Maths M.C.Q

3. 2m - I

Solution:

Let the lower limit = k

Mid-point = m

Upper limit = l

Mid-point $=\frac{\text{(upper limit + lower limit)}}{2}$

$\text{m}=\frac{(\text{k+l})}{2}$

$2\text{m}=\text{k}+\text{l}$

$\text{k}=2\text{m}-\text{l}$

Therefore, lower limit = 2m - l

3. 7

Solution:

Mean of 5 observations = 11

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

$\Rightarrow11=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$

$\Rightarrow11=\frac{5\text{x}+20}{5}$

$\Rightarrow55=5\text{x}+20$

$\Rightarrow5\text{x}=35$

$\Rightarrow\text{x}=7$

2. 21

Solution:

The given data is in ascending order.

Here, n is 10, which is an even number. Thus, we have:

Median = mean of $\Big(\frac{\text{n}}{2}\Big)\text{th}$ and $\Big(\frac{\text{n}}{2}+1\Big)\text{th}$ observations $=\frac{1}{2}$ (5th observation + 6th observation)

$=\frac{1}{2}(\text{x}+2+\text{x}+4)=(\text{x}+3)$

= 24

Also, x + 3 = 24

⇒ x = 21

3. 55

Solution:

Class mark $=\frac{\text{( upper limit + Lower limit ) }}{2}$

$=\frac{50+60}{2}$

$=\frac{110}{2}$

So class mark is 55.

2. $\frac{2\text{x}+3\text{y}}{5}$

Solution:

Average is equal to the sum of all the values in the data set divided by the number of values in the data set.

Average $=\frac{\text{x}+\text{x}+\text{y}+\text{y}+\text{y}}{5}$

Average $=\frac{2\text{x}+3\text{y}}{5}$

2. 53kg.

Solution:

Let the weight of the 6^th boy be x kg.

$\frac{51+45+49+46+44+\text{x}}{6}=48$

$\Rightarrow51+45+49+46+44+\text{x}=48\times6$

$\Rightarrow235+\text{x}=288$

$\Rightarrow\text{x}=53\text{kg}$

So, the weight of the 6^th boy is 53kg.

1. 9

Solution:

Mean of first five observations $=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}=11$

⇒ 5x + 20 = 55

⇒ x = 7

⇒ First three numbers are 7, 9, 11

$\text{Mean}=\frac{7+9+11}{3}=\frac{27}{3}=9$

4. 34

Solution:

$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$

⇒ a + b + c + d + e = 140 ...(1)

Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$

⇒ a+ c + e = 72 ...(2)

Subtracting equation (2) from (1), we have

b + d = 68

$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$

2. 36

Solution:

Mean of first 13 observation = 32

$\therefore$ Sum of all first 13 observation = (32 × 13) = 416

Mean of last 13 observation = 40

$\therefore$ Sum of all last 13 observation = (40 × 13) = 520

Mean of 25 observation = 36

$\therefore$ Sum of all first 25 observation = (36 × 25) = 900

Hence, 13^th observation = 416 + 520 - 900 = 36

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

Solution:

We have $\bar{\text{x}}$ is the mean of x₁, x₂, ......, x_n, and $\bar{\text{y}}$ is the mean of y₁ , y₂ , ... , y_n

So, $\bar{\text{x}}=\frac{(\text{x}_1+\text{x}_2+\text{x}_3+\ ....\ +\text{x}_\text{n})}{\text{n}}$

$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+\ .....\ +\text{x}_\text{n}=\text{n}\bar{\text{x}}$

And $\bar{\text{y}}=\frac{(\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n})}{\text{n}}$

$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n}=\text{n}\bar{\text{y}}$

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

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

2. 0

Solution:

Given that $\bar{\text{x}}$ is the mean of x₁, x₂ x₃, ..., x_n.

$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+_{\dots}+\text{x}_\text{n}}{\text{n}}=\bar{\text{x}}$

$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+_{\dots}+\text{x}_\text{n}=\text{n }\bar{\text{x}}\ \dots\text{(i)}$

Consider,

$\big(\text{x}_1-\bar{\text{x}}\big)+\big(\text{x}_2-\bar{\text{x}}\big)+\big(\text{x}-\bar{\text{x}}\big)+_{\dots}+\big(\text{x}_\text{n}-\bar{\text{x}}\big)$

$=\big(\text{x}_1+\text{x}_2+\text{x}_3+_{\dots}+\text{x}_\text{n}\big)-\text{n }\bar{\text{x}}$

$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}\ \dots$ (from (i))

$=0$

2. 42kg

Solution:

Arranging the numbers in ascending order, we have:

31, 35, 36, 38, 40, 44, 45, 52, 55, 60

Here, n is 10, which is an even number. Thus, we have:

Median = Mean of $\Big(\frac{\text{n}}{2}\Big)\text{th}$ obervation & $\Big(\frac{\text{n}}{2}+1\Big)\text{th}$ obervation Median Weight = mean of the weights of $\Big(\frac{10}{2}\Big)\text{th}$ student & $\Big(\frac{10}{2}+1\Big)\text{th}$ student = mean of the weights of 5th student & 6th student$=\frac{1}{2}(40+44)=42$

Hence, the median weight is 42kg.

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

Solution:

Required mean $=\frac{(\text{ax}_1+\text{ax}_2+_{\dots}+\text{ax}_\text{n})+(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ \dots,\frac{\text{x}_\text{n}}{\text{a}})}{2\text{n}}$

$=\frac{1}{2}\bigg[\frac{\text{a}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}+\frac{\frac{1}{2}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}\bigg]$

$=\frac{1}{2}\Big[\text{a}\bar{\text{x}}+\frac{1}{\text{a}}\bar{\text{x}}\Big]$

$=\Big[\text{a}+\frac{1}{\text{a}}\Big]\frac{\bar{\text{x}}}{2}$

4. 16

Solution:

$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$

$\Rightarrow34+\text{x}=50$

$\Rightarrow\text{x}=16$

2. Mode = 3 median - 2 mean

Solution:

The empirical relationship between the three measures of central tendencies is 3 Median = Mode + 2 Mean.

The relationship is as per observation. A distribution in which the values of mean, median and mode coincide (i.e., mean = median = mode) is called symmetrical

Distribution. conversely, when the values of mean, median, mode are not equal, the distribution is called asymmetrical or skewed.

Knowing any two values, the third can be computed by this formula

3 median = 2 mean + mode

2 mean = 3 median – mode

$\text{Mean}=\frac{1}{2}(3 \text{ median - mode})$

3. $\frac{\displaystyle\sum_{\text{i=1}}^{\text{n}}\text{n}_\text{i}\text{x}_\text{i}}{\displaystyle\sum_{\text{i=1}}^{\text{n}}\text{n}_\text{i}}$

Solution:

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.

4. 212, 237

Solution:

For 1st class mark

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

1st class mark = 212

Add the class size to get the class mark of the next class

2nd class mark = 212 + 25

2nd class mark = 237

Hence, the class mark of first two intervals are 212 and 237.

4. 4.25

Solution:

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.

The first 4 prime numbers are 2, 3, 5, 7

So, mean is $\frac{17}{4}=4.25$

2. Is increased by 5

Solution:

Is increased by 5

Then old mean x old $=\frac{\displaystyle\sum_{\text{i}=1}^{\text{n}}\text{x}_{\text{i}}}{\text{n}}$

Now, adding 5 in each observation, the new mean becomes

$\overline{\text{x}}_\text{New}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+....+(\text{x}_\text{n}+5)}{\text{n}}$

$\Rightarrow\overline{\text{x}}_\text{New}=\frac{(\text{x}_1 + \text{x}_2+....+\text{x}_\text{n})+5\text{n})}{\text{n}}$

$\Rightarrow\overline{\text{x}}_\text{New}=\frac{\displaystyle\sum_{\text{i}=1}^{\text{n}}\text{x}_{\text{i}}}{\text{n}}+5=\overline{\text{x }}{\text{old}}+5$

$\Rightarrow\overline{\text{x}}_\text{New}=\overline{\text{x}}_\text{old}+5$

2. 34

Solution:

Prime numbers between 30 and 40 are 31 and 37.

Mean $=\frac{31+37}{2}$

Mean $=34$

3. 6

Solution:

The observations in ascending order can be written as:

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

Median $=\Big(\frac{9+1}{2}\Big)\text{th}$ term = 5th term = 6