MATHS M.C.Q (1 Marks)

4. 4.5

Solution:

Factors of 10 are, 1, 2, 5, 10

Hence required mean $ =\frac{1+2+5+10}{4}$

$ =\frac{18}{4}=4.5$

1. 16.25

Solution:

Factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

$\text{mean}=\frac{\text{sum}}{\text{count}}$

$\text{Mean}=\frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}​$

$ \text{mean}=\frac{195}{12}$

$ \text{Mean}=16.25$

1. 6.25

Solution:

For first 10 overs, run rate = 3.2

⇒ Runs scored = 3.2 × 10 = 32

Total runs to be scored = 282

Runs Left to be scored in 4040 overs $ = \frac{282-32}{40}$

$ =\dfrac{250}{40}$

Required Run-Rate in remaining overs = 6.25

2. 1

Solution:

Mean of the series: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25

$ \text{mean} = \frac{\text{Sum}}{\text{Number of observations}}$

$\text{mean} = \frac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}$

$ \text{mean} = \frac{297}{10} = 29.7$

2. 36 years

Solution:

Average age of the two brother = 9 years

$ \therefore$ Age of two brother = 9 × 2 = 18 years

If their mother age is included then the average age is increased by 9

$ \therefore$ Average age of 3 = 9 + 9 = 18 years

Now Total age of three = 3 × 18 = 54 Years

$ \therefore$ Mothers age = 54 - 18 = 36 years.

1. l = 1.25, k = -5

Solution:

Given, $\text{w}_\text{i}=\text{lx}_\text{i}+\text{k},\ \bar{\text{x}}_\text{i}=48,\text{ sx}_\text{i}=12,\text{ w}_\text{i}=55$ and $\text{sw}_\text{i}=15$

Then, $\bar{\text{w}}_\text{i}=\text{l}\bar{\text{x}}_\text{i}+\text{k}$ $\big[\text{where }\bar{\text{w}}_\text{i}\text{ is mean w}_\text{i}{'\text{s}}\text{ and }\bar{\text{x}}_\text{i}\text{ is mean of x}_\text{i}{'\text{s}}\big]$

$\Rightarrow55=\text{l}\times48+\text{k}\ ...(\text{i})$

Now, $\text{SD of w}_\text{i}=\text{l}(\text{SD of x}_\text{i})$

$\Rightarrow15=\text{l}\times12$

$\Rightarrow\text{l}=\frac{15}{12}=12.5$

From Eq. (i) we get k = 55 - 1.25 × 48 = -5

1. 178

Solution:

The lines of 5 bulbs are given by

1357, 1090, 1666, 1494, 1623

$\therefore\ \text{Mean}=\frac{1357+1090+1666+1494+1623}{5}$

$\Rightarrow\bar{\text{x}}=\frac{7230}{5}=1446$

$\therefore\ \text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{890}{5}=178$

3. 14

Solution:

Mean of 20 observations = 12.5 Sum of 20 observations = 12.5 × 20 = 250

Since 15 was misread as,

15 New sum = 250 - (-15) + 15 = 280

The correct $\text{mean} = \dfrac{280}{20} = 14$

4. 30
5. Solution:

Mean is given by $ \text{mean}=\dfrac{\text{sum of the elements}}{\text{total number of elements}}$

$=\text{mean}=\frac{45+35+20+30+15+25+40​}{7}$

$=\frac{210}{7}$

$ =30$

1. 24

Solution:

Arranging is ascending order the values are

$\frac{\text{x}}{5},\frac{\text{x}}{4},\frac{\text{x}}{3},\frac{\text{x}}{2},\text{x}$

Middle value $ =\frac{\text{x}}{3}\Rightarrow\frac{\text{x}}{3}=\text{x}=24$

4. 41

Solution:

Let the average score till 11th innings be x according to question $ ⇒ \frac{11\text{x}+63}{12}$

= x + 2 ⇒ 11x + 63 = 12x + 24 ⇒ x = 39

12th inning average ⇒ 39 + 2 = 41

4. $ \displaystyle \frac{\text{M}}{50}$

Solution:

Given $ \text{mean}= \frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}...........\text{x}_{50}}{50}$

$ \text{mean} =\frac{\frac{\text{x}_{1}}{50}+\frac{\text{x}_{2}}{50}+...........\frac{\text{x}_{50}}{50}}{50}$

$ =\frac{\text{x}_1+\text{x}_2+....\text{x}_{50}}{50\times50}$

$ \displaystyle \frac{\text{M}}{50}$

1. 11

Solution:

$ 1\ +\ 2+\ 3\ +....\text{n}=\frac{\text{n}\times(\text{n}\ +\ 1)}{2} $

$ \text{Then u}=\frac{\text{n}\times(\text{n}+1)}{2\times \text{n}}$

$ ∴\text{u}=\frac{(\text{n}+1)}{2}​$

But Given: $ \text{u}=\dfrac{(\text{n}+7)}{3}$

Thus, $ \text{u}=\frac{(\text{n}+1)}{2}=\frac{(\text{n}+7)}{3}​$

Solving above we get,

n=11

3. 50.9

Solution:

Mean of 200 observations = 50

Sum of 200 observations = 50 × 200 = 10000

After replacing the misread observation 92 to 192 and 8 to 88

Sum of 200200 observations = 10000 - 92 + 192 - 8 + 88 = 10180

$ \text{New mean} = \frac{10180}{200}​ = 50.9$

3. 252500

Solution:

Here, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$

$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$

$\Rightarrow{\sum\text{x}_\text{i}}=5000$

$\therefore\ \text{SD}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$

$\Rightarrow\ 5=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{5000}{100}\Big)^2}$

$\Rightarrow25=\frac{\sum\text{x}_\text{i}^2}{100}=2525$

$\therefore\ {\sum\text{x}_\text{i}^2}=2525\times100=252500$

1. 8.25

Solution:

The first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Multiplying each number by −1, we get

-1, -2, -3, -4, -5, -6, -7, -8, -9, -10

Adding 1 to each of these numbers, we get

0, -1, -2, -3, -4, -5, -6, -7, -8, -9

Now,

 $\sum\text{x}_\text{i}=$ 0 + (-1) + (-2) + (-3) + (-4) + (-5) + (-6) + (-7) + (-8) + (-9) = -45

$\sum\text{x}_\text{i}^2=$ 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285

$\therefore$ Variance of the obtained numbers

$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$

$=\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2$

$=28.5-20.25$

$=8.25$

1. 37.5

Solution:

Average of remaining 48 numbers

$= \frac{(50 \times38 )- 55 - 45}{48} = 37.5$

3. 24

Solution:

Numbers between 6 and 34 divisible by 5 are 10, 15, 20, 25, 30.

Required average = $\frac {10+15+20+25+30}{5} = \frac {100}{5} $

= 20

3. 50

Solution:

$ \text{Mean} = \frac { \text{Sum of observations}}{\text{Total number of observations}} = \dfrac {600}{12} = 50$

3. $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$

Solution:

Therefore are 2n + 1 terms.

⇒ N = 2n + 1

$\sum\text{x}_\text{i}=\text{a}+\text{a}+\text{d}+\text{a}+2\text{d}+\text{a}+3\text{d}+...+\text{a}+2\text{nd}$

$=(2\text{n}+1)\text{a}+\text{d}(1+2+3+...+2\text{n})$ [a + a + a + ...(2n + 1)times = (2n + 1)a]

$=(2\text{n}+1)\text{a}+\frac{2\text{n}(2\text{n}+1)\text{d}}{2}$ $\Big[$Sum of the first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2},$ but here we are considering$\Big]$

$=(2\text{n}+1)\text{a}+(2\text{n}+1)\text{nd}$

$=(2\text{n}+1)(\text{a}+\text{nd})$

$\overline{\text{X}}=\frac{(2\text{n}+1)(\text{a}+\text{nd})}{(2\text{n}+1)}$

$=\text{a}+\text{nd}$

$\sum\big|\text{x}_\text{i}-\overline{\text{X}}\big|=\text{nd}+(\text{n}-1)\text{d}+(\text{n}-2)\text{d}\\+...+\text{d}+0+\text{d}+2\text{d}+3\text{d}+...+\text{nd}$

$=\text{d}(\text{n}+(\text{n}-1)+(\text{n}-2)+...+1)\\ \ +0+\text{d}(1+2+3+....+\text{n})$

$=\frac{\text{dn}(\text{n}+1)}{2}+\frac{\text{dn}(\text{n}+1)}{2}$ $\Big\{\because1+2+3+....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big\}$

$=\text{n}(\text{n}+1)\text{d}$

Mean deviation about the mean $=\frac{\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|}{\text{N}}$

$=\frac{\text{n}(\text{n}+1)\text{d}}{(2\text{n}+1)}$

4. 2.87

Solution:

We know that the standard deviation of first n natural number is $\sqrt{\frac{\text{n}^2-1}{12}}.$

$\therefore$ Standard deviation of first 10 natural numbers

$=\sqrt{\frac{10^2-1}{12}}$

$=\sqrt{\frac{99}{12}}$

$=\sqrt{8.25}$

$=2.87$

Hence, the correct answer is option (d).

1. Var (Y) = a² Var (X)

Solution:

$\text{Var}(\text{x})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}$ where Mean $\Big(\overline{\text{X}}\Big)=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$

$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$ and $\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$

We have,

$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$

$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$

$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{ax}_\text{i}+\text{b}}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}+\frac{\text{nd}}{\text{n}}$

$=\text{a}\overline{\text{X}}+\text{b}$

$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big\{\text{ax}_\text{i}+\text{b}-\big(\text{a}\overline{\text{X}}+\text{b}\big)\Big\}^2}{\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{ax}_\text{i}-\text{a}\overline{\text{X}}\big)^2}{\text{n}}$

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

$=\text{a}^2\text{Var}(\text{X})$

3. $ \displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{6}$

Solution:

The first natural numbers are 1,2,3,......nTheir square are1$1 ^2,2^2,3^2......\text{n}^2$

$ \text{Mean}=\dfrac{1^2+2^2+3^2+........+\text{n}^2}{\text{n}}$

$ =\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}$

$\therefore$ Mean=n12 + 22 + 32 +........+ n2

$\therefore$ square of n natural numbers is$ =\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$

$ \text{Mean}=\dfrac{(\text{n}+1)(2\text{n}+1)}{6}$ 

3. $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$

Solution:

It is given that x₁, x₂, ...,x_n be n observations and $\overline{\text{X}}$ be their arithmetic mean.

The standard deviation is given observations is $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}.$

Also,

$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}=\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$

Hence, the correct answers are options (c) and (d).

Disclaimer: For option (c) to be the only correct answer, option (d) should be different from the given value.

4. None of these

Solution:

Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$

$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$

$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$

$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$

$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$

$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$

$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$

$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$

$\sigma=\sqrt{\text{Variance}(\text{X})}$

$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$

4. both B and C

Solution:

Median is the middle most value of a series.

So when the series has odd number of elements then,

median can be calculated easily but when the series has even number of elements then,

The series has two middle values, so

median is calculated either by taking out the average of both the

value or the median is the$ \frac{(\text{N}+1)}{ 2}$ th element of the series.

2. $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

Solution:

$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$

$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$

$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$

$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$

$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$

We know:

$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\sigma^2$

$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$

$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$

$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$

$\text{SD of Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$

$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$

$\text{ac}<0$

$\Rightarrow\text{a}<0\text{ or }\text{c}<0$

$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$

$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

2. $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$

Solution:

The mean deviation for n observations x₁, x₂, ...,x_n from their mean $\overline{\text{X}}$ is $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$

Disclaimer: There is some printing error in option (b) given in the question. The answer would be option (b) if it given as $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$

3. Frequency of the class preceding the modal class

Solution:

In the formula for mode of a grouped data,

$\text{ mode} =\text{l}+\left \{\frac {\text{f}_1-\text{f}_0}{2\text{f}_2-\text{f}_0-\text{f}_2}\right \}\times\text{h}$ where symbols have their usual meaning

f₀ represents Frequency of the class preceding the modal classwhere

f = Frequency,

1 = Lowest value of the modal range,

f₁​= Frequency of modal class,

f₂ Frequency of class succeeding the modal class and

f₀ = Frequency of class preceding the modal class.Hence, option C is correct.

3. $ \frac{8\text{th}\ \text{observation}\ +\ 9\text{th}\ \text{observation}}{2}$

Solution:

For even number of observations median is the mean of $ \frac{\text{n}}{2}$ 

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

So, median of 16 observation$= \frac{8\text{th}\ \text{observation}\ +\ 9\text{th}\ \text{observation}}{2}$

3. $\overline{\text{x}}= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }​$

Solution:

$\text{ A Median} = \text{L} + \begin{pmatrix} \frac{N}{2} - \text{c.f.} \end{pmatrix} \frac{\text{h}}{\text{f}} , ​ ​ ​ $ 

$ \text{mode}=\text{L}+\Big(\frac{\text{f}\text{m}-\text{f}1}{\text{f}\text{m}-\text{f}1-\text{f}2}\Big)\text{h}$ Where,

L is lower boundary of median class,

N is sum of frequencies, c.f. is cumulative frequency,

h is width of classes, f is frequency of mode class,

$ \text{f}_\text{m}$ is frequency of modal class,

$ \text{f}_\text{1}$ is frequency of pre-modal class,

$ \text{f}_\text{2}$ is frequency of post-modal class Mean is

given $= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }​$ Thus, Mean given in option C is incorrect.

4. 71

Solution:

Wrong average =70

70 $ \therefore$ Wrong sum of weight of 20 students = 20 × 70

$ \therefore$ Correct sum of weights of 20 students

= 1400 - wrong weight + correct weight

= 1400 - 70 + 90 = 1420

$∴ \text{Correct mean} = \frac{1420}{20}=71 \text{kg.}$

3. 252500

Solution:

Let $\overline{\text{x}}$ and $\sigma$ be the mean and standard deviation of 100 observations, respectively.

$\therefore\overline{\text{x}}=50,\ \sigma=5$ and n = 100

$\text{Mean},\ \overline{\text{x}}=50$

$\Rightarrow\frac{\sum\text{x}_\text{i}}{100}=50$

$\Rightarrow\sum\text{x}_\text{i}=5000\ ...(1)$

Now,

Standard deviation, $\sigma=5$

$\Rightarrow\sqrt{\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{\sum\text{x}_\text{i}}{100}\Big)^2}=5$

$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{5000}{100}\Big)^2=25$ [From (1)]

$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}=25+2500=2525$

$\Rightarrow{\sum\text{x}_\text{i}^2}=252500$

Thus, the sum of all squares of all the observations is 252500.

Hence, the correct answer is option (c).

3. 17.5

Solution:

Arranging the given data in ascending order 4, 12, 13, 16, 19, 28, 31,

The middle terms are 16, 19 Hence median

$ =\frac{16 \ +\ 19}{2}$

$ =17.5$

1. 10

Solution:

$ ∑ \text{fx} = 15\text{p} + 1110; ∑ \text{f} = 50 + \text{p}$

$ \text{x} = 21$

$ 21=\frac{15_p\ +\ 1110}{50 \ +\ p}$

$ ⇒ 21\text{p} − 15\text{p} = 1110 − 1050$

$ \Rightarrow\text{p}=\frac{60}{6}=10$