Maths M.C.Q

2. 51

Solution:

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

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

$\Rightarrow\sum\text{x}_\text{i}=50\times100=5,000$

If one of the observation which was 50 is resplaced by 150, then

$\sum\text{x}_\text{i}=5,000-50+150=5100$

Then, the resulting mean $=\frac{5100}{100}=51$

2. 93 marks

Solution:

Let, Vihaan obtains x marks in the fourth test.

So,

$\frac{92+85+78+\text{x}}{4}=87$

$\frac{255+\text{x}}{4}=87$

255 + x = 348

x = 348 - 255

x = 93 marks

1. 1, 3, 3, 3, 5

Solution:

For the data 2, 2, 2, 2, 4 of 5 numbers, we have

Mean $=\frac{2+2+2+2+4}{5}=\frac{12}{5}=2.4$

Median $=\Big(\frac{5+1}{2}\Big)^\text{th}$ Value = 3rd Value = 2

Since, 2 occurs maximum number of times, Mode = 2

Mean ≠ Median

For the data 1, 3, 3, 3, 5 of 5 numbers, we have

Mean $=\frac{1+3+3+3+5}{5}=\frac{15}{5}=3$

Median $=\Big(\frac{5+1}{2}\Big)^\text{th}$ Value = 3^rd Value = 3

Since, 3 occurs maximum number f times, Mode = 3

Mean = Median = Mode

1. 5cm

Solution:

Use unitary method

0.25cm - 100 people

So 1cm - 400 people

So for 2000 people:

$\frac{2000}{400}=5\text{cm}$

4. 38

Solution:

Let x₁, x₂, x₃, x₄, and x₅ be five numbers and one of the excluded number be x₅,

Given, mean of the numbers = 30

$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}{5}=30$

$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}=150$

$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}=150-\text{x}_5$

On dividing both sides by 4, we get

$\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}{4}=\frac{150-\text{x}_5}{4}\ \dots(\text{i})$

Given, mean of the numbers = 28

$\therefore\ \frac{150-\text{x}_5}{4}=28$ [from Eq. (i)]

$\Rightarrow 150 -\text{x}_5=112$

$\Rightarrow \text{x}_5=150 -112$

$\Rightarrow\text{x}_5=38$

Hence, the excluded number is 38.

4. $\text{k}\overline{\text{X}}$

Solution:

Let us take n observations X₁, ...., X_n

If $\overline{\text{X}}$ be the mean of the n observations then we have

$\overline{\text{X}}=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$

$\Rightarrow\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}=\text{n}\overline{\text{X}}$

Multiply a constant k to each of the observations then observations becomes kXi......kX_n

If $\overline{\text{Y}}$ be the mean of the new observations, then we have

$\overline{\text{Y}}=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{kX}_{\text{i}}$

$=\frac{\text{k}}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$

$={\text{k}}\cdot=\frac{1}{\text{n}}\sum_{\text{i}=1}^{\text{n}}\text{X}_{\text{i}}$

$=\text{k}\overline{\text{X}}$

2. 7

Solution:

Let x and y be the uppel and lower and lower class limit in a frequency distribution.

Now, mid value of a class $\frac{(\text{x}+\text{y})}{2}=10$ [given]

⇒ x + y = 20 ...(i)

Also, given that, width of class x - y = 6 ...(ii)

On adding Eqs. (i) and (ii), we get

2x = 20 + 6

⇒ 2x = 26

⇒ x = 13

On putting x = 13 in Eq. (i), we get

13 + y = 20

⇒ y = 7

Hence, the lower limit of the class is 7.

3. 17.3

Solution:

Average of upper limit &lower limit of a class interval is called its mid point.

For the class interval 12.3 - 22.3

$\text{Mid - point }=\frac{12.3+22.3}{2}$

$\text{Midpoint is }:17.3$

3. $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}_2}$

Solution:

If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the mean of n group with n₁, n₂, ...., n_n number of observation respectively, then the mean $\bar{\text{x}}$

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

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

Hence, the mean $\bar{\text{x}}$ of all group taken together is gien up $=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$

2. 29

Solution:

Arranging the points in an ascending order,

We have:

0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56,

Here, n = 11, Which is odd

$\therefore\ $median score = value of $=\frac{1}{2}(11+1)^\text{th}$ term

= value of $\Big(\frac{1}{2}\times12\Big)^\text{th}$ term

= value of 6^th term

= 29

2. 25

Solution:

Now,

Mean $=\frac{303+9\text{p}}{41+\text{p}}$

Given, 

Mean = 8

$\therefore\frac{303+9\text{p}}{41+\text{p}}=8$

⇒ 303 + 9p = 328 + 8p

⇒ p = 25

4. 2M² – 1

Solution:

Given ,$\frac{\text{x}+\frac{1}{\text{x}}}{2}=\text{M}$

Taking square on both sides

$\bigg(\frac{\text{x}+\frac{1}{\text{x}}}{2}\bigg)^2=(\text{M})^2$

$\bigg(\text{x}+{\frac{1}{\text{x}}}\bigg)^2=(2\text{M})^2$

$\bigg(\text{x}^2+2+{\frac{1}{\text{x}^2}}\bigg)=(2\text{M})^2$

$\bigg(\text{x}^2+{\frac{1}{\text{x}}}\bigg)^2={4\text{M}^2-2}$

Divide by 2 on both sides to get mean

$\bigg(\frac{\text{x}^2+\frac{1}{\text{x}^2}}{2}\bigg)^2=2\text{M}^2-1$

1. $\text{l}+\bigg(\frac{\frac{\text{n}}{2}-\text{cf}}{\text{f}}\bigg)\times{\text{h}}$

Solution:

Where, l = lower limit of the median class

F = frequency of the median class

CF = cumulative frequency of the class preceding the median class

N = number of observations

H = size of the class interval (assuming all class sizes to be equal)

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

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. Sum of first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2}$

So, mean of first n natural numbers is $\frac{\frac{\text{n}(\text{n}+1)}{2}}{\text{n}}=\frac{(\text{n}+1)}{2}$

4. 2m - I

Solution:

Let the lower limit = k

Midpoint = m

Upper limit = I

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

$\text{m}=\frac{\text{(K + I)}}{2}$

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

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

Therefore, lower limit = 2m - I

3. 53kg

Solution:

Mean weight of six boys = 48kg

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

We know:

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

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

$=\frac{235+\text{x}}{6}$

Given,

Mean = 48kg

$=\frac{235+\text{x}}{6}=48$

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

Hence, the weight of the 6th boy is 53kg.

2. 18, 20, 22

Solution:

Let the three consecutive even integers be- x, x + 2, x + 4

$\frac{\text{x}+\text{x}+2\text{x}+4}{3}=20$

$\frac{3\text{x}+6}{3}=20$

$3\text{x}+6=60$

$3\text{x}=54$

$\text{x}=18$

So the three numbers are 18, 20, 22.

3. 42kg

Solution:

Arranging the points in an ascending order,

We have:

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

Here, n = 10, Which is even

$\therefore\ $median = mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{10}{2}+1\Big)^\text{th}$ terms

= mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{12}{2}\Big)^\text{th}$ term

= mean of 5^th and 6^th term

$=\frac{1}{2}(40+44)$

$=\frac{1}{2}\times84$

$=42\text{kg}$

1. 47 and 37

Solution:

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is $\frac{\text{l+m}}{2}$ and the class-size is (l - m).

Given that the mid-value of the class is 42 and the class-size is 10. Therefore, we have two equations

$\frac{\text{l+m}}{2}=42$

⇒ l + m = 84,

m - l = 10

Adding the above two equations, we have

(l + m) + (m - l) = 84 + 10

⇒ l + m + m - l = 94

⇒ 2m = 94

⇒ l = 37

Substituting the value of m in the first equation, we have

l + 47 = 84

⇒ l = 84 - 47

⇒ l = 37

Hence, the upper and lower limits of the class are 47 and 37 respectively. Thus, the correct choice is (a).

4. Mean

Solution:

3 Median = 2 Mean + Mode

⇒ 3 Median – Mode = 2 Mean

$\Rightarrow\text{Mean}=\frac{3}{2}\text{ Median }-\frac{\text{Mode}}{2}$

$\Rightarrow\text{Mean}+\frac{3}{2}\text{ Median }-\frac{\text{Mode}}{2}-\text{Mode = Mean}$

$\Rightarrow\text{Mean}+\frac{3}{2}\text{(Median - }\text{Mode) = Mean}$

4. 3

Solution:

From the given frequency distribution table:

$3.5=\frac{2\times1+3\times2+4\text{x}+5\times4+6\times5}{2+3+4+5+6}$

$3.5=\frac{2+6+4\text{x}+20+30}{20}$

70 + 4x + 58

4x = 12

x = 3

1. $\text{k}\overline{\text{X}}$

Solution:

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

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

if each observation is multiplied by k, then

$\text{New Mean},\overline{\text{X'}}=\frac{(\text{Sum of all observations})\text{k}}{\text{n}}$

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

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

Solution:

For the data A: 2, 3, 7, 1, 3, 2, 3, of 7 numbers, we have

Mean $=\frac{2+3+7+1+3+2+3}{7}=\frac{21}{7}=3$

Arranging the data A in ascending orer we have 

A: 1, 2, 2, 3, 3, 3, 7

Median $=\Big(\frac{7+1}{2}\Big)^\text{th}$ Value

= 4^th value = 3

Since, 3 occurs maximum number of times, Mode = 3

Mean = Median = Mode

2. 36

Solution:

Given, Mean of 25 observations = 36

$\therefore$ Sum of 25 observations = 36 × 25 = 900

Now, the mean of first 13 observations = 32

$\therefore$ Sum of first 13 observations = 13 × 32 = 416

and the Mean of last 13 observations = 40

$\therefore$ Sum of last 13 observations = 40 × 13 = 520

So, 13th observation = (Sum of last 13 observations + Sum of first 13 observations) - (Sum of 25 observations)

= (520 + 416) - 900 = 936 - 900 = 36

Hence, the 13th observation is 36.

3. $\frac{55.2\text{km}}{\text{h}}$

Solution:

Sum of all the recorded speeds is

48 + 52 + 57 + 55 + 42 + 39 + 60 + 49 + 53 + 47 = 502

Because of the error, $\frac{5\text{km}}{\text{h}}$ in each case

The sum increases by 50 i.e., 552

So the average speed of 10 vehicle is $\frac{55.2\text{km}}{\text{h}}$

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

Solution:

Given that:

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

$\frac{5\text{x}+25}{5}=9$

5x + 25 = 45

5x = 20

x = 4

So the last three numbers are 9, 11, 14

So there mean is $11\frac{1}{3}$

2. 21

Solution:

The data is given to be in an ascending order,

Here, n = 10, which is even

$\therefore\ $Median = mean of $\Big(\frac{10}{2}\Big)^\text{th}$ and $\Big(\frac{10}{2}+1\Big)^\text{th}$ term

= mean of 5^th and 6^th terms

= $\frac{1}{2}\Big[(\text{x}+2)+(\text{x}+4)\Big]$

$=\frac{1}{2}\big[2\text{x}+6\big]$

But median = 24

$\Rightarrow\frac{1}{2}\big[2\text{x}+6\big]=24$

$\Rightarrow2\text{x}+6=48$

$\Rightarrow2\text{x}=42$

$\Rightarrow\text{x}=21$