MATHS M.C.Q (1 Marks)

3. 408

Solution:

⇒ Let total number of employees be

x ⇒ Average salary of total employee

= Rs. 70 = Average of 12 employees = Rs. 400 = Rs. 400 ⇒ Average of remaining employees

$ \text{Rs}.60∴ 70=\frac{400\times 12+(\text{x}-12)\times 60}{\text{x}}$

$ ∴ 70\text{x}=4800+60\text{x}-720$

$ ∴ 70\text{x}=4080+60\text{x}$

$ ∴ 10\text{x}=4080\therefore\text{x}=408$

Total number of employees are 408.

2. 12 years

Solution:

⇒ The average age of 66 students is 11 years.

⇒ Sum of age of 66 students = 6 × 11 = 66

⇒ When two more students of age 14 and 16 added to 6 students then,

total students will become 8 ⇒ Sum of age of 8 students = 66 + 14 + 16 = 96

⇒ Required average $ =\dfrac{96}{8}=12\ \text{years}$

3. -1

Solution:

Mean of the given distribution is,

$ =\frac{(\text{x}+2)+(2\text{x}+3)+(3\text{x}+4)+(4\text{x}+5)}{4} = \text{x} +2$

= 4(x + 2) + (2x + 3) + (3x + 4) + (4x + 5) ​= x + 2, (given)

$=\frac{10\text{x}+14}{4} = \text{x}+2$

= 10x + 14 = 4x + 8 ⇒ x = -1

1. 8.25

Solution:

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

On multiplying each number by -1, we get -1, -2, -3, -4, -5, -6, -7, -8, -9, -10 On adding 1 in each number.

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

$\therefore\ \sum\text{x}_\text{i}=-\frac{9\times10}{2}=-45$

and $\sum\text{x}^2_\text{i}=0^2+(-1)^2+(-2)^2+\ ....\ +(9)^2=\frac{9\times10\times19}{6}=285$

$\text{SD}=\sqrt{\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2}=\sqrt{\frac{285}{10}-\frac{2025}{100}}$

$=\sqrt{\frac{2850-2025}{100}}=\sqrt{8.25}$

Now, $\text{variance}=(\text{SD})^2=\big(\sqrt{8.25}\big)^2=8.25$

3. 25%

Solution:

Standard deviation is expressed in the following manner:

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

$=\sqrt{\frac{1530}{10}-(12)^2}$

$=\sqrt9$

$=3$

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

$=\frac{3}{12}\times100$

$=25%$

4. 50

Solution:

Mean of 8 numbers=25

$ \therefore\ \text{A.M}=\frac{\sum \text{x}}{\text{n}}$

​$ \Rightarrow 25=\dfrac{\sum \text{x}}{8}$

$ \Rightarrow∑\text{x}=25×8=200$

If each number is multiply by 2 then new sum

$ =200\times 2=400$

$ ∴ \text{New mean}=8400​=50$

1. a = 1.25, b = -5

Solution:

It is given that y_i = ax_i + b for i = 1, 2, 3, ..., n, where a and b are constants.

$\overline{\text{x}_\text{i}}=48$ and $\sigma_{\text{x}_\text{i}}=12$

$\overline{\text{y}_\text{i}}=55$ and $\sigma_{\text{y}_\text{i}}=15$

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

$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\sum(\text{ax}_\text{i}+\text{b})}{\text{n}}$

$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\text{a}\frac{\sum\text{x}_\text{i}}{\text{n}}+\frac{\sum\text{b}}{\text{n}}$

$\Rightarrow\overline{\text{y}_\text{i}}=\text{a}\overline{\text{x}_\text{i}}+\text{b}$

$\Rightarrow55=48\text{a}+\text{b}\ ...(1)$

Now,

Standard deviation of y_i = Standard deviation of ax_i + b

$\Rightarrow\sigma_{\text{y}_\text{i}}=\text{a}\times\sigma_{\text{x}_\text{i}}$

$\Rightarrow15=12\text{a}$

$\Rightarrow\text{a}=\frac{15}{12}=1.25$

Putting a = 1.25 in (1), we get

b = 55 - 48 × 1.25 = 55 - 60 = -5

Thus, the values of a and b are 1.25 and -5, respectively.

Hence, the correct answer is option (a).

2. 6

Solution:

Given set M = {x, 2x, 4x}

Average (arithmetic mean) of the numbers in set M is 14.

Value of x will be,

$ \Rightarrow \frac{\text{x}\ +\ 2\text{x}\ +\ \text{4x}}{3}$

$ \Rightarrow7\text{x}=42$

$ \Rightarrow\text{x}=6$

3. 17

Solution:

No. of test in the month Jan 2000 = 4 Total Marks obtained in 4 test

= 16 + 14 + 18 + 20

$ 68\therefore \text{A.M}=\frac{\sum \text{x}}{\text{n}}=\frac{68}{4}=17$

3. $\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$

Solution:

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

$\overline{\text{Y}}=\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\frac{\text{a}\sum\text{X}+\text{nb}}{\text{c}}}{\text{n}}$

$=\frac{\text{a}\sum\text{X}}{\text{nc}}+\frac{\text{nb}}{\text{nc}}$

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

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

$=\sigma^2$

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

$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\Big)}{\text{n}}$

$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\Big)^2}{\text{n}}$

$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\big(\text{x}_1-\overline{\text{X}}\big)^2}{\text{n}}$

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

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

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

3. 8.6

Solution:

$\text{ Mean} = \frac{9+11+12+4+7}{5}$

$ ​\text{Mean} = \dfrac{43}{5}= 8.6$

1. 4

Solution:

Median of length drawn to the side, $\text{c}= \frac{1}{2}\sqrt{\left (2(\text{a}^{2}+\text{b}^{2})-\text{x}\text{c}^{2}\right)}$
$ = \frac{1}{2}\sqrt{\left (2(5^{2}+3^{2})-2^{2}\right)} $

$ = \frac{1}{2}\sqrt{\left (2(34)-2^{2}\right)}$

$ = \frac{1}{2}\sqrt{\left (68-4\right)} $

$ = \frac{1}{2}\sqrt{64} $

$ = \frac{8}{2}$

$=4$

4. 5.6

Solution:

The first five prime numbers are 2, 3, 5, 7, 11

$\text{Mean}=\frac{\text{sum of the terms}}{\text{no. of terms}}$

$\text{Mean}=\frac{2+3+5+7+11}{5}$

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

2. 40

Solution:

In this question one day attendance not givenGiven attendance as per Answer.

are 40, 42, 30, 35, 45, 44, 41, 38, 44 and 41 Then mean

$ =\frac{40+42+30+35+45+44+41+38+44+41}{10}$

$ =\frac{400}{10}=40$

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

Solution:

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

3. $\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$

Solution:

The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$

2. 10 × (11)!

Solution:

$ \sum(\text{r}^2+1)\text{r}!=\sum[\text{r}(\text{r }+1) -(\text{r}-1)\text{r}!$

$ =\sum\limits^{10}_\text{r=1}[\text{r}(\text{r }+1)!-(\text{r}-1)\text{r}!]$

$= (1×2!−0×1!)+(2×3!−1×2!)+......+(10×11!−9×10!)=10×11!​ $

2. 2.57

Solution:

Observations are fiven by 3, 10, 10, 4, 7, 10, and 5

$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$

$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$

3. 62

Solution:

The series in ascending order is: 34, 37, 53, 55, x, x + 2, 77, 83, 89 and 

The series has 10 numbers, even numbers.

Hence, the median will be the mean of the two middle numbers:

median = mean of 5^th and 6^th terms

$63=\frac{\text{x}\ + \ \text{x}\ +\ 2}{2}$

$1260=2\text{x}+2$

$ 2\text{x}=124$

$ \text{x}=62$

3. 120

Solution:

Mean of 200 observations = 150

Sum of 200 observations = 200 × 150 = 30000

Mean of 300 observations = 100

Sum of 300 observations = 300 × 100 = 30000

Total Sum = 30000 + 30000 = 60000

Number of observations = 100 + 200 = 500

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

$ = \frac{60000}{500} = 120$

2. x₃

Solution:

The five consecutive odd numbers are $ \text{x}_1+ \text{x}_1+2, \text{x}_1 + 4,\text{x}_1 +6,\text{x}_1 +8$

$ \therefore \text{mean}=\frac{\text{x}_1 \ + \ \text{x}_1 \ +\ 2+\text{x}_1 \ +\ 4\ +\ \text{x}_1 \ +\ 6\text{x}_1 \ +\ 8}{6}$

$ =\frac{5\text{x}_1\ +\ 20}{5}$

$ =\text{x}_1\ +\ 4=\text{x}_3$

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

Solution:

The given observations are 6, 5, 9, 13, 12, 8, 10.

Now,

$\sum\text{x}_\text{i}=$ 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63

$\sum\text{x}_\text{i}^2=$ 36 + 25 + 81 + 169 + 144 + 64 + 100 = 619

$\therefore$ Standard deviation of the observations, $\sigma$

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

$=\sqrt{\frac{1}{7}\times619-\Big(\frac{1}{7}\times63\Big)^2}$

$=\sqrt{\frac{619}{7}-81}$

$=\sqrt{\frac{619-567}{7}}$

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

Hence, the correct answer is option (d).

2. 29.55 cm

Solution:

Incorrect total of 20 measurement = 20 × 56 = 1120

Correct total = 1120 - 64 + 61 = 1117

$ ∴ \text{Correct average} = \displaystyle \frac{1117}{20} = 55.85\text{cm}$

2. 13

Solution:

Given observations x, x + 2, x + 4, x + 6, x + 8

$\Rightarrow\frac{5\text{x}+ 20}{5}=11$

⇒ x = 7

So, the observations are 7, 9, 11, 13, 15

Req. mean $ =\frac{11+13+15}{3}=13$

2. 24, 12, 6

Solution:

Let the third number be x Then Second number = 2x

and first number = 4x

Sum of the reciprocals of these 3 numbers $ =\frac{1}{\text{4x}}+\frac{1}{2\text{x}}+\frac{1}{\text{x}}=\frac{1+2+4}{4\text{x}}$

$= \frac{7}{\text{4x}}$

Given, $\frac{7}{4\text{x}}=3\times \frac{7}{72}$

$=4\text{x}=24= \text{x}=6 $

Therefore, the three numbers are,

4 × 6, 2 × 6, 6 i.e. 24, 12, 6.

1. 82

Solution:

Average $ = \frac{20\left ( 80 \right )\ +\ 5\left ( 90 \right )}{25}$

$=\frac{1600\ +\ 450}{25}$

$=\frac{2050}{25}=82$

3. ks

Solution:

Here, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$

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

$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$

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

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

$=\text{K}.\text{S}$

1. 8.6

Solution:

N = 10

$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$

$=\frac{494}{10}$

$=49.4$

Mean deviation from the mean $=\frac{88.8}{10}$

$= 8.88$

Disclaimer: No option is matching the answer.

3. 35 years

Solution:

Let the age of each student be x years

Then, the age of teacher will be (x + 20) years

$\text{Mean age} =\frac{\left (\text{x}+20 \right )+3\text{x}}{4}$

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

⇒ x = 15

Hence, age of the teacher = 35 years

2. 10

Solution:

Given, set $$f = 4, 5, 11, 13, 16, 18, x Average of set

f = 11

$\Rightarrow \dfrac{4+5+11+13+16+18+\text{x}}{7}=11$

$\Rightarrow \frac{67+\text{x}}{7}=11$

$\Rightarrow 67+{\text{x}}=77$

$\Rightarrow \text{x}=77-67=10$

1. 8.25

Solution:

The given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

If 1 is added to each number, then the new numbers obtained are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Now,

 $\sum\text{x}_\text{i}=$ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65

$\sum\text{x}_\text{i}^2=$ 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505

$\therefore$ Variance of the numbers so obtained

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

$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$

$=50.5-42.25$

$=8.25$

4. 63Kg

Solution:

$ \displaystyle \frac{54+59+\text{x}+53+73+49+50+58+45}{9}=56$
On simplification $ \text{x} = 63$

4. 8.25

Solution:

Given numbers are 1, 2, 3,4, 5, 6, 7, 8, 9 and 10

If 1 is added to each number, then observations will be 2, 3,4, 5, 6,7, 8, 9, 10 and 11

$\therefore\ \sum\text{x}_\text{i}=2+3+4+\ ....\ +11$

$=\frac{10}{2}\big[2\times2+9\times1\big]=5[4+9]=65$

and $\sum\text{x}^2_\text{i}=2^2+3^2+4^2+5^2+\ .....\ +11^2=(1^2+2^2+3^2+\ .....\ +11^2)-(1^2)$

$=\frac{11\times12\times23}{6}-1=505$

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

$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$

$=50.5-(6.5)^2$

$=50.5-42.35$

$=8.25$

2. 26

Solution:

Mean of 55 numbers = 18

Sum of these 55 numbers = 18 × 5 = 90

Let number that has been excluded be x New mean = $ \dfrac{90-\text{x}}{4} = 16$

Solving this, we get 90 - x = 64

x = 26

2. 25

Solution:

Given observations 23, 28, 13, 16, 20 No. of observations are 5 Mean of

observations $ \dfrac{23+28+13+16+20}{5}$

$ =\dfrac{100}{5}$

$ =20$