[5 marks sum]

Question 15 Marks

Find the mean and median of the data $: 35, 48, 92, 76, 64, 52, 51, 63$ and $71.$If $51$ is replaced by $66,$ what will be the new median?

Answer

Let $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3, ...., x_n$.
Mean $=\frac{x_1+x_2+x_3+\ldots \ldots \ldots+x_n}{n}$
Therefore,
Mean of given data $=\frac{35+48+92+76+64+52+51+63+71}{9}$
$=\frac{552}{9}$
$=61.33$
Let us rewrite the given data in ascending order:
Thus, we have
$35, 48, 51, 52, 63, 64, 71, 76, 92$
There are $9$ observations, which is odd.
Therefore, median = $\left(\frac{n+1}{2}\right)^{\text {th }}$ Observation
$\Rightarrow$ Median$=\left(\frac{9+1}{2}\right)^{\text {th }}$Observation
$\Rightarrow$ Median$=\left(\frac{10}{2}\right)^{\text {th }}$ Observation
$\Rightarrow$ Median $= 5^{th}$ Observation
$\Rightarrow$ Median $= 63.$
If $51$ is replaced by $66$, the new set of data in ascending order is: $35, 48, 52, 63, 64, 66, 71, 76, 92$
Since median $= 5^{th}$ observation,
We have a new median $= 64.$

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Question 25 Marks

Find the mean of the following data: $30, 32, 24, 34, 26, 28, 30, 35, 33, 25.(i)$ Show that the sum of the deviations of all the given observations from the mean is zero.$(ii)$Find the median of the given data.

Answer

Let $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3,,\dots ...., x_n$
Mean $=\frac{x_1+x_2+x_3+\ldots \ldots+x_n}{n}$
Therefore,
Mean of given data = $\frac{30+32+24+34+26+28+30+35+33+25}{10}$
$=\frac{297}{10}$
$=29.7$
$(i)$ Let us tabulate the observations and their deviations from the mean

Observation $x_i$	Deviation$x_i-\bar{x}$
$30$	$0.3$
$32$	$2.3$
$24$	$-5.7$
$34$	$4.3$
$26$	$-3.7$
$28$	$-1.7$
$30$	$0.3$
$35$	$5.3$
$33$	$3.3$
$25$	$-4.7$
Total	$0$

From the table, it is clear that the sum of the deviations from the mean is Zero.
$(ii)$ Consider the given data :
$30, 32, 24, 34, 26, 28, 30, 35, 33, 25.$
Let us rewrite the above data in ascending order
$24, 25, 26, 28, 30, 30, 32, 33, 34, 35.$
There are 10 observations, which is even.
Therefore,
Median $=\frac{1}{2}\left[\left(\frac{n}{2}\right)^{\text {th }}\right.$ term $+\left(\frac{n}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}[\left(\frac{10}{2}\right)^{\text {th }}$term$+(\frac{10}{2}+1)^{\text {th }}$term$]$
$=\frac{1}{2}[(5)^{\text {th }}$term$+\left(\frac{10}{2}+1\right)^{\text {th }}$term$]$
$=\frac{1}{2}[5^{\text {th }}$term$+(5+1)^{\text {th }}$term$]$
$=\frac{1}{2}[5^{\text {th }}$term$+6^{\text {th }}$term$]$
$=\frac{1}{2}[30+30]$
$=\frac{1}{2}[60]$
$=30 .$

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Question 35 Marks

The mean of $10$ numbers is $24$. If one number is included, the new mean is $25$. Find the included number.

Answer

Let $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3, ,\dots...., x_n$
Mean of given data $=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$
Given that mean of $10$ numbers is $24.$
That is,
$\frac{x_1+x_2+x_3+\ldots+x_{10}}{10}=24$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_{10} = 10 \times 24$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_{10} = 240$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_{10} + x_{11} = 240 + x_{11},\dots ....(1)$
Also, given that mean of $11$ number is $25.$
That is,
$\frac{x_1+x_2+x_3+\ldots+x_{10}+x_{11}}{11}=25$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_{10} + x_{11} = 11 \times 25$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_{10} + x_{11} = 275 ,\dots....( 2 )$
From equations $(1)$ and $(2)$, we have :
$x_1 + x_2 + x_3 + ,\dots... + x_{10} + x_{11} = 240 + x_{11} = 275$
$240 + x_{11} = 275$
$\Rightarrow x_{11} = 275 - 240 = 35$

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Question 45 Marks

The mean of $6$ numbers is $42$. If one number is excluded, the mean of the remaining number is $45$. Find the excluded number.

Answer

Let $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3, ,\dots...., x_n$
Mean of given data $=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$
Given that mean of $6$ number is $42.$
That is,
$\frac{x_1+x_2+x_3+\ldots+x_6}{6}=42$
$\Rightarrow x_1 + x_2 + x_3 + ,\dots... + x_6 = 6 x 42$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 252 - x_6,\dots...(1)$
Also, given that the mean of $5$ number is $45.$
That is,
$\frac{x_1+x_2+x_3+x_4+x_5}{5}=45$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 5 x 45$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 225 .,\dots...(2)$
From equation $(1)$ and $(2)$, we have
$x_1 + x_2 + x_3 + x_4 + x_5 = 252 - x_6 = x_1 + x_2 +x_3 + x_4 + x_5 = 225$
$252 - x_6 = 225$
$\Rightarrow x_6 = 252 - 225 = 27$

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Question 55 Marks

The heights $($in $cm)$ of the volleyball players from team $A$ and team $B$ were recorded as:Team $A:180, 178, 176, 181, 190, 175, 187;$Team $B:174, 175, 190, 179, 178, 185, 177;$Which team had a greater average height?Find the median of team $A$ and team $B.$

Answer

Total number of players in each team $= 7$
Mean height of team $ A=\frac{\text { Sum of the height of players of the team A }}{\text { Total number of team A players }}$
$=\frac{180+178+176+181+190+175+187}{7}$
$=\frac{1267}{7}$
$= 181 \ cm$
Mean height of team $B =\frac{\text { Sum of the height of players of the team B }}{\text { Total number of team B players }}$
$=\frac{174+175+190+179+178+185+177}{7}$
$=\frac{1258}{7}$
$= 179.7 \ cm$
Thus, team $A$ has greater average height.
Median of team $A$:
Arranging heights in ascending order, we get
$175, 176, 178, 180, 181, 187, 190$
Total number of observations $= n = 7 ($odd$)$
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ Observation $=\left(\frac{7+1}{2}\right)^{\text {th }}$ Observation $=4^{\text {th }}$ Observation $=180 \ cm$
Median of team $B:$
Arranging heights in ascending order, we get
$174, 175, 177, 178, 179, 185, 190$
Total number of observations $= n = 7 ($odd$)$
$\therefore$ Median = $=\left(\frac{n+1}{2}\right)^{\text {th }}$ Observation $=\left(\frac{7+1}{2}\right)^{\text {th }}$ Observation $=4^{\text {th }}$ Observation $=178 \ cm$

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Question 65 Marks

The mean weight of $60$ students in a class is $40 \ kg$. The mean weight of boys is $50 \ kg$ while that of girls is $30 \ kg$. Find the number of boys and girls in the class.

Answer

Total number of students $= 60$
Mean weight of $60$ students $= 40$
Let the number of boys $= x$
Then, the number of girls $= 60 - x$
Mean weight of boys $=\frac{\text { Total weight of boys }}{\text { Total number of boys }}$
$\Rightarrow 50=\frac{\text { Total weight of boys }}{x}$
$\Rightarrow$ Total weight of boys $= 50x$
Mean weight of girls $=\frac{\text { Total weight of girls }}{\text { Total number of girls }}$
$\Rightarrow 30=\frac{\text { Total weight of girls }}{60-x}$
$\Rightarrow$ Total weight of girls $= 30( 60 - x )$
Now,
Mean weight of $60$ students $=\frac{\text { Total weight of boys }+ \text { Total weight of girls }}{\text { Total number of students }}$
$\Rightarrow 40=\frac{50 x+30(60-x)}{60}$
$\Rightarrow 2400 = 50x + 1800 - 30x$
$\Rightarrow 20x = 600$
$\Rightarrow x = 30$
$\Rightarrow 60 - x = 60 - 30 = 30$
Hence, the number of boys is $30$ and the number of girls is also $30.$

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Question 75 Marks

Find the mean and median of all the positive factors of $72.$

Answer

Let us find the factors of $72:$
$72 = 1 \times 72$
$= 2 \times 36$
$= 3 \times 24$
$= 4 \times 18$
$= 6 \times 12$
$= 8 \times 9$
$= 9 \times 8$
$= 12 \times 6$
$= 18 \times 4$
$= 24 \times 3$
$= 36 \times 2$
$= 72 \times 1$
Therefore, the data set is $:1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$
Mean of the above data set $=\frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}$
$=\frac{195}{12}$
$=16.25$
Since the number of observation is $12,$ which is even,
median is given by
Median$=\frac{1}{2}[\left(\frac{n}{2}\right)^{\text {th }}$term$+(\frac{n}{2}+1)^{\text {th }}$term$]$
$=\frac{1}{2}[(\frac{12}{2})^{\text {th }}$term$+(\frac{12}{2}+1)^{\text {th }}$term$]$
$=\frac{1}{2}[6^{\text {th }}$term$+7^{\text {th }}$term$]$
$=\frac{1}{2}[8+9]$
$=\frac{1}{2} \times 17$
$=8.5$

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MATHEMATICS [5 marks sum]

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