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} $