Tag: maths

Questions Related to maths

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = $

  1. $-\frac{4}{3}p$

  2. 5q

  3. -7r

  4. 5q-7r


Correct Option: D
Explanation:

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = \frac{1}{3}(-2p+6q-9r)-\frac{1}{3}(-2p -9q +12r) $
$=\frac{1}{3} [\left (-2p+6q-9r)-(-2p -9q +12r) \right]$
$=\frac{1}{3} (-2p + 6q - 9r + 2p + 9q -12r)$
$=\frac{1}{3} (15q - 21r)$
$=(5q - 7r)$

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$

  1. $(a+y)^{2}$

  2. $\frac{3a}{16}$

  3. $\frac{9}{16}(a+y)^{2}$

  4. 1


Correct Option: C
Explanation:

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left (\frac{3}{4}(a+y) \right) \right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{4}(a+y)\right ]$
= $ \frac{3}{4}(a+y) \left [ \frac{3}{4}(a+y)\right ]$
= $ \frac{9}{16}(a+y)^2$

$-84\times 29+365=$?

  1. $2436$

  2. $2801$

  3. $-2801$

  4. $-2071$

  5. None of these


Correct Option: D
Explanation:

Given Exp. $=-84\times (30-1)+365$
$=-(84\times 30)+84+365$
$=-2520+449$
$=-2071$

In a class, there are 18 boys who are over 160 em tall. If these constitute three-fourths of the boys and the total number of boys is two-third of the total number of students in the class, what is the number of girls in the class?

  1. 6

  2. 12

  3. 18

  4. 24


Correct Option: B
Explanation:

Total number of boys $\displaystyle=18\div\frac{3}{4}=24$

Total number of students $\displaystyle=24\times\frac{3}{2}=36$

$\therefore$ Number of girls $=36-24=12$

$35+15\times 1.5=$?

  1. $85$

  2. $51.5$

  3. $57.5$

  4. $5.25$

  5. None of these


Correct Option: C
Explanation:

Given Exp.$=35+15\times \cfrac{3}{2}=35+\cfrac{45}{2}=35+22.5=57.5$

$(800\div 64)\times (1296\div 36)=$?

  1. $420$

  2. $460$

  3. $500$

  4. $540$

  5. None of these


Correct Option: E
Explanation:

Given Exp. $=\cfrac { 800 }{ 64 } \times \cfrac { 1296 }{ 36 } =450$

$9+\cfrac { 3 }{ 4 } +7+\cfrac { 2 }{ 17 } -\left( 9+\cfrac { 1 }{ 15 }  \right) =$?

  1. $7+\cfrac { 719 }{ 1020 } $

  2. $9+\cfrac { 817 }{ 1020 } $

  3. $9+\cfrac { 719 }{ 1020 } $

  4. $7+\cfrac { 817 }{ 1020 } $

  5. None of these


Correct Option: D
Explanation:

Given the sum $=9+\cfrac { 3 }{ 4 } +7+\cfrac { 2 }{ 17 } -\left( 9+\cfrac { 1 }{ 15 }  \right) $
$=(9+7-9)+\left( \cfrac { 3 }{ 4 } +\cfrac { 2 }{ 17 } -\cfrac { 1 }{ 15 }  \right) $
$=7+\cfrac { 765+120-68 }{ 1020 } $
$=7+\cfrac { 817 }{ 1020 } $

${\rm{1}}\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ is equal to____________.

  1. ${\rm{1}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$

  2. ${\rm{1}}{{\rm{0}}^3}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$

  3. ${\rm{1}}{{\rm{0}}^{ - 2}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$

  4. ${\rm{1}}{{\rm{0}}^{ - 3}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$


Correct Option: D
Explanation:

We have,

$1\ g/cm^3$

Since,
$1000\ g=1\ kg$
$1\ g=10^{-3} \ kg$

So,

$1\ g/cm^3 =10^{-3}\ kg/cm^3$

Hence, this is the answer.

A spherical bowl  with diameter  $24$ m. How much it weigh  (approx. in kg)  (if $1$ gm $= 1$ cubic cm).

  1. $7216473.6 kg$

  2. $216473.6 kg$

  3. $721473.6 kg$

  4. None of these


Correct Option: A

Weight of $1$ mango is $250$ gms. Then identify the weight of $16$ mangos. (Write your answer in kilograms)

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: D
Explanation:
the weight of $1$ mango $=250 $  $gram$

weight of $16$ mangoes $=250 \times 16 = 4000$  $gram$

We know that

$1$  $kg =1000$  $gram$

$1$  $gram =\dfrac1{1000}$  $kg$

Given That, we have to convert $4000$  $gram$  to   $kg$

$4000$  $gm =\dfrac{1}{1000} \times 4000$  $kg$

                    $ =\dfrac{4000}{1000} $  $kg$

                   $=4$  $kg$

So, option $D$ is correct