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Question 1: The mean of 100 observations is 50. If one observation 50 is replaced by 150, what will be the resulting mean?
Solution: Original Sum $= 100 \times 50 = 5000$. New Sum $= 5000 - 50 + 150 = 5100$. New Mean $= 5100 / 100 = 51$.
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Question 2: The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Solution: If each observation is decreased by $k$, the mean decreases by $k$. New Mean $= 20 - 5 = 15$.
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Question 3: There are 50 students in a class, of which 40 are boys and the rest are girls. The average weight of the class is 44 kg and the average weight of the girls is 40 kg. Find the average weight of the boys.
Solution: Total weight $= 50 \times 44 = 2200$ kg. Girls $= 10$, Total weight of girls $= 10 \times 40 = 400$ kg. Weight of boys $= 2200 - 400 = 1800$ kg. Avg weight of boys $= 1800 / 40 = 45$ kg.
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Question 4: The mean of 6, 4, 7, $p$ and 10 is 8. Find the value of $p$.
Solution: Sum $= 6+4+7+p+10 = 27+p$. Mean $= (27+p)/5 = 8 \Rightarrow 27+p = 40 \Rightarrow p = 13$.
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Question 5: The median of the observations 11, 12, 14, 18, $x+2$, $x+4$, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of $x$.
Solution: $n=10$. Median $= (5^{th} + 6^{th})/2 = (x+2 + x+4)/2 = (2x+6)/2 = x+3$. Given $x+3 = 24 \Rightarrow x = 21$.
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Question 6: The mean of 100 items was found to be 30. If at the time of calculation two items were wrongly taken as 32 and 12 instead of 23 and 11, find the correct mean.
Solution: Incorrect Sum $= 100 \times 30 = 3000$. Correct Sum $= 3000 - (32+12) + (23+11) = 3000 - 44 + 34 = 2990$. Correct Mean $= 2990 / 100 = 29.9$.
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Question 7: Find the mean of the first 10 prime numbers.
Solution: Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Sum $= 129$. Mean $= 129/10 = 12.9$.
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Question 8: The mean weight of a class of 35 students is 45 kg. If the weight of the teacher be included, the mean weight increases by 500g. Find the weight of the teacher.
Solution: Total weight (students) $= 35 \times 45 = 1575$ kg. New mean $= 45.5$ kg. Total weight (with teacher) $= 36 \times 45.5 = 1638$ kg. Teacher's weight $= 1638 - 1575 = 63$ kg.
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Question 9: If the mode of the following data is 14, find the value of $k$: 10, 12, 14, 15, 15, 14, 12, 10, 15, $k$.
Solution: Current counts: 10(2), 12(2), 14(2), 15(3). For mode to be 14, it must have the highest frequency. So $k$ must be 14 (making count of 14 equal to 3, but usually mode is unique or defined clearly, here if $k=14$, counts are 14(3) and 15(3) making it bimodal. If question implies unique mode 14, then $k$ must be 14 and perhaps one 15 is removed or similar. Assuming standard problem type, $k=14$ makes it a mode).
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Question 10: The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Solution: Sum of 5 $= 5 \times 18 = 90$. Sum of 4 $= 4 \times 16 = 64$. Excluded number $= 90 - 64 = 26$.