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The average of eleven numbers is 80. The average of the first four numbers is 74.5 and that of the next four numbers is 82.5. The nineth number is 6 more than the tenth number, and the tenth number is 6 less than the eleventh number. What is the average of the tenth and eleventh numbers?

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 = $=\frac{80+86}{2}=83$

The average of 12 number is 39 . The average of last 5 number is 35 , and that of the first four number is 40 . The fifth number is 10 less than the sixth number and 7 more than the seventh number. The average of the fifth and sixth number is:

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

$=\frac{5^{\text {th }} \text { number }+6^{\text {th }} \text { number }}{2}$

$\Rightarrow \frac{30+40}{2}=35$

The average weight of 15 students in a class increase by $1.5 \mathrm{~kg}$ when one of the students weighting $45 \mathrm{~kg}$ is replaced by a new student. What is the weight (in $\mathrm{kg}$ ) of the new student?

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**

The average weight of the 9 persons in boat is increased by $1 \frac{1}{3} \mathrm{~kg}$ when one of the crew of weight $60 \mathrm{~kg}$ is replaced by a new person. The weight of new person is:

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 $=60+12=72 \mathrm{~kg}$

The average weight of 15 crewman in a boat increased by $\frac{1}{3} \mathrm{~kg}$ when one of the crewmen whose weight is $55 \mathrm{~kg}$ is replaced by new man. What is the weight of that new man.

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 $=55+5=60 \mathrm{~kg}$

The average of 50,70,40,20, a, and b is 50 and the average of 40,45,50,55,40, c, and d is 55 . What is the average of a, b, c, and d ?

$\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, $\frac{a+b+c+d}{4}=\frac{120+155}{4}=\frac{275}{4}=68.75$

In the first 10 overs of a cricket game, the run rate was only 6.4. What should be the average run rate in the remaining 40 overs to reach the target of 264 runs?

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 $=8+4=12$

The average weight of 5 persons increases by 2.8 Kg when a new person comes in place of one of them weighing 38 Kg. The weight of the new person is

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 $=53.5-55=1.5 \mathrm{~kg}$

The average weight of 24 persons in a group is 72 kg. If 3 persons with average weight 78 kg leave the group and 4 persons of average weight 75 kg join the group, then what will be the average weight (in kg ) of the persons in the group now?

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

The average weight of 6 persons increases by 1.5 kg when a new person comes in place of one of them who was weighing 65 kg. What is the weight of the new person?

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 $=65+6 \times 1.5=65+9=74 \mathrm{Kg}$