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

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}}$