Averages are the cheapest marks in the quantitative section and the biggest trap in data interpretation. The cheap marks come from questions that only look like averages: "the average of five numbers is 24, and four of them are known" is a subtraction, not a statistics problem. The trap is the same word doing two different jobs — a weighted mean and an average of averages are not the same number.
Averages also quietly power rate questions. Average speed is a weighted average, and nearly every candidate who loses marks there loses them to the belief that averaging two speeds is the same as finding the average speed.
Think of an average as a way of sharing a total evenly.
The second form is the one that solves questions. It says that an average question is almost always about a total: if you know the mean and the count, you know the sum without ever adding anything up.
Two consequences worth holding:
| Measure | Definition | Watch out for |
|---|---|---|
| Mean | Sum divided by count | Sensitive to extremes: one huge value moves it far |
| Median | Middle value of the ordered list; with an even count, the mean of the two middle values | You must sort first, and an even count needs a half-step |
| Mode | The value that occurs most often | A set can have no mode, one mode, or several |
| Range | Largest minus smallest | Not a measure of centre — it is a crude spread |
The same fact expressed as distances: the combined mean divides the gap between the two group means in the inverse ratio of the group sizes. Two groups of 30 and 20 students average 60 and 70; the combined mean sits 30/50 of the way from 70 towards 60, which is 64. The bigger group wins the tug of war.
For two equal distances covered at speeds a and b, this simplifies to the harmonic mean 2ab ÷ (a + b). For unequal distances or unequal times, always go back to totals.
Total = 24 × 5 = 120. Known sum = 18 + 30 + 22 + 20 = 90. The fifth number is 120 − 90 = 30.
Time out = 60 ÷ 40 = 1.5 h; time back = 60 ÷ 60 = 1 h; total 120 km in 2.5 h = 48 km/h. The shortcut 2ab ÷ (a + b) gives (2 × 40 × 60) ÷ 100 = 48, and the wrong answer 50 is always among the options.
(30 × 60 + 20 × 70) ÷ 50 = (1,800 + 1,400) ÷ 50 = 3,200 ÷ 50 = 64. Note that the answer is nearer 60, because there are more students there.
Old total = 180. Removing 27 leaves 153 over 11 numbers, which is 13.9. Removing an above-average value pulls the mean down — a five-second sanity check before you compute.
A batsman's average after 10 innings is 30. He scores 80 in the eleventh. What is his new average?
What is the median of 7, 3, 9, 5, 11, 4?
A shop sells 40 kg of rice at Rs 100 per kg and 60 kg at Rs 150 per kg. What is the average price per kg of the mixture?
The mean of five numbers is 18. Four are 12, 20, 16 and 22. What is the fifth?
Without looking back: write the two forms of the mean relationship, and the average speed for equal distances at 30 km/h and 60 km/h.
The answer: mean = total ÷ count, equivalently total = mean × count; and 2ab ÷ (a + b) = 3,600 ÷ 90 = 40 km/h.
An average is a total divided by a , so the total equals the mean multiplied by the count. A combined mean from two groups of different sizes is called a mean, and it sits nearer the group. For equal distances at 40 and 60 km/h the average speed is km/h.
A data interpretation table shows three departments with average salaries of Rs 60,000 (40 staff), Rs 80,000 (25 staff) and Rs 100,000 (5 staff). A question asks for the overall average salary. Another asks whether the overall average is above or below Rs 80,000. Answer both without a calculator.
Exact answer: (40 × 60,000 + 25 × 80,000 + 5 × 100,000) ÷ 70 = (2,400,000 + 2,000,000 + 500,000) ÷ 70 = 4,900,000 ÷ 70 = Rs 70,000.
The comparison question is answered by shape, not arithmetic: 40 of the 70 staff sit in the Rs 60,000 department, so the weighted mean must be dragged to the low side — below 80,000, and in fact 70,000. The lever rule gives this in five seconds: the mean sits 40/70 of the way from 80,000 towards 60,000, which is a drop of about 11,400 from 80,000.
Note the distractor that this question is designed to catch: the unweighted average of 60,000, 80,000 and 100,000 is 80,000, which is the option for candidates who forget that departments differ in size.
You can now handle totals, weighted groups and average speed. The next lesson specialises that last idea into its own question family — distance, speed and time — which is one of the most reliably repeatable five-mark blocks in the quantitative section.