MATHS M.C.Q (1 Marks)

2. 21

Solution:

Let in common no. in each group is X

Then Member of each group is 2X, 3X and 5X

Total of three group = (2X + 3X + 5X) 12 = 120x

And total of Two group = (2X + 3X)3 = 15X (given mean of two group is 3)

Then total of third group = 120X - 15X = 115X

Mean of third group $ =\frac{115\text{X}}{5\text{X}}=21$

3. 5.5

Solution:

Numbers are 1, 2, 3,..10

$\text{Sum of the numbers} = {\frac{\text{n}(\text{n}+1)}{2}}= \frac{10\times11}{2} = 55$

$\text{Mean} = \frac{55}{10} =5.5$

4. 44

Solution:

Given $\sum\text{x}_\text{i}^2=18000,\ \sum\text{x}_\text{i}=960$ and n = 60

$\therefore$ Variance

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

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

$=300-256$

$=44$

Hence, the correct answer is option (d).

1. 25

Solution:

Mean of 1010 observations = 25

Sum of 1010 observations = 25 × 10 = 250

After removing an observation with a value = 25

New sum = 250 - 25 = 225

New $ \text{mean} = \frac{225}{9} = 25$

2. 11.5

Solution:

 $ \text{mean}=\frac{6+10+11+11+13+13+13+16+12}{8}=11.6$

2. 2.57

Solution:

The given observations are 3, 10, 10, 4, 7, 10, 5.

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

Now,

Mean deviation from mean, MD

$=\frac{\sum|\text{x}_\text{i}-7|}{7}$

$=\frac{|3-7|+|10-7|+|10-7|+|4-7|+|7-7|+|10-7|+|5-7|}{7}$

$=\frac{4+3+3+3+0+3+2}{7}$

$=\frac{18}{7}$

$=2.57$

Hence, the correct answer is (b).

1. 5.8

Solution:

Values of x are: 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5and11.1

$\text{Mean}=\frac{\ \text{Sum}}{\text{Count}}​$

$\text{Mean}=\frac{9.8+5.4+3.7+1.7+1.8+2.6+2.8+8.6+10.5+11.1}{10}​$

$ \text{Mean}=\frac{58}{10}$

$\text{Mean}=5.8$

2. 89.85kg

Solution:

Difference in weight = 87 - 78 = 9kg

$ \therefore$ Correct average weight $ = 89.4 +\frac {9}{20}$

= 89.4 + 0.45 = 89.85kg

2. 13 years

Solution:

Total age of 15 students = (15 × 15)years = 225years

Total age of 5 students = (5 × 14)years = 70years

Total age of other 9 students = (9 × 16)years = 144years

$ \therefore$ Age of the 15th student = 225 - (70 + 144) = 225 - 214 = 11 years.

2. $ 0$

Solution:

Both the summations runs over $ \text{p,q,r} ∴ \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - (\text{p}^2 +\text{q}^2+\text{r}^2) - (\text{p}^2 +\text{q}^2+\text{r}^2)$

= 0 Hence, the answer is 0.

3. $\text{V}=\sigma^2$

Solution:

The variance is the square of the standard deviation.

2. 11.15

Solution:

Mean of 3 nos = 10 Total of 3 nos is 10 × 3 = 30 Mean of other 4 nos is 12 

Total of 4 nos is 12 × 4 = 48

total of 4 + 3 = 7

numbers is 48 + 30 = 78

Mean of 7 numbers is $ \cfrac{78}{7} = 11.15.$

4. 4

Solution:

If a set of observations, with SD σσ, are multiplied with a non-zero real number a, then SD of the new observations will be $|\text{a}|\sigma.$
Dividing the set of observations by -2 is same as multiplying the observations by $\frac{1}{-2}.$

New $\text{S.D.}=\Big|-\frac{1}{2}\Big|\times8$

$=\frac{8}{2}$

$=4$