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Conversion of units - class-VII

Description: conversion of units
Number of Questions: 26
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Tags: perimeter and area of rectilinear figures area and volume metric system perimeter and area maths measures and motion revision
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A square field is 40m long. The perimeter of the square field is (in cm)

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $14000 \, cm$

  2. $16000 \, cm$

  3. $12000 \, cm$

  4. $8000 \, cm$

Show answer
Correct Option: B

By converting the $0.0287m^2$ into the $cm^2$, the answer will be

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $28.7 \, cm^2$

  2. $2870 \, cm^2$

  3. $287 \, cm^2$

  4. None of these

Show answer
Correct Option: C

$10^{-5} \, m^2=? \, hectare$

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $1$

  2. $10$

  3. $10^{-2}$

  4. $10^{-1}$

Show answer
Correct Option: A

A rectangular field is 40m long and 30m wide. perimeter of rectangular field is

maths revision unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $120$m

  2. $130$m

  3. $140$m

  4. $150$m

Show answer
Correct Option: C

If $1 \, m^2$ of a certain plot in a certain area costs Rs. $1500,$ how much does $1$ hectare cost?

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $15 \times 10^5$

  2. $15 \times 10^6$

  3. $15 \times 10^7$

  4. $15 \times 10^4$

Show answer
Correct Option: B

By converting the 80.2km into the hectare, the answer will be

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. 0.08020ha

  2. 8020ha

  3. 802ha

  4. 0.802ha

Show answer
Correct Option: B

By converting the 4.8mm into the cm, the answer will be

maths revision unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. 0.048cm

  2. 0.48cm

  3. 48cm

  4. 480cm

Show answer
Correct Option: A

$500.2867 m^2= \, ? hectare$

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $5002.867$

  2. $5002867$

  3. $500.2867$

  4. $0.05002867$

Show answer
Correct Option: D

$10^{-10} \, cm^2=? \, hectare$

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $10^{-28}$

  2. $10^{-8}$

  3. $10^{-18}$

  4. $10^{18}$

Show answer
Correct Option: C

By converting the $4.8mm^{2}$ into the $m^2,$ the answer will be

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $4.8 \times 10^{-7} m^2$

  2. $4.8 \times 10^{-6} m^2$

  3. $4.8 \times 10^{-8} m^2 $

  4. $4.8 \times 10^{-5}m^2$

Show answer
Correct Option: B

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

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
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