Tag: metric system

Questions Related to metric system

Consider the railway platform which is in square shape having side length $2\ km$. Then area of the platform is $4$ ____.

maths perimeter and area converting between units of areas the metric system metric system

  1. $m$

  2. $km$

  3. $km^{2}$

  4. $m^{2}$

Show answer
Correct Option: C
Explanation:

Area of square platform$=2\times 2=4 km^2$

The area of a square field is $30\dfrac {1}{4}m^2$. Calculate the length of the side of the square.

maths perimeter and area converting between units of areas the metric system metric system

  1. $5\dfrac {1}{3}m$

  2. $5\dfrac {1}{2}m$

  3. $5\dfrac {2}{5}m$

  4. $5\dfrac {1}{4}m$

Show answer
Correct Option: B
Explanation:

Let side of square field be $x$.


Then area of square field $= x^2$


According to question

$x^2 = 30 \dfrac{1}{4} m^2$

$x^2 = \dfrac{121}{4} m^2$

$x = \dfrac{\sqrt{121}}{\sqrt{4}} m$

$x = \dfrac{11}{2}$

Hence side of square field is $5 \dfrac{1}{2} m$

Option (B)

The area of a square field is $80\dfrac {244}{729}$ square metres. Find the length of each sides field.

maths perimeter and area converting between units of areas the metric system metric system

  1. $8\dfrac {25}{27}m$

  2. $8\dfrac {24}{27}m$

  3. $8\dfrac {26}{27}m$

  4. $8\dfrac {22}{27}m$

Show answer
Correct Option: C
Explanation:

Given area of square field $= 80 \dfrac{244}{729} m^2$


Let us assume that side of square field is x 


Then area of field $= x^2$

According to question

$x^2 = 80 \dfrac{244}{729} m^2$

$x^2 = \dfrac{58,564}{729} m^2$

$x = \dfrac{\sqrt{58,564}}{\sqrt{729}} m$

$x = \dfrac{242}{27} m$

Hence side of field is $8 \dfrac{26}{27} m$.

Option (C)

The area of a square field is $325\ m^2$. Find the approximate length of one side of the field. (upto 2 places of decimals) (in $m^2$)

maths perimeter and area converting between units of areas the metric system metric system

  1. $19.03$

  2. $18.02$

  3. $18.03$

  4. $17.03$

Show answer
Correct Option: C
Explanation:

We know area of any square is its side x side 

Let us assume that side of square field is x 
Then area of square field $= x^2$
According to question

$x^2 = 325 m^2$

$\Rightarrow x^2 = 325$

$x = \sqrt{325}$

So $x = 18.03$

Hence side of square field is $18.03 m$

option (C)

The area of a square playground is $256.6404$ square metres. Find the length of one side of the playground.

maths perimeter and area converting between units of areas the metric system metric system

  1. $16.04$ metres

  2. $16.02$ metres

  3. $16.06$ metres

  4. $16.08$ metres

Show answer
Correct Option: B
Explanation:

Let side of square play ground be x

Then area of square play ground will be $x^2$
According to question

$x^2 = 256.6404 m^2$

$x = \sqrt{256.6404} m$

$x = \dfrac{\sqrt{2566404}}{\sqrt{10000}} m$

$= \dfrac{1602}{100}$

Hence side of square is $16.02 m$

Option (B)

By converting the $5.6 m^2$ into the $cm^2$, the answer will be

maths perimeter and area converting between units of areas the metric system metric system

  1. $0.0056cm^2$

  2. $5600cm^2$

  3. $56000cm^2$

  4. $560cm^2$

Show answer
Correct Option: B

Which would be weight closest to 800 kg?

maths metric system converting between units of areas the metric system units of area and their conversions

  1. Feather

  2. Ball

  3. Cow

  4. None of these

Show answer
Correct Option: C
Explanation:

Weight of feather is very low about $0.082$ grams.


Weight of ball is also normally about $150$ grams.

And it's well known that cow is heavy and it weighs about $800$ kg. In fact weight of a bull(male) is about $1100$ kg.

Subtract $2\ \text{kg}\ 54\ \text{g}$ from $12\ \text{kg}\ 530\ \text{g}$.

maths metric system converting between units of areas the metric system units of area and their conversions

  1. $10\ \text{kg}\ 476\ \text{g}$

  2. $104\ \text{kg}\ 76\ \text{g}$

  3. $1\ \text{kg}\ 476\ \text{g}$

  4. $1047\ \text{kg}\ 6\ \text{g}$

Show answer
Correct Option: A
Explanation:
We know $1$ kg $=1000$ gm
(i) We have to subtract   $2$ kg  $54$ gm from $12$ kg $530$ gm
$2$ kg $54$ gm $ = 2$ kg $+$ $54$ gm $=A $    .......(1)
$12$ kg $530$ gm $ = 12$ kg $+$ $530$ gm $=B $     .......(2)
Now as per the question, we have to subtract $A$ from $B$
$B-A = [12$ $\text{kg}$ $530$  $\text{gm}$] $-$[$2$ $\text{kg}$ $54$ $\text{gm}$ $]$
Subtract like terms 
$B-A=\left[12 \ \text{kg} -2 \ \text{kg}\right]+\left[530 \ \text{gm}-54 \ \text{gm}\right]$

$B-A = 10  $ kg $+ $ $476$ gm
So, option A is correct

Subtract $21\ kg\ 370\ g$ from $37\ kg\ 675\ g$ without conversion into gram.

maths metric system converting between units of areas the metric system units of area and their conversions

  1. $15\ kg\ 305\ g$

  2. $15\ kg\ 470\ g$

  3. $16\ kg\ 305\ g$

  4. $16\ kg\ 300\ g$

Show answer
Correct Option: C
Explanation:

Let $  B = $   $37$ $kg$  $675$  $gm$ 

$A = $   $21$ $kg$  $370$  $gm$ 
(i)  Subtract   $21$ $kg$  $370$  $gm$ from $37$ $kg$  $675$  $gm$

$21$ $kg$  $370$  $gm$ $ = 21kg + 370gm$  $=A $   ...................(1)

$37$ $kg$  $675$  $gm$ $ = 37kg + 675gm$  $=B $   ...................(2)


Now as per the question we have to subtract A from B

$B-A = [37$ $kg$  $+ $  $675$  $gm$] $ - $ [$21$ $kg$ $+ $  $370$  $gm$]

$B-A = [37$ $kg$ $- $ $21$  $kg$] $+  $ [$675$ $gm$ $-$ $370$  $gm$]


$B-A = 16  $  $kg $ $+  $  $305$ $gm$ 

$16  $  $kg $  $305$ $gm$ is the answer.

A truck was loaded with $482\ kg\ 100\ g$ of pumpkins and $307\ kg\ 432\ g$ of watermelons. Find the total weight carried by the truck.

maths metric system converting between units of areas the metric system units of area and their conversions

  1. $78\ kg\ 953\ g$

  2. $789\ kg\ 532\ g$

  3. $89\ kg\ 532\ g$

  4. $780\ kg\ 432\ g$

Show answer
Correct Option: B
Explanation:
Let $482$ $kg$  $100$  $gm$ $ = 482kg + 100gm$  $=A =  $  Weight of  Pumpkins
$307$ $kg$  $432$  $gm$ $ = 307kg + 432gm$  $=B =  $  Weight of  Watermelons
(i)  Given that, we have to add $482$ $kg$  $100$  $gm$   to $307$ $kg$  $432$  $gm$
$ 482kg + 100gm$  $=A $   ...... (1)
$  307kg + 432gm$  $=B $   ...... (2)


Now as per the question we have to add A and B

$A+B = [482$ $kg$  $100$  $gm$] $+ $ [$307$ $kg$  $432$  $gm$] 
$A+B = [482$ $kg$ $+$  $307$  $kg$] $+ $ [$100$ $gm$ $+$  $432$  $gm$]

$A+B = 789  $  $kg$ $+ $   $532$ $gm$ 
The Total weights carried  in Truck is  $ 789  $  $kg$   $532$ $gm$
Hence, option $B$ is correct