Maths M.C.Q

2. 7

Solution:

Given,

Mid value of the class = 10

Width of each class = 6

Now,

Let the lower limit be x.

We know,

Upper limit = Lower limit + class size

 = x + 6

Also,

Mid value $=\frac{\text{x}+\text{x}+6}{2}$

$=\frac{2\text{x}+6}{2}=\text{x}+3$

$\Rightarrow\text{x}+3=10$

$\Rightarrow\text{x}=7$

Thus, the lower limit is 7.

1. $\frac{\text{x}+\text{y}}{2}$

Solution:

Since x¯ and y¯ are two numbers, though being means, their arithmetic mean is given by:

$\text{z}=\frac{\text{x and y}}{2}$

3. 6

Solution:

First, we arrange the given numbers in ascending order is,

3, 4, 4, 5, 6, 7, 7, 7 and 12

Here, n = 9

Since, n is odd, so we use the formula for median,

Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$

$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put n = 9]

$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$

$=5^{\text{th}}\text{observation}=6$ 

4. 8

Solution:

Arrange the given data in ascending order.

7, 7, 7, 8, 9, 10, 13

For odd number (n) of observation median $=\Big[\frac{(\text{n+1)}}{2}\Big]\text{th}$ term,

Here n = 7 so median $=\Big[\frac{{(7+1)}}{2}\Big]\text{th}$ term = 4th term that is 8,

Hence median = 8

3. 6

Solution:

Arranging the points in an ascending order,

We have:

3, 4, 4, 5, 6, 7, 7, 7, 12

Here, n = 9, Which is odd

$\therefore\ $median score = value of $\frac{1}{2}(9+1)^\text{th}$ terms

= value of $\Big(\frac{1}{2}\times10\Big)^\text{th}$ term

= value of 5^th term

= 6

2. $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$

Solution:

Mean of ax₁, ax₂, ..., ax_n, is $\text{a}\bar{\text{x}}$

Mean of $\frac{\text{x}_1}{\text{a}},\ \frac{\text{x}_2}{\text{a}},\ ...,\ \frac{\text{x}_\text{n}}{\text{a}}$ is $\frac{1}{\text{a}}\bar{\text{x}}$

so the their mean is $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}.$

4. 55.5

Solution:

upper limit - lower limit = class width

upper limit - lower limit = 10

$\frac{\text{(upper limit + lower limit)}}{2}=\text{mid}-\text{value}$

upper limit + lower limit = 2 × 60.5

upper limit + lower limit = 121

By solving the above two equations, we get

upper limit = 65.5

Lower limit = 55.5

4. Mean, Median and Mode of A are equal.

Solution:

$\text{Mean of A}=\frac{1+2+2+3+3+3+7}{7}$

$=\frac{21}{7}=3$

$\text{Mean of B}=\frac{3+5+5+7+8+9+12}{7}$

$=\frac{49}{7}=7$

$\text{Mean of C}=\frac{2+3+4+4+4+7+11}{7}$

$=\frac{35}{7}=5$

Mode of A = 3; Median of A = 3

Mode of B = 5; Median of B = 7

Mode of C = 4; Median of C = 4

(Mean of A = 3) $\neq$ (Mode of C = 4)

(Mean of C = 5) $\neq$ (Median of B = 4)

(Median of B = 7) $\neq$ (Mode of A = 3)

Mean of A = 3, Mode of A = 3, Median of A = 3

1. $\text{x}+\frac{\text{n}+1}{2}$

Solution:

If the first item is increased by 1, second by 2, third by 3 and so on,

So the new sequence thus formed is x₁ + 1, x₂ + 2, x₃ + 3 ...... x_n + n

So the new mean is increased by $\frac{\text{n}+1}{2}$

So the new mean is $\text{x}+\frac{\text{n}+1}{2}$

Where x is the mean of first n numbers.

4. 56.5

Solution:

Given, n = 50, then mean $\bar{\text{x}}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$

Then, $\bar{\text{x}}=\frac{1}{50}\times\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}$

$\Rightarrow\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}=50\bar{\text{x}}$

Now, subtract each observation from 53, we get a new mean say $\bar{\text{x}}_\text{new}$

$\therefore\ \bar{\text{x}}_\text{new}=\frac{(-\text{x}_1+53)+(-\text{x}_2+53)\ +\ .....\ +(-\text{x}_{50}+53)}{50}$

$\Rightarrow-3.5=\frac{-(\text{x}_1+\text{x}_2\ +\ .....\ +\text{x}_{50})+(53+53\ +\ ....\ +50\text{times})}{50}$

$\Rightarrow-3.5\times50=(-\text{x}_1+\text{x}_2+\ ....\ +\text{x}_{50})+53\times50$

$\therefore$ Mean of 50 observation $=\frac{1}{50}\sum\limits^{50}_{\text{i}=1}\text{x}_{\text{i}}$

$=\frac{1}{50}\times2852=56.5$ $\Bigg[\because\text{ mean}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}\Bigg]$

Hence, the mean of given number is 56.5

3. $11\frac{1}{3}$

Solution:

Mean of 5 observations = 9

$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$\Rightarrow9=\frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}$

$\Rightarrow9=\frac{5\text{x}+18}{5}$

$\Rightarrow45=5\text{x}+18$

$\Rightarrow5\text{x}=20$

$\Rightarrow\text{x}=4$

So, the mean of the last three observations

$=\frac{\text{x}+5+\text{x}+7+\text{x}+10}{3}$

$=\frac{4+5+4+7+4+10}{3}$

$=\frac{34}{3}$

$=11\frac{1}{3}$

2. 2m - l

Solution:

Let x and y be the lower and upper class limit of a continuous frequency distribution.

Now, mid-point of a class $=\frac{(\text{x}+\text{y})}{2}=\text{m}$ [given]

⇒ x + y = 2m = x + l = 2m

$\big[\therefore$ y = l = upper class limit (given)$\big]$

⇒ x = 2m - l

Hence, the lower class limit of the class is 2m - l.

2. $\frac{1}{2}(\bar{\text{x}}+\bar{\text{y}})$

Solution:

Since $\bar{\text{z}}$ is the mean of x₁, x₂, ..., x_n, y₁, y₂, ..., y_n,

$\bar{\text{z}}=\frac{(\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n})+(\text{y}_1+\text{y}_2+{\dots}\text{y}_\text{n})}{2\text{n}}$

Since $\bar{\text{x}}$ is the mean of x₁, x₂, ..., x_n,

$\bar{\text{x}}=\frac{\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n}}{\text{n}}$

$\Rightarrow\text{x}_1+\text{x}_2+_{\dots}+\text{x}_\text{n}=\text{n}\bar{\text{x}}$

Since $\bar{\text{y}}$ is the mean of y₁, y₂, ..., y_n,

$\bar{\text{y}}=\frac{\text{y}_1+\text{y}_2+_{\dots}+\text{y}_\text{n}}{\text{n}}$

$\Rightarrow\text{y}_1+\text{y}_2+_{\dots}+\text{y}_\text{n}=\text{n}\bar{\text{y}}$

so,

$\bar{\text{z}}=\frac{\text{n}\bar{\text{x}}}{2\text{n}}+\frac{\text{n}\bar{\text{y}}}{2\text{n}}=\frac{\text{n}\bar{\text{x}}+\text{n}\bar{\text{y}}}{2\text{n}}$

$=\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$

3. 2

Solution:

Adjusted Frequency $=\Big(\frac{\text{Frequncey of the class }}{\text{widthof the class}}\Big)\times5$

Therefore adjusted frequencey of $25-45=\frac{8}{20}\times5=2$

3. 26

Solution:

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.

If the number of observation is even (as in this example), then the median is the average of two middle numbers

Hence, the average of 25 and 27 is $\frac{25+27}{2}=26$

So the median is 26.

2. $\overline{\text{X}}+\text{k}$

Solution:

$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$

$=\frac{\text{Sum of all observations}}{\text{n}}$

Now if k is aded to each observation

$\text{New Mean},\overline{\text{X'}}=\frac{\text{Sum of all observations}+\text{nk}}{\text{n}}$

$=\frac{\text{Sum of all observations}}{\text{n}}+\text{k}$

$\Rightarrow\overline{\text{X}'}=\overline{\text{X}}+\text{k}$

1. 6.5 - 12.5

Solution:

Class mark $=\frac{\text{upper limit+lower limit}}{2}$

2 × 9.5 = upper limit + lower limit

19 = upper limit + lower limit

Class size = upper limit - lower limit

6 = upper limit - lower limit

Solving above two equations we get,

2 × upper limit = 25

upper limit = 12.5

Hence, lower lower limit = 12.5 - 6 = 6.5

So, the class interval is 6.5 - 12.5.

1. 0.6

Solution:

First five prime numbers are 2, 3, 5, 7, 11

Mean is $\frac{2+3+5+7+11}{5}$

$\frac{28}{5}=5.6$

Median is 5

So the difference between them is 5.6 - 5 = 0.6

2. 38

Solution:

Let a, b, c, d and e are five numbers, then

Mean $=\frac{\text{(a + b + c + d + e)}}{5}=30$

⇒ (a + b + c + d + e) = 150 ..... (1)

Now Let the number excluded be a 

Then new mean $=\frac{\text{(b + c + d + e)}}{4}=28$

⇒ (b + c + d + e) = 112

Putting this value in (1)

⇒ a + 112 = 150

⇒ a = 150 - 112 = 38

$\therefore$ Excluded number = 38

3. $11\frac{1}{3}$

Solution:

Given that, the mean of the observation x, x + 3, x + 5, x + 7 and x + 10 is 9.

$\therefore\ \frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}=9$

$\Rightarrow {\text{5x}}+25=45$

$\Rightarrow\text{5x}=20$

$\Rightarrow \text{x}=4$

$\therefore$ Last three odservations are x + 5 = 4 + 5 = 9, x + 7 = 4 + 7 = 11

and x + 10 = 4 + 10 = 14

So, the mean of last three observations $=\frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3}$

Hence, the mean of last three observation is $11\frac{1}{3}.$

3. 13

Solution:

Let the lower limit be x.

Class interval = 4

Upper limit = x + 4

Now, mid - value of a class $=\frac{\text{x}+4+\text{x}}{2}=\text{x}+2=15(\text{given)}$

x = 13 = lower limit

4. $2\text{m}-\text{L}$

Solution:

Mid - value $=\frac{\text{Lower limit + Upper limit}}{2}$

$\Rightarrow\text{m}=\frac{\text{L+U}}{2}$

⇒ U = 2m - L

$\therefore$ Upper class boundary of the class = 2m – L

2. 99

Solution:

$\text{Mean}=81=\frac{\text{Sum of seven numbers}}{7}$

⇒ Sum of seven numbers = 81 × 7 = 567

Let the discared number be x.

⇒ Sum of 6 numbers = 567 - x

Now, mean of remaining 6 numbers $=\frac{567-\text{x}}{6}=78$

⇒ 567 - x = 468

⇒ x = 99

So, discarded number is 99

2. 105

Solution:

$\text{Class mark }=\frac{\text{Upperlimit + Lowerlimit}}{2}$

$\Rightarrow \text{Class mark}=\frac{90+120}{2}=\frac{210}{2}=105$

3. $\frac{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}}$

Solution:

Sum of the terms $=\text{n}_1\bar{\text{x}_1}+\text{n}_2\bar{\text{x}}_2+\ _{\dots},+\text{n}_\text{n}\bar{\text{x}}_\text{n}$

Number of terms $=\text{n}_1+\text{n}_2+\ _{\dots}+\text{n}_\text{n}$

Required mean $=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits_{\text{i}=1}^\text{n}\text{n}_\text{i}}$

2. $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$

Solution:

Given, mean of x₁, x₂, ......, x_n is $\bar{\text{x}}$

$\therefore\ \sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}=\text{n}\bar{\text{x}}$

Now, let the mean of $\Big(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)$ is $\bar{\text{z}}$

Then, $\bar{\text{z}}=\frac{(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n})+\Big(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)}{\text{n}+\text{n}}$

$=\frac{\text{a}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})+\frac{1}{\text{a}}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\cdot\text{n}\bar{\text{x}}}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big){\bar{\text{x}}}}{2}$

3. 40

Solution:

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.

For even number of observations, the median is calculated as an average of two middle numbers.

For the given example 30 and x are in the middle and the median is 35.

So,

$35=\frac{30+\text{x}}{2}$

$70=30+\text{x}$

$\text{x}=40$

3. 3.5

Solution:

The given data is 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5 and 6.

Arranging the given data in ascending order, we have

-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6

Here, the number of observation n = 14, which is an even number.

Hence, the median is

$\frac{\Big(\frac{\text{n}}{2}\Big)^\text{th}\text{ observation }+{\Big(\frac{\text{n}}{2}+1\Big)}^\text{th}\text{ observation }}{2}$

$=\frac{\Big(\frac{14}{2}\Big)^\text{th}\text{ observation }+{\Big(\frac{14}{2}+1\Big)}^\text{th}\text{ observation }}{2}$

$=\frac{7^\text{th}\text{ observation }+8^\text{th}\text{ observation }}{2}$

$=\frac{2+5}{2}$

$=\frac{7}{2}$

$=3.5$

2. 17.5

Solution:

Class mark is the mid-value of each class interval.

For class interval 15 - 20

Class mark $=\frac{15+20}{2}=\frac{35}{2}=17.5$