Tag: maths

Questions Related to maths

Nehal purchased $7$ kg $200$ g of sugar, $9$ kg $395$ g of rice. What is the total weight which Nehal carried?

  1. $10$ kg $375$ g

  2. $16$ kg $595$ g

  3. $20$ kg $495$ g

  4. $24$ kg $765$ g


Correct Option: B
Explanation:
Total weight carried$=\left( 7.200+9.395 \right) ㎏=16.595㎏$
$=16㎏$ $595 gm$

Chris sold 112 kg 342 g of newspapers and 195 kg of magazines. Find the total quantity of articles sold.

  1. $200$ kg $143$ g

  2. $256$ kg $246$ g

  3. $307$ kg $342$ g

  4. $384$ kg $342$ g


Correct Option: C
Explanation:
Total quantity of articles sold$=\left( 112.342+195 \right) ㎏=307.342㎏$
$=307$ ㎏ $342$ gm

How many spherical bullets can be made out of a solid cube of lead whose edge measures $44\space cm$, each bullet being $4\space cm$ in diameter.

  1. $2451$

  2. $2541$

  3. $2304$

  4. $2536$


Correct Option: B
Explanation:

Number of spherical bullets formed $ = \dfrac {Volume  of 

cube}{Volume   of   one   spherical  bullet} $





Volume of a cube of edge a $ = {a}^{3} $





 Volume of a sphere of radius 'r' $ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $





As the diameter of the sphere is $4$ cm, its radius r $ = 2$ cm





Hence, number of spherical bullets formed $ = \dfrac {44 \times 44 \times 44}{\dfrac {4}{3}

\times \dfrac {22}{7} \times 2 \times 2 \times 2} = 2541 $

A hemispherical tank of radius $1\displaystyle\frac{3}{4}m$ is full of water. It is connected with a pipe which empties it at the rate of $7\space litres$ per second. How much time will it take to empty the tank completely?

  1. $26.74\space min.$

  2. $26.54\space min.$

  3. $26.4\space min.$

  4. $26\space min.$


Correct Option: A
Explanation:

Suppose the pipe takes $x$ seconds to empty the tank. 

Then, 
The volume of the water that flows out of the tank in $x$ seconds =Volume of the hemispherical tank

The volume of the water that flows out of the tank $x$ in seconds= Volume of the hemispherical shell of radius $175cm$

$\Rightarrow 7000x=\cfrac { 2 }{ 3 } \times \cfrac { 22 }{ 7 } \times 175\times 175\times 175$

$\Rightarrow x=\cfrac { 2 }{ 3 } \times \cfrac { 22 }{ 7 } \times \cfrac { 175\times 175\times 175 }{ 7000 } =1604.16seconds$

$\Rightarrow x=\cfrac { 1604.16 }{ 60 } =26.74\quad minutes$

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, then its mean is

  1. 40

  2. 30

  3. 20

  4. none of these


Correct Option: A
Explanation:

Given $\sigma = 20$, coefficient of variation $=50$ %
We know coefficient of variation $=\cfrac{\sigma }{\bar{x}}\times 100=50$
$\Rightarrow \bar{x} = 2\times \sigma = 40$

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250 $. Then Co-efficient of variation is

  1. $10\%$

  2. $40\%$

  3. $50\%$

  4. None


Correct Option: A
Explanation:
$\sum(x-\overline{x})^2=250$, $\overline{x}=50$
$\Rightarrow$  Standard deviation $(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$
$\Rightarrow$  Coefficient of variation $=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$
                                             $=\dfrac{\sigma}{Mean}\times 100$

                                             $=\dfrac{5}{50}\times 100$

                                             $=10\%$

The Coefficient of Variation is given by:

  1. $\dfrac{Mean}{\ Standard \ \ deviation } \times 100$

  2. $\dfrac{\ Standard \ \ deviation }{Mean}$

  3. $\dfrac{Standard \ \ deviation }{Mean }\times 100$

  4. $\dfrac{Mean}{Standard \ Deviation}$


Correct Option: C
Explanation:

The coefficient of variation (CV) is a standardized measure of dispersion 

. It is defined as the ratio of the standard deviation to the mean.
$CV\quad =\quad \cfrac { \sigma  }{ Mean }\times100 $

If mean of a series is 40 and variance 1486, then coefficient of variation is 

  1. $0.9021$

  2. $0.9637$

  3. $0.8864$

  4. $0.9853$


Correct Option: B
Explanation:

If mean of the given dist. be $\bar{x}$ and S.D be $\sigma $
then given $\bar{x} = 40, \sigma^2 = 1486$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, the its mean is

  1. 40

  2. 30

  3. 20

  4. None of these


Correct Option: A
Explanation:

We know if a distribution having mean $\bar{x}$ and standard deviation $\sigma$
then coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$
$\therefore \cfrac{20}{\bar{x}}\times 100=50\Rightarrow \bar{x} = 40$
Hence required mean is $=40$

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of varition is

  1. 10%

  2. 40%

  3. 50%

  4. None of these


Correct Option: A
Explanation:

Given $\displaystyle \Sigma \left ( x _{i}-\overline{x} \right )^{2}=250$,$n=10,\overline{x}=50$

Now, $\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x _{i}-\overline{x} \right )^{2}}$

$= \sqrt{\dfrac{1}{10}\times 250}=5$ 
Hence coefficient of variation $\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$%