Find the harmonic mean of 2 and 4.
Answer: B
The harmonic mean (H.M.) of n numbers is the reciprocal of the arithmetic mean of their reciprocals.
For two numbers a and b, H.M. = \(\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}\).
H.M. = \(\frac{2 \times 2 \times 4}{2+4} = \frac{16}{6} = \frac{8}{3}\).
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The average of the first 10 multiples of 3 is:
Answer: B
The first ten multiples of 3 run from 3 to 30 in equal steps, so the average is the average of the extremes.
\[ \frac{3 + 30}{2} = 16.5 \]
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A student's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of students in the class is:
Answer: C
The extra marks added to the total sum = 83 - 63 = 20.
Let N be the number of students. The increase in the total sum (20) caused the average to increase by 1/2.
Increase in average = \(\frac{\text{Increase in sum}}{\text{Number of students}}\)
\(\frac{1}{2} = \frac{20}{N}\)
N = 20 × 2 = 40.
There are 40 students in the class.
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For which of the following is the median and mean the same?
Answer: A
The mean and median are the same for a data set that is symmetrically distributed.
The set {1, 2, 3, 4, 5} is an arithmetic progression, which is a symmetric distribution.
Mean = (1+2+3+4+5)/5 = 15/5 = 3.
Median = The middle value, which is 3.
All other sets are skewed.
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Find the weighted mean of the following data: Value(x) = {10, 20}, Weight(w) = {2, 3}.
Answer: B
\[ \bar{x} = \frac{\sum wx}{\sum w} = \frac{10(2) + 20(3)}{2 + 3} \]
\[ = \frac{80}{5} = 16 \]
The plain average would be 15; the heavier weight on 20 pulls it up.
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The average of a non-zero number and its square is 5 times the number. The number is:
Answer: A
Let the number be x.
The average of the number and its square is \(\frac{x + x^2}{2}\).
This is equal to 5 times the number, so \(5x\).
\(\frac{x + x^2}{2} = 5x\)
\(x + x^2 = 10x\)
Since x is non-zero, we can divide by x:
1 + x = 10
x = 9.
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If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
Answer: B
Let's count the frequency of each number in the data set (excluding x):
Currently, the data is multimodal with modes 15, 16, and 17. For the mode of the entire set to be 15, the value of x must be 15. This would make the frequency of 15 equal to 3, which would be the highest frequency.
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The mean of 11 numbers is 35. If the mean of the first 6 numbers is 32 and that of the last 6 numbers is 37, find the 6th number.
Answer: B
Total sum of 11 numbers = 11 × 35 = 385.
Sum of the first 6 numbers = 6 × 32 = 192.
Sum of the last 6 numbers = 6 × 37 = 222.
The 6th number is included in both the 'first 6' and 'last 6' groups. So, if we add the sum of the first 6 and the last 6, the 6th number is counted twice.
Sum of (first 6 + last 6) = 192 + 222 = 414.
This sum is equal to (Sum of all 11 numbers) + (6th number).
414 = 385 + (6th number)
6th number = 414 - 385 = 29.
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For a data set, the mean is 25 and the median is 22. What is the approximate value of the mode?
Answer: A
\[ \text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 3(22) - 2(25) \]
\[ = 66 - 50 = 16 \]
Mean above median indicates a right skew, so the mode should be the smallest of the three - a check on the sign.
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The average weight of 8 men is increased by 1.5 kg when one of the men, who weighs 65 kg, is replaced by a new man. The weight of the new man is:
Answer: C
The total increase in weight for the group of 8 men is 8 × 1.5 kg = 12 kg.
This increase is the difference between the new man's weight and the old man's weight.
Weight of new man = Weight of old man + Total increase
Weight of new man = 65 kg + 12 kg = 77 kg.
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