1). If the sum of 11 consecutive natural numbers is 2761, then the middle number is:
A). 249 
B). 250 
C). 251 
D). 252 
Correct Answer: 251
Let the first natural number be x
According to the question
x+x+1x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10=2761
=> 11x + 55 = 2761
x = \( \Large \frac{2761  55}{11} = 246 \)
Middle number = x + 5 = 246 + 5 = 251 
2). The mean weight of 9 items is 15. If one more item is added to the series, the mean 16. The value of 10th item is
A). 35 
B). 30 
C). 25 
D). 20 
Correct Answer: 25 Total weight of 9 items = 15 x 9 = 135
and total weigth of 10 items = 16 X 10 = 160
Therefore, weight of 10 items = 160  135 = 25

3). The average of the square of the number 0, 1, 2, 3, 4.........n is:
A). \( \Large \frac{1}{2}n \left(n+1\right) \) 
B). \( \Large \frac{1}{6}n \left(2n+1\right) \) 
C). \( \Large \frac{1}{6} \left(n+1\right) \left(2n+1\right) \) 
D). \( \Large \frac{1}{6}n \left(n+1\right) \) 
Correct Answer: \( \Large \frac{1}{6}n \left(2n+1\right) \)
Mean: \( \Large \frac{0^{2}+1^{2}+2^{2}+3^{2}+....+n^{2}}{ \left(n+1\right) } \)
= \( \Large \frac{n \left(n+1\right) \left(2n+1\right) }{6 \left(n+1\right) } = \frac{1}{6}n \left(2n+1\right) \) 
4). If the average of number 148, 146, 144, 142,..... in AP be 125, then the total numbers in the series will be
A). 18 
B). 24 
C). 30 
D). 48 
Correct Answer: 24
Given series is 148, 146, 144, 142 .....whose first term and common difference is
\( \Large a = 148,\ d = \left(146  148\right) = 2 \)
\( \Large S_{n}=\frac{n}{2}\left[ 2a+ \left(n1\right)d \right] = 125 \)
=> \( \Large 125n = \frac{n}{2}\left[ 2 \times 148+ \left(n1\right) \times \left(2\right) \right] \)
=> \( \Large n^{2}24n = 0 => n \left(n24\right)=0 \)
n = 24 
5). The mean of the values of 1, 2, 3, ..., n with respectively frequencies x, 2x, 3x, ..., nx is
A). \( \Large \frac{n}{2} \) 
B). \( \Large \frac{1}{3} \left(2n+1\right) \) 
C). \( \Large \frac{1}{6} \left(2n+1\right) \) 
D). \( \Large \frac{n}{3} \) 
Correct Answer: \( \Large \frac{1}{3} \left(2n+1\right) \)

6). The mean of 12 items is \( \Large \bar{X} \) . If the first term is increased by 1, second by 2 and so on, then the new mean is
A). \( \Large \overline{X}+n \) 
B). \( \Large \overline{X}+\frac{n}{2} \) 
C). \( \Large \overline{X}+\frac{n+1}{2} \) 
D). none of these 
Correct Answer: \( \Large \overline{X}+\frac{n+1}{2} \)

7). If the mean of n observation \( \Large 1^{2},\ 2^{2},\ 3^{2},........n^{2} \), is \( \Large \frac{46}{11}n \), then n is equal to
A). 11 
B). 12 
C). 23 
D). 22 

8). In a moderately skewed distribution the values of mean and median are 5 and 6 respectively. The value of mode in such a situation is approximately equal to:
A). 8 
B). 11 
C). 16 
D). none of these 
Correct Answer: 8
Given that, mean = 5, median = 6
For a moderately skewed distribution,
We have Mode = 3 median  2 mean
=> Mode = 3(6)  2(5) = 8. 
9). The value of mean, median and mode coincides, then the distribution is
A). Positive skewness 
B). Symmetrical distribution 
C). Negative skewness 
D). All of these 
Correct Answer: Symmetrical distribution
If mean, median and mode coincides, then there is a symmetrical distribution. 
10). The algebraic sum of the deviation of 20 observations measured from 30 is 2. Then mean of observations is
A). 28.5 
B). 30.1 
C). 30.5 
D). 29.6 
Correct Answer: 30.1 Give that, \( \Large \sum^{20}_{i=1} \left(x_{i}30\right)=2 \)
\( \Large \sum^{20}_{i=1}x_{i}\sum^{20}_{i=1} \left(30\right)=2 \)
\( \Large \overline{x} = \frac{20.30}{20} + \frac{2}{30}=30+0.1=30.1 \)
