Sum = Average × Count

  • Average = sum ÷ count, so sum = average × count and count = sum ÷ average. Nearly every question in this chapter starts by turning averages into sums.
  • 12 numbers with average 17.5 add up to 12 × 17.5 = 210.
  • A group with total 1,350 and average 45 has 1,350 ÷ 45 = 30 members.
  • The average always lies between the smallest and the largest value.
  • The average keeps the unit of the data: an average of weights in kg is in kg.

Consecutive Numbers and APs

  • Evenly spaced numbers (an AP): average = (first + last) ÷ 2, which is the middle term.
  • Odd count: the average is the middle number. Even count: it is halfway between the two middle numbers.
  • 5 consecutive odd numbers with average 41 are 37, 39, 41, 43 and 45.
  • 8 consecutive even numbers with average 27: the 4th and 5th are 26 and 28, so they run from 20 to 34.
  • Largest − smallest = (count − 1) × step.

Averages of Number Series

  • First n natural numbers: n+12\frac{n + 1}{2}. First n odd numbers: n. First n even numbers: n + 1.
  • Squares of the first n natural numbers: (n+1)(2n+1)6\frac{(n + 1)(2n + 1)}{6}. Cubes: n(n+1)24\frac{n(n + 1)^2}{4}.
  • First n multiples of k: k(n+1)2\frac{k(n + 1)}{2}.
  • Multiples of k between two limits: find the first and last multiple inside, then take (first + last) ÷ 2.
  • Primes are not evenly spaced, so list them and add. The first 5 primes: 28 ÷ 5 = 5.6.

Missing Number and Changing Every Value

  • Missing number = average × count − sum of the known numbers.
  • Add or take away k from every value: the average goes up or down by k.
  • Multiply or divide every value by k: the average is multiplied or divided by k.
  • Both together: average 15, every value × 4 and then − 3, new average = 4 × 15 − 3 = 57.
  • k added to only some values: the total rises by k × (how many changed); divide by the full count.

Days and Overlapping Groups

  • Mon to Wed vs Tue to Thu: the shared days cancel, so Thu − Mon = 3 × (difference of the averages).
  • First k and last k of n numbers, when 2k is more than n: the middle ones count twice. Their sum = first-k sum + last-k sum − total.
  • Average of 9 is 30, first 5 is 27 and last 5 is 34: the 5th = 135 + 170 − 270 = 35.
  • Pairwise averages: add the three pair sums and halve for A + B + C.

One Added or Removed

  • One joins and the average rises by d: newcomer = old average + new count × d.
  • One leaves and the average falls by d: the one who left = old average + (count left behind) × d.
  • When the average moves the other way, use minus instead of plus.
  • Several join or leave: their sum = new total − old total, or old total − new total.
  • 20 numbers with average 30; one more makes it 31: new number = 30 + 21 × 1 = 51.

Replacement

  • The count stays the same, so the total changes by count × change in average.
  • New person = old person + count × rise, or old person − count × fall.
  • 6 people, one of 58 kg replaced and the average goes up by 2 kg: new person = 58 + 6 × 2 = 70 kg.
  • Two replaced: the new pair’s sum = the old pair’s sum ± count × change.

Wrong Entries

  • Correct total = wrong total − wrong value + right value.
  • Change in average = (right value − wrong value) ÷ count.
  • Digits reversed: the error is 9 × (difference of the digits). 63 written as 36 is 27 short.
  • A value left out: add it and raise the count by 1. A value counted twice: take it out once and lower the count by 1.
  • 25 numbers with average 40, 82 read as 28: correct average = 40 + 54 ÷ 25 = 42.16.

Cricket and Bowling Averages

  • Batting average = total runs ÷ innings.
  • Runs needed in the next innings = new average × new innings − runs so far, or new average + old innings × rise.
  • A score s in the nth innings raises the average by d: new average = s − (n − 1) × d.
  • Bowling average = runs given ÷ wickets. A lower bowling average is better.
  • 400 runs for 16 wickets is 25 a wicket; 5 more wickets for 20 runs makes it 420 ÷ 21 = 20.

Target Marks

  • Marks needed = target average × total tests − marks so far.
  • Shortcut: target + number of earlier tests × shortfall per test.
  • Average 62 in 4 tests and a target of 65 in 5: 65 + 4 × 3 = 77.
  • With several tests left, divide the marks needed by the number of tests left to get the average needed.
  • If the marks needed are more than the paper’s maximum, the target cannot be reached.

Weighted Average and Group Sizes

  • Combined average = n1a1+n2a2n1+n2\frac{n_1 a_1 + n_2 a_2}{n_1 + n_2}. It lies between the two averages, nearer the bigger group.
  • Group sizes from the averages: n1:n2=(a2−A):(A−a1)n_1 : n_2 = (a_2 - A) : (A - a_1), the distances swapped.
  • Boys 40, girls 55, class 46: boys : girls = 9 : 6 = 3 : 2.
  • One group’s average from the overall one: (overall total − other group’s total) ÷ its count.
  • The same balancing idea is alligation (Chapter 6).

Weighted Percentages and Salaries

  • Never average percentages from papers with different maximum marks. Change them into marks first.
  • 40% of 60 and 70% of 140: (24 + 98) ÷ 200 = 61%, not 55%.
  • Average salary across grades = total pay ÷ total staff.
  • Different percent rises with the same head count: % rise in the average = total rise ÷ old total pay × 100.
  • Everyone gets the same ₹k rise: the average rises by ₹k.

The Average of Averages Trap

  • The plain average of two averages is right only when the groups are the same size.
  • 10 students averaging 80 and 30 averaging 60: class average = (800 + 1,800) ÷ 40 = 65, not 70.
  • Prices: equal quantities, take the plain average. Equal money spent at ₹30 and ₹45 a kg gives ₹36 a kg, not ₹37.50.

Average Ages Over Time

  • After t years every age, and so the average, rises by t; t years ago it was t less.
  • A baby born since: bring the old members’ total up to today; baby = new total − that.
  • Teacher included: teacher = old average + new count × rise.
  • Same average as t years ago after one of n members is replaced: the new member is n × t years younger.
  • Average at the youngest’s birth: remove the youngest, then take the youngest’s age off each of the others.

Other Age Problems

  • The difference between two ages never changes.
  • Sum and difference given: younger = (sum − difference) ÷ 2.
  • Children born at equal gaps: the average is the middle child’s age. 5 children 2 years apart with average 10: the youngest is 6.
  • Digits reversed: the ages differ by 9 × (difference of the digits) and add up to 11 × (sum of the digits).