Maths M.C.Q (1 Marks)

$82 = \frac{{(27 + x) + (31 + x) + (89 + x) + (107 + x) + (156 + x)}}{5}$

==> $82 \times 5 = 410 + 5x$ ==> $410 - 410 = 5x$

==> $x = 0$

Required mean is,

$\bar x = \frac{{130 + x + 126 + x + 68 + x + 50 + x + 1 + x}}{5}$

$\bar x = \frac{{375 + 5x}}{5}$

$ = \frac{{375 + 0}}{5}$$ = \frac{{375}}{5}$= $75.$

i.e., $3 = \frac{{5 + 8 + 18 + 4f}}{{15 + f}}$

==> $45 + 3f = 31 + 4f$

==> $45 - 31 = f$ ==> $f = 14$.

==> $\sum\limits_{i = 1}^{20} {{x_i} - 20 \times 30 = 20} $

==> $\sum\limits_{i = 1}^{20} {{x_i} = 620} $.

Mean = $\frac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} $

$= \frac{{620}}{{20}} = 31$.

$ = \frac{{1 + 1 + 1 + 1 + ..... + 1}}{{\frac{{n(n + 1)}}{2}}}$

$ = \frac{n}{{\frac{{n(n + 1)}}{2}}} = \frac{2}{{n + 1}}$.

==> ${x_1} + {x_2} + ..... + {x_{10}} = 45$

$\frac{{{x_{11}} + {x_{12}} + ..... + {x_{40}}}}{{30}} = 3.5$

==> ${x_{11}} + {x_{12}} + ..... + {x_{40}} = 105$

${x_1} + {x_2} + ..... + {x_{40}} = 150$

$\frac{{{x_1} + {x_2} + ..... + {x_{40}}}}{{40}} $

$= \frac{{150}}{{40}}$$ = \frac{{15}}{4}$.

i.e., $2.6 = \frac{{1 \times 4 + 2 \times 5 + 3 \times y + 4 \times 1 + 5 \times 2}}{{4 + 5 + y + 1 + 2}}$

or $31.2 + 2.6y = 28 + 3y$ or $0.4y = 3.2$

==> $y = 8$.

$\frac{{\sum\limits_{i = 1}^n {({x_i} + 2i)} }}{n} = \frac{{\sum\limits_{i = 1}^n {{x_i}} + 2\sum\limits_{i = 1}^n i }}{n} = \frac{{n\bar x + 2(1 + 2 + ...n)}}{n} = \frac{{n\bar x + 2\frac{{n(n + 1)}}{2}}}{n} = \bar x + (n + 1)$

$ = \frac{{n\bar x + 2\frac{{n(n + 1)}}{2}}}{n} = \bar x + n + 1$.

Sum of items added$ = 60 + 70 + 80 = 210$

Sum of items replaced$ = 40 + 20 + 50 = 110$

New sum $ = 4900 + 210 - 110 = 5000$

Correct mean$ = \frac{{5000}}{{100}} = 50$.

After one number excluded

Sum of total number = $16 × 4 = 64$

Then, excluded number is $90 -64 = 26.$

Sum of weight of $6$ students

$= 52 + 58 + 55 + 53 + 56 + 54 = 328\ kg$

$\therefore $ Weight of seventh student = $385 -328 = 57\ kg.$

Sum of items added = $19 +31 = 50$

Sum of items replaced = $91+ 13 = 104$

New sum = $4500 - 104 + 50$ = $4446$

New mean$ = \frac{{4446}}{{100}}$ = $44.46$

Here, total number of items is $41$

$i.e.$, an odd number.

Hence, the median is $\frac{{41 + 1}}{2}^{th}$

$i.e.$, $21^{st}$ item.

From cumulative frequency table, we find that median

$i.e.$, $21^{st}$ item is $155$,

(All items from $16$ to $22^{nd}$ are equal, each $155$).

$\bar x = \frac{{0.{q^n} + 1.\frac{n}{1}{q^{n - 1}}p + 2.\frac{{(n)(n - 1)}}{{2!}}{q^{n - 2}}{p^2} + .....n.{p^n}}}{{{q^n} + \frac{n}{1}{q^{n - 1}}p + \frac{{n(n - 1)}}{2}{q^{n - 2}}{p^2} + ..... + {p^n}}}$

$ = \frac{{0.{^n}{C_0}{q^n}{p^0} + {{1.}^n}{C_1}\,{q^{n - 1}}p + ..... + n.{\,^n}{C_n}{q^0}{p^n}}}{{^n{C_0}{q^n}{p^0}{ + ^n}{C_1}{q^{n - 1}}{p^1} + .....{ + ^n}{C_n}{q^{n - n}}{p^n}}}$

$ = \frac{{\sum\limits_{r = 0}^n r {.^n}{C_r}{q^{n - r}}{p^r}}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{\sum\limits_{r = 1}^n r .\frac{n}{r}\,{\,^{n - 1}}{C_{r - 1}}{q^{n - r}}.p.\,{p^{r - 1}}}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{np\left( {\sum\limits_{r = 1}^n {^{n - 1}{C_{r - 1}}{p^{r - 1}}{q^{(n - 1) - (r - 1)}}} } \right)}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{np{{(q + p)}^{n - 1}}}}{{{{(q + p)}^n}}} = np$, . $[\because q + p = 1]$

$n = \sum f = 159$. Here $n = 159$, which is odd

$\therefore $Median number$ = \frac{1}{2}(n + 1) = \frac{1}{2}(159 + 1) = 80$,

which is in the class $30-40$. (see the row of cumulative frequency $95$, which contains $80$).

Hence median class is $30-40$.

$\therefore$  We have,

$l$ = Lower limit of median class = $30$

$f$ = Frequency of median class = $45$

$C$ = Total of all frequencies preceding median class = $50$

$i$ = Width of class interval of median class = $10$

$\therefore$  Required median $ = l + \frac{{\frac{N}{2} - C}}{f} \times i$ $ = 30 + \frac{{\frac{{159}}{2} - 50}}{{45}} \times 10 = 30 + \frac{{295}}{{45}} = 36.55$.

$\mathrm{x}_{10+\mathrm{n}}=148+(\mathrm{n}-1)(-2)=\mathrm{x}_{17}=136, \mathrm{x}_{18}=134$

hence median $=135$

Here, total number of items is $41$ i.e., an odd number.

Hence, the median is $\frac{41+1}{2}$ th i.e., $21^{\text {st }}$ item.

From cumulative frequency table, we find that median

i.e., $21^{\text {st }}$ item is $155,$

(All items from $16$ to $22^{\text {nd }}$ are equal, each $155$ ).

$\frac{685}{2}=342.5$

$40+\frac{342.5-296}{180} \times 10$

$40+\frac{46.5}{18}=40+2.8=42.6$

$=\frac{\mathrm{x}_{1}+1+\mathrm{x}_{2}+2+\mathrm{x}_{3}+3+\mathrm{x}_{4}+4+\ldots+\mathrm{x}_{10}+10}{10}$

$=\frac{\sum x_{i}+\frac{10 \times 11}{2}}{10}=3+\frac{11}{2}=\frac{17}{2}$

 Mean deviation$ = \frac{{\Sigma |{x_i} - \bar x|}}{n}$

$ = \frac{{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}}{5}$

$ = \frac{{2 + 1 + 0 + 1 + 2}}{5}$$ = \frac{6}{5} = 1.2$.

No. of observations $=3 \mathrm{n}$

mean $(\bar X) = \frac{{n \times a + 2n \times ( - 2a)}}{{3n}} =  - a$

$\therefore $ Mean deviation about mean $ = \frac{{\Sigma \left| {{{\rm{x}}_1} - {\rm{\bar x}}} \right|}}{{3{\rm{n}}}}$

$\frac{\mathrm{n} \times 2 \mathrm{a}+2 \mathrm{n} \times \mathrm{a}}{3 \mathrm{n}}=\frac{4 \mathrm{a}}{3}$

Mean deviation $=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\mathrm{a}\right|}{10}=30$

$\frac{2(4.5 x+3.5 x+2.5 x+1.5 x+.5 x)}{10}=30$

$\frac{2(12.5)|\mathrm{x}|}{10}=30$

$|x|=12$

$S.D.$ = $\sigma $ = $\sqrt {\frac{1}{n}\sum {x_i^2 - (\bar x} {)^2}} $

=$\sqrt {\frac{1}{5}(1 + 4 + 9 + 16 + 25) - 9} $=$\sqrt {11 - 9} = \sqrt 2 $.

Hence, variance = $\frac{1}{n}\Sigma {({x_i} - \overline x )^2}$

$ = \frac{1}{5}\{ {(2 - 6)^2} + {(4 - 6)^2} + {(6 - 6)^2} + {(8 - 6)^2} + {(10 - 6)^2}\} $

$ = \frac{1}{5}\left\{ {(16 + 4 + 0 + 4 + 16} \right\}$$ = \frac{1}{5}\left\{ {40} \right\}$ $ = 8$.

Both sequence have same standard deviation.

Note:

If $1^{\text {st }}$ sequence is $x _{ i }$ and $2^{\text {nd }}$ sequence is $y _{ i }$,

$y _{ i }=10+ x _{ i } \Rightarrow \overline{ y }=10+\overline{ x }$

So, $\overline{ y }- y _{ i }=\overline{ x }- x _{ i }$

$ = \frac{{n(n + 1)\;(2n + 1)}}{{6n}} - {\left( {\frac{{n(n + 1)}}{{2n}}} \right)^2} $

$= \frac{{{n^2} - 1}}{{12}}$.

$ = \frac{{21}}{6} = \frac{7}{2}$

$S.D.$ $ = \sigma = \sqrt {\frac{1}{n}\Sigma x_i^2 - {{(\bar x)}^2}} $

$ = \sqrt {\frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) - \frac{{49}}{4}} $

$ = \sqrt {\frac{{91}}{6} - \frac{{49}}{4}} $

$ = \sqrt {\frac{{182 - 147}}{{12}}} $

$ = \sqrt {\frac{{35}}{{12}}} $.

Mean $ = 4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4$

==> $x + y = 11$ .....$(i)$

and variance = $5. 2$

==> $\frac{{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}}}{5} - {({\rm{mean}})^2} = 5.2$

$41 + {x^2} + {y^2} = 5[5.2 + {(4)^2}]$

$41 + {x^2} + {y^2} = 106$

${x^2} + {y^2} = 65$.....$(ii)$

Solving $(i)$ and $(ii)$ for $x$ and $y$, we get

$x = 4,y = 7$ or $x = 7,y = 4$.

Hence, new variance $ = {5^2} \times 9$$ = 225$.

${\sigma ^2} = \frac{{\sum {f_i}d_i^2}}{{\sum {f_i}}} - {\left( {\frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}$

$= \frac{{900}}{{10}} - {\left( {\frac{{ - 30}}{{10}}} \right)^2}$

${\sigma ^2} = 90 - 9 = 81$

==> $\sigma$ = $9$.

Corrected $\bar x = 7990/200$$ = 39.95$

Incorrect $\Sigma {x^2} = n\,[{\sigma ^2} + {\bar x^2}] = 200[{15^2} + {40^2}] = 365000$

Correct $\Sigma {x^2} = 365000 - 2500 + 1600$$ = 364100$

Corrected $\sigma = \sqrt {\frac{{364100}}{{200}} - {{(39.95)}^2}} $

$ = \sqrt {(1820.5 - 1596)} $$ = \sqrt {224.5} = 14.98$.

$\bar x = $ combined mean $ = \frac{{5 \times 8 + 3 \times 8}}{{5 + 3}}$ $ = \frac{{64}}{8} = 8$

Combined variance $ = \frac{{{n_1}(\sigma _1^2 + D_1^2) + {n_2}(\sigma _2^2 + D_2^2)}}{{{n_1} + {n_2}}}$,

where ${D_1} = {\bar x_1} - \bar x$, ${D_2} = {\bar x_2} - \bar x$

Now, ${D_1} = 8 - 8;\,\,{D_2} = 8 - 8 = 0$

Combined variance $ = \frac{{5(18) + 3(24)}}{{5 + 3}}$ $ = \frac{{90 + 72}}{8}$ $ = \frac{{162}}{8}$ = $20.25$.

Then, mean $ = 4.4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4.4$

==> $x + y = 13$.....$(i)$

and variance $= 8.24$

==> $\frac{{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}}}{5}$ -${({\rm{mean}})^2} = 8.24$

==> $41 + {x^2} + {y^2} = 5\,\{ {(4.4)^2} + 8.24\} $

==> ${x^2} + {y^2} = 97$.....$(ii)$

Solving $(i)$ and $(ii)$ for $x$ and $y$, we get

$x = 9,\,\,y = 4$ or $x = 4,y = 9$.

${\sigma ^2} = \frac{{\sum {{f_i}} d_i^2}}{{\sum {{f_i}} }} - {\left( {\frac{{\sum {{f_i}} {d_i}}}{{\sum {{f_i}} }}} \right)^2}$

$=\frac{900}{10}-\left(\frac{-30}{10}\right)^{2}$

$\sigma^{2}=90-9=81 $

$\Rightarrow \sigma=9$

Variance on doubling each observation

$=2^{2} \times 16=64$

Std. deviation $=\sqrt{\operatorname{var}}=8$

$\sigma^{2}=\frac{6 \times 24+3 \times 36}{6+3}+\frac{6 \times 3}{(6+3)^{2}}(11-14)^{2}$

$\sigma^{2}=\frac{144+108}{9}+\frac{18}{81} \times 9=28+2=30$

varience $=\frac{1^{2}+3^{2}+6^{2}+7^{2}+9^{2}+10^{2}}{6}-\left(\frac{1+3+6+7+9+10}{6}\right)^{2}$

$=10$

$\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2} = 100} $

new observations are $2 \mathrm{x}_{1}, 2 \mathrm{x}_{2}, \ldots \ldots, 2 \mathrm{x}_{20}$

Their mean $=\overline{\mathrm{x}}_{1}=\frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{20}\right)}{20}=2 \overline{\mathrm{x}}$

Now, variance$ = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - 2\bar x} \right)}^2}} $

$ = \frac{1}{{20}} \times 4\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2}} $

$ = \frac{1}{{20}} \times 4 \times 100 = 20$

average $\mathrm{m}_{1}=60, \mathrm{m}_{2}=40$

$\sigma_{1}=4, \sigma_{2}=6$

Standard deviation of combined series

$\sigma=\sqrt{\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(m_{1}-m_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$

$=\sqrt{\frac{10 \times 16+10 \times 36}{10+10}+\frac{10 \times 10(60-40)^{2}}{(10+10)^{2}}}$

$=\sqrt{8+18+100}=\sqrt{126}=11.2$

= $75 + 80 + 85$= $240$

If the marks of another subjected is added, then the marks will be $ \ge $ $240$ out of $400$

Minimum average marks $ = \frac{{240}}{4} = 60\% $,

[When marks in the fourth subject = $0$].

Given, $\bar x = 30$, ${\bar x_1} = 32$, $\overline {{x_2}} = 27$

Let ${n_1} + {n_2} = 100$ and ${n_1}$ denotes men, ${n_2}$ denotes women for this ${n_2} = 100 - {n_1}$

$30 = \frac{{32{n_1} + (100 - {n_1})27}}{{100}}$

==> $30 = \frac{{32{n_1} + 2700 - 27{n_1}}}{{100}}$

==> $3000 - 2700 = 32{n_1} - 27{n_1}$

==>$300 = 5{n_1}$ ==>${n_1} = 60$

So, ${n_2} = 40$

Hence, the percentage of women in the group is $40$.

$ = \frac{3}{{\frac{1}{{30}} + \frac{1}{{25}} + \frac{1}{{50}}}} \ km/hr$.

$40 + \frac{1}{2} = \frac{{35 \times 40 + w}}{{35 + 1}}$

==> $36 \times 40 + 36 \times \frac{1}{2} = 35 \times 40 + w$

==> $w = 58$

Weight of the teacher = $58\ kg$.

Total marks obtained

$= (40 × 50) + (35 × 60) + (45 × 55) + (42 × 45)$

$= 8465$

Overall average of marks per students ,

$ = \frac{{8465}}{{162}} = 52.25$.

Given, $\bar x = 500$,${\bar x_1} = 510$,${\bar x_2} = 460$

Let ${n_1} + {n_2} = 100$ and ${n_1}$ denotes male, ${n_2}$ denotes female for this ${n_2} = 100 - {n_1}$

$500 = \frac{{510{n_1} + (100 - {n_1})460}}{{100}}$

==> $50000{\rm{ }} = 510{n_1} + 46000 - 460{n_1}$

==> $50000{\rm{ }} - 46000 = 50{n_1}$

==> $4000 = 50{n_1}$

==> ${n_1} = \frac{{4000}}{{50}} = 80$.

Hence, the percentage of male employees in the factory is $80$.

Time taken for first half journey is, ${t_1} = (d/{v_1})$ and time taken for rest half journey is, ${t_2} = (d/{v_2})$

$\therefore$ ${V_{av}} = \frac{{2d}}{{(d/{v_1}) + (d/{v_2})}}$$ = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}$.

Let the given $11$ distinct positive integers are in increasing order

$x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$, so $x_{11}$ is largest of these integers and the median is $x_6$.

Now, median of first $10$ numbers is

$\frac{x_6+x_6}{2}=m$ (Let)

Now, we have to replace largest number $x_{11}$ by $m$ and then increasing order will be

$x_1, x_2, x_3, x_4, x_5, m, x_6, x_7, x_8, x_9, x_{10}$

$m < x_6$ as $x_5 < \frac{x_5+x_6}{2} < x_6$

So, median decreases.

Total number of students $=50$

Average marks of student $=47.5$

$\therefore$ Total marks of students

$=50 \times 47.5=2375$

Now, the student get integer marks Hence, the maximum number of students we will divide total mark by $48$.

$\frac{2375}{48}=49$

Four digits number which is divisible by $7$ are $1001,1008,1015, \ldots .$ $9996 .$

Hence, total number of such numbers $=1286$

$Median=\frac{\left(\frac{N}{2}\right)^{\text {th }} observatio+\left(\frac{N}{2}+1\right)^{\text {th }}\,observation}{2}$

$Median=\frac{\left(\frac{1286}{2}\right)^{\text {th }}\,observation+\left(\frac{N}{2}+1\right)^{\text {th }}\,observation}{2}$

$=\frac{643^{th}+644^{th}}{2}$

$=\frac{(1001+(642)7)+(1001)+(643)7)}{2}$

$=\frac{2(1001)+7(642+643)}{2}$

$=\frac{2(1001)+7(1285)}{2}$

$=1001+4497.5=5498.5$

Given, average income of $x$ people $=P$ and average income of $y$ people $=Q$

$\therefore$ Total income of people in two villages are $P_x$ and $Q_y$ respectively.

One person moves from first village to second village.

Then, number of people in first village

$=x-1$ and second village $=y+1$.

Average income $=P^{\prime}$ and $Q^{\prime}$

$\therefore$ Total income $=P^{\prime}(x-1)$ and $Q^{\prime}(y+1)$

Total income in both cases are same

$\therefore P x+Q y=P^{\prime}(x-1)+Q^{\prime}(y+1)$

$\Rightarrow P x-P^{\prime}(x-1)=Q^{\prime}(y+1)-Q y$

$\Rightarrow x\left(P-P^{\prime}\right)+P^{\prime}=y\left(Q^{\prime}-Q\right)+Q^{\prime}$

$\therefore P^{\prime} \neq P$ and $Q^{\prime} \neq Q$

Hence, option $(c)$ is correct.