Tag: converting between units of areas

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Questions Related to converting between units of areas

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

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

  1. $5\dfrac {1}{3}m$
  2. $5\dfrac {1}{2}m$
  3. $5\dfrac {2}{5}m$
  4. $5\dfrac {1}{4}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

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

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

  1. $8\dfrac {25}{27}m$
  2. $8\dfrac {24}{27}m$
  3. $8\dfrac {26}{27}m$
  4. $8\dfrac {22}{27}m$
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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 $m^2$)

  1. $19.03$
  2. $18.02$
  3. $18.03$
  4. $17.03$
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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

  1. $16.04$ metres
  2. $16.02$ metres
  3. $16.06$ metres
  4. $16.08$ metres
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

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

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

  1. $10\ \text{kg}\ 476\ \text{g}$
  2. $104\ \text{kg}\ 76\ \text{g}$
  3. $1\ \text{kg}\ 476\ \text{g}$
  4. $1047\ \text{kg}\ 6\ \text{g}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $15\ kg\ 305\ g$
  2. $15\ kg\ 470\ g$
  3. $16\ kg\ 305\ g$
  4. $16\ kg\ 300\ g$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  1. $78\ kg\ 953\ g$
  2. $789\ kg\ 532\ g$
  3. $89\ kg\ 532\ g$
  4. $780\ kg\ 432\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
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