Tag: maths

Questions Related to maths

State true or false.
$10cm$ in kilometre is $0.0001km$

  1. True

  2. False


Correct Option: A
Explanation:

As we know,


$1$ m $= 100$ m


$1$ km $= 1000$ m

Given, $10$ cm

So, $10$ cm $= \dfrac{10}{100}$ m$= 0.1$ m

$0.1$ m = $\dfrac{0.1}{1000}$ km $= 0.0001$ km

So, $10$ cm $= 0.0001$ km

State true or false.

$9$ paise is equal to Rs.$0.09$.

  1. True

  2. False


Correct Option: A
Explanation:

As we know,

1 Rs = 100 paisa

Given, 9 paisa
So, 9 paisa = $\frac{9}{100}$ Rs
= 0.09 Rs

Fill in the banks: 
$1765$ grams $=................... kg$

  1. $1.765$

  2. $1.65$

  3. $1.55$

  4. $1.25$


Correct Option: A
Explanation:

$1$ kg $= 1000$ gm


So, $1765$ grams $= \dfrac{1765}{1000}$ kg


$= 1.765$ kg 

Let $D$ represent a repeating decimal. If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures which do repeat themselves, then the incorrect expression is

  1. $D = P.QQQ....$

  2. $10^{r}D = P.QQQ...$

  3. $10^{r + s}D = PQ .QQQ .....$

  4. $10^{r}(10^{s} - 1)D = Q(P - 1)$

  5. $10^{r} . 10^{2s}D = PQ.QQQ...$


Correct Option: D
Explanation:

$D = .PQQQ .... = .a _{1} ...a _{r}b _{1} ...b _{x}b _{1} ...b _{s} ...$ So $(a), (b), (c)$ and $(e)$ are all correct choices. To check that $(d)$ is incorrect, we have
$10^{r + s}D - 10^{r}D = PQ - P. \therefore 10^{r}(10^{s} - 1)D = P(Q - 1)$.

$2+\dfrac {1}{2}$ In the decimal form

  1. $2.5$

  2. $1.5$

  3. $3.5$

  4. $2$


Correct Option: A
Explanation:

$2+\dfrac{1}{2}$

As $\dfrac{1}{2}=0.5$
So $2+0.5=2.5$

If the standard deviation of $x _{1},x _{2},.....x _{n}$ is 3.5, then the standard deviatiuon of $-2x _{1}-3,-2x _{2}-3....,-2x _{n}-3$ is

  1. -7

  2. -4

  3. 7

  4. 1.75


Correct Option: C

If $\sigma$ $f _i$ $x _i$  = 20 and $\sigma$ $f _i$ = 4, what is the mean of the data.

  1. $\dfrac{1}{5}$

  2. $80$

  3. $16$

  4. $5$


Correct Option: A
Explanation:
$\sigma fixi=20$
and $\sigma fi=4$

Hence, Mean $=\dfrac{\sigma fi}{\sigma fixi}=\dfrac{4}{20}=\dfrac{1}{5}$

Hence Option $A$ is correct

The variance of the data $6,\ 8,\ 10,\ 12\,,14\,,\ 16,\ 18,\ 20,\ 22,\ 24$ is

  1. $15$

  2. $20$

  3. $30$

  4. $33$


Correct Option: D
Explanation:
Mistake :$14$ is not given
Mean $\bar x=\dfrac{6+8+10+12+14+16+18+20+22+24}{10}=\dfrac{150}{10}=15$
Variance$=\dfrac{1}{n} \sum\limits _{i=1}^n(x _{i}-\bar x)^2$
$\implies \dfrac{1}{10}((6-15)^2+(8-15)^2+(10-15)^2+(12-15)^2+(14-15)^{2}+(16-15)^2+(18-15)^2+(20-15)^2$
$+(22-15)^2+(24-15)^2$

$\implies \dfrac{81+49+25+9+1+1+9+25+49+81}{10}$

$\implies \dfrac{330}{10}=33$

The variate x and u are related by $\displaystyle u= \frac{x-a}{h}$ then correct relation between $\displaystyle \sigma _{x}:and:\sigma _{u}$

  1. $\displaystyle \sigma _{x}= h\sigma _{u}$

  2. $\displaystyle \sigma _{x}= h+\sigma _{u}$

  3. $\displaystyle \sigma _{u}= h\sigma _{x}$

  4. $\displaystyle \sigma _{u}= h+\sigma _{x}$


Correct Option: A
Explanation:

Given $\displaystyle u =\frac{x}{h}-\frac{a}{h}$
Since,S.D. is not depend on change of origin but it is depend on change of scale.
$\displaystyle \therefore \sigma _{u}=\frac{\sigma _{x}}{h}$
$\Rightarrow h\sigma _{u}=\sigma _{x}$

Standard deviation of a collection of data is $2\sqrt{2}$. If each value in a data set  is multipled by $3$, then the standard deviation of the new data is.

  1. $\sqrt{12}$

  2. $4\sqrt{2}$

  3. $6\sqrt{2}$

  4. $9\sqrt{2}$


Correct Option: C
Explanation:
The standard deviation would also be multiplied by $3$.
Because the mean would also be $3x$ larger, the differences from the mean would be $3x$ larger too.
It is the same idea as if you were looking at your data set through an enlarging lens- everything would be $3x$ bigger, not only the data values, but also the mean, the differences from the mean, but just everything!
$\therefore$ the standard deviation becomes $2\sqrt{2}\times 3=6\sqrt{2}$