Tag: units of area and their conversions

Questions Related to units of area and their conversions

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

A shopkeeper purchased $392\ kg\ 500\ g$ of orange. Later on, he found that $56\ kg\ 460\ g$ of oranges were rotten. Find the quantity of oranges in good condition.

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

  1. $330\ kg\ 400\ g$

  2. $336\ kg\ 4\ g$

  3. $336\ kg\ 40\ g$

  4. $33.6\ kg\ 40\ g$

Show answer
Correct Option: C
Explanation:
Quantity of oranges in good condition$=\left( 392.500-56.460 \right) ㎏$
$=336.040ℊ\Rightarrow$ $336$ ㎏ $40$ gm

Add $95\ kg\ 45\ g$ and $45\ kg\ 300\ g$.

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

  1. $14\ kg\ 34\ g$

  2. $140\ kg\ 300\ g$

  3. $14\ kg\ 345\ g$

  4. $140\ kg\ 345\ g$

Show answer
Correct Option: D
Explanation:

We know that, $gram$  is abbreviated as $gm$

$1$  $kg =1000$  $gram$


(i) To add   $95$ $kg$  $45$  $gm$  to $45$ $kg$  $300$  $gm$

$95$ $kg$  $45$  $gm$ $ = 95kg + 45gm$  $=A $   ...................(1)

$45$ $kg$  $300$  $gm$ $ = 45kg + 300gm$  $=B $   ...................(2)


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

$A+B = [95$ $kg$  $45$  $gm$] $+ $ [$45$ $kg$  $300$  $gm$]

$A+B = [95$ $kg$ $+$  $45$  $kg$] $+ $ [$45$ $gm$ $+$  $300$  $gm$]


$A+B = 140  $  $kg$ $+ $   $345$ $gm$ 

So, Option $D$ is correct

If Raina weighs $54\ kg\ 43\ g$ and Rohit weighs $60\ kg\ 760\ g$. Then, the sum of weights of Raina and Rohit is :

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

  1. $114\ kg\ 803\ g$

  2. $115\ kg\ 19\ g$

  3. $110\ kg\ 703\ g$

  4. $104\ kg\ 803\ g$

Show answer
Correct Option: A
Explanation:
Let $54$ $kg$  $43$  $gm$ $ = 54kg + 43gm$  $=A =  $  Weight of  Raina

$60$ $kg$  $760$  $gm$ $ = 60kg + 760gm$  $=B =  $  Weight of  Rohit

(i) We have to add $54$ $kg$  $43$  $gm$ to $60$ $kg$  $760$  $gm$

$ 54kg + 43gm$  $=A $   ...................(1)

$  60kg + 760gm$  $=B $   ...................(2)


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

$A+B = [54$ $kg$  $43$  $gm$] $+ $ [$60$ $kg$  $760$  $gm$]

$A+B = [54$ $kg$ $+$  $60$  $kg$] $+ $ [$43$ $gm$ $+$  $760$  $gm$]


$A+B = 114  $  $kg$ $+ $   $803$ $gm$ 

The sum of weights of Raina and Rohit is  $ 114  $  $kg$   $803$ $gm$
Hence, Option $A$ is correct

Average weight of $25$ persons is increased by $1$ kg when one man weighing $60$ kg is replaced by a new person. Weight of new person is

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

  1. $50$ kg

  2. $61$ kg

  3. $86$ kg

  4. $85$ kg

Show answer
Correct Option: D
Explanation:

Total weight increased $=1\times 25=25 kg$
$\therefore$ weight of new person is $60+25=85 kg$

If $A$ weighs $43\ kg\ 234\ g$ and $B$ weighs $56\ kg\ 450\ g$. Then the difference between the weights of $A$ and $B$ is :

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

  1. $12\ kg\ 811\ g$

  2. $13\ kg\ 216\ g$

  3. $13\ kg\ 306\ g$

  4. $10\ kg\ 216\ g$

Show answer
Correct Option: B
Explanation:
The weight of $A = 43\ kg\ 234\ g$

the weight of $A = 56\ kg\ 450\ g$
(i) Difference between weights of $A$ and $B$ is obtained by subtracting $43$ $kg$  $234$  $gm$ from $56$ $kg$  $450$  $gm$

$43$ $kg$  $234$  $gm$ $ = 43kg + 234gm$  $=A $   .... (1)

$56$ $kg$  $450$  $gm$ $ = 56kg + 450gm$  $=B $   ..... (2)


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

$B-A = [56\ kg+ 450\ gm] - [43\ kg+234\ gm$

$B-A = [56\ kg-43\ kg]+ [450\ gm -234\ gm] $


$B-A = 13\ kg\ 216\ gm$
So, Option $B$ is correct

Add $17\ kg,13\ kg\ 940\ g$ and $15\ kg\ 65\ g$.

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

  1. $40\ kg\ 65\ g$

  2. $4\ kg\ 650\ g$

  3. $46\ kg\ 50\ g$

  4. $46\ kg\ 5\ g$

Show answer
Correct Option: D