MATHS M.C.Q (1 Marks)

3. 7

Solution:

$\text{ Average} = \displaystyle \frac{2 + 4 + 6 + 8 + 10}{5} = 6$

3. 15

Solution:

The first five multiples of 7 are 7, 14, 21, 28 and 35.

$ \text{Required mean }= \dfrac{7+14+21+28+35}{7}=\dfrac{105}{7}=21$

2. 74%

$ \text{Average}=\frac{20.80+30.70}{20+30}=74$

3. 18

Solution:

Average of 15 numbers 15 × 18

= 270 Average of first 8 number is 19

$\therefore$ Sum of first 8 number=19 × 8 = 152 = 19 × 8 = 152

Average of first 8 number = 17

$\therefore$ Sum of 8 number = 8 × 17 = 136 = 8 × 17 = 136

∴ 8th number is = (152 + 136) - 270 ⇒ 288 - 270 - 8

1. s

Solution:

The given observations are a, b, c, d, e.

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

$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$

Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$

Now, consider the observations a + k, b + k, c + k, d + k, e + k.

New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$

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

$=\frac{5\text{m}+5\text{k}}{5}$

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

$\therefore$ New standard deviation

$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$

$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}}$ [Using (1)]

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$

$=\text{s}$

Hence, the correct answer is option (a).

2. 7.8

Solution:

Required average $ = \frac{3 + 5 + 7 + 11 + 13}{5} = \frac{39}{5} =7.8$

1. s

Solution:

Given observations are a, b, c d and e.

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

$\sum\text{x}_\text{i}={\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}={5}\text{m}$

Now, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$

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

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

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$

$=\text{s}$

1. 4500

Solution:

Total production has to be 4375 × 12 = 52500

First three months production is 4000 × 3 = 12000

Total production has to be in remaining 9 months = 52500 - 12000 = 40500

Average production per month in remaining 9 months $ = \frac{40500}{9} = 4500$

1. 81

Solution:

Given that $\sigma_\text{c}=5$

We know that $\text{C}=\frac{5}{9}(\text{F}-32)$

$\Rightarrow\text{F}=\frac{9\text{C}}{5}+32$

$\therefore\ \sigma_\text{F}=\frac{9}{5}\sigma_\text{c}=\frac{9}{5}\times5=9$

$\therefore\ \sigma^2_{\text{F}}=(9)^2=81$

4. 44

Solution:

We know that variance $(\sigma^2)\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2$

$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2=300-256=44$

1. 65

Solution:

Total students = 100 Average marks

=72 Total marks of the class = 72 × 100 = 7200

Total marks of the boys = 70 × 75 = 5250

Total marks of the girls = 7200 = 5250 = 1950

Average marks of the girls $ = \dfrac{1950}{30}=65$ hence, option B is correct.

1. 10%

Solution:

We have:

$\overline{\text{X}}=50,\ \text{n}=10$

$\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2=250$

$\therefore\text{SD}=\sqrt{\text{Variance of X}}$

$=\sqrt{\frac{\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}}$

$=\sqrt{\frac{250}{10}}$

$=5$

Using $\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$

$\Rightarrow\text{CV}=\frac{5}{50}\times100$

$=10\%$

2. $ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )^{2}}{4}$

Solution:

First n natural numbers are 1,2,3,4,......,n

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

Sum of the cubes of n natural numbers $ =\left(\frac{\text{n}(\text{n}+1)}{2}\right)^2$

$ \therefore \text{Mean}=\left(\dfrac{\text{n}(\text{n}+1)}{2}\right)^2\times\frac{1}{\text{n}}$

$ =\dfrac{\text{n}(\text{n}+1)^2}{4}$

2. 33

Solution:

By definition of average,

$ =\cfrac{60+40+10+40+4+70+30+10}{8}$

$ =\cfrac{264}{8}=33$

3. 926

Solution:

Formula,

$ \cfrac{\sum {\text{x }} }{N}=\cfrac{864+874+884+1000+1008}{5}$

$ =\frac{4630}{5}$

$ =926$

4. 78

Solution:

$\text{Mean}=\frac{75+120+12+50+70.5+140.5}{6}$

​= 78 The average daily sale is therefore the mean = 78.

3. 80

Solution:

Let the number of boys and girls be x and y.

$ \therefore 52\text{x}+42\text{y}=50(\text{x}+{y})$

$ \Rightarrow 52\text{x}+42\text{y}=50\text{x}+50\text{y}$

$ \Rightarrow 2\text{x}=8{\text{y}} = \text{x }{ =4y} $

$ \therefore$Total number of students in class

= x + y = 4y + y = 5y

$ \therefore$ Required % of boys

$=\frac { 4\text{y} }{ 5\text{y} } \times 100=80%$

4. x + 5

Solution:

By definition

$ \text{Average} =\cfrac{\text{x}+(\text{x}+3)+(\text{x}+4)+((\text{x}+8)+(\text{x}+10)}{5}$

$ =\cfrac{5\text{x}+25}{5}$

$ ​=(\text{x}+5)$

1. 0

Solution:

Here, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ and $\bar{\text{x}}_2=25$

$\therefore\ \text{CV}_1=\frac{\sigma_1}{\bar{\text{x}}_1}\times100$

$\Rightarrow50=\frac{\sigma_1}{30}\times100$

$\therefore\ \sigma_1=\frac{30\times50}{100}=15$

and $\text{CV}_2=\frac{\sigma_2}{\bar{\text{x}}_2}\times100$

$\Rightarrow60=\frac{\sigma_2}{25}\times100$

$\therefore\ \sigma^2=\frac{60\times25}{100}=15$

Now, $\sigma_1-\sigma_2=15-15=0$ 

3. 10

Solution:

We have,

$= \frac{7+8+\text{x}+11+14}{5}$

$ = \text{x} = 40+\text{x} = 5\text{x} = \text{x} = 10$

3. 144

Solution:

The mean of 100 numbers is 45

$ ∴$ Sum of all 100 numbers = 100 × 45 = 4500

The mean of last 99 numbers is 44

$ ∴$Sum of all last 99 numbers = 99 × 44 = 4356

⇒ The first number = 4500 - 4356 = 144

3. 3.5

Solution:

First six natural numbers = 1, 2, 3, 4, 5, 6

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

$\text{Mean} = \frac{1 + 2 + 3 + 4 + 5 +6}{6} = \dfrac{21}{6}$

$ =3.5$

2. 24 years

Solution:

let presently the member be x_1, x_2, x₃.....x₈

So the average age$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$

......1Now the average age of all the members 3 years ago

$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$

​if x₁ the younger member is replaced by the older member y₁ ​then,

$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$

$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
$ ⇒\text{x}1​=\text{y}1​+24\Rightarrow \text{x}_1-\text{y}_1=24 $

4. 2.87

Solution:

We know that SD of first n natural numbers $\sqrt{\frac{\text{n}^2-1}{12}}$

Here, $\text{n}=10$

$\therefore\ \text{SD}=\sqrt{\frac{(10)^2-1}{12}}=\sqrt{\frac{99}{12}} $

$=\sqrt{8.25}=2.87$

3. 4.5

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 by taking out the average of both the value.

The given series is first arranged into ascending; 1, 2, 2, 3, 4, 5, 6, 7, 8, 9

$\text{N} = 10$

$ \text{median}= \frac{(10+1)}{2}$

$ \text{th term} = \frac{11}{2}$

$ \text{th term} = 5.5 $

$ = \frac{( \text{value of 5th term }+ \text{value of 6th term)}}{2}$

$ = \frac{(4+5)}{2} = \frac{9}{2} = 4.5$

2. 161

Solution:

⇒ Sum of the observation = 23 × 7 = 161

If each observation is multiplied by 23 then the sum is also multiplied by 23.

New sum = 23 × 161 = 3703

⇒ New mean $ =\frac{3703}{23}=161$

3. 1.2

Solution:

$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$

$=\frac{25}{5}$

$=5$

Taking the absolute value of deviation of each term from the mean, we get:

$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$

$=\frac{2+1+0+1+2}{5}$

$=\frac{6}{5}$

$=1.2$

3. $ \displaystyle 4\frac{3}{4}$

Solution:

$= \displaystyle 1\frac{1}{6}+2\frac{1}{3}+6\frac{2}{3}+ 8\frac{5}{6}$

$=\frac{7}{6}+ \frac{7}{3 }+ \frac{20}{3 }+\frac{53}{6}$

$ = \frac{6+7+14+40+53}{6} ​ $

$= \frac{114}{6}$

$ 19\therefore \text{Average}=\frac{19}{4}=4\dfrac{3}{4}$

1. $\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$

3. 20 years

Solution:

Let the average age of the staff

= x Age of the new teacher

= y According to the questionNew age of the staff reduced by

$ 2 \text{years}\Rightarrow \dfrac{20\text{x}-60+\text{y}}{20}$

$\text{ x}-2⇒20\text{x}−60+\text{y}​$

$ \Rightarrow 20\text{x}-60+\text{y}$

$ =20(\text{x}-2)⇒20\text{x}−60+\text{y}$

$ ⇒20\text{x}−60+\text{y}$

$ \Rightarrow\text{y}=60-40=20$

$ ⇒\text{y}=60−40$

= 20 Hence the age of the new teacher is 20 years.

3. 12.67

Solution:

$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$

$\text{M}_\text{e}=50$

$\therefore\ \text{MD}=\frac{114}{9}=12.67$

1. 15.15

Solution:

Mean of 20 observatios = 15

Sum of 20 observations = 15 × 20 = 300

Correct sum = 300 + 8 + 4 - 3 - 6 = 300

Correct mean $ = \dfrac{303}{20} = 15.15$

4. $\text{M.D.}=\frac{\sum\text{f}\ |\text{d}|}{\sum\text{f}}$

1. $\sqrt{\frac{52}{7}}$

Solution:

Given data are 6, 5, 9, 13, 12, 8 and 10

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

$=\sqrt{\frac{619}{7}-\Big(\frac{63}{7}\Big)^2}=\sqrt{\frac{4333-396}{49}}$

$=\sqrt{\frac{396}{49}}=\sqrt{\frac{52}{7}}$

4. 40

Solution:

Given:Average age of 30 students = 9years.

Total age of 3030 students = 9 × 30 = 270 years. Teachers age included.

So, average age of 30 students + one teacher = 10years.

⇒ Total age of 30 students + one teacher = 10 × 31 = 310 years.

$ \therefore$ age of teacher = 310 - 270 = 40 years.