Maths M.C.Q. [1 Marks Each]

 Total marks of first batch $= 55 \times 50 = 2750$
total marks of second batch $= 60 \times 55 = 3300$
total marks of third batch $= 45 \times 60 = 2700$
total number of students in three batch $= 55 + 60 + 45 = 160$
$\therefore$ average marks of all the students $ = \frac{2750+3300+2700}{160} = \frac{8750}{160} = 54.6875$

Here,
Highest observation $= 21$
Lowest observation $= 4$
Range = Highest observation - Lowest observation
$= 21 - 4 = 17$

 The number on the cards are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$
The even numbers on the cards are $2, 4, 6, 8, 10,$
$\therefore $ Probability of getting an even numbered card $=\frac{\text{Number of even numbered card}}{\text{Number of cards with numbers from 1 to 10}}=\frac{5}{10}=\frac{1}{2}$
Hence, the correct option is $(c).$

$\text{Average} = \frac{0.3+0.03+0.003}{3}$
$ = \frac{0.333}{3} = 0.111$

Sum of $10$ observations $= 10 × 15 = 150$
Sum of $11$ observations $= 150 + 15 = 165$
Number observations $= 11$
Mean of $11$ observations $= \frac{165}{11}=15$
Thus, the new mean is $15$
Hence, the correct option is $(d).$

Required mean $ = \frac{1+2+3+ ......... +}{\text{n}}$
$ = \frac{\text{n}}{2} = \frac{(2+(\text{n-1}).(1)}{2} = \frac{{\text{n}}+{1}}{2}$

 $\text{mean} = \frac{\text{Total of sample values}}{\text{sample size}}$

 The possible numbers on a dice are $1, 2, 3, 4, 5, 6.$
There is only one even prime number which is $2.$
$\therefore$ Probability of getting an even prime $ =\frac{\text{Number of even prime numbers}}{\text{Number of all possible outcomes on the dice}}=\frac{1}{6}$
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 $(a).$

Given,
$M = 1, 2, 3, 4, 5, 6, 7$ Each number in set $N$ is generated by dividing each number in set M by $2$
$N = 0.5, 1, 1.5, 2, 2.5, 3, 3.5$ Arithmetic mean of numbers of set N is
$ = \frac{0.5+1+1.5+2+2.5+3+3.5}{7} = \frac{14}{7} = {2}$

 The question tells us that the mean of $x, y$ and $z$ is $y.$
$\text{i.e.} = \frac{\text{x+y+z}}{3} = \text{y}$
$\text{i.e.} = \text{x+z} = 2\text{y}$

$\text{class marks} = \frac{\text{upperlimit + lower limit}}{2}\Rightarrow \frac{120+90}{2} = \frac{210}{2} = {105}$

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 $(b).$

$\text{Average score}=\frac{\text{Sum of scores in all exams}}{\text{Total number of exams}}$
$\therefore$ Average score in first three examination $=\frac{97+73+88}{3}=\frac{258}{3}=86$
Also, average score in four examination $=\frac{97+73+88+80}{4}=\frac{338}{4}=84.5$
Hence, average score is decreased by $(86 - 84.5) = 1.5$

Let the marks for annual marks be $x.$
Given: - Marks of first unit test $= 35$
Marks of second unit test $= 21$ average of $3$ marks $= 30$
$\Rightarrow \frac{{\text{x}} + { 35 } +{ 21 }} {3} = {30}$
$\Rightarrow x + 35 + 21 = 90$
$\Rightarrow x + 56$
$\Rightarrow x = 34$ Marks he should get is $x = 34$

 Mean is the average of the values of the data set. Given the height of the girls are
$140, 142, 135, 133, 137, 150, 148, 138$ The total number of girls is $8$
$\therefore$ mean $ = \frac{140+142+135+133+137+150+148+138}{8}$
$\Rightarrow $ mean $ = \frac{1123}{8} = {140.375}$

Let the mean be $\bar{\text{x}}$ According to the question,
$\bar{\text{x}} - {10} = {60}{\text{%}} \text{ of } \bar{\text{ x}}$
$\bar{\text{x}} = {25}$
Now, each observation is increased by $5.$
$\therefore$ New mean $ = \bar{\text{x}}+5$
$= 25 + 5 = 30.$

We have, $33, 38, 48, 33, 34, 34, 33$ and $24.$
On arranging the data in ascending order, we get $24, 33, 33, 33, 34, 34, 38$ and $48.$
Here, $33$ occurs more frequently, i.e. $3$ times.
Mode of data $= 33$
Note: Mode is the observation that occurs most frequently in the data.

 Mean $ = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10}{10} = {5.5}$

Given, the average of $3, 4,$ and $x$ is $2$
we have to find $x$
$\Rightarrow \frac{{3+4+}{\text{x}}}{3} = 2$
$\Rightarrow{7}+{\text{x}} = {6}$
$\text{x} = -{1}$

Arranging the numbers $22, 21, 23, 24, 21, 20, 23, 26$ and $26$ in increasing order, we get
$20, 21, 21, 22, 23, 23, 24, 26, 26$
Here, the frequencies $21, 23$ and $24$ is $2$
So, for $23$ to be the mode of the data, the value of $x$ should be $23$
Hence, the correct option is $(c).$

The first five prime numbers are: $2, 3, 5, 7, 11$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{2+3+5+7+11}{5}$
$=\frac{28}{5}$
$=5.6$
Thus, the mean of first five prime number is $5.6$
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 $(c).$

$\frac{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + {\text{e}}}{5} = {28}$
$\frac{{\text{a}} + {\text{c}} + {\text{e}}}{3} = 24$
$\frac{{\text{b + }}{\text{d}}} {2} = ?\Rightarrow\text{a} + {\text{b}} + {\text{c}} + {\text{d}}+{\text{e}} = {140}$
$ = \text{a} + {\text{c}}+ {\text{e}} = 72$
$\Rightarrow\text{b} + {\text{d}} = {68}$
$\therefore \frac{{\text{b}} + {\text{d}}}{2} = {34}$

$\frac{\text{Sum}}{\text{Total}} = {20}$
$\frac{\text{Sum}}{5}= {20}$
Sum $= 100$
Let no. excluded be $x$
$\frac{\text{New sum}}{\text{Total}} = {15}$
New Sum $= 60$
Excluded $N = 100 - 60 = 40$

Sum of number $= 152$
Required mean $ = \frac{152}{8} = {19}$

Since mean $ = \frac{0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8+}{16}$
$\Rightarrow $ mean $ = \frac{64}{16}$
$\Rightarrow $ mean $ = {4}$

Given data $31, 45, 84, 23, 67$ the sum of observations $31 + 45 + 84 + 23 + 67 = 240$
the mean $\frac{240}{5} = {48}$

Required mean $ =\frac{ \text{a+}(\text{a+d})+(\text{a+2d})+ ..... +{(\text{a+2nd}})}{\text{2n+1}}$
$ = \frac{{\text{2n+1}}}{2} = \frac{{\text{2a}+(\text{2n+1}-1)\text{d}}}{\text{2n+1}} = \text{a + nd}$

The given data is $16, 29, 60, 18, 27$ the sum of observations is $16 + 29 + 60 + 18 + 27 = 150$
the mean is given as $\frac{150}{5} = {30}$

Probability of getting $5=\frac{14}{80}=\frac{7}{40}$
Therefore
Probability of not getting $5=1-\frac{7}{40}=\frac{33}{40}$
Hence, the correct option is $(c).$

Here $x, y$ and $z$ be three observations.
We know that, $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{\text{x+y+z}}{3}$

 Here, highest observation $= +20$ and lowest observation $= -17$
As we know,
Range $=$ Highest observation $–$ Lowest observation $= +20 - (-17) = 20 + 17 = 37$

 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 $(a).$

 $\text{A.M} = \frac{\sum\text{x}}{\text{n}}\Rightarrow{8} = \frac{6+10+{\text{x}}+12}{4}$
$\Rightarrow{32} = {6+10} + {\text{x}} + {12} \Rightarrow = {32}-{28}$
$\Rightarrow{\text{x}} = {4}$

 Probability of occurrence of event $E$ is $P(E) = \frac{\text{N(E)}}{\text{N(S)}}$
Where $n(E)$ is no. of cases favorable to event $E$ and $n(S)$ is total no. of cases.

 $\text{Average} = \frac{2+4+6+8+10}{5} = {6}$

Mean of $p, q$ and $r =$ Mean of $q, 2r$ and $s$
$\frac{\text{p+q+r}}{3}=\frac{\text{q+2r+8}}{3}$
$\Rightarrow\text{p + q + r}=\text{q + 2r }+8$
$\Rightarrow\text{p}=\text{r + s}$
Hence, the correct option is $(d).$

 Mean weight $= 21\ kg$
Number of students $= 21$
Sum of weights of $21$ students $= 21 × 21 = 441$
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 $(b).$

 Given data $35, 34, 25, 52, 27, 23, 24, 36$ No of observations $8$ sum of observations $35 + 34 + 25 + 52 + 27 + 23 + 24 + 36 = 256$
Mean is $\frac{256}{8} = {8}$

 Here, $\frac{50}{100}=\frac{1}{2}$
So, if the the probability of an event is $\frac{1}{2},$ then the number of favourable cases has to be $50.$
Hence, the correct option is $(d).$