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

Mean $ =\frac{ {\text{Total weight of 9 apples}}}{\text{total no of apples}}$
$\frac{50+60+65+62+67+70+64+45+48}{9} \Rightarrow\frac{531}{9} = 59$

Given three year ago the average age of $5$ family member is $17$ year.
Then present age of six members family $= 17 × 6 = 102$ yearsAnd present age of five members
family $= (17 + 3) × 5 = 100$ years so age of baby to day $= 102 - 100 = 2$ years.

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

Range is the difference between the smallest value and largest value of the data set.
given data set is $14, 12, 17, 18, 16, x$ and Range is given as $20$ In the given set the smallest value is $12$ and the largest value is $18.$
$\therefore$ the range is $18 - 12 = 6 \neq$ hence, $18$ is not the largest value. Variable $x$ is the largest value.
$\therefore$ range is $x - 12 = 20$
$\Rightarrow x = 20 + 12 = 32$

The probability of getting neither green nor red ball is equal to the probability of getting blue balls.
Number of blue balls $= 2$
Total number of balls $= 4 + 4 + 2 = 10$
Therefore
Probability of getting neither green nor red ball $=\frac{2}{10}=\frac{1}{5}$
Hence, the correct option is $(d).$

The prime numbers between $20$ and $30$ are $23, 29$ mean of the data set is the average of values in the data set.
$\therefore$ the mean of the prime numbers is $\frac{23 + 9}{2} = \frac{52}{2} = {36}$

Range $=$ largest score $-$ smallest score Smallest score $= 13$ and range is $44,$ so $x$ is must be largest score. sum of digits of $x$ is $5 + 7 = 12.$

Total number of outcomes = Total number of students $= 5$
Number of possible outcomes = Students participating in a quiz $= 2$
$\therefore\text{Probability}=\frac{\text{Number of possible outcomes}}{\text{Total number of outcomes}}=\frac{2}{5}$
to To find percentage, we havemultiply it by hundred $=\frac{2}{5}\times100=40\%$

Given, mean $= 7$ and data is $6, 8, 5, x, 4$
$\therefore \frac{6+8+5+\text{x}+{4}}{{5}} = 7$
$\Rightarrow{\text{x+23}} = 35$
$\Rightarrow{\text{x}} = 12$

Given observations $22, 16, 19, 12, 26, 32, 87, 58$ No. of observations 8 sum of observations.
$22 + 16 + 19 + 12 + 26 + 32 + 87 + 58 = 272$
Mean is given as $\frac{272}{8} = {34}$

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

Three bowling scores of tournament are $167, 178$ and $186.$
average score will be $ = \frac{167+178+186}{3} = {177}$

The mean is same like average. Add all and then divide by the number of colleges. Average
$ = \frac{(370+310+380+340+310)}{5} = \frac{1710}{5} = {342}$
$\therefore$ the mean of these monthly fees is $342.$

Consider the given series. $1, 2, 4, 8, 16, ..... ,2n.$
$ A.M = \frac{1+2+4+8+16+ ..... +2{\text{n}}}{\text{n+1}}$
$ A.M =\frac{ {2}^{0}+{2}^{1}+{2}^{2}+{2}^{3}+{2}^{4}+ ...... +{2}{\text{n}}}{\text{n+1}}$
$A.M = \frac{ \frac{{1}({2}^{\text{n+1}} -1)}{2-1}}{\text{n+1}}$
$A.M = \frac{{2}^{\text{n+1}}-{1}}{\text{n+1}}$

The average of $11, 12, 13, 14$ and $x$ is $13$
$\therefore\frac{11+12+13+14+{\text{x}}}{{5}} = 13$
$\therefore 50+{\text{x}} = 65{\text{x}} = 15$

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

$\Rightarrow 16x = 18x - 14$
$\Rightarrow x = 7$

 $ = \frac{4\times1.5+10\times2+6\times1}{20}$
$ = \frac{32}{20} = {1.6}{\text{m}}$

 Average amount of interest paid by the Company during the given period.
$ = \frac{\text{ Rs. } \big[23.4 + 32.5 + 41.6 + 36.4 + 49.4\big]}{5} \text{lakhs}$
$ =\frac{ \text{ Rs. }\big[183.3\big]}{5}\text{lakhs}$
$= Rs. 36.66$ lakhs

 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 $(a).$

 Here, the observations are $5, 7, x, 10, 5$ and $7$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow7=\frac{5+7+\text{x}+10+5+7}{6}$
$\Rightarrow\text{x}+34=42$
$\Rightarrow\text{x}=42-34=8$
Hence, the correct option is $(c).$

 The consecutive odd numbers from $1$ to $13 = 3, 5, 7, 9, 11$
thus required average.
$\frac{ = {3}^{2} + {5}^{2} +{7}^{2}+{9}^{2}+{11}^{2}} {5}$
$ = \frac{9+25+49+81+121}{5} = \frac{285}{5} ={57}$

$A.M$ of numbers $ = \frac{\text{sum of these number }}{\text{their number}}$
$ = \frac{7 + 6 + 5 + 3 + 8 + 6 + 7}{7} = \frac{42}{7} = {6}$

Since, Mean $ = {\frac{1+2+3+ ...... +{\text{n}}}{\text{n}}}$
$\Rightarrow$ Mean $= \frac{[\text{n}({\text{n+1}})]}{2}$
$ = [{\frac{1+2+3+ ...... +{\text{n}}}{\text{n}}}= \frac{\text{n}({\text{n+1}})}{2}]$
$\Rightarrow$ Mean $= \frac{\text{n}({\text{n+1}})}{2{\text{n}}}$
$\Rightarrow$ Mean $= \frac{\text{n+1}}{2}$

By definition,
$\text{Average} =\frac{ \text{x}+(\text{x+3})+({\text{x+6}})+({\text{x+9}})+({\text{x+12}})}{5}$
$\frac{{5}+{30}}{5} = {\text{x+6}}$

Sum of eight numbers $= 38.4 \times 8 = 307.2$
$\therefore$ Total sum = Average \times Number of items
sum of seven numbers $= 39.2 \times 7 = 274.4$
$\therefore$ Eight number $= 307.2 - 274.4 = 32.8$

 If $x = 10$ then the observation $3x = 30$
and if $x = 15$ then the observation $3x = 45$
with values between $30$ to $45,$ the Mean, Median and Quartile Deviation will change.
but the range will not change as the highest value among the set of observations $48$
and least value is $8$ and does not depend on the value of $x$

 The given observations are: $96, 104, 121, 134, 142, 149, 153$ and $161$
Mean $ = \frac{\text{Sum}}{\text{Total}}$
Mean $ = \frac{96+104+121+134+142+149+153+161}{8}$
Mean $= 132.5$

mean $ = \frac{\text{Sum}}{\text{Total}} = {-7}\frac{\text{Sum}}{10}$
$= −7$ sum$= -705$ is added to every $10$ no. mean $ = \frac{-70 + 50}{10} $
$= -2$ since Total added $= 50$

 The total no. of observations are $100$ The mean of those observations are $30$
So The sum of observations is $30 \times 100 = 3000$ the wrong observations are $32$ and $12$ those are to be subtracted from sum of observations
So, $3000 - 32 - 12 = 2956$ the correct observations are $23$ and $11$ those are to be added
so, $2956 + 23 + 11 = 2990$ the mean is given by.
$ = \frac{2990}{100} = {29.9}$

No. of observations 5 sum of observations $36 + 40 + 32 + 48 + 44 = 200$ mean $\frac{200}{5} = {40}$

The average maximum temp from $12th$ sept to $18th$ sept is $35$ and The average maximum temp from $13th$ sept to $19th$ sept is $34$ then total temp from $12th$ sept to $18th$ sept $= 35 \times 7 = 245$ and total temp from $13th$ sept to $19th$ sept $= 34 \times 7 = 238$ then diff $= 245 - 238 = 7$ so Maximum temperature on $12th$ is $7^\circ $ less than that for $19th$ September. $77$

If there are three people each pays $\frac{\text{x}}{3}$
if there are four people each pays $\frac{\text{x}}{4}$
the difference $ = \frac{\text{x}}{3} - \frac{\text{x}}{4} = \frac{{4}{\text{x-3x}}}{12} = \frac{\text{x}}{12} $

 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 $(d).$

 Let there be $100$ students who took the examination.
The number of students who wrote answers in Hindi and in English are thus $40$ and $60$ respectively.
$\therefore$ average marks of all the candidates
$= (40 × 74 + 60 × 77) ÷ 100$
$= 7580 ÷ 100 = 75.8$

$ n = 25$
$u = 78.4 ($initial wrong mean$)$
$x = n × u = 25 × 78.4 = 1960$
Now $69$ was read as $96,$ so
$= 1960 - 96 + 69 = 1987$
$\therefore {\text{u'}} = \frac{1987}{25}$
thus $u = 79.48$

Average is the sum of the data values divided by the total values. Given that $13, 15, 11, 9, 8$ are the ages of $5$ children.
$\therefore$ average age $ = \frac{13 + 15 + 11 + 9 + 8}{5} = \frac{56}{5} = {11.2}$

Given observations: $994, 996, 998, 1000, 1002$
$\text{mean} = \frac{\text{sum}}{\text{number of observations}}$
$\text{mean} = \frac{{994+996+998+1000+1002}}{5}$
$\text{mean} = \frac{4990}{5}$
$\text{mean} = 998$

 Mean $ = \frac{2 + 3 + 4 + 5 +6}{5} = \frac{20}{5} = {4}$
Median is the middle most number $= 4$

We know that average of a bunch of numbers is the sum of the numbers divided by how many numbers in the bunch.
We can set up an equation for the average from the given question,
$\Rightarrow \frac{12+{\text{a}}+6+8+2+14}{6} = {6}$
$\Rightarrow\frac{{42}+{\text{a}}}{6} = {6}$
$\Rightarrow{42}+{\text{a}} = {36}$
$\text{a} = -{6}$

The first five natural numbers are: $1, 2, 3, 4, 5$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$=\frac{1+2+3+4+5}{5}$
$=\frac{15}{5}$
$=3$
Thus, the mean of first five natural number is $3$
Hence, the correct option is $(c).$

After one value is removed:
Since all of the values are Integers, the sum here must be an integer. Sum $= ($number$) × ($average$).$
Since the average $ = {35}\frac{7}{17},$ and the sum must be an integer, the number of integers must be a multiple of $17.$
For any evenly spaced set,
average = median. After one of the consecutive integers is removed, most of the remaining set will still be evenly spaced.
As a result, the average of the remaining set $ ={37}\frac{7}{17}$ will still be close to the median. Implication: The number of integers $= 4 × 17 = 68$ with the result that $ = {35}\frac{7}{17},$
will be close to the median of the 68 mostly consecutive integers.
$\therefore{\text{sum}} = {68}\times{35}\frac{7}{17} = {2408}$
Original set:
Since $68$ integers remain after one of the integers is removed, the original set contains $69$ integers. Sum of the first n positive integers $ = \frac{(\text{n})({\text{n+1}})}{2}$
$\therefore{\text{sum}} = {69}\times\frac{70}{2} = {2415}.$
Removed integer = original sum - sum after one integer is removed $= 2415 - 2408 = 7.$

Mean $ = \frac{\text{Sum of observation}}{\text{Total}} = \frac{{5}\times{4}}{2(5)} = 2$ Sum of $1st\ 5$ whole no.
$ = \frac{5(4)}{2} = {10}$
$(0, 1, 2, 3, 4)$ First $5$ whole no.

Sum of $5$ numbers $= 27 \times 5 = 135$ when one of the numbers is excluded.
sum of remaining $4$ numbers $= 4 \times 25 = 100$ Excluded number $= 135 - 100 = 35.$

Let the $20$ numbers be $a_1​, a_2​, .... ,a_{20}$
Given that average of $20$ numbers is zero.
$\therefore\frac{ {\text{a}}_1+{\text{a}}_2+...+{\text{a}}_{20}}{20}$
$⇒ a_1​ + a_2​ + ... +a_{20}​=0$
$⇒ a_1​ + a_2​ + ... +a_{19}= -a_{20}$
$\therefore$ at the most $19$ numbers can be greater than zero.

Total amount spent on shirts is $5 \times 450 = 2250$
amount on trousers is $4 \times 750 = 3000$ and on shoes $12 \times 750 = 9000$
hence total amount spent is $2250 + 3000 + 9000 = 14250$
so average per article is $\frac{14250}{21} = 678.57$

Average is the sum of the data values divided by the total values.The first $4$ prime numbers are $2, 3, 5, 7$
$\therefore$ average of first $4$ prime numbers $ = \frac{2 + 3 + 5 + 7}{4} = \frac{17}{4} = {4.25}$

 Required average
$ = \frac{\big(50.25\times{16} + 45.15 \times{8}\big)}{16 + 8}$
$ =\frac{ \big(804 + 361.20\big)}{24}$
$ = \frac{1165.20}{24}$
$ = {48.55}$

Given, Rahul and $34$ mean score in $5$ test was $84$. And his means score in the first $4$ test of these test was $87.$ Then total score in $5$ tests $= 84 × 5 = 420$ And total score in $4$ tests $= 87 × 4 = 348$ So, the score in fifth test $= 420 - 348 = 72.$

$\text{mean}\frac{\text{sum of the terms}}{\text{total number of terms}}$
$\Rightarrow \frac{{2+4+6+8+}{\text{x+y}}}{6} = {5}$
$\frac{{20+}{\text{x+y}}}{6} = {5}$
$20 + x + y = 5 (6)$
$x + y = 30 - 20 = 10$

$ = \frac{33.5+30.4+25.6+31.5+29}{5} = \frac{150}{5} = {30}$

Let the grade he got on final test be $x$ The average of all four grades is $90.$
So, we have $ = \frac{{92+89+86+}{\text{x}}}{4} = {90}$
$\Rightarrow x = 360 - 267 = 93$

Since the weight of each student increases by $3\ kgs,$ their mean will also increase by $3\ kgs,$ hence new mean becomes $235 + 3 = 238\ kg$

 The sum of two numbers is twice their average.
$x + y = 100 $
$y + z = 160$
$x = 100 - y $
$z = 160 - y$
substitute these expressions for $z$ and $x : z - x = (160 - y) - (100- y) = 160 - y - 100 + y= 160 - 100 = 60$
alternatively, pick smart numbers for $x$ and $y.$
Let $x = 50$ and $y = 50 ($this is an easy way to make their average equal to $50).$
since the average of $y$ and $z$ must be $80,$ you have $z = 110$
$\therefore z - x = 110 - 50 = 60$

Given is that mean age of all males is $15$ years and age of all females is $19$ years in a colony, when we find the mean age mm of the combined group of males and females in colony, we can conclude that mean age will be more than $15$ but less than. so $15 < m < 19$

We know that Mean $ = \frac{\text{sum of observations}}{\text{total numer of observations}}$
$\therefore$ Mean of $10$ natural numbers $ = \frac{\text{sum of 10 natural number}}{10}$
We know that the sum of $′n′$ natural numbers $ = \frac{\text{n}{(\text{n+1})}}{2}$
$\therefore$ Sum of $10$ natural numbers $ = \frac{{10}\times{11}}{2} = {55}$
Mean of natural numbers $ = \frac{55}{10} = \frac{11}{2}$

The first $50$ natural numbers are $1, 2, 3, ..... ,50$
We know that, the sum of first nn natural numbers is $ = \frac{\text{n}({\text{n+1}})}{2}$
$\therefore$ sum of first $50$ natural numbers is $\frac{{50}{(51)}}{2} = {25}(51) = {1275}$
the average of first $50$ natural numbers is $\frac{1275}{50} = {25.5}$

Let the numbers be $x - 2, x, x + 2, x + 4$ according to the question,
$ = \frac{\text{x-2+x+x+2+x+4}}{4} = {15}$
$4x + 4 = 60$
$4x = 56$
$⇒ x = 14$
hence, the second highest number $i.e. x + 2 = 14 + 2 = 16$

$\frac{\text{(x + 2)}\times{60}+\text{x}\times{120}+\text{(x - 2)}\times{180}}{{\text{x + 2}}+{\text{x + x}}-{2}} = {100}$
$\frac{{60}\text{x}+{120}+{120}{\text{x}}+{180}{\text{x}} - {360}}{\text{3x}} = {100}$
$360x - 240 = 300x$
$60x = 240$
$x = 4$

Let the average age of the whole team by $x$ years.
$\therefore 11x - (26 + 29) = 9(x - 1)$
$⇒ 11x - 9x = 46$
$⇒ 2x = 46$
$⇒ x = 23.$
So, average age of the team is $23$ years.

 Let the man spends $Rs. x$ on purchasing each kind of pens at their respective rates. Total cost $= x + x + x = 3x$
Now, number of pens bought of $Rs. 5$ will be $\frac{\text{x}}{5}$
Number of pens bought of $Rs. 10$ will be $\frac{\text{x}}{10}$
Number of pens bought of $Rs. 20$ will be $\frac{\text{x}}{20}$
So, total number of pens bought $ = \frac{\text{x}}{5}+\frac{\text{x}}{10}+\frac{\text{x}}{20}$
Average cost $ =\frac{ \text{Total cost}}{\text{No. of total pens}}= \frac{ \text{x+x+x}}{\frac{\text{x}}{5}+\frac{\text{x}}{10}+\frac{\text{x}}{20}}$
$ = \frac{{3}}{\frac{1}{5}+\frac{1}{10}+\frac{1}{20}} = \frac{60}{7}$

 $\text{Mean} = \frac{\text{x + x + 2 + x + 4 + x + 6 + x + 8}}{8} $
$= 205x = 80x = 16$

 First $726$ natural numbers $= 1, 2, 3, 4 .... 726$
sum of $726$ numbers $= 1 + 2 + 3 + 4 .... 72$
this forms an $AP,$ with first term $= 1,$ last term $= 726$ and number of terms $= 1$
$\text{sum of AP} = \frac{\text{n}}{{2}} (\text{a + 1})$
$\text{sum of AP} = \frac{726}{2} (1+726)$
$\text{sum of AP} =263901$
$\text{mean} = \frac{\text{sum}}{\text{number of turns}} = \frac{263901}{726} = 363.5$

 The factor of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$
$\therefore{\text{A.M}} = \frac{\sum{\text{x}}}{\text{n}}$
$\Rightarrow\frac{1+2+3+4+6+8+12+24+}{8}\Rightarrow\frac{60}{8} = \frac{15}{2}$

 Given $3$ Quantities: $12 - n, 12, 12 + n$ Average of the above $3$ Quantities
$ = \frac{\text{sum of the above 3 quantities}}{\text {number of quantities}} $
$ = \frac{{12 }-{\text{n}} +12+12+{\text{n}}}{3}$
$ = \frac{{12}\times{3}}{3} = {12}$
$\therefore$ average of the Quantities listed above is $12.$

 $\frac{2+4+6+8+10+12+}{6} = \frac{42}{6} = 7$

Mean of n observations $= 12$
Sum of observations $= 132$
$\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow12=\frac{132}{\text{n}}$
$\Rightarrow\text{n}=\frac{132}{12}=11$
Thus, the value of $n$ is $11$
Hence, the correct option is $(c).$

Let $a,b$ and $c$ are three numbers $ = \frac{\text{a + b + c}}{3} = {60}$
$a + b + c = 180$
and $a = \frac{1}{4} \text{(b+c)} = \frac{1}{4} (180) - {\text{a}}$
$4a = 180 - a$
$\text{a} = \frac{180}{5} = {36}$

The first four natural numbers are $1, 2, 3, 4$
$\text{A.M} = \frac{\sum{\text{x}}}{\text{n}} = \frac{{1}^{3} +{2}^{3} + {3}^{3}+{4}^{3}}{4}$
$\frac{1+4+27+64}{4} = \frac{100}{4} = {25}$

 There are $52$ cards in a standard deck. There are four different suits Diamonds (red), Clubs (black), Hearts (red), and Spades (black) each containing $13$ cards.
$\therefore $ Number of red cards $($favourable outcomes$) = 13 + 13 = 26$
Therefore
Probability of getting a red card $=\frac{26}{52}=\frac{1}{2}$
Hence, the correct option is $(b).$

 We know that sum of $1$ to $n$ numbers $ = \frac{\text{n}(\text{n+1})}{1}$
Then sum of $1$ to $100 = \frac{{100}(100+1)}{2} = \frac{{100}\times{101}}{2} = \frac{10100}{2} = 5050$
Then average of $1$ to $100$ positive numbers $ = \frac{5050}{100} = 50.5$

 Given observations $25, 37, 84$ No .of observations $3$ The mean of
observation $\frac{25+37+84}{3}=\frac{146}{3}=48.6$

 Mean of $5$ number $= 27.$ If $1$ of the numbers is excluded,
$\therefore M′ = M - 2$
$= 27 - 2$
$= 25 × 4$
$= 100M = 27 × 5$
$= 135$ Excluded number $= 135 - 100 = 35.$

Required mean $ = \frac{130+126+68+50+1}{5} = \frac{375}{5} = {75}$

 Mean of $20$ observation $= 12.5$
sum of $20$ observations $= 12.5 × 20 = 250$
Wrong observation is $-15$ Correct observation is $15$ So,
sum of $20$ observation correctly $= 250 + 15 + 15 = 280$
first $15$ is to eliminate the wrong observation and the other $15$ for adding the correct observation in the data.
correct mean $ = \frac{280}{20} = {14}$

Total numbers $= 50$
Mean of numbers after subtracting $53$ from each $= 3.5$
Sum of numbers after subtracting $53$ from each $= 3.5 × 50 = 175$
Sum of the original numbers $= 175 + 53 × 50 = 2825$
Mean of the original numbers $ = \frac{2825}{50} = {56.5}$

 Total marks obtained by all students $= 10 \times 75 + 12 \times 60 + 8 \times 40 + 3 \times 30 = 1880$
The total number of students $= 10 + 12 + 8 + 3 = 33$ Average of their score
$ = \frac{1880}{33} = 56.96969697 ($aproximately$)$

 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 $(a).$

 Number of all possible outcomes $= 100$
Number of head obtained $= 59$
Number of tail obtained $($favourable outcomes$) = 100 - 59 = 41$
Therefore
Probability of getting a tail $\frac{41}{100}$
Hence, the correct option is $(b).$

 Given the average age of two brothers is $9$ years
then total age $= 9 \times 2 = 18$ years
If mothers age included the average age $= 9 + 9 = 18$ years
then total age of mother and two brothers $= 18 \times 3 = 54$ years
so age of mother $= 54 - 18 = 36$ years.

 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 $(d).$

 Given data is $56, 15, 48, 49, 52, 57, 30, 51, 42, 50$ No. of observations $10$ sum of observations is
$56 + 15 + 48 + 49 + 52 + 57 + 30 + 51 + 42 + 50 = 450$
mean of the data is $\frac{450}{10} = {45}$

 Given: the mean of the observations $7, 8, 9, 11$ and $x$ is $10$
Mean of observations $=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow10=\frac{7+8+9+11+\text{x}}{5}$
$\Rightarrow35 + \text{x} = 50$
$\Rightarrow\text{x} = 50 - 35 = 15$
Thus, the value of $x$ is $15$
Hence, the correct option is $(b).$

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 $(b).$

Let the five numbers be $A, B, C, D, E$
Given, $A + B + C + D + E = 555$
$\frac{\text{A+B}}{2}$ = 75
$\Rightarrow A + B = 150$ and $C = 115$
$\therefore 150 + 115 + D + E = 555$
$\Rightarrow D + E = 555 -265 = 290$
$\Rightarrow $ Required Average $\frac{\text{D+E}}{2} = \frac{290}{2}= 145$

Mean $ =\frac{ \text{Sum of observations}}{\text{Total number of observations}} = \frac{5+6+8+9+12+13+17}{7} = \frac{70}{7} = {10}$

Tossing a coin, either we get a head $(H)$ or a tail $(T).$
So, the probability of getting a head is $\frac{1}{2}$
Hence, the correct option is $(a).$

Number of boys $= 70$
Average marks of boys $= 75$
Total marks of boys $= 70 \times 75 = 5250$
Total marks of the class $= 72 \times 100 = 7200$
Total marks of girls $= 1950$
Average of the girls $ = \frac{1950}{30} = {65}$

$\Rightarrow $ Factors of $2$ are $1, 2, 3, 4, 6, 8, 12, 24$
$\Rightarrow $ The observations are $1, 2, 3, 4, 6, 8, 12, 24$
$\therefore$ Number of observations = 8 Arithmetic mean $ = \frac{\text{Sum of observations}}{\text{Number of observations}}$
$\therefore$ Arithmetic mean$ = \frac{1 + 2 + 3 + 4 + 6 + 8 + 12 + 24}{8}$
$\therefore$ Arithmetic mean$ = \frac{60}{8}$
$\therefore$ Arithmetic mean $=7.5$

 Distances jumped by $10$ athletes $($in $cm) 205, 200, 275, 260, 259, 199, 252, 239, 228$ and $281$
Mean $ = \frac{205+200+275+260+259+199+252+239+228+281}{10}$
Mean $ = \frac{2398}{10}$
Mean $= 239.8$

 Given odd numbers between $18$ and $28$ are $19, 21, 23, 25, 27.$
$\text{Average} = \frac{19+21+23+25+27}{5} = \frac{115}{5} = 23$

 Mean $ = \frac{{1 }\times{ \text{ x + 2 }}\times{ 2 }{\text{ x + 3 }}\times{ 3}{\text{x}} ...... }{\text{x + 2x + 3x .....}}$
$ = \frac{{1 }\times{ \text{ x + 2 }}\times{ 2 }{\text{ x + 3 }}\times{ 3}{\text{x}} ...... }{{\text{x}}({1+2+3 ......})}$
$=\frac{\frac{\text{n(n+1})({2}\text{n+1})}{6}} {\frac{\text{n(n+1})}{2}}$
Mean $\Rightarrow \frac{\text{n(n+1})({2}\text{n+1})}{6} \times \frac{\text{n(n+1})}{2}$
Mean $ = \frac{\text{2n+1}}{3}$

Included value $31 \times 11 - 32.6 \times 10 = 15$

 The mean of $10, 15, 19, 30, 43, 69$ and $x$ is $x.$
$\therefore\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$
$\Rightarrow\text{x}=\frac{10+15+19+30+43+69+\text{x}}{7}$
$\Rightarrow\text{x}+186=7\text{x}$
$\Rightarrow\text{x}=\frac{186}{6}=31$
Thus, the observations are $10, 15, 19, 30, 43, 69$ and $31$
Arranging the numbers $10, 15, 19, 30, 43, 69$ and $31$ in increasing order, we get
$10, 15, 19, 30, 31, 43, 69$
Thus, the median is $30$
Hence, the correct option is $(c).$

Factors of $10$ are: $1, 10, 2, 5$
$\text{mean of factors of 10} = \frac{\text{sum}}{\text{count of number}}$
$\text{mean} = \frac{1+10+2+5}{4}$
$\text{mean} = \frac{18}{4}$
$\text{mean} = 4.5$

The observations are: $x + 3, x + 5, x + 7, x + 9, x + 11$
$\text{Mean} = \frac{\text{Sum}}{{\text{Number of observations}}} = \frac{\text{x+3,x+5,x+7,x+9,x+11}}{5}$
$ = \frac{5\text{x+}{35}}{5} = \text{x+7}$

 Given data $6, 8, 10, 12, 14, 16, 18$
the sum of observations is $6 + 8 + 10 + 12 + 14 + 16 + 18 = 84$
mean is given as $\frac{84}{7} = {7}$

 Sum of $100$ numbers $= 100 × 44 = 4400$
Sum of $104$ numbers $= 104 × 50 = 5200$
Sum of $4$ numbers $= 5200 - 4400 = 800$
$\therefore$ Average of four new numbers $ = \frac{800}{4} = 200$

 We know that, median is the middle most observation.
For finding the median of the data firstly,
we arrange the data in ascending order, i.e. Ascending order
$i.e\ 3, 3, 4, 4, 5, 6, 7.$
$n = 7 ($odd$)$
$\therefore$ Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)^{\text{th}}$
observation = Value of $\Big(\frac{7+1}{2}\Big)^{\text{th}}$ observation
$= 4th$ observation $= 4$

 Average of the expression $= \frac{\text{sum of expression}}{\text{total number of expressions}}$
$\Rightarrow \frac{{2}{\text{x}}+4+{5}{\text{x}} - 1 -{\text{x}} +{3}}{3}$
$\Rightarrow \frac{{6}{\text{x}}+{6}}{3}$
$\Rightarrow\frac{{3}({2}{\text{x}}+{2})}{3} = {2}{\text{x}}+{2}$

$\text{Probability}=\frac{\text{Number of possible outcomes}}{\text{Total number of outcomes}}$
Total number of cards in set $(d) = 6$
Number of possible outcomes $= 2 [\because 2$ aces in every set, given$]$
So, probability $=\frac{2}{6}$

First $8$ whole numbers are $0, 1, 2, 3, 4, 5, 6, 7$
Mean $ = \frac{0+1+2+3+4+5+6+7}{8} = \frac{28}{8} = {3.5}$

$\frac{5+0+6+}{5}\frac{1}{4}+8\frac{3}{4}$
$ = \frac{20}{5} = {4}$

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 $(a).$

Required average $ = \Big(\frac{{67}\times{2+35+}\times{2+6}\times{3}}{2+2+3}\Big) = \Big(\frac{134+70+18}{7}\Big) = \frac{227}{7}$
$ = {31}\frac{5}{7}{\text{ years}}$

Average of Number $ = \frac{\text{sum of all number}}{\text{total number of number}}$
$\frac{201+207+210+213+204}{5} = > \frac{1035}{5} = {207}$

The data in arranging order is: $5, 7, 9, 10, 11$
As the number of observations is odd $(5),$ the median is the middle term which is $9$
Hence, the correct option is $(b).$

$ \Rightarrow $ Runs scored by Sachin in fist innings $= 80$
$\Rightarrow $ Runs scored by Sachin in fistinnings $= 120$
$\Rightarrow $ Total runs scored in two innings by Sachin $= 80 + 120 = 200$
$\Rightarrow $ Average scored of Sachin in each innings
$ = \frac{200}{2} = {100}$

 Mean of remaining set $= \frac{{50}\times 38 - ( 45+55)}{50 - 2} = \frac{1800}{48} = {37.5}$

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 $(d).$

Arithmetic mean, $A = \frac{\text{S}}{\text{N}} = \frac{6+{\text{p}}+12+8+9}{5} = {9}$
$\frac{{35+}{\text{p}}}{5} = {9}$
$35 + p = 45$
$p = 45 - 35 = 10$

$\frac{{11}+{12}+{13}+{14}\text{ - x}}{5} = {13}$
$x = 65 - 50$
$x = 15$

 If a train covers a certain distance at $x\ kmph$ and equal distance at $y\ kmph$ than average speed $ =\frac{2\text{xy}}{\text{x+y}}$
here $x = 30\ km$ per hour
$y = 60\ km$ per hour
$\therefore$ average speed $\frac{2\times60\times30}{30+60}$
$\Rightarrow\frac{360}{90} = 40\ km $ per hour

First $n$ odd natural numbers are $1, 3, 5, ..., (2n - 1)$
so, the required mean $ = \frac{1+3+5+ .... +({2}{\text{n-1})}}{\text{n}}$
$ = \frac{\text{n}}{2}\frac{[{1+2}{\text{n-1]}}} {\text{n}} = \frac{{\text{n}}^{2}}{\text{n}} = {\text{n}}$

Let the distance of his office $= x\ km$
$\therefore$ Required Average speed
$ = \frac{\text{Total Distance}}{\text{Total time taken}} = \frac{\text{x}}{40} + \frac{\text{x}}{60} = {48}\frac{\text{km}}{\text{hour}}$

 The number on the faces of a dice are $1, 2, 3, 4, 5,$ and $6.$
$\therefore$ Number of possible outcomes $= 6.$
Hence, the correct option is $(a).$

 Numbers are $= 80, 90, 100, 110, 120, 130$
Average $ = \frac{80 + 90 + 100 + 110 + 120 + 130}{6} = > \frac{630}{6} = {105}$