Section 9, arithmetic
Averages
Every averages question gets easy once you convert averages into totals: sum = average × count. CAT’s favourite twist is optimisation, the largest or smallest possible value under an average and a few constraints.
How CAT asks it
Members join or leave a group; a wrong entry is corrected; two groups’ averages change between two tests; averages of terms in a progression; and “what is the maximum possible value of…” questions with integer or distinctness constraints.
Concepts you must know
- Sum = average × count. Work in sums, not averages.
- A member joins: new member = old average + (change in average × new count).
- Replacement: incoming − outgoing = change in average × count.
- Weighted average = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂). It lies between a₁ and a₂, closer to the bigger group.
- Average of an AP = (first + last) ÷ 2 = the middle term.
- Adding k to every value adds k to the average; multiplying every value by k multiplies the average by k.
- Ages: everyone gets one year older each year, so a family’s total age rises by members × years.
- Optimisation rule: to maximise one value, push all the others as low as the constraints allow, and vice versa.
Thought process
- Convert every average into a total.
- Write what changed between the “before” and “after” totals.
- For max/min questions, list the constraints (integer? distinct? positive? an upper limit?) and push the other values to their extremes.
Patterns CAT repeats
Numbered so your error log can point at them: “Pattern 2”.
- 1Joining, leaving and replacement.
- 2A wrong entry corrected.
- 3Weighted average of groups: sections, boys and girls.
- 4Two groups across two tests: use totals, and watch for data that does not matter.
- 5Optimisation: the max/min of one value, or of the median, given an average and constraints.
- 6Averages of AP terms or of numbers in a range with a property.
- 7Cricket: runs needed to lift a batting average.
- 8Ages over time, with members added.
Worked examples
Attempt each one before opening its solution, then compare methods.
Example 1, joining
The average weight of 24 students is 50 kg. Including the teacher raises the average by 1 kg. Find the teacher’s weight.
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Teacher = 50 + 1 × 25. Check: 25 × 51 = 1,275 = 1,200 + 75. Answer: 75 kg.
Example 2, replacement
The average age of 8 men rises by 2 years when two men aged 20 and 24 are replaced by two women. Find the women’s average age.
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The total rises by 8 × 2 = 16, so the women total 44 + 16 = 60. Answer: 30 years.
Example 3, wrong entry
The average of 40 numbers was found to be 35, but 83 had been misread as 38. Find the correct average.
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The total rises by 45, so the average rises by 45/40 = 1.125. Answer: 36.125.
Example 4, weighted
Section A has 30 students averaging 60 and section B has 20 averaging 75. Find the overall average.
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(1,800 + 1,500) ÷ 50. Answer: 66.
Example 5, two tests
A class has 30 boys and 20 girls. In Test 1 the girls averaged 6 marks more than the boys. In Test 2 the girls’ average fell by 2 while the class average rose by 1. By how much did the boys’ average change?
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Class total change = 50 × 1 = +50. Girls’ total change = 20 × (−2) = −40. So the boys’ total rose by 90, which is 90/30 = 3 per boy. The “6 more” was never needed; CAT often adds such data. Answer: up by 3.
Example 6, optimisation, maximum
The average of 5 distinct positive integers is 20. What is the largest possible value of the largest one?
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Sum = 100. Make the other four as small as possible: 1 + 2 + 3 + 4 = 10. Answer: 90.
Example 7, optimisation, median
The average of 7 distinct positive integers is 12. What is the largest possible median?
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Sum = 84. The smallest three are 1, 2, 3. With median m, the top three are at least m + 1, m + 2, m + 3. So 6 + m + 3m + 6 = 84 and m = 18. Check: 1, 2, 3, 18, 19, 20, 21 sums to 84. Answer: 18.
Example 8, AP average
Find the average of all three-digit numbers that leave remainder 2 when divided by 7.
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They form an AP from 100 (= 98 + 2) to 996 (= 994 + 2). Average = (100 + 996) ÷ 2. Answer: 548.
Example 9, cricket
A batter averages 40 after 20 innings. How many runs must she score in the 21st innings to raise her average to 42?
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40 + 2 × 21. Check: 42 × 21 = 882 = 800 + 82. Answer: 82.
Example 10, ages over time
Four years ago, a family of 5 had an average age of 30. A baby has been born since, and the 6 members now average 28.5 years. How old is the baby?
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The original 5 now total 150 + 5 × 4 = 170. The family of 6 totals 6 × 28.5 = 171. Answer: 1 year.
Traps
- Averaging two averages when the groups differ in size.
- Forgetting the extra year per person per year in age questions.
- Ignoring “distinct” or “positive” in optimisation; those words decide the answer.
Practice set
Timed, about 2.5 minutes per question. Log every miss in the error log.
- 1
The average of 11 numbers is 50. The first 6 average 49 and the last 6 average 52. Find the 6th number.
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56: 294 + 312 − 550, since the 6th number is counted in both groups.
- 2
The average of 5 consecutive odd numbers is 61. Find the largest.
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65: the middle number is 61.
- 3
50 students average 40 marks. 10 students averaging 52 leave. Find the new average.
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37: (2,000 − 520) ÷ 40.
- 4
The average of a, b, c is 20 and of b, c, d is 25. If d = 30, find a.
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15: (a + b + c) − (b + c + d) = 60 − 75, so a = d − 15.
- 5
The mean of 6 distinct positive integers is 15. What is the largest possible value of the largest one?
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75: 90 − (1 + 2 + 3 + 4 + 5).
- 6
11 players average 26 years. Without the captain, the rest average 25. Find the captain’s age.
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36: 286 − 250.