RS AGGARWAL CLASS 9 CHAPTER 18 MEAN, MEDIAN AND MODE OF UNGROUPED DATA EXERCISE 18D

  EXERCISE 18D

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Question 1:

Find the mode of the following items.
0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6

Answer 1:

On arranging the items in ascending order, we get:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Clearly, 6 occurs maximum number of times.
∴ Mode = 6

Question 2:

Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20

Answer 2:

On arranging the values in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
Clearly, 25 occurs maximum number of times.
∴ Mode = 25

Question 3:

Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9

Answer 3:

On arranging the shoe sizes in ascending order, we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Clearly, 9 occurs maximum number of times.
∴ Mode = 9

Question 4:

A cricket player scored the following runs in 12 one-day matches:
50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.
Find his modal score.

Answer 4:

On arranging the runs in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Clearly, 50 occurs maximum number of times.
∴ Modal score = 50

Question 5:

If the mean of the data 3, 21, 25, 17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x, find the mode of the data.

Answer 5:

We know that,

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

The given data is 3, 21, 25, 17, (x + 3), 19, (x – 4).

Mean of the given data = $\frac{3+21+25+17+\left(x+3\right)+19+\left(x-4\right)}{7}$
$$\begin{gathered} \Rightarrow 18\left(7\right)=84+2x \\ \Rightarrow 126-84=2x \\ \Rightarrow 2x=42 \\ \Rightarrow x=21 \end{gathered}$$

Hence, the value of x is 21.

Now, the given data is 3, 21, 25, 17, 24, 19, 17
Arranging this data in ascending order:
3, 17, 17, 19, 21, 24, 25

Here, 17 occurs maximum number of times.
∴ Mode = 17

Hence, the mode of the data is 17.

Question 6:

The numbers 52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 have been arranged in an ascending order and their median is 55. Find the value of x and hence find the mode of the given data.

Answer 6:

Arranging the given data in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57

Number of terms = 9 (odd)

$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 55={\left(\frac{9+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 55={\left(5\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 55=2x+1 \\ \Rightarrow 2x=55-1 \\ \Rightarrow 2x=54 \\ \Rightarrow x=27 \end{gathered}$$

Hence, the value of x is 27.

Arranging the given data in ascending order:
52, 53, 54, 54, 55, 55, 55, 56, 57

Here, 55 occurs maximum number of times.
∴ Mode = 55

Hence, the mode of the data is 55.

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Question 7:

For what value of x is the mode of the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value of x, find the median.

Answer 7:

Given: Mode = 25
∴ 25 occurs maximum number of times.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, x + 3, 25, 26, 27, 40

∴ x + 3 = 25
⇒ x = 25 − 3
⇒ x = 22

Hence, the value of x is 22.

Arranging the given data in ascending order:
15, 20, 22, 23, 24, 25, 25, 26, 27, 40

Number of terms = 10 (even)

$$\begin{gathered} \therefore \mathrm{Median}=\mathrm{mean} \mathrm{of} \left[{\left(\frac{n}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{n}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ =\mathrm{mean} \mathrm{of} \left[{\left(\frac{10}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{10}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ =\mathrm{mean} \mathrm{of} \left[{\left(5\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(6\right)}^{\mathrm{th}} \mathrm{term}\right] \\ =\mathrm{mean} \mathrm{of} \left[24 \mathrm{and} 25\right] \\ =\frac{24+25}{2} \\ =\frac{49}{2} \\ =24.5 \end{gathered}$$

Hence, the median is 24.5 .

Question 8:

The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47 have been arranged in an ascending order and their median is 45. Find the value of x. Hence, find the mode of the above data.

Answer 8:

Arranging the given data in ascending order:
42, 43, 44, 44, (2x + 3), 45, 45, 46, 47

Number of terms = 9 (odd)

$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 45={\left(\frac{9+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 45={\left(5\right)}^{\mathrm{th}} \mathrm{term} \\ \Rightarrow 45=2x+3 \\ \Rightarrow 2x=45-3 \\ \Rightarrow 2x=42 \\ \Rightarrow x=21 \end{gathered}$$

Hence, the value of x is 21.

Arranging the given data in ascending order:
42, 43, 44, 44, 45, 45, 45, 46, 47

Here, 45 occurs maximum number of times.
∴ Mode = 45

Hence, the mode of the data is 45.

MULTIPLE CHOICE QUESTIONS

Question 1:

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
(a) 5
(b) 6
(c) 7
(d) 8

Answer 1:

(c) 7

Mean of 5 observations = 11
We know:
$$\begin{gathered} \mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations} } \\ \Rightarrow 11=\frac{x+x+2+x+4+x+6+x+8}{5} \\ \Rightarrow 11=\frac{5x+20}{5} \\ \Rightarrow 5x+20=55 \\ \Rightarrow 5x=35 \\ \Rightarrow x=7 \end{gathered}$$

Question 2:

If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(a) $10\frac{1}{3}$
(b) $10\frac{2}{3}$
(c) $11\frac{1}{3}$
(d) $11\frac{2}{3}$

Answer 2:

(c) 11$\frac{1}{3}$
Mean of 5 observations = 9
We know:
$$\begin{gathered} \mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow 9=\frac{x+x+3+x+5+x+7+x+10}{5} \\ \Rightarrow 9=\frac{5x+25}{5} \\ \Rightarrow 5x+25=45 \\ \Rightarrow 5x=20 \\ \Rightarrow x=4 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{last} \mathrm{three} \mathrm{observations} \mathrm{are} (4+5), (4+7) \mathrm{and} (4+10), \mathrm{i}.\mathrm{e}., 9, 11 \mathrm{and} 14. \\ \mathrm{Now}, \\ \mathrm{Mean} \mathrm{of} \mathrm{the} \mathrm{last} \mathrm{three} \mathrm{terms}=\frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3} \end{gathered}$$

Question 3:

If $\overline{x}$ is the mean of $x_1, x_2, x_3, ..., x_n,$ then $\sum _{i=1}^n(x_i - \overline{x}) = ?$
(a) −1
(b) 0
(c) 1
(d) n − 1

Answer 3:

(b) 0

$$\begin{gathered} If \overline{x } \mathrm{is} \mathrm{the} \mathrm{mean} \mathrm{of} x_1, x_2, x_3, x_{4,...}{x}_{n,} \mathrm{then} \mathrm{we} \mathrm{have}: \\ \sum _{i=1}^n{x}_i=\overline{x } \\ \mathrm{O}r, \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x} }=0 \end{gathered}$$

Question 4:

If each observation of the data is increased by 8, then their mean
(a) remains the same
(b) is decreased by 8
(c) is increased by 5
(d) becomes 8 times the original mean

Answer 4:

(b) is decreased by 8

$\mathrm{Let} \mathrm{the} \mathrm{numbers} \mathrm{be} x$₁, x_2,...x_n.
$\mathrm{Hence}, \mathrm{mean} = \frac{x_1+x_2+....+x_n}{n}$

Now the new numbers after decreasing every number by 8 : (x₁−8) , (x₂−8)...,(x_n−8)

$\mathrm{New} \mathrm{Mean} = \frac{\left(x_1-8\right)+\left(x_2-8\right)+....+\left(x_n-8\right)}{n}$

           $= \frac{x_1+x_2+....+x_n-8n}{n}$

           $$\begin{gathered} = \frac{x_1+x_2+....+x_n}{n}-8 \\ \therefore \mathrm{New} \mathrm{mean} = \mathrm{mean} - 8 \end{gathered}$$          

Hence, mean is decreased by 8.

Question 5:

The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
(a) 52 kg
(b) 52.8 kg
(c) 53 kg
(d) 47 kg

Answer 5:

(c) 53 kg

Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.

$$\begin{gathered} \mathrm{We} \mathrm{know}: \\ \mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations} }{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations} } \\ =\frac{51+45+49+46+44+x}{6} \\ =\frac{235+x}{6} \\ \mathrm{Given}: \\ \mathrm{Mean}=48 \mathrm{kg} \\ \Rightarrow \frac{235+x}{6}=48 \\ \Rightarrow 235+x=288 \\ \Rightarrow x=53 \\ \mathrm{Hence}, \mathrm{the} \mathrm{weight} \mathrm{of} \mathrm{the} 6\mathrm{th} \mathrm{boy} \mathrm{is} 53 \mathrm{kg}. \end{gathered}$$

Question 6:

The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
(a) 38.6
(b) 39.4
(c) 39.8
(d) 39.2

Answer 6:

(b) 39.4

Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = $\left(39\times 50\right)=1950$
Correct sum = (1950 + 43 $-$ 23) = 1970
$\therefore \mathrm{Mean}=\frac{1970}{50}=39.4$