How to Use These Questions
The 35 questions below jump between the four topics, so a cricket average can come straight after a family-age question. First decide what kind of change it is: one group gaining or losing a member, two groups put together, or ages moving with time. After that the working is usually one or two lines.
Do each one on paper with a timer running, at about 36 to 45 seconds a question. If one runs past a minute, look up its fast route in Shortcuts & Speed Methods and redo it the next day.
About a fifth of the questions are on averages of numbers: number series, overlapping groups and pair averages. About a quarter are on changes to a group: people joining, leaving or being replaced, wrong entries, cricket and target marks. Weighted and combined averages take the biggest share, close to 30%, since recent CGL papers ask them often. The last quarter is average ages.
The [Tier 2] questions add one more step, such as finding n from the average of squares, different raises for two groups, a bowler's wickets, or ages with reversed digits. Tier 1 students should still work through them, as each rests on a method from the topic sections.
Solved Examples
Question 1
In a class of 45 students, the boys average 64 marks and the girls average 71 marks in a test. The class average is 68.2. How many girls are in the class?
Answer. I use the see-saw. The boys are 68.2 − 64 = 4.2 below the class average and the girls are 71 − 68.2 = 2.8 above it. The group sizes go the other way round, so boys : girls = 2.8 : 4.2 = 2 : 3. That's 5 units for 45 students, so 1 unit = 9 and there are 27 girls (and 18 boys). Check: 1,152 + 1,917 = 3,069 = 45 × 68.2.
Question 2
Find the average of all the prime numbers between 100 and 130.
Answer. There's no formula for primes, so I just list them: 101, 103, 107, 109, 113 and 127. That's 6 primes adding up to 660, so the average is 660 ÷ 6 = 110. The usual slips are 111 (3 × 37), 119 (7 × 17) and 121 (11 × 11), which all look prime but aren't.
Question 3
The average weight of 50 students is 46 kg. Six students leave, and the average weight of the rest goes up by 0.75 kg. What is the average weight of the six who left?
Answer. The average went up, so the six who left were lighter than 46 kg. The 44 who stay now average 0.75 kg above 46, which is 33 kg extra in all, so the six leavers were 33 kg short of 6 × 46 = 276 kg. They weigh 243 kg together, which is 40.5 kg each on average. Check: 2,300 − 44 × 46.75 = 2,300 − 2,057 = 243.
Question 4
Three years ago, the average age of a family of 5 was 26 years. A baby has been born since then, and today the average age of the 6 members is 24.5 years. How old is the baby?
Answer. The five older members have each gained 3 years, so today they average 29 and total 5 × 29 = 145. The family of 6 totals 6 × 24.5 = 147. The baby is the difference, 2 years.
Question 5
A shopkeeper buys 25 kg of rice at ₹42 a kg, 35 kg at ₹48 a kg and 40 kg at ₹51 a kg. What is his average cost per kg?
Answer. The quantities are different, so I can't just average the prices. With ₹48 as the base: 25 × (−6) + 35 × 0 + 40 × (+3) = −150 + 120 = −30. Over 100 kg that's −0.30, so the average cost is ₹47.70 a kg.
Question 6
The average of 11 numbers is 42. The average of the first five is 39 and the average of the last five is 46. Find the sixth number.
Answer. This time the two groups don't overlap. The first five and the last five leave out only the 6th number, so 6th number = full total − both groups = 462 − 195 − 230 = 37.
Question 7
In a committee of 8 members, one member aged 63 is replaced by a new member, and the average age of the committee falls by 3.5 years. How old is the new member?
Answer. The count stays at 8, so the total age falls by 8 × 3.5 = 28 years. All of that comes from the swap, so the new member is 28 years younger than the one who left: 63 − 28 = 35 years.
Question 8
140 candidates took an exam and their average was 44 marks. The candidates who passed averaged 53 and those who failed averaged 32. How many candidates passed?
Answer. The ones who passed are 9 above 44, and the ones who failed are 12 below. The sizes go the opposite way, so passed : failed = 12 : 9 = 4 : 3. 7 units = 140, so 1 unit = 20 and 80 candidates passed. Check: 4,240 + 1,920 = 6,160 = 140 × 44.
Question 9
Four brothers and sisters were born 3 years apart, and their present ages add up to 50 years. How old is the eldest?
Answer. Their ages go up in equal steps, so the average, 50 ÷ 4 = 12.5, sits halfway between the two middle children. With a gap of 3 they are 11 and 14, so the ages are 8, 11, 14 and 17, and the eldest is 17 years old.
Question 10
A batsman has a certain average after some innings. In his next innings he scores 132 runs, and his average goes up by 4 to 52. How many innings has he played now, including this one?
Answer. Before this innings his average was 48. Out of the 132 runs, 52 count for this innings at the new average, and the other 80 are extra. Those 80 runs lifted each old innings by 4, so there were 80 ÷ 4 = 20 old innings. With this one he has played 21 innings. Check: 20 × 48 + 132 = 1,092 = 21 × 52.
Question 11
In an office, the average monthly salary of all the staff is ₹31,500. The 14 officers average ₹54,000 and the other workers average ₹27,000. How many people work in the office?
Answer. The officers are 22,500 above ₹31,500 and the workers are 4,500 below. So officers : workers = 4,500 : 22,500 = 1 : 5. With 14 officers there are 70 workers, which makes 84 people in all.
Question 12
The average of A and B is 38, the average of B and C is 49, and the average of A and C is 45. Find B.
Answer. Adding the three pair averages gives A + B + C straight away: 38 + 49 + 45 = 132. (Each pair total is double its average, and each letter turns up in two pairs, so the doubles cancel.) A + C = 2 × 45 = 90, so B = 132 − 90 = 42. Check: A = 34 and C = 56 fit all three pairs.
Question 13
A husband and wife had an average age of 23 years when they married 9 years ago. Now, with their two children, the average age of the family is 19.5 years. The children were born 4 years apart. How old is the younger child?
Answer. The couple are 9 years older each, so they now total 46 + 18 = 64. The family of 4 totals 4 × 19.5 = 78, so the children add up to 14. With a gap of 4 they are 9 and 5, so the younger one is 5 years old.
Question 14
A student scores 80% in a paper of 100 marks, 66% in a paper of 150 marks and 59% in a paper of 200 marks. What is his overall percentage?
Answer. I turn everything into marks: 80 + 99 + 118 = 297 out of 450. That's 297 ÷ 450 = 66%. The plain average of the three percentages is about 68.3%, which is wrong because the papers carry different maximum marks.
Question 15
Ten numbers were added and their average found. Later it was noticed that one two-digit number had been written with its digits reversed, which made the average 2.7 more than it should have been. The digits of the correct number add up to 11. What is the correct number?
Answer. The total went up by 10 × 2.7 = 27. A number and its reverse differ by 9 × (difference of the digits), so the digits differ by 3. They add up to 11, so they are 4 and 7. The wrong entry was the bigger one, so the correct number is 47 (written as 74).
Question 16
The average age of 8 men goes up by 2 years when two of them, aged 35 and 41, are replaced by two women. What is the average age of the two women?
Answer. The total age rises by 8 × 2 = 16 years, and all of it comes from the swap. The two men total 76, so the two women total 76 + 16 = 92, and their average is 46 years.
Question 17
Class A has 30 students with an average of 64 marks. When class A is put together with class B, which averages 72, the combined average is 67. How many students are in class B?
Answer. Class A is 3 below 67 on each of 30 students, a shortfall of 90. Each class B student is 5 above 67, so it takes 90 ÷ 5 = 18 students to balance it. Check: 1,920 + 1,296 = 3,216 = 48 × 67.
Question 18
The average of five consecutive odd numbers is 61. If the next three odd numbers are also included, what is the new average, and what is the largest number in the list?
Answer. The five numbers are 57 to 65. Three more terms with a step of 2 move the middle forward by = 3, so the new average is 64. The list now runs 57, 59, … , 71, so the largest is 71. Check: = 64.
Question 19
The average weight of a group of 14 people is 56 kg. When P joins them, the average becomes 57 kg. Then Q, who is 6 kg lighter than P, also joins. What is the average weight of the group now?
Answer. P = new average + old count × rise = 57 + 14 × 1 = 71 kg, so Q is 65 kg. Q is 8 kg above the current average of 57, and that 8 is shared by 16 people, so the average goes up by 0.5 to 57.5 kg.
Question 20
Sections A, B and C have 30, 40 and 50 students. The average marks of A and B are 58 and 64, and the average of all 120 students is 62. Find the average of section C.
Answer. With 62 as the base: A gives 30 × (−4) = −120 and B gives 40 × (+2) = +80, which leaves −40. Section C must make up +40 across 50 students, which is 0.8 each, so C averages 62.8. Check: 1,740 + 2,560 + 3,140 = 7,440 = 120 × 62.
Question 21
The average age of a family of 7 is 26 years, and the youngest member is 5. What was the average age of the family at the time the youngest was born?
Answer. Without the youngest, the other six total 182 − 5 = 177 today. Five years ago each of them was 5 years younger, so they totalled 177 − 30 = 147. The family then had only these six, so the average was 147 ÷ 6 = 24.5 years.
Question 22 [Tier 2]
The average of the squares of the first n natural numbers is 105. Find n.
Answer. The average is , so (n + 1)(2n + 1) = 630. Here 2n + 1 is roughly twice n + 1, so is close to 630, is close to 315, and n + 1 is about 18. Trying n = 17: 18 × 35 = 630, which fits. So n = 17.
Question 23
A batsman's average in 36 innings is 42. His highest score is 150 runs more than his lowest. If these two innings are left out, the other 34 innings average 40. What is his highest score?
Answer. All 36 innings total 1,512 and the other 34 total 1,360. So highest + lowest = 152, and highest − lowest = 150. The highest is = 151 (and the lowest is just 1).
Question 24
A man spends ₹1,800 on rice at ₹24 a kg, and another ₹1,800 on rice at ₹40 a kg. What is his average cost per kg?
Answer. The money is equal, not the kg, so I can't just average the prices. He gets 1,800 ÷ 24 = 75 kg and 1,800 ÷ 40 = 45 kg, so 120 kg for ₹3,600, which is ₹30 a kg. The plain average, ₹32, is the trap. The formula = 30 agrees.
Question 25
Four years ago, the average age of a family of 5 was 34 years. Since then a bride has joined the family, and today the average age of the 6 members is 36.5 years. How old is the bride?
Answer. The five are 4 years older each, so they average 38 and total 190. The six total 6 × 36.5 = 219, so the bride is 29 years old.
Question 26 [Tier 2]
The average monthly salary of 40 employees is ₹30,300. The 12 managers average ₹45,000. The managers get a 10% raise and everyone else gets a 5% raise. What is the new average salary?
Answer. All 40 total 12,12,000 and the managers total 5,40,000, so the other 28 total 6,72,000. The managers' raise is 10% of 5,40,000 = 54,000, and the others' is 5% of 6,72,000 = 33,600. That's 87,600 extra shared by 40 people, or 2,190 each, so the new average is 30,300 + 2,190 = ₹32,490. A flat 7.5% on ₹30,300 is the trap, because the managers' bigger raise is on bigger salaries.
Question 27
A student's average marks in 7 subjects are 58. How many marks must he get in the 8th subject to raise his average to 62?
Answer. Target + subjects done × gap = 62 + 7 × 4 = 90. Check: 8 × 62 − 7 × 58 = 496 − 406 = 90.
Question 28
What is the average of all the numbers between 300 and 600 that are divisible by 13?
Answer. 13 × 24 = 312 is the first one after 300, and 13 × 46 = 598 is the last one before 600. They form an AP, so the average is = 455. Averaging the limits, = 450, is the trap.
Question 29
The average age of 35 students and their teacher is 15 years. If the teacher's age is left out, the average falls by 1 year. How old is the teacher?
Answer. Without the teacher, the 35 students average 14, so the teacher lifts each of them by 1, from 14 to 15. He has to bring 15 for himself plus 35 × 1 for the students, so the teacher = 15 + 35 × 1 = 50 years. Check: 36 × 15 − 35 × 14 = 540 − 490 = 50.
Question 30 [Tier 2]
A batch of 60 students has an average of 52 marks. The top 20 students average 18 marks more than the other 40. What is the average of the top 20?
Answer. The other 40 are twice as many as the top 20, so the batch average sits one-third of the way from the lower group to the higher one. That puts it 18 ÷ 3 = 6 above the lower average. So the other 40 average 46, and the top 20 average 46 + 18 = 64. Check: 1,280 + 1,840 = 3,120 = 60 × 52.
Question 31 [Tier 2]
A bowler's average is 11.9 runs per wicket. In his next match he takes 5 wickets for 27 runs, and his average improves by 0.5. How many wickets has he taken in all, including this match?
Answer. The new average is 11.4. At that rate the 5 new wickets would "cost" 5 × 11.4 = 57 runs, but he gave only 27, a saving of 30. That saving makes every old wicket 0.5 cheaper, so he had 30 ÷ 0.5 = 60 wickets before this match. In all he has 65 wickets. Check: 60 × 11.9 + 27 = 741 = 65 × 11.4.
Question 32
The average age of A and B is 32 years. Five years ago, A was twice as old as B. How old is A now?
Answer. A + B = 64 now, so five years ago their ages added up to 54. Then A was twice B, so B was 54 ÷ 3 = 18 and A was 36. Now A is 41 years old (and B is 23).
Question 33
The average temperature of a week was 28°C. The average of the first four days was 26.5°C, and the average of the last four days was 30.25°C. What was the temperature on the fourth day?
Answer. The fourth day is in both groups. With differences from 28: the first four days are 1.5 below each (−6) and the last four are 2.25 above each (+9). So the fourth day = 28 − 6 + 9 = 31°C. The totals give the same: 106 + 121 − 196 = 31.
Question 34
The average marks of 25 students are 54. If the topper and the lowest scorer are left out, the remaining 23 average 53. The topper scored 98. How many marks did the lowest scorer get?
Answer. All 25 total 1,350 and the other 23 total 1,219. So the topper and the lowest scorer have 131 between them, and the lowest scorer got 131 − 98 = 33.
Question 35 [Tier 2]
The present ages of a father and his son have the same two digits in reverse order. The average of their ages is 33, and the father is 36 years older than the son. How old is the father?
Answer. I write the father's age as 10a + b and the son's as 10b + a. Their sum is 11(a + b) = 66, so a + b = 6. Their difference is 9(a − b) = 36, so a − b = 4. That gives a = 5 and b = 1, so the father is 51 and the son is 15.