[3 marks sum]

Question 13 Marks

If the mean of 8, 10, 7, x + 2 and 6 is 9, find the value of x.

Answer

The mean $8,10,7, x+2$ and 6 is 9
$ \begin{aligned} & \therefore \frac{8+10+7+\mathrm{x}+2+6}{5}=9 \\ & \Rightarrow \frac{\mathrm{x}+33}{5}=9 \\ & \Rightarrow \mathrm{x}+33=9 \times 5 \\ & \Rightarrow x=45-33 \\ & \Rightarrow x=12 \end{aligned} $

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Question 23 Marks

The pocket expenses (per day) of Anuj, during a certain week, from Monday to Saturday were ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25. Find the mean pocket expenses per day.

Answer

The pocket expenses (per day) during a certain week are: ₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60 and ₹87.25
$\therefore$ Mean of pocket expenses per day $=\frac{85.40+88.00+86.50+84.75+82.60+87.25}{6}$ (Here n $=6)$
$ =\frac{514.5}{6} $
$ =\text₹ 85.75 $
$\therefore$ Anuj expenses per day $=\text₹ 85.75$

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Question 33 Marks

The mean of a certain number of observations is 32. Find the resulting mean, if the observation is decreased by 20%.

Answer

Observation is decreased by $20 \%$.
The new mean of observation $=32-20 \%$ of the 32
$ \begin{aligned} & =32-\frac{20 \times 32}{100} \\ & =32-6.4 \\ & =25.6 \end{aligned} $

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Question 43 Marks

Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.

Answer

Mean of $x+3, x+5, x+7, x+9$ and $x+11$
$ \begin{aligned} & \therefore \text { Mean }=\frac{(\mathrm{x}+3)(\mathrm{x}+5)(\mathrm{x}+7)(\mathrm{x}+9)(\mathrm{x}+11)}{2} \ldots \ldots . .(\text { Here } \mathrm{n}=5) \\ & =\frac{5 \mathrm{x}+35}{5} \\ & =\frac{5(\mathrm{x}+7)}{5} \\ & =\mathrm{x}+7 \end{aligned} $

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Question 53 Marks

Find the mean of all factors of 10.

Answer

Factors of $10=1,2,5,10$
$ \text { Mean }=\frac{\text { Sum of observations }}{\text { No. of observations }} $
Sum of observations $=1+2+5+10=18$
No. of observations $=4$
Mean $=\frac{18}{4}=\frac{9}{2}=4.5$

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Question 63 Marks

Find the median of 3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9 and 7.6.

Answer

$3.6,9.4,3.8,5.6,6.5,8.9,2.7,10.8,15.6,1.9$ and 7.6.
Arranging in ascending order: $1.9,2.7,3.6,3.8,5.6,6.5,7.6,8.9,9.4,10.8,15.6$
Here, number of terms $=11$ which is odd
$ \begin{aligned} & \therefore \text { Median }=\frac{\mathrm{n}+1}{2}=\frac{11+1}{2}=6 \text { th term } \\ & =6 \text { yh term }=6.5 \end{aligned} $
Hence, median $=6.5$

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Question 73 Marks

Find the median of 9, 3, 20, 13, 0, 7 and 10

Answer

$9,3,20,13,0,7$ and 10
Arranging in ascending order: $0,3,7,9,10,13,20$
Here, number of terms $=7$ which is odd
$\therefore$ Median $=\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}$ th term $=4$ th term $=9$
Hence, median $=9$

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Question 83 Marks

Find the median of 5, 7, 9, 11, 15, 17, 2, 23 and 19.

Answer

$5,7,9,11,15,17,2,23$ and 19
Arranging in ascending order: 2, 5, 7, 9, 11, 15, 17, 19, 23
Here, number of terms $=9$ which is odd
Median $=\frac{\mathrm{n}+1}{2}=\frac{9+1}{2}$ th term $=5$ th term $=11$
Hence, median $=11$

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Question 93 Marks

The mean of 5 numbers is 27. If one new number is included, the new mean is 25. Find the included number.

Answer

Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On including an observation the mean of 6 observation = 25 × 6 = 150
⇒ Included observations = Total Mean of 6 observations – Total mean of 5 observations
= 150 − 135 = 15

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Question 103 Marks

The mean of five numbers is 27. If one number is excluded, the mean of the remaining numbers is 25. Find the excluded number.

Answer

Mean of 5 observations = 27
Total sum of 5 observations = 27 × 5 = 135
On excluding an observation, the mean of remaining 6 observations = 25
⇒ Total of remaining 4 observations = 25 × 4 = 100
⇒ Included observation = Total mean of 5 observations – Total mean of 4 observations
= 135 − 100 = 35

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Question 113 Marks

The mean of marks scored by 100 students was found to be 40, later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.

Answer

Mean of 40 observations $=100$
Total sum of 40 observations $=100 \times 40=4000$
Incorrect total of 40 observation is $=4000$
Correct total of 40 observations $=4000-83+53=3970$
$\therefore$ Correct mean $=\frac{3970}{100}=39.70$

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Question 123 Marks

If the mean of 6, 4, 7, p and 10 is 8, find the value of p.

Answer

The mean of $6,4,7, p$ and 10 is 8
$ \begin{aligned} & \therefore \text { Mean }=\frac{6+4+7+p+10}{5}=8 \\ & \Rightarrow 27+p=40 \\ & \Rightarrow p=40-27 \\ & \Rightarrow p=13 \end{aligned} $

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MATHS [3 marks sum]

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