| I. Average = [Sum of observations / Number of observations] | |
| II. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed during the whole journey is [2xy / x + y] kmph. |
| 1. | Find the average of all prime numbers between 30 and 50. |
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| Sol. There are five prime numbers between 30 and 50. They are 31, 37, 41, 43 and 47. ∴ Required average = [(31 + 37 + 41 + 43 + 47) / 5] = 199/5 = 39.8. |
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| 2. | Find the average of first 40 natural numbers. |
| Sol. Sum of first n natural numbers = 40 * 41 / 2 = 820. ∴ Required average = 820 / 40 = 20.5. |
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| 3. | Find the average of first 20 multiples of 7. |
| Sol. Required average = 7(1+2+3 ... + 20) / 20 = [7 * 20 * 21 / 20 * 2] = [147 / 2] = 73.5. | |
| 4. | The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53kg is replaced by a new man. find the weight of the new man. |
| Sol. Total weight increased = (1.8 * 10)kg = 18kg. ∴ Weight of the new man = (53 + 18)kg = 71kg. |
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| 5. | A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. find his average after 17th inning. |
| Sol. Let the average after 17th inning = x. Then, average after 16th inning = (x - 3) ∴ 16(x - 3) + 87 = 17x or x = (87 - 48) = 39. |
| 7. | The average runs of a cricket player of 10 innings was 32. How many runs must he makes ih his next innings so as to increase his average of runs by 4? |
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| 8. | The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is |
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| 9. | The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? |
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