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Conversion of units - class-VII

Description: conversion of units
Number of Questions: 26
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Tags: revision measures and motion maths perimeter and area metric system area and volume perimeter and area of rectilinear figures
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Add $17\ kg,13\ kg\ 940\ g$ and $15\ kg\ 65\ g$.

  1. $40\ kg\ 65\ g$

  2. $4\ kg\ 650\ g$

  3. $46\ kg\ 50\ g$

  4. $46\ kg\ 5\ g$


Correct Option: D

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 :

  1. $12\ kg\ 811\ g$

  2. $13\ kg\ 216\ g$

  3. $13\ kg\ 306\ g$

  4. $10\ kg\ 216\ g$


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

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

  1. $50$ kg

  2. $61$ kg

  3. $86$ kg

  4. $85$ kg


Correct Option: D
Explanation:

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

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

  1. $114\ kg\ 803\ g$

  2. $115\ kg\ 19\ g$

  3. $110\ kg\ 703\ g$

  4. $104\ kg\ 803\ g$


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

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

  1. $14\ kg\ 34\ g$

  2. $140\ kg\ 300\ g$

  3. $14\ kg\ 345\ g$

  4. $140\ kg\ 345\ g$


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

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.

  1. $330\ kg\ 400\ g$

  2. $336\ kg\ 4\ g$

  3. $336\ kg\ 40\ g$

  4. $33.6\ kg\ 40\ g$


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

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. $780\ kg\ 432\ g$


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

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. $16\ kg\ 300\ g$


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.

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


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

Which would be weight closest to 800 kg?

  1. Feather

  2. Ball

  3. Cow

  4. None of these


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.

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

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


Correct Option: B

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

  1. $10^{-28}$

  2. $10^{-8}$

  3. $10^{-18}$

  4. $10^{18}$


Correct Option: C

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

  1. $0.0056cm^2$

  2. $5600cm^2$

  3. $56000cm^2$

  4. $560cm^2$


Correct Option: B

$500.2867 m^2= \, ? hectare$

  1. $5002.867$

  2. $5002867$

  3. $500.2867$

  4. $0.05002867$


Correct Option: D

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

  1. 0.048cm

  2. 0.48cm

  3. 48cm

  4. 480cm


Correct Option: A

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

  1. 0.08020ha

  2. 8020ha

  3. 802ha

  4. 0.802ha


Correct Option: B

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

  1. $15 \times 10^5$

  2. $15 \times 10^6$

  3. $15 \times 10^7$

  4. $15 \times 10^4$


Correct Option: B

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

  1. $120$m

  2. $130$m

  3. $140$m

  4. $150$m


Correct Option: C

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

  1. $1$

  2. $10$

  3. $10^{-2}$

  4. $10^{-1}$


Correct Option: A

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

  1. $28.7 \, cm^2$

  2. $2870 \, cm^2$

  3. $287 \, cm^2$

  4. None of these


Correct Option: C

A square field is 40m long. The perimeter of the square field is (in cm)

  1. $14000 \, cm$

  2. $16000 \, cm$

  3. $12000 \, cm$

  4. $8000 \, cm$


Correct Option: B

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


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)

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. $17.03$


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


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 $30\dfrac {1}{4}m^2$. 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. $5\dfrac {1}{4}m$


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)

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

  1. $m$

  2. $km$

  3. $km^{2}$

  4. $m^{2}$


Correct Option: C
Explanation:

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

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