Maths MCQ

Mean $= 11$
Number of observations $= 7$
Sum of observations $= 11 \times 7 = 77$
Sum of new observations $= 2 \times 77 = 154$
Mean of new observations $=\frac{154}{7}=22$
Hence, the correct option is $(c).$

Mean of five numbers $= 4$
Sum of five numbers $= 5 \times 4 = 20$
$\text{New mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{20+1+1+1+1+1}{5}$
$=\frac{25}{5}$
$=5$
Thus, the new mean is $5$
Hence, the correct option is $(b).$

The relation between Mean, Median and Mode is Mode $-$ Mean $= 3 ($Median $-$ Mean$).$
Hence, the correct option is $(d).$

Mean of data $= 15$
Sum of observations $= 195$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations(n)}}$
$\Rightarrow15=\frac{195}{\text{n}}$
$\Rightarrow\text{n}=\frac{195}{15}=13$
Thus, the number of observations is $13$
Hence, the correct option is $(a).$

Arranging the given data in increasing order, we get
$8, 9, 9, 10, 11, 11, 12$
As the number of observations is odd $(7),$ the median is the middle term which is $10$
Hence, the correct option is $(a).$

The first seven even natural numbers are: $2, 4, 6, 8, 10, 12, 14$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{2+4+6+8+10+12+14}{7}$
$=\frac{56}{7}$
$=8$
Thus, the mean of first seven even natural number is $8$
Hence, the correct option is $(b).$

Median divides the data into two equal parts. Since, the number of observations is $11$,
so after arranging in increasing or decreasing order, the number of observations to the left of the median is five.
Thus, the required number of observations is $5$
Hence, the correct option is $(a).$

The first six multiples of $5$ are: $5, 10, 15, 20, 25, 30$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{5+10+15+20+25+30}{6}$
$=\frac{105}{6}$
$=17.5$
Thus, the mean of first six multiples of $5$ is $17.5$
Hence, the correct option is $(c).$

Arranging the data in ascending order, we get
$7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12$
Here, $10$ has the maximum frequency $(4)$
Hence, the correct option is $(a).$

Mean weight $= 21\ kg$
Number of students $= 21$
Sum of weights of $21$ students $= 21 \times 21 = 441$
Sum of weights of $20$ students left $= 441 - 21 = 420$
Mean of remaining students $=\frac{420}{20}=21\text{kg}$
Hence, the correct option is $(b).$

Sum of $10$ observations $= 95$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{95}{10}$
$=9.5$
Thus, the mean is $9.5$
Hence, the correct option is $(a).$

Arranging the numbers $10, 12, 6, 18$ in ascending order, we get
$6, 10, 12, 18$
Thus, for $10$ to be the median of the data, $x < 6$ or $6\leq\text{x}\leq10$
Hence, the correct option is $(d).$

The mean of $9, 10, 15, x, 6, 8$ and $12$ is $11$
$\therefore\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow11=\frac{9+10+15+\text{x}+6+8+12}{7}$
$\Rightarrow\text{x}+60=77$
$\Rightarrow\text{x}=77-60=17$
So, the observations are $9, 10, 15, 17, 6, 8$ and $12$
Arranging the the observations in increasing order, we get
$9, 10, 15, 17, 6, 8, 12$ or $6, 8, 9, 10, 12, 15, 17$
Thus, the median is $10$
Hence, the correct option is $(b).$

Given: the mean of the observations $20, 42, 35, 45$ and $x.$
Mean of observations $=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow37=\frac{20+42+35+45+\text{x}}{5}$
$\Rightarrow142+\text{x}=185$
$\Rightarrow\text{x}=185-142=43$
Thus, the value of $x$ is $43$
Hence, the correct option is $(a).$

Arranging the data $9, 6, 3, 4, 9, 8, 6, 4, 6$ in ascending order, we get
$3, 4, 4, 6, 6, 6, 8, 9, 9$
Since the mode of the data is $6$,
so the value of $x$ cannot be $4$ or $9.$
Hence, the correct option is $(d).$