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Clearly,

Sum of 11 numbers =

Also, Sum of first eight numbers =

 So, Sum of last three numbers = 880 – 628 = 252

Now, Let 10^th number = x

 then, 9^th number = x + 6

 and, 11^th number = x + 6

 ⇒ x+6 + x +x+6 = 252

 ⇒ 3x + 12 = 252

⇒ 3x = 240

⇒ x = 80

⇒ 9^th number =11^th number =  86

⇒ Required Average of the tenth and eleventh numbers =  $=\frac{80+86}{2}=83$

Given, 

$12 \times 39=468$

$\left.\begin{array}{l}5 \times 35=175 \\ 4 \times 50=200\end{array}\right]+=375$

Sum of 5^th, 6^th and 7^th number = 468 - 375 = 93

3x-27= 93

$3 x=120$

$x=40$

According to the question,

Required Average

$=\frac{5^{\text {th }} \text { number }+6^{\text {th }} \text { number }}{2}$
$\Rightarrow \frac{30+40}{2}=35$

Let the weight of 1 student $=x \mathrm{~kg}$
the weight of 15 students $=15 x \mathrm{~kg}$
The weight of new student $=\mathrm{y} \mathrm{kg}$
ATQ,
$15 x-45+y=15(x+1.5)$
$15 x-45+y=15 x+22.5$
$y=67.5 \mathrm{~kg}$

ALTERNATE METHOD

ATQ,
Average weight of the 9 boatsmen increased
by $=1 \frac{1}{3}$ years
Total increased in weight $=9 \times \frac{4}{3}=12 \mathrm{~kg}$
Weight of new man $=60+12=72 \mathrm{~kg}$

ATQ,
Average weight of the 15 crewmen increased by $=\frac{1}{3} \mathrm{~kg}$
$\therefore$ Total increase in weight $=15 \times \frac{1}{3}=5 \mathrm{~kg}$
Weight of oldman $=55 \mathrm{~kg}$
Weight of new man $=55+5=60 \mathrm{~kg}$

$\begin{aligned}
&\frac{50+70+40+20+a+b}{6}=50 \\
&a+b=300-180 \\
&a+b=120 \\
&\frac{40+45+50+55+40+c+d}{7}=55 \\
&c+d=385-230
\end{aligned}$
$c+d=155$
So, $\frac{a+b+c+d}{4}=\frac{120+155}{4}=\frac{275}{4}=68.75$

Let the run rate for the last 40 overs was $\mathrm{x}$, then
$
\begin{aligned}
&10 \times 6.4+x \times 40=264 \\
&40 x=264-64 \\
&40 x=200 \\
&x=5
\end{aligned}
$

Let there are $\mathrm{n}$ observations, then
Sum of $\mathrm{n}$ observations $=54 \times n=54 n$
Now, each observation is increased by 8 , So
New sum of observations $=54 n+8 \times n=54 n+8 n=62 n$
It is then divided by 2 , So
Result $=\frac{62 n}{2}=31 n$
Hence, Mean $=\frac{31 n}{n}=31$

Sum of all 15 numbers $=15 \times 55.4=831$
Sum of the first 6 numbers $=6 \times 50=300$
Sum of the next 4 numbers $=4 \times 60.5=242$
So,
Sum of remaining 5 numbers $=831-(300+242)=831-542=289$
Hence, Average of remaining 5 numbers $=\frac{289}{5}=57.8$

Increase in average $=0.5$
Increase in total $=92-78$
$=14$
Total no of students $=\frac{14}{0.5}$
$=28$

The sum of weight of $\mathrm{X}$ and $\mathrm{Y}$ is $=40 \times 2=80 \mathrm{~kg}$
The sum of weight of $\mathrm{Y}$ and $\mathrm{R}$ is $=62 \times 2=124 \mathrm{~kg}$
The sum of weight of $\mathrm{X}$ and $\mathrm{R}$ is $=56 \times 2=112 \mathrm{~kg}$
Adding (1),(2) and (3), we get
$
\begin{aligned}
&2(\mathrm{X}+\mathrm{Y}+\mathrm{R})=316 \\
&(\mathrm{X}+\mathrm{Y}+\mathrm{R})=158
\end{aligned}
$
Putting the value of (2) in eq. (4)
$
\begin{aligned}
&(\mathrm{X}+124)=158 \\
&\mathrm{X}=158-124=34 \mathrm{~kg}
\end{aligned}
$

Let $x$ be the average after $17^{\text {th }}$ innings
According to question
$
\begin{aligned}
&\frac{17 x+80}{18}=x+4 \\
&\Rightarrow 17 x+80=18 x+72
\end{aligned}
$
$
\Rightarrow \mathrm{x}=8
$
So, average after the $18^{\text {th }}$ innings $=8+4=12$

Overall increase in total weight $=$ increase in avg weight $\times$ no of persons
$
\begin{aligned}
&=2.8 \times 5 \\
&=14 \mathrm{Kg}
\end{aligned}
$
Weight of new person $=38 \mathrm{Kg}+14 \mathrm{Kg}$
$
=52 \mathrm{Kg}
$

Average of 19 students $=74$
one student were recorded as 65 in place of 56
So that total marks will be 9 less
So the actual average $=74-\frac{9}{19}$
$=74-0.5$
$=73.5$

Increase in total marks of all 30 students $=65-56=9$ marks
Decrease in total marks of all 30 students $=84-48=36$ marks
Overall change $=+9-36$
$=27$ decrease
Now the correct avg of all students $=76-\frac{27}{30}$
$=75.1$

Prime no between 25 and $45=29,31,37,41,43$
Average of these numbers $=\frac{29,+31+37+41+43}{5}$
$
\begin{aligned}
&=\frac{181}{5} \\
=& 36 \frac{1}{5}
\end{aligned}
$

$\begin{aligned} \text { Required average } &=\frac{20 \times 152+15 \times 168}{35} \\ &=\frac{3040+2520}{35} \\ &=\frac{5560}{35}=158.85 \simeq 159 \end{aligned}$

$
\begin{aligned}
\text { Required average } &=\frac{56 \times 55-12 \times 60+6 \times 52.5}{56-12+6} \\
&=\frac{3080-720+315}{50} \\
&=\frac{2675}{50}=53.5 \mathrm{~kg}
\end{aligned}
$
Average weight of the remaining students decreases by $=53.5-55=1.5 \mathrm{~kg}$

$\begin{aligned} \text { Required average } &=\frac{24 \times 72-3 \times 78+4 \times 75}{24-3+4} \\ &=\frac{1728-234+300}{25} \\ &=\frac{1794}{25}=71.76 \end{aligned}$

If the incoming person was of the same weight as the outgoing person, there would be no change in the average weight, but due to the arrival of the new person, the average weight increased by 1.5 kg, which means that its weight was equal to that of the outgoing person. Also increases the weight of all by 1.5 kg

therefore

Weight of the new person $=65+6 \times 1.5=65+9=74 \mathrm{Kg}$