Average

This is where the chapter starts. The questions are short, and they build the habit of thinking in totals.

1. Direct average. Add the values and divide by how many there are. When the values follow a regular pattern (like 15, 25, 35, 45), the middle value is the average. With an even count, take the mid-point of the two middle values: 25 + 352 = 30. No adding needed.

2. Missing value. "The average of 6 numbers is 20. Five of them are given. Find the sixth." Required total = 6 × 20 = 120, so the missing value = 120 − (sum of the five). The same thinking answers "what must he score in the next test to make his average 80": required total − present total.

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3. Change in every value. This is the most scoring part of the type.

• If every value is increased or decreased by k, the average also goes up or down by k.
• If every value is multiplied or divided by k, the average is also multiplied or divided by k.
• If only some values change, find the net change in the total and divide it by the number of values: change in average = Net change in totaln.

Radhika's 4 marks have an average of 50. Add 3 to each and the average becomes 53. Double each and it becomes 100. If only one mark goes up by 12, the total goes up by 12, so the average goes up by 124 = 3.

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4. Values linked by a relation. Sometimes the numbers are given as ratios or percentages of each other, such as "the first is 20% more than the second". Take the simplest one as x (or in parts), write the others in terms of it, and use Sum = Average × Number.

5. Average and the sign of the numbers. If the average of n numbers is zero, the positives and negatives cancel. At most (n − 1) of them can be negative, because one large positive number can balance all the rest. In the same way, at most (n − 1) can be positive.

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When numbers go up by the same step every time (consecutive numbers, consecutive odd or even numbers, multiples of a number, any AP), you do not need to add anything to find the average.

The rule: for numbers in AP (equal gaps), Average = Middle term = First term + Last term2.

Why does this work? Pair the first with the last, the second with the second-last, and so on. Every pair has the same sum (first + last), so every pair has the same average, and that is also the middle value.

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• Odd count (say 7 numbers): the average is exactly the 4th number, the one in the middle.
• Even count (say 8 numbers): the average is the mid-point of the 4th and 5th numbers. So an even count of consecutive odd numbers has an even average, and the other way round.

Standard results (first n terms):

• First n natural numbers: n + 12
• First n odd numbers (1, 3, 5, …): n
• First n even numbers (2, 4, 6, …): n + 1
• First n multiples of k: k(n + 1)2
• Squares 1², 2², …, n²: (n + 1)(2n + 1)6
• Cubes 1³, 2³, …, n³: n(n + 1)²4

Finding the numbers from the average. If 5 consecutive odd numbers have an average of 41, the middle one is 41, so the numbers are 37, 39, 41, 43, 45. Smallest = Average − 2 × 2 and largest = Average + 2 × 2. In general, for k numbers with common difference d: smallest = Average − (k − 1)d2 and largest = Average + (k − 1)d2.

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Shifting the set. If the same block of consecutive numbers moves forward by m places, the average moves forward by m × d. If numbers are added at the ends, find the new middle.

Radhika writes 9 consecutive even numbers with an average of 30. The middle (5th) number is 30, so the numbers run from 22 to 38. She then adds the next two even numbers, 40 and 42. Now the 11 numbers run from 22 to 42, and the new average = 22 + 422 = 32.

For numbers that are not in AP, such as primes, there is no shortcut. List them and add.

Squares and cubes of the first n numbers. The numbers 1², 2², 3², … do not go up by equal steps, so the mid-point rule is not allowed here. Use the sum formulas and divide by n:

Average of 1², 2², …, n² = (n + 1)(2n + 1)6

Average of 1³, 2³, …, n³ = n(n + 1)²4

Check with n = 5. The squares add up to 1 + 4 + 9 + 16 + 25 = 55, so the average is 11, and 6 × 116 = 11 ✓. The cubes add up to 225, so the average is 45, and 5 × 364 = 45 ✓. For the squares of the first n even numbers, multiply the first result by 4, because (2k)² = 4k². For the squares of the first n odd numbers the average is (2n − 1)(2n + 1)3.

Removing terms from an AP. If you take the first term out of an AP with common difference d, what is left is again an AP, and its first term has moved up by d. So the average goes up by d2. Taking out the last term pulls the average down by d2. In general, removing k terms from the beginning raises the average by kd2, and removing k terms from the end lowers it by kd2. If you remove the same number of terms from both ends, the average does not change.

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The multiples of 5 from 5 to 50 have the average 27.5. Remove 5, and the multiples 10 to 50 have the average 30, which is 27.5 + 52. Remove 50 instead, and 5 to 45 average 25 = 27.5 − 2.5. This works only when the removed terms are the ones at the ends of the series.

When two or more groups are put together, the combined average is not the simple average of the group averages. Each group pulls the result towards itself, and the bigger the group, the stronger the pull. That is why it is called a weighted average.

Combined average = n₁a₁ + n₂a₂ + n₃a₃ + …n₁ + n₂ + n₃ + …

Here n is the size (weight) of a group and a is its average. The weights can be actual counts, a ratio, percentages or fractions, because only their proportion matters. "30% are women" just means weights of 30 : 70.

In Radhika's coaching class, 30 boys average 60 marks and 20 girls average 75. Combined average = 30 × 60 + 20 × 7550 = 1800 + 150050 = 66. It is closer to 60 because the boys' group is bigger.

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The reverse questions are asked more often:

• The overall average and one group are known, and you need the other group's average: other total = overall total − known total.
• Both group averages and the overall average are known, and you need the number in a group. Here the balance rule is the quickest method.

Balance rule for two groups. The overall average A lies between the two group averages. Members of group 1 are (A − a₁) below it and members of group 2 are (a₂ − A) above it. The two sides must balance:

n₁(A − a₁) = n₂(a₂ − A) ⇒ n₁n₂ = a₂ − AA − a₁

Say officers earn 50,000, clerks earn 20,000 and the overall average is 30,000. Officers are 20,000 above and clerks are 10,000 below, so officers : clerks = 10,000 : 20,000 = 1 : 2. This is the same alligation cross you will use in Mixture and Alligation.

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Calendar questions, such as "a month of 30 days beginning with a Sunday", are weighted averages too. Count how many days of each kind the month has and use those counts as the weights.

"A teacher joins and the average rises by 1." "One student leaves and the average falls by 0.5." Such questions can always be done with totals: new total − old total. The deviation idea turns them into one-line sums, and it helps to see why.

A new member joins. The old group has n members with average A. The newcomer joins and the average rises by t, so the new average is A + t for (n + 1) people. For this to happen, the newcomer must bring:

• A, his own fair share, so that the old average is not disturbed, and
• an extra t for each of the (n + 1) people now in the group.

So, New member = A + (n + 1) × t. If the average falls by t, then New member = A − (n + 1) × t.

Radhika's class of 24 students has an average weight of 40 kg. When the teacher joins, the average becomes 41 kg. Teacher = 40 + 25 × 1 = 65 kg. By totals: 25 × 41 − 24 × 40 = 1025 − 960 = 65 ✓.

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A member leaves. There are n members with average A. One leaves and the average of the remaining (n − 1) falls by t. The person who left was above the average, and his extra was what kept the other (n − 1) people t higher. So the person who left = A + (n − 1) × t. If the average rises after he leaves, he was below the average: A − (n − 1) × t.

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A group joins or leaves. If m people join and the average rises by t, the total of the new people = mA + (n + m)t, so their average = A + (n + m)tm. Leaving works the same way with (n − m).

Finding the original number. Sometimes the newcomer's value is given and the number of people is asked. Use the same formula backwards: 36,000 = 24,000 + (n + 1) × 400 ⇒ n + 1 = 30, so there were 29 people before.

In a replacement the number of people does not change. Someone goes out and someone else comes in his place. So the whole change in the total comes from the difference between the newcomer and the person who left.

Change in total = New member − Old member
Change in average = New member − Old membern

Turn it around and you get the formula behind nearly every replacement question:

New member = Old member + n × (change in average)

Use + if the average increases and − if it decreases.

Radhika's team of 8 has some average weight. A player of 52 kg is replaced and the average goes up by 1.5 kg. The total went up by 8 × 1.5 = 12 kg, so the new player weighs 52 + 12 = 64 kg. The average itself was never needed.

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Two or more replaced. If k people are replaced by k new people:
Total of the new k = Total of the old k + n × (change in average)

Finding n. If both values are known: n = New − Oldchange in average.

Mixed cases. When some people leave and a different number join (2 leave, 3 join), the count changes. Go back to totals: new total = old total − (leavers) + (joiners), and divide by the new count.

Same average after replacement. If the average does not change, the newcomer's value equals the value of the person replaced. For groups, the two totals are equal.

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A value was copied wrongly, or a mark was changed after re-checking, or the digits of a number got reversed. In every such case only one thing has happened: the total has changed by (correct value − wrong value). Everything else stays the same.

Correct average = Wrong average + Correct value − Wrong valuen

If the correct value is bigger, the correct average is higher. If it is smaller, the correct average is lower.

Radhika's class of 40 has an average of 55. One mark was entered as 73 instead of 37. The total is 36 too high, so the correct average = 55 − 3640 = 55 − 0.9 = 54.1.

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Several wrong entries. Add up all the corrections (each one is correct − wrong, with its sign) and divide by n once.

Finding n. If the error is known and the change in average is given, n = Error in totalChange in average. For example, a mark of 45 written as 85 raised the average by 0.8, so n = 400.8 = 50.

Digits interchanged. A two-digit number 10a + b becomes 10b + a, a change of 9(a − b). So when the digits are reversed, the total changes by 9 × (difference of the digits). Examiners like this twist.

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Extra marks in some subjects. "If he had got 12 more in Maths and 8 more in Science, his average would have risen by 2." The total rises by 20 = 2 × n, so n = 10 subjects. It is the same idea.

"The average of 13 numbers is 40. The average of the first 7 is 38 and of the last 7 is 43. Find the 7th number." Look at the positions. The first 7 are positions 1–7 and the last 7 are positions 7–13. Together they cover all 13 numbers, but the 7th number is counted twice.

So, (sum of first 7) + (sum of last 7) = (sum of all 13) + 7th number.

Common number = Sum of first group + Sum of last group − Sum of all

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Always count positions before you start. Three situations come up:

• The two groups overlap in one number (6 + 6 out of 11). The common number is the answer, from the formula above.
• The two groups overlap in two numbers. For example, the first 17 and the last 16 days of a 31-day month share days 16 and 17. The formula gives the sum of those two, and another condition splits them.
• The two groups leave a gap (first 7 and last 7 out of 15). The middle number left out = Total − first group − last group.

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Radhika noted the temperature for a week. The first 4 days averaged 30 °C, the last 4 days 33 °C, and the whole week 31 °C. Days 1–4 and days 4–7 share day 4, so day 4 = 4 × 30 + 4 × 33 − 7 × 31 = 120 + 132 − 217 = 35 °C.

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When extra conditions are given ("the 9th number is 5 more than the 10th"), first find the total of the unknown numbers from the sums. Then use the conditions to split that total.

Here a few numbers are linked through the averages of their pairs or groups: "the average of A, B, C is 40, of A and B is 45, and of B and C is 38". Turn every statement into a sum, then add or subtract the equations. Often you will not need to find every number.

Three numbers, two pairs. A + B + C = 3 × (average of the three). A + B and B + C come from the pair averages. Then
B = (A + B) + (B + C) − (A + B + C)
because B is the common member of both pairs.

All three pairs given. (A + B) + (B + C) + (C + A) = 2(A + B + C). So the sum of the three numbers is half the sum of the three pair sums. After that, each number = total − the pair that does not contain it.

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Difference of two pair averages. l + m2 − m + n2 = l − n2. The common member m cancels, so l − n = 2 × (difference of the averages).

Two overlapping groups of three. The averages of (1st, 2nd, 3rd) and of (2nd, 3rd, 4th) are given. Subtract the sums: 1st − 4th = 3 × (difference of the averages).

"Average of any two added to the third". For numbers a, b, c with sum S: a + b2 + c = S − c2 + c = S + c2. Adding all three results gives 3S + S2 = 2S. So S = sum of the three results2.

Radhika's three bags weigh a, b and c. The first two average 12 kg, the last two average 15 kg, and all three average 13 kg. The middle bag b = 24 + 30 − 39 = 15 kg.

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Ages have one property that ordinary numbers do not: every year, every person becomes one year older. So if the same n people are looked at t years later, every age goes up by t, the total goes up by n × t, and the average goes up by exactly t.

• Same group, t years later: average + t. Same group, t years earlier: average − t.
• The total changes by n × t, where n is the number of people who were there for those t years.

Bring everything to the present first. "3 years ago the average age of a husband and wife was 30" means their present total = 2 × 30 + 2 × 3 = 66. "2 years from now the average of the three will be 25" means the present total = 3 × 25 − 3 × 2 = 69. So the child is 69 − 66 = 3 years old. Turn each statement into a present total, then compare.

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Birth of a member. Before a baby was born, the family had one member less. To find the average at the birth of the youngest (now k years old), take the present total, remove the youngest (k), and remove k years from each of the others.

Radhika's family of 4 has a present average age of 30 years, and Radhika is the youngest at 10. At her birth the other 3 were each 10 years younger, so the total was 120 − 10 − 3 × 10 = 80 and the average was 803 = 2623 years.

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Average unchanged after some years. If a team's average today equals its average t years ago, some older members were replaced by younger ones. In t years the total should have grown by n × t. Since it did not, the newcomers are together n × t years younger than the people they replaced.

Ratio questions. "The average age of a mother and daughter is 40 and their ages are in the ratio 3 : 1." The total is 80, split in the ratio as 60 and 20. For their ages after k years, add k to each and simplify the ratio.

Cricket questions are average questions told in cricket words. In these questions, batting average means total runs ÷ number of innings, so Total runs = Average × Innings.

Runs needed in the next innings. A batsman has an average of A after n innings. To raise it by t in the (n + 1)th innings, he must score his old average plus t for each of the (n + 1) innings:

Runs needed = A + (n + 1) × t

This is the joining formula of Type 4, with the new innings as the new member. If his average falls by t, the runs = A − (n + 1) × t.

Radhika's cousin averages 30 in 14 innings. To make it 32, he needs 30 + 15 × 2 = 60 in the 15th innings. By totals: 15 × 32 − 14 × 30 = 480 − 420 = 60 ✓.

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Finding the number of innings. If he scored R in the last innings and his average rose by t to the new average B, then R = B + n × t, where n is the number of earlier innings. Or use R − old average = (n + 1) × t.

Highest and lowest scores left out. Total of all innings − total of the remaining innings = Highest + Lowest. If "Highest − Lowest" is also given, solve the two together: Highest = Sum + Difference2.

Run rate. Run rate = RunsOvers. Required run rate for the remaining overs = Target − Runs already scoredOvers left.

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Bowling average (runs conceded per wicket) works the same way: Runs given = Average × Wickets.

Not-out innings. In the questions so far every innings counted as one in which the batsman got out, so innings and dismissals were the same. In real cricket the batting average is runs divided by the number of times the batsman was dismissed. An innings in which he remains not out adds runs to the total but adds nothing to the divisor.

Batting average = Total runsInnings − Not outs

Radhika's cousin has played 11 innings, 1 of them not out, and his average is 42. His total is 42 × (11 − 1) = 420 runs. In the next innings he scores 90 and stays not out. The total becomes 510 and the dismissals are still 10, so the new average is 51010 = 51. Had he been out for 90, the average would have been 51011 ≈ 46.4. So decide first whether the new innings is an "out" or a "not out", and only then divide.

Average speed is the one place where "average" does not mean the simple average of the given values.

Radhika goes 60 km to her college at 30 km/h and comes back at 60 km/h. Going takes 2 h and returning takes 1 h. Average speed = 1203 = 40 km/h, not 30 + 602 = 45. She spends more time at the slower speed, so the slower speed gets more weight.

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Equal distances at speeds x and y: Average speed = 2xyx + y, the harmonic mean. For three equal distances: 3xyzxy + yz + zx.

Equal times at speeds x and y: Average speed = x + y2. Only in this case is the simple average correct.

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Average price with equal spending. The same thinking works with money. If someone spends the same amount every year on petrol at prices p₁, p₂, p₃ per litre, the quantity bought each year is amountprice, and the average price = Total moneyTotal litres = 31p₁ + 1p₂ + 1p₃. Equal spending behaves like equal distance.

This type is kept basic here. Trains, boats, relative speed and the harder forms of average speed are covered in the Time, Speed and Distance chapters.

A journey in three or more parts. The rule never changes: average speed = total distance ÷ total time. Only the way of getting the time of each part changes. If the parts are given as fractions of the distance (the first half, the next quarter, the last quarter), choose a total distance that every fraction divides, find the time for each part and add. If the parts are given as times, the distance of each part is speed × time.

Radhika cycles 120 km. She covers the first half at 30 km/h, the next 13 at 20 km/h and the last 16 at 40 km/h. The three distances are 60, 40 and 20 km, and the times are 2 h, 2 h and 0.5 h, so the total time is 4.5 hours. Average speed = 1204.5 = 2623 km/h. The simple average of the three speeds, which is 30, is wrong, because she spends different times at the three speeds.