2. Find his previous batting average.
A. 159 B. 157 C. 161 D. 179
43. 2 years ago, the ratio of the ages of Sanjay and Rahul was 3 : 4. 10 years from now, the ratio of their ages will be 7 : 8. What is the sum of their present ages?
A. 29 years B. 45 years C. 25 years D. 21 years
44. A batsman scores 125 runs in his 19th innings and thus raises his average by 4. Find his previous batting average.
A. 45 B. 53 C. 49 D. 68
45. A mother was 20 years old when her daughter was born. The mother is now five times as old as the daughter. What will be the mother's age after 8 years?
A. 32 years B. 41 years C. 28 years D. 33 years
46. A batsman scores 123 runs in his 8th innings and thus raises his average by 1. Find his previous batting average.
A. 114 B. 115 C. 123 D. 116
47. The average age of 10 workers is recorded. When a worker aged 52 years is replaced by a new one, the average age falls by 4 years. Find the age of the new worker.
A. 12 years B. 40 years C. 48 years D. 92 years
48. The average age of 4 workers is recorded. When a worker aged 58 years is replaced by a new one, the average age rises by 1 years. Find the age of the new worker.
A. 59 years B. 4 years C. 54 years D. 62 years
49. The ages of a father and his son are in the ratio 3 : 7 and the product of their ages is 1029. What is the difference between their ages?
A. 28 years B. 196 years C. 42 years D. 4 years
50. The ages of a mother and her daughter are in the ratio 4 : 5 and the product of their ages is 180. What is the difference between their ages?
A. 1 years B. 3 years C. 6 years D. 9 years
Chapter 7 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. A. Let the ages be 3x and 5x. Difference = 2x = 18 ⇒ x = 9. So Deepak = 27 years and Vikas = 45 years. After 5 years, Vikas will be 45 + 5 = 50 years.
APAR26-07-33 | Ratio and difference: age after n years | Easy
2. B. Let the present ages be 3x and 4x. Then (3x + 2)/(4x + 2) = 10/13 ⇒ 39x + 26 = 40x + 20 ⇒ 1x = 6 ⇒ x = 6. Present ages: Anita = 18, Pooja = 24; difference = 6, sum = 42. (Note that the difference of ages never changes with time.)
APAR26-07-31 | Ratio now and ratio later | Easy
3. B. Using the standard formula for the average of first 97 even numbers: average = n+1 = 97+1 = 98. Memorising these standard formulas saves time versus listing and adding every term.
APAR26-07-08 | Average of standard sequences | Easy
4. B. Sum of ratio terms = 2 + 3 + 5 = 10; one part = 50/10 = 5 years. Vikas's present age = 2 × 5 = 10 years; after 4 years = 10 + 4 = 14 years.
APAR26-07-36 | Three ages in a ratio with their sum | Easy
5. C. Rekha's present age = 45 − 10 = 35 years. Let the ages be 7x and 8x: 7x = 35 ⇒ x = 5, so Suresh's present age = 8 × 5 = 40 years. (Applying the ratio to 45 directly gives 51.43, which is wrong.)
APAR26-07-35 | Ratio and one future age | Easy
6. D. Total of all 8 = 8 × 77 = 616. Total of known 7 = 7 × 83 = 581. Missing value = 616 − 581 = 35.
APAR26-07-03 | Missing value from average | Easy
7. A. Change in total = n × k = 5 × 2 = 10. New value = old value + 10 = 55 + 10 = 65.
APAR26-07-07 | Average change on replacement | Easy
8. C. Let Pooja's present age be x, so Vikas's is 49 − x. 2 years ago: (49 − x − 2) = 2(x − 2) ⇒ 47 − x = 2x − 4 ⇒ 3x = 51 ⇒ x = 17. So Pooja = 17 years and Vikas = 49 − 17 = 32 years. (Splitting 49 directly in the ratio 2 : 1 gives 32.67, which ignores the 2-year shift.)
APAR26-07-34 | Sum of ages and a past multiple | Easy
9. A. Sum of a group of numbers = average value × number of terms = 34 × 27 = 918.
APAR26-07-05 | Average from sum (direct) | Easy
10. B. Using the standard formula for the average of first 63 odd numbers: average = n = 63 = 63. Memorising these standard formulas saves time versus listing and adding every term.
APAR26-07-01 | Average of standard sequences | Easy
11. A. New average = 44 − 1 = 43. New member's value = newAvg × (n+1) − oldAvg × n = 43 × 26 − 44 × 25 = 18.
APAR26-07-02 | Average change on member added | Easy
12. B. Change in total = n × k = 5 × 2 = 10. New value = old value + 10 = 46 + 10 = 56.
APAR26-07-06 | Average change on replacement | Easy
13. A. Sum of a group of numbers = average value × number of terms = 430 × 15 = 6450.
APAR26-07-04 | Average from sum (direct) | Easy
14. D. New average = 29 + 3 = 32. New member's value = newAvg × (n+1) − oldAvg × n = 32 × 10 − 29 × 9 = 59.
APAR26-07-09 | Average change on member added | Easy
15. D. Let Geeta's present age be x, so Anita's is 51 − x. 3 years ago: (51 − x − 3) = 2(x − 3) ⇒ 48 − x = 2x − 6 ⇒ 3x = 54 ⇒ x = 18. So Geeta = 18 years and Anita = 51 − 18 = 33 years. (Splitting 51 directly in the ratio 2 : 1 gives 34, which ignores the 3-year shift.)
APAR26-07-32 | Sum of ages and a past multiple | Easy
16. C. Let the daughter's present age be x and the mother's be y. 6 years ago: y − 6 = 5(x − 6). 7 years hence: y + 7 = 3(x + 7). Subtracting, 2x = 38 ⇒ x = 19, y = 71. So the daughter is 19, the mother is 71, and their sum is 90. (Forgetting to shift both ages by 6 in the first equation is the usual slip.)
APAR26-07-43 | Two conditions: past multiple and future multiple | Medium
17. A. Let the son's age be x; the father's age is 6x. After 9 years: 6x + 9 = 3(x + 9) ⇒ 3x = 18 ⇒ x = 6. So the son is 6, the father is 36, and the sum is 42. (Adding 9 only to the father's side, i.e. 6x + 9 = 3x, is the common mistake.)
APAR26-07-38 | Present multiple and future multiple | Medium
18. D. Let the daughter's present age be x and the mother's be y. 3 years ago: y − 3 = 3(x − 3). 12 years hence: y + 12 = 2(x + 12). Subtracting, 1x = 18 ⇒ x = 18, y = 48. So the daughter is 18, the mother is 48, and their sum is 66. (Forgetting to shift both ages by 3 in the first equation is the usual slip.)
APAR26-07-39 | Two conditions: past multiple and future multiple | Medium
19. D. Change in total = n × k = 4 × 5 = 20. New value = old value + 20 = 60 + 20 = 80.
APAR26-07-18 | Average change on replacement | Medium
20. C. Change in total = n × k = 5 × 1 = 5. New value = old value − 5 = 36 − 5 = 31.
APAR26-07-14 | Average change on replacement | Medium
21. C. New average = 60 − 1 = 59. New member's value = newAvg × (n+1) − oldAvg × n = 59 × 13 − 60 × 12 = 47.
APAR26-07-11 | Average change on member added | Medium
22. C. New average = 43 + 2 = 45. New member's value = newAvg × (n+1) − oldAvg × n = 45 × 22 − 43 × 21 = 87.
APAR26-07-16 | Average change on member added | Medium
23. C. Change in total = n × k = 11 × 1 = 11. New value = old value − 11 = 42 − 11 = 31.
APAR26-07-17 | Average change on replacement | Medium
24. C. New average = score − (n−1)×k = 154 − 10×7 = 84. Previous average = new average − k = 84 − 7 = 77.
APAR26-07-15 | Cricket batting average | Medium
25. C. Multiplying every value by k multiplies the average by k too: new average = 73 × 5 = 365.
APAR26-07-22 | Average under uniform transformation | Medium
26. D. Total of all 8 = 8 × 21 = 168. Total of known 7 = 7 × 22 = 154. Missing value = 168 − 154 = 14.
APAR26-07-19 | Missing value from average | Medium
27. C. P+Q+R = 300, Q+R+S = 153, so P − S = 300 − 153 = 147. With P + S = 199, adding gives 2P = 346, so P = 173.
APAR26-07-20 | Average of overlapping groups | Medium
28. C. Change in total = n × k = 8 × 1 = 8. New value = old value + 8 = 17 + 8 = 25.
APAR26-07-10 | Average change on replacement | Medium
29. D. New average = 37 + 3 = 40. New member's value = newAvg × (n+1) − oldAvg × n = 40 × 12 − 37 × 11 = 73.
APAR26-07-21 | Average change on member added | Medium
30. A. Adding a constant k to every value adds k to the average too: new average = 21 + 10 = 31.
APAR26-07-24 | Average under uniform transformation | Medium
31. B. Multiplying every value by k multiplies the average by k too: new average = 53 × 4 = 212.
APAR26-07-13 | Average under uniform transformation | Medium
32. C. Let the ages be 4x and 5x; 1x = 5 ⇒ x = 5, so the ages are 20 and 25. After 8 years: 28 : 33 = 28 : 33. (Adding 8 to the ratio terms themselves, 12 : 13, is the classic error — the ratio terms are not the ages.)
APAR26-07-41 | Ratio and difference: ratio after n years | Medium
33. B. Let the daughter's present age be x and the mother's be y. 3 years ago: y − 3 = 7(x − 3). 15 years hence: y + 15 = 3(x + 15). Subtracting, 4x = 48 ⇒ x = 12, y = 66. So the daughter is 12, the mother is 66, and their sum is 78. (Forgetting to shift both ages by 3 in the first equation is the usual slip.)
APAR26-07-46 | Two conditions: past multiple and future multiple | Medium
34. C. Let Priya's present age be x, so Pooja's is 78 − x. 3 years ago: (78 − x − 3) = 5(x − 3) ⇒ 75 − x = 5x − 15 ⇒ 6x = 90 ⇒ x = 15. So Priya = 15 years and Pooja = 78 − 15 = 63 years. (Splitting 78 directly in the ratio 5 : 1 gives 65, which ignores the 3-year shift.)
APAR26-07-40 | Sum of ages and a past multiple | Medium
35. A. Let Meena's present age be x, so Neha's is 27 − x. 3 years ago: (27 − x − 3) = 2(x − 3) ⇒ 24 − x = 2x − 6 ⇒ 3x = 30 ⇒ x = 10. So Meena = 10 years and Neha = 27 − 10 = 17 years. (Splitting 27 directly in the ratio 2 : 1 gives 18, which ignores the 3-year shift.)
APAR26-07-37 | Sum of ages and a past multiple | Medium
36. B. Let the son's age be x; the father's age is 5x. After 4 years: 5x + 4 = 4(x + 4) ⇒ 1x = 12 ⇒ x = 12. So the son is 12, the father is 60, and the sum is 72. (Adding 4 only to the father's side, i.e. 5x + 4 = 4x, is the common mistake.)
APAR26-07-42 | Present multiple and future multiple | Medium
37. D. P+Q+R = 198, Q+R+S = 165, so P − S = 198 − 165 = 33. With P + S = 149, adding gives 2P = 182, so P = 91.
APAR26-07-12 | Average of overlapping groups | Medium
38. C. Let the ages be 2x and 3x; 1x = 7 ⇒ x = 7, so the ages are 14 and 21. After 6 years: 20 : 27 = 20 : 27. (Adding 6 to the ratio terms themselves, 8 : 9, is the classic error — the ratio terms are not the ages.)
APAR26-07-45 | Ratio and difference: ratio after n years | Medium
39. B. Let Anita's present age be x, so Meena's is 58 − x. 7 years ago: (58 − x − 7) = 3(x − 7) ⇒ 51 − x = 3x − 21 ⇒ 4x = 72 ⇒ x = 18. So Anita = 18 years and Meena = 58 − 18 = 40 years. (Splitting 58 directly in the ratio 3 : 1 gives 43.5, which ignores the 7-year shift.)
APAR26-07-44 | Sum of ages and a past multiple | Medium
40. B. New average = score − (n−1)×k = 95 − 18×1 = 77. Previous average = new average − k = 77 − 1 = 76.
APAR26-07-23 | Cricket batting average | Medium
41. B. By alligation, men:women = (overall−low):(high−overall) = (23−19):(26−23) = 4:3 = 4:3. Wait — ratio men:women = 3:4. With total 77, one unit = 11, so men = 33, women = 44.
APAR26-07-29 | Weighted average — find group size | Difficult
42. A. New average = score − (n−1)×k = 199 − 19×2 = 161. Previous average = new average − k = 161 − 2 = 159.
APAR26-07-28 | Cricket batting average | Difficult
43. C. Let the ages 2 years ago be 3x and 4x, so the present ages are 3x + 2 and 4x + 2. After 10 years: (3x + 12)/(4x + 12) = 7/8 ⇒ 24x + 96 = 28x + 84 ⇒ x = 3. Present ages: Sanjay = 11, Rahul = 14; sum 25, difference 3. (Using 3x and 4x as the present ages is the trap — they are the ages 2 years ago.)
APAR26-07-49 | Two conditions: ratio in the past and ratio in the future | Difficult
44. C. New average = score − (n−1)×k = 125 − 18×4 = 53. Previous average = new average − k = 53 − 4 = 49.
APAR26-07-25 | Cricket batting average | Difficult
45. D. The difference of their ages is always 20 years. If the daughter is x now, the mother is 5x, so 5x − x = 20 ⇒ x = 20/4 = 5. The daughter is 5 and the mother is 25. After 8 years: daughter 13, mother 33. (Dividing 20 by 5 instead of 4 is the usual error.)
APAR26-07-50 | Age at birth of the child and present multiple | Difficult
46. B. New average = score − (n−1)×k = 123 − 7×1 = 116. Previous average = new average − k = 116 − 1 = 115.
APAR26-07-30 | Cricket batting average | Difficult
47. A. Change in total = n × k = 10 × 4 = 40. New value = old value − 40 = 52 − 40 = 12.
APAR26-07-27 | Average change on replacement | Difficult
48. D. Change in total = n × k = 4 × 1 = 4. New value = old value + 4 = 58 + 4 = 62.
APAR26-07-26 | Average change on replacement | Difficult
49. A. Let the ages be 3x and 7x. Product = 21x² = 1029 ⇒ x² = 49 ⇒ x = 7. Ages: 21 and 49; sum = 70, difference = 28. (Taking x = 49 instead of √49 is the usual slip.)
APAR26-07-48 | Ratio and product of ages: sum or difference | Difficult
50. B. Let the ages be 4x and 5x. Product = 20x² = 180 ⇒ x² = 9 ⇒ x = 3. Ages: 12 and 15; sum = 27, difference = 3. (Taking x = 9 instead of √9 is the usual slip.)
APAR26-07-47 | Ratio and product of ages: sum or difference | Difficult