The Change × Count Idea
When one person joins or leaves a group, you can always work with totals: old total = old count × old average, new total = new count × new average, and the difference is the person who joined or left. That never fails. There is also a faster way to see it, and SSC's options are built to catch people who use it carelessly.
Joining. 24 students have an average weight of 52 kg. A new student joins and the average becomes 52.4 kg.
With totals: 25 × 52.4 − 24 × 52 = 1310 − 1248 = 62 kg.
The fast way: the newcomer has to bring himself up to the new average, 52.4 kg, and then add 0.4 kg for each of the 24 old students to lift them to 52.4 kg as well. So he weighs 52.4 + 24 × 0.4 = 62 kg. Starting from the old average instead, he supplies 0.4 kg for all 25 people, himself included: 52 + 25 × 0.4 = 62 kg. Both lines give the same answer, as long as you pair the new average with the old count, or the old average with the new count.
When the average falls. A team of 18 workers has an average daily wage of ₹640. When a trainee joins, the average falls by ₹5. The trainee earns 640 − 19 × 5 = ₹545 (or 635 − 18 × 5 = ₹545).
Leaving. 30 students average 46 kg. One leaves and the average becomes 45.6 kg, a fall of 0.4 kg. The one who left was taking more than his share: he weighed 46 + 29 × 0.4 = 57.6 kg. Check with totals: 30 × 46 − 29 × 45.6 = 1380 − 1322.4 = 57.6 kg.
Here n is the count before the change, A the old average, B the new average and d the size of the change:
| What happens | Value of that person (from A) | Value of that person (from B) |
|---|---|---|
| joins, average rises by d | ||
| joins, average falls by d | ||
| leaves, average rises by d | ||
| leaves, average falls by d |
The last column is the easiest to remember: the person's value is the new average, plus or minus d for each of the n people who were there at the start.
Which way does it go? If the average goes up when someone joins, he is above both averages; if it goes down, he is below both. If someone leaves and the average goes up, he was below it. Use this to strike out options before calculating.
Key Point: A change of d in the average of N people means the total changed by N × d. Every question in this section comes back to that.
One Value Added or Removed: Every Variant
Recent papers ask this in many forms. All of them come from the same equation, but knowing the forms by sight saves time.
1. New average after adding a known value. 14 numbers average 41, and 71 is added.
The new value is 30 above the average, and that 30 is shared by 15 numbers.
2. New average after removing a known value. 20 numbers average 45, and 26 is removed. The removed value was 19 below the average. Taking it out spreads that 19 over the other 19 numbers, so each goes up by 19 ÷ 19 = 1 and the new average is 46.
3. The removed value, when the new average is given. The average of 25 numbers is 52; after one is removed it becomes 51. The removed number is 52 + 24 × 1 = 76.
4. The original count. In every one-value case (taking both differences as sizes, without signs),
The average of some numbers is 28. When 52 is included, it becomes 30. The new number is 52 − 30 = 22 above the new average, and 22 ÷ 2 = 11, so there were 11 numbers at first.
For removal: the average of some numbers is 46, and when 74 is removed it falls to 44. The removed number is 74 − 44 = 30 above the new average, and 30 ÷ 2 = 15, so there were 15 numbers at first (14 are left). Check: 15 × 46 = 690, and 690 − 74 = 616 = 14 × 44.
5. The original average. There are 17 numbers; when 90 is added the average rises by 2.5. Then 90 = A + 18 × 2.5, so A = 45.
6. The average does not change. Then the value added or removed equals the average. In the same way, adding a number 10 above the average of n numbers raises the average by , and removing a number 10 above it lowers the average by .
7. One added, then one removed. A set of 9 numbers averages 47. 80 is added and then 35 is removed. The total goes from 423 to 468 and the count is back to 9, so the average is 52. When the count ends where it started, this is just replacement (see Replacement below): the average changes by (80 − 35) ÷ 9 = 5.
Most wrong answers here come from using n where n + 1 (or n − 1) is needed, so check which count goes with which average before you choose.
Several Values Added or Removed at Once
When several people join or leave together, treat them as one person carrying the whole group's total. The one-person formulas then work with k places in place of 1.
k join and the average changes. 36 students average 48 kg. Four new students join and the average rises by 0.6 kg. With totals: 40 × 48.6 − 36 × 48 = 1944 − 1728 = 216 kg for the four, an average of 54 kg.
The change × count form: the newcomers must fill their own 4 places at the old average (4 × 48 = 192 kg) and also supply 0.6 kg to all 40 people (24 kg). Total 216 kg.
Use a minus sign if the average falls. Dividing by k gives their average: 48 + 40 × 0.6 ÷ 4 = 54 kg.
k join with known values. 18 members average 54 kg, and two new members of 58 kg and 66 kg join. New average = (972 + 124) ÷ 20 = 54.8 kg. Or, faster: the newcomers are 4 and 12 kg above 54, which is 16 kg extra, shared by 20 people: 54 + 0.8 = 54.8 kg.
One of two newcomers known. 16 members average 57 kg. Two join and the average becomes 58 kg. One of them weighs 64 kg. The two together weigh 18 × 58 − 16 × 57 = 1044 − 912 = 132 kg, so the other weighs 68 kg.
k leave. 45 workers average ₹16,200 a month. Five leave, and the average of the remaining 40 is ₹16,000. The five took away 45 × 16,200 − 40 × 16,000 = ₹7,29,000 − ₹6,40,000 = ₹89,000, an average of ₹17,800.
In change × count form, the average fell by ₹200 for the 40 who stayed, so the leavers were above the old average by 40 × 200 = ₹8,000 between them: 5 × 16,200 + 8,000 = ₹89,000.
with plus when the average falls after they leave and minus when it rises.
Some leave and some join. A hostel mess has 40 boarders paying an average of ₹3,150 a month. Six boarders paying an average of ₹3,450 leave and eight new ones paying an average of ₹2,850 join. New total = 1,26,000 − 20,700 + 22,800 = ₹1,28,100 for 42 boarders, an average of ₹3,050.
Finding how many joined. If the newcomers' average is known but their number is not, the question is a weighted average of two groups; the Weighted Average section handles it with the "distance from the overall average" method.
Replacement: One or Two People Replaced
In a replacement, one person goes out and another comes in, so the count does not change. Only the total changes, and by exactly the difference between the two people:
One person replaced, average rises. The average weight of 8 rowers goes up by 1.5 kg when one rower of 62 kg is replaced by a new one. The total rose by 8 × 1.5 = 12 kg, so the new rower weighs 62 + 12 = 74 kg.
One person replaced, average falls. The average age of the 12 members of a board falls by 2 years when a 58-year-old member retires and a new member takes the seat. The total fell by 24 years, so the new member is 58 − 24 = 34 years old. This is why an older member being replaced pulls the average age down. (Questions where the average age stays the same as it was some years ago, because of a replacement, are in the Average Ages section.)
Two replaced. The average weight of 10 players rises by 2.5 kg when two players of 64 kg and 71 kg are replaced by two new players. The total rose by 25 kg, so the new pair weigh 135 + 25 = 160 kg together, an average of 80 kg. The 25 kg is shared by two people, so their average is 67.5 + 12.5 = 80, not 67.5 + 2.5.
Finding the size of the group. When a 54 kg person is replaced by a 72 kg person, the average weight of a group rises by 1.5 kg. The total rose by 18 kg, so there are 18 ÷ 1.5 = 12 people.
A chain of joining and replacing. The average weight of A, B and C is 62 kg. When D joins them, the average of the four becomes 60.5 kg. Then E, who is 4 kg heavier than D, replaces A, and the average of B, C, D and E becomes 59.75 kg.
- A + B + C = 186, so D = 4 × 60.5 − 186 = 56 kg and E = 60 kg.
- B + C + D + E = 4 × 59.75 = 239, so B + C = 239 − 116 = 123.
- A = 186 − 123 = 63 kg.
A check: replacing A by E took the four-person total from 242 to 239, a fall of 3, and E − A = 60 − 63 = −3.
Replacement and joining together. If the count changes as well, go back to totals. 20 members average 50 kg; one of 45 kg is replaced by one of 66 kg and another person of 55 kg joins. New total = 1000 + 21 + 55 = 1076 for 21 people, an average of about 51.24 kg.
[SSC Tip] In a replacement the count stays n. The two classic slips are using n + 1 (as if someone had only joined) and adding the change in average directly to the old person's weight, which gives 62 + 1.5 = 63.5 kg in the rowing case.
Wrong Entries and Misread Values
A mistake in one value changes the total by (wrong value − correct value), so the average is off by that amount divided by n:
A number copied as another. The average of 50 observations was found to be 64. Later it turned out that 86 had been read as 68. The total was 18 short, so the correct average is 64 + 18 ÷ 50 = 64.36.
Several wrong entries. 46 and 29 were recorded as 64 and 23. The net error is (64 + 23) − (46 + 29) = 87 − 75 = +12, so the recorded total was 12 too high. With 25 observations and a recorded average of 52, the correct average is 52 − 12 ÷ 25 = 51.52.
Digits reversed. A two-digit number with digits a and b is 10a + b; reversed, it is 10b + a. The difference is 9(a − b), always a multiple of 9.
- In a class of 30, a mark of 38 entered as 83 raises the total by 45 and the average by 45 ÷ 30 = 1.5.
- Going backwards, if the average of a class of 30 went up by 1.5, the total rose by 45 = 9 × 5, so the digits of the mark differ by 5. The mark could be 16, 27, 38 or 49; the question has to give one more fact to fix it.
Finding the number of items. A mark of 56 was entered as 81, and the class average went up by 0.5. The total rose by 25, so there are 25 ÷ 0.5 = 50 students.
A value left out. Two different slips are possible.
- The value is missing from the total, but the division is still by the right count. 12 numbers were averaged as 34 because 54 was left out of the sum. The correct total is 12 × 34 + 54 = 462, and the average is 38.5.
- The value is missing completely, so the division was by one less. The average of 11 numbers came out as 34 because the 12th number, 46, was forgotten. Then the correct average is 34 + (46 − 34) ÷ 12 = 35.
A value counted twice. A list of 15 numbers was averaged as 50, but 80 had been counted twice, so 16 entries were divided by 16. The true total is 16 × 50 − 80 = 720, and the correct average is 720 ÷ 15 = 48.
[SSC Tip] First decide whether the recorded total was too high or too low. If a big number was entered as a small one, the correct average is higher than the recorded one, and two of the four options usually go the wrong way.
Excluding the Highest or the Lowest
Questions that drop the highest or the lowest value give you totals to compare with the whole.
The average of 11 results is 60. Leaving out the highest, the average of the other 10 is 57; leaving out the lowest, it is 62.
- whole total: 11 × 60 = 660
- highest = 660 − 10 × 57 = 90
- lowest = 660 − 10 × 62 = 40
- range (highest − lowest) = 50
A shortcut for the range: highest − lowest = 10 × (62 − 57) = 50, because the two groups of 10 differ only in which extreme they contain.
Both left out together. A list of 15 numbers averages 48. With the highest and lowest both dropped, the other 13 average 47. Then highest + lowest = 15 × 48 − 13 × 47 = 720 − 611 = 109. You need one more fact (their difference, or one of them) to find each.
Highest exceeds lowest by a given amount. A set of 20 readings averages 45. The highest is 38 more than the lowest, and without both of them the other 18 average 44. Then
- highest + lowest = 900 − 792 = 108
- highest − lowest = 38
- highest = (108 + 38) ÷ 2 = 73, and lowest = 35
When does dropping a value leave the average unchanged? Only when that value equals the average. Dropping the highest and lowest together leaves the average unchanged when their mean equals the average, for example 30 and 70 in a list averaging 50.
Quick checks. Dropping the highest pulls the average down; dropping the lowest pulls it up. The highest can never be below the original average, and the lowest can never be above it. If an option breaks any of these, it is wrong.
Dropping several. If the top 3 of 25 scores average 92 and the class average is 64, the other 22 average (1600 − 276) ÷ 22 = 60.18, about 60.2. The idea is the same as one value: take the dropped total away from the whole total, then divide by the new count.
The same three lines solve the batsman's highest and lowest scores in the cricket part below.
Write the three totals (all, without highest, without lowest) in a column before subtracting, and multiply the reduced group's average by n − 1, not n.
Cricket: Batting Average and Run Rate
A batsman's average is total runs divided by innings played. (In real cricket you divide by the number of times he was out; SSC questions use innings unless they mention "not out".) Every batting question is a one-person-joins question in disguise: each new innings is a new member joining the list of scores.
Runs needed in the next innings. After 14 innings a batsman averages 36. To lift his average to 38 in the 15th innings, he needs 36 + 15 × 2 = 66 runs. In totals, he must go from 504 to 570.
Average after the latest innings. A batsman scores 92 in his 21st innings and his average goes up by 2.5. That 92 is 20 × 2.5 = 50 more than the new average, so the new average is 92 − 50 = 42. Check: before, 20 innings at 39.5 make 790 runs; 790 + 92 = 882, and 882 ÷ 21 = 42.
Runs in the latest innings. After his 16th innings a batsman's average is 47, and that innings raised it by 2. He scored 47 + 15 × 2 = 77 in it.
When the average falls. A batsman scores just 9 in his 26th innings and his average falls by 1.4. Then the new average is 9 + 25 × 1.4 = 44.
How many innings? A batsman's average rises by 2 after he scores 85 in his latest innings, and it is now 51. Then 85 − 51 = 34 = (innings before) × 2, so he had played 17 innings and the latest was his 18th.
Highest and lowest scores. A batsman's average over 40 innings is 51. His highest score is 153 more than his lowest, and without these two innings his average is 49.5.
- all 40 innings: 2040 runs; the other 38: 38 × 49.5 = 1881
- highest + lowest = 159, and highest − lowest = 153
- highest = 156, lowest = 3
[Tier 2] Not-out innings. A batsman made 1,125 runs in 30 innings and was not out in 5 of them. His batting average is 1125 ÷ 25 = 45, not 1125 ÷ 30 = 37.5.
Run rate. A team's run rate is its average number of runs per over. In the first 10 overs of a 50-over match the run rate was 3.8, so 38 runs came. To reach 278, the team needs 240 runs in the remaining 40 overs, a run rate of 6. The trap answer 278 ÷ 50 = 5.56 forgets the runs already made and the overs already used.
Cricket: Bowling Average
A bowler's average is the runs he gives away per wicket:
Here lower is better. A bowler with an average of 13.6 who takes 6 wickets for 33 runs in a match gives away only 5.5 runs per wicket in that match, well below his average, so his average falls.
Wickets taken before the last match. A bowler's average is 13.6 runs per wicket. In his last match he takes 6 wickets for 33 runs, and his average drops by 0.6, to 13. If he had w wickets before:
A faster line: at the new average the 6 new wickets would have cost 6 × 13 = 78 runs, but he gave only 33, a saving of 45 runs. That saving lowered the average of his 75 earlier wickets by 0.6 each, since 45 ÷ 0.6 = 75.
So the wickets before the last match = (wickets in the match × new average − runs in the match) ÷ fall in average.
Wickets taken in the last match. A bowler has 70 wickets at an average of 22, so he has given away 1,540 runs. In his next match he gives away 56 runs and his average becomes 21. If he took k wickets, then 1540 + 56 = 21(70 + k), so 70 + k = 76 and k = 6.
Average rising. If a bowler gives away more per wicket in a match than his average, his average goes up. A bowler has 28 wickets at 20 (560 runs). A match of 2 wickets for 76 runs makes it 636 ÷ 30 = 21.2.
Total wickets after the match. The same equation answers it; just add the match wickets: 75 + 6 = 81 in the first case.
Bowling average versus economy. Economy is runs per over, a different measure. Questions that give runs per over are run-rate questions (see Batting Average and Run Rate above), not bowling-average questions.
[SSC Tip] "Improved his average" in bowling means the average went down. Students who add the change instead of subtracting it get the usual wrong option. Check: if the match figures (runs ÷ wickets) are below his old average, the new average must be lower.
Target Questions: Marks Needed and the Minimum Score
A target question asks what is still needed to reach a set average. Work out the total the target demands, subtract what is already there, and the rest is what is needed.
Marks needed in the last test. Kavya's average in 5 tests is 66. To make her average 71 over 6 tests she needs 6 × 71 − 5 × 66 = 426 − 330 = 96. The shortcut: 71 + 5 × 5 = 96, that is, the target itself plus 5 marks to make up for each of the 5 earlier tests.
When the target cannot be reached. Kunal averages 78 in 4 tests out of 100 and wants 85 over 5 tests. He needs 85 + 4 × 7 = 113, more than the maximum of 100, so it cannot be done. The best he can get is (312 + 100) ÷ 5 = 82.4. SSC sometimes asks for this highest possible average.
How many more tests at a fixed score. A student averages 64 in 9 tests and then scores 80 in every test after that. How many more tests does she need to reach an average of 68? She is 9 × (68 − 64) = 36 marks short, and each new test brings 80 − 68 = 12 marks above the target. So she needs 36 ÷ 12 = 3 more tests. Check: 576 + 240 = 816 = 12 × 68.
Monthly sales target. A shopkeeper's sales for 5 months were ₹42,300, ₹38,900, ₹45,600, ₹40,200 and ₹44,500, a total of ₹2,11,500. For an average of ₹43,000 over 6 months he needs ₹2,58,000 in all, so the 6th month must bring ₹46,500. With deviations from 43,000 it is quicker: −700, −4,100, +2,600, −2,800 and +1,500 add up to −3,500, so the 6th month needs 43,000 + 3,500 = ₹46,500.
Minimum average in the remaining papers. To pass, a student needs an average of at least 60 over 8 papers. She has 410 marks in the first 6. She needs 480 − 410 = 70 more in 2 papers, so at least 35 per paper on average.
Not letting the average fall. A batsman averaging 41 must score at least 41 in his next innings to keep his average from falling; any score below the current average pulls it down, and any score above it pulls it up.
Weighted percentages across papers with different maximum marks are in the Weighted Average section.
Solved Examples
Question 1
The average weight of 27 students in a class is 41 kg. When a new student joins, the average weight becomes 41.5 kg. What is the weight of the new student? (a) 54.5 kg (b) 55.5 kg (c) 55 kg (d) 41.5 kg
Answer. The newcomer must be at the new average, 41.5 kg, and also give 0.5 kg to each of the 27 old students. So he weighs 41.5 + 27 × 0.5 = 41.5 + 13.5 = 55 kg. The answer is (c) 55 kg. With totals: 28 × 41.5 − 27 × 41 = 1162 − 1107 = 55. Using 27 with the old average (41 + 13.5 = 54.5) is the usual slip.
Question 2
The average monthly salary of 15 employees of a firm is ₹28,400. When a trainee joins, the average falls by ₹600. What is the trainee's salary? (a) ₹18,800 (b) ₹19,400 (c) ₹27,800 (d) ₹18,200
Answer. The average fell, so the trainee earns less than both averages. From the old average he is short by ₹600 for each of the 16 people, himself included: 28,400 − 16 × 600 = 28,400 − 9,600 = ₹18,800. The answer is (a) ₹18,800.
Question 3
The average of 21 numbers is 64. When one number is removed, the average of the remaining numbers is 62.5. What is the number that was removed? (a) 95.5 (b) 92.5 (c) 34 (d) 94
Answer. The average fell when it left, so the removed number was above the average. Removed = old average + (remaining count) × fall = 64 + 20 × 1.5 = 94. The answer is (d) 94. With totals: 21 × 64 − 20 × 62.5 = 1344 − 1250 = 94.
Question 4
The average weight of a group of people is 48 kg. When a person weighing 78 kg joins the group, the average weight goes up by 2 kg. How many people were in the group at first? (a) 15 (b) 14 (c) 13 (d) 39
Answer. The new average is 50 kg. The newcomer is 78 − 50 = 28 kg above it, and that 28 kg went as 2 kg to each of the people already there. So there were 28 ÷ 2 = 14 people. The answer is (b) 14. Check: 14 × 48 + 78 = 750 = 15 × 50. Using the old average, 30 ÷ 2 = 15, gives the count after he joined.
Question 5
The average weight of 40 students is 52 kg. Five new students join and the average weight of all the students rises by 0.5 kg. What is the average weight of the five new students? (a) 56.5 kg (b) 56 kg (c) 52.5 kg (d) 74.5 kg
Answer. The five must fill their own places at 52 kg (260 kg) and also add 0.5 kg for all 45 students (22.5 kg). So together they weigh 282.5 kg, and their average is 282.5 ÷ 5 = 56.5 kg. The answer is (a) 56.5 kg.
Question 6
The average monthly salary of 50 workers in a unit is ₹21,000. Eight workers are transferred out, and the average salary of the remaining workers becomes ₹20,600. What was the average salary of the eight workers who left? (a) ₹21,400 (b) ₹23,500 (c) ₹23,100 (d) ₹18,900
Answer. Total before: 50 × 21,000 = ₹10,50,000. Total after: 42 × 20,600 = ₹8,65,200. The eight took away ₹1,84,800, which is ₹23,100 each on average. The answer is (c) ₹23,100. The average fell when they left, so they must have earned more than ₹21,000.
Question 7
The average weight of 22 members of a team is 61 kg. Two new members join, and the average weight becomes 62 kg. If one of the new members weighs 70 kg, what is the weight of the other? (a) 74 kg (b) 54 kg (c) 78 kg (d) 76 kg
Answer. Total now: 24 × 62 = 1488 kg. Total before: 22 × 61 = 1342 kg. So the two new members weigh 146 kg together, and the other one weighs 146 − 70 = 76 kg. The answer is (d) 76 kg.
Question 8
The average weight of 9 members of a rowing crew increases by 2.4 kg when one of them, who weighs 58 kg, is replaced by a new member. What is the weight of the new member? (a) 60.4 kg (b) 79.6 kg (c) 82 kg (d) 34.4 kg
Answer. The count stays 9, so the total went up by 9 × 2.4 = 21.6 kg. The new member weighs 58 + 21.6 = 79.6 kg. The answer is (b) 79.6 kg. Adding just 2.4 gives 60.4 kg.
Question 9
The average weight of 11 players falls by 2 kg when two players weighing 78 kg and 82 kg are replaced by two new players. What is the average weight of the two new players? (a) 69 kg (b) 78 kg (c) 138 kg (d) 58 kg
Answer. The total fell by 11 × 2 = 22 kg. The two who left weighed 160 kg, so the two new players weigh 160 − 22 = 138 kg together, and their average is 69 kg. The answer is (a) 69 kg.
Question 10
When a person weighing 47 kg in a group is replaced by a person weighing 68 kg, the average weight of the group increases by 0.75 kg. How many people are there in the group? (a) 21 (b) 27 (c) 28 (d) 29
Answer. The total went up by 68 − 47 = 21 kg, and that raised the average by 0.75 kg. So the group has 21 ÷ 0.75 = 28 people. The answer is (c) 28.
Question 11
The average age of 10 employees in a department falls by 3 years when a 57-year-old employee retires and a new employee is appointed in his place. What is the age of the new employee? (a) 54 years (b) 30 years (c) 24 years (d) 27 years
Answer. Replacement keeps the count at 10, so the total age fell by 10 × 3 = 30 years. The new employee is 57 − 30 = 27 years old. The answer is (d) 27 years. Using 11 people gives 24 years.
Question 12
The average weight of A, B and C is 70 kg. When D joins them, the average weight of the four becomes 68 kg. If E, who weighs 5 kg more than D, replaces A, the average weight of B, C, D and E becomes 67.5 kg. What is the weight of A? (a) 67 kg (b) 69 kg (c) 62 kg (d) 71 kg
Answer. A + B + C = 210, and A + B + C + D = 272, so D = 62 kg and E = 67 kg. Next, B + C + D + E = 4 × 67.5 = 270, so B + C = 270 − 62 − 67 = 141. Then A = 210 − 141 = 69 kg. The answer is (b) 69 kg. Check: swapping A for E changed the four-person total from 272 to 270, and E − A = 67 − 69 = −2.
Question 13
The average of 45 observations was found to be 72. Later it was noticed that the value 53 had been copied as 98. What is the correct average? (a) 71 (b) 73 (c) 70 (d) 69
Answer. The recorded total was 98 − 53 = 45 too high. Spread over 45 observations, the average was 1 too high, so the correct average is 72 − 1 = 71. The answer is (a) 71. A big number replaced a small one, so the correct average must be lower.
Question 14
In a class, a student's marks were entered as 85 instead of 58. Because of this, the average marks of the class went up by 0.75. How many students are in the class? (a) 27 (b) 30 (c) 36 (d) 40
Answer. The total went up by 85 − 58 = 27 marks, and this raised the average by 0.75. So there are 27 ÷ 0.75 = 36 students. The answer is (c) 36.
Question 15
The average of 18 numbers was calculated as 46. Later it was found that the number 72 had been left out while adding, though the total was still divided by 18. What is the correct average? (a) 54 (b) 49 (c) 46 (d) 50
Answer. The total was 72 short, so the average was 72 ÷ 18 = 4 short. The correct average is 46 + 4 = 50. The answer is (d) 50. In totals: 18 × 46 = 828, the correct total is 828 + 72 = 900, and 900 ÷ 18 = 50.
Question 16
The average of 13 values is 54. If the highest value is left out, the average of the rest is 51.5, and if the lowest value is left out, the average of the rest is 57. What is the difference between the highest and the lowest values? (a) 84 (b) 66 (c) 18 (d) 71.5
Answer. Total of all 13 = 702. Without the highest, 12 × 51.5 = 618, so the highest is 84. Without the lowest, 12 × 57 = 684, so the lowest is 18. The difference is 84 − 18 = 66. The answer is (b) 66. Shortcut: 12 × (57 − 51.5) = 66. Using 13 instead of 12 gives 71.5.
Question 17
A batsman has an average of 39 runs after 18 innings. How many runs must he score in his 19th innings to raise his average to 42? (a) 96 runs (b) 93 runs (c) 99 runs (d) 42 runs
Answer. He needs the target, 42, plus 3 runs for each of the 18 innings already played: 42 + 18 × 3 = 96. In totals: 19 × 42 − 18 × 39 = 798 − 702 = 96. The answer is (a) 96 runs.
Question 18
A batsman scores 108 runs in his 25th innings and thereby increases his average by 3 runs. What is his average after the 25th innings? (a) 33 (b) 39 (c) 36 (d) 105
Answer. The 108 runs supplied the new average for this innings plus 3 runs for each of the 24 earlier innings. So the new average is 108 − 24 × 3 = 36. The answer is (c) 36. Check: before, 24 innings at 33 make 792 runs; 792 + 108 = 900 = 25 × 36. 33 is his average before this innings.
Question 19
A batsman's average in 44 innings is 56 runs. His highest score exceeds his lowest score by 168 runs. If these two innings are excluded, his average in the remaining 42 innings is 54 runs. What is his highest score? (a) 196 (b) 168 (c) 14 (d) 182
Answer. All 44 innings: 44 × 56 = 2464 runs. The other 42: 42 × 54 = 2268 runs. So highest + lowest = 196, and highest − lowest = 168. Adding the two lines, 2 × highest = 364, so the highest is 182 (and the lowest is 14). The answer is (d) 182.
Question 20
A bowler's average is 16.5 runs per wicket. In his next match he takes 5 wickets for 35 runs, and his average falls by 0.5. How many wickets had he taken before this match? (a) 95 (b) 90 (c) 70 (d) 85
Answer. Let the earlier wickets be w. Then , so 0.5w = 80 − 35 = 45 and w = 90. The answer is (b) 90. Quick line: at the new average the 5 wickets would cost 80 runs, he saved 45, and 45 ÷ 0.5 = 90. 95 is his total after the match.
Question 21
Aman's average in the first 7 tests is 71. How many marks does he need in the 8th test to make his average 74? (a) 95 (b) 98 (c) 92 (d) 77
Answer. Target total: 8 × 74 = 592. Present total: 7 × 71 = 497. He needs 592 − 497 = 95. Shortcut: 74 + 7 × 3 = 95. The answer is (a) 95.