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Define weight and units of weight - class-V

Description: define weight and units of weight
Number of Questions: 19
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Tags: how much does it weigh? weight how big? how heavy? maths decimals
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A shopkeeper purchased $346$ kg $500$ g of orange. Later on, he found that $109$ kg $300$ g of oranges were rotten. Find the quantity of oranges in good condition.

  1. $150$ kg $670$ g

  2. $204$ kg $600$ g

  3. $237$ kg $200$ g

  4. $250$ kg $940$ g


Correct Option: C
Explanation:

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

$1$  $kg =1000$  $gram$
Total  weight of  Oranges purchased $ = B = $   $346$ $kg$  $500$  $gm$ 
Total weight of rotten oranges $=A = $   $109$ $kg$  $300$  $gm$ 
(i) We have to subtract $109\ kg$ $300\ gm$ from $346$ $kg$  $500$  $gm$ to get the weight of good conditioned oranges

$109$ $kg$  $300$  $gm$ $ = 109kg + 300gm$  $=A $   ...................(1)

$346$ $kg$  $500$  $gm$ $ = 346kg + 500gm$  $=B $   ...................(2)


$B-A = [346\ kg+500\ gm]-[109\ kg +300\ gm]$
$ = [346\ kg-109\ kg] +  $ $[500$ $gm$ $-$ $300$  $gm$]
$= 237  $  $kg $ $+  $  $200$ $gm$ 

Therefore, $237  $  $kg $  $200$ $gm$   of oranges are in good condition.

How much heavier is the toffee packet which has mass $1$ kg $345$ g in comparison to $1$ kg of chocolates?

  1. $1$ kg $345$ g

  2. $345$ g

  3. $1$ kg

  4. $1$ kg $234$ g


Correct Option: B
Explanation:

Weight of first toffee packet $=1.345\ kg$ 


Weight of second toffee packet $=1\ kg$ 

Required difference
$=1.345-1$

$=0.345\ kg$

$=345\ gm$

Hence, this is the answer.

If $15$ bananas measure $1.2$ kg, then find the weight of one banana. (Give yous answer in grams)

  1. $80$ gms

  2. $70$ gms

  3. $60$ gms

  4. $50$ gms


Correct Option: A
Explanation:

Total number of Bananas $=15$


Total mass of 15 Bananas  $=1.2 kg$

Total mass of $1$ banana $=\dfrac{1.2}{15} $   $kg$

We know that

$1$  $kg =1000$  $gram$

$1$  $gm =\dfrac1{1000}$  $kg$

we have to convert $\dfrac{1.2}{15}$  $kg$  to   $grams$


$\dfrac{1.2}{15}$  $kg =\dfrac{1.2}{15} \times 1000$  $gm$

                $=80$  $gm$

So, Option $A $ is correct 

A hollow iron pipe is $21\,cm$ long and its external diameter is $8\,cm$. If the thickness of the pipe is $1\,cm$ and iron weighs $8\,g/c{m^3}$, then the weight of pipe is :

  1. $3.6\,kg$

  2. $3.696\,kg$

  3. $36\,kg$

  4. $36.9\,kg$


Correct Option: B
Explanation:

Volume of iron $=\pi(R^2-r^2)\times h$


$=\pi(4^2-3^2)\times 21$

$=\cfrac{22}{7}\times(16-9)\times 21$

$=462cm^2$

Weight $=\cfrac{8\times 462}{1000}kg=3.696kg$

Working together, pipes $A$ and $B$ can fill an empty tank in $10\ hours$. they worked together for $4$ hours and then $B$ stopped and $A$ continued filling the tank till was full. It took a total of $13\ hours$ to fill the tank. How long would it take $A$  to fill the empty tank alone?

  1. $13\ hours$

  2. $15\ hours$

  3. $17\ hours$

  4. $18\ hours$


Correct Option: B

A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all  round. Find the weight, in the kilogram, of 2 meters of the pipe if 1 $\displaystyle cm^{3}$ of the metal weighs 7.7 g.

  1. 12.31 kg

  2. 19.31 kg

  3. 13.31 kg

  4. 14.31 kg


Correct Option: C
Explanation:

Inner diameter=5 cm. radius=2.5cm
thickness =5mm=0.5cm
outer radius=2.5+0.5=3cm
length=2m=200cm
volume of pipe=
$V=\Pi { (R }^{ 2 }-{ r }^{ 2 })h$
$V=3.14{ (3 }^{ 2 }-{ 2.5 }^{ 2 })200$
$V=22/7
(9-6.25)200$
$V=12100/7{ cm }^{ 3 }$
 1 $cm^3$ of the metal weighs 7.7 $g$.
$ So, \cfrac{12100}{7} cm^{3} \hspace {1 mm} weighs =7.7
\cfrac{12100}{7} gm =13310 \hspace {1 mm} gm = 13.31 kg$
 

The average weight of three boys $P$, $Q$ and $R$ is $\displaystyle 54\frac {1}{3}$ $Kg$, while the average weight of three boys $Q$, $S$ and $T$ is $53$ $kg$. What is the average weight of $P$, $Q$, $R$, $S$ and $T$?

  1. $52.4$ $kg$

  2. $53.2$ $kg$

  3. $53.8$ $kg$

  4. Data inadequate


Correct Option: D
Explanation:

As per question, 
$P+Q+R=54\frac { 1 }{ 3 } $---------eq 1
$Q+S+T=53 $-----------eq2
Here is 5 variable and 2 euqation.Data inadequate to solve the question.
Answer (D) 
Data inadequate

How many smaller solid balls of radius 2 cm can be made by melting a solid sphere of radius 8 cm?

  1. 128

  2. 512

  3. 64

  4. 32


Correct Option: C
Explanation:

Number of balls made $ = \dfrac {Volume  of   sphere } {Volume  of 

each  spherical  ball} $
Volume of a sphere of radius 'r' $ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $

Hence, number of balls made $ =\dfrac { \dfrac { 4 }{ 3 } \pi \times  { 8 }^{ 3 } }{ \dfrac { 4 }{ 3 } \pi \times { 2 }^{ 3 } } = 64  $

If $\square =2$ and $\triangle =4$. Then $6\times \square - 3\times \triangle =$ ?

  1. $0$

  2. $1$

  3. $3$

  4. $2$


Correct Option: A
Explanation:
$\Box =2,\triangle =4,$, then $6\times \Box -3\times \triangle =?$
$\Rightarrow 6\times 2-3\times 4=12-12=0$

Weight of one balloon is $3$ gms and of one string is $4$ gms. Then what is the combined weight of $3$ balloons and $6$ strings ?

  1. $9$ gms

  2. $24$ gms

  3. $33$ gms

  4. $41$ gms


Correct Option: C
Explanation:

Total weight$=\left( 3\times 3 \right) +\left( 6\times 4 \right) =33 gm$

A shopkeeper sold 87 kg 356 g of wheat on Saturday and 106 kg 278 g of wheat on Sunday. Find the total weight of wheat sold on both the days.

  1. $193$ kg $634$ g

  2. $165$ kg $534$ g

  3. $149$ kg $445$ g

  4. $125$ kg $356$ g


Correct Option: A
Explanation:
Total weight of wheat sold$=\left( 87.356+106.278 \right) ㎏$
$=193.634㎏$
$=193 ㎏ 634 gm$

Granny purchased $8$ kg $593$ g of sweets and snacks for the occasion. Out of which $6$ kg $368$ g were consumed. What quantity of sweets and snacks were left?

  1. $1$ kg $225$ g

  2. $2$ kg $225$ g

  3. $1$ kg $100$ g

  4. $2$ kg $347$ g


Correct Option: B
Explanation:
Quantity of sweets and snacks left$=\left( 8.593-6.368 \right) ㎏=2.225㎏$
$=2$ ㎏ $225$ gm

Southee purchased $8$ kg $428$ g of grain and Miln purchased $4$ kg $634$ g more. What quantity of grain did Jasmine and Melissa purchase?

  1. $13$ kg $405$ g

  2. $13$ kg $534$ g

  3. $13$ kg $62$ g

  4. $13$ kg $302$ g


Correct Option: C
Explanation:
Southee purchased $8.428\ kg$ grain and Miln purchased $4. 634\ kg$ 

Quantity purchased by Southee and Miln$=\left( 8.428+4.634 \right) ㎏$
$=13.062$ ㎏ $=13㎏62$ gm

Kaley weight $48$ kg $405$ g and Nancy weights $34$ kg $560$ g. Who weighs less and by how much?

  1. Kaley weighs less by $15$ kg $345$ g

  2. Nancy weighs less by $15$ kg $345$ g

  3. Kaley weighs less by $13$ kg $845$ g

  4. Nancy weighs less by $13$ kg $845$ g


Correct Option: D
Explanation:
Weight of Nancy is less than Kaley's weight by
$\left( 48.405-34.560 \right) =13.845㎏$
$13#$ ㎏ $845$ gm

Sara bought $10$ apples and $5$ bananas and Sudha bought $6$ apples and $9$ bananas. If one apple weighs $250$ g whereas $1$ banana weighs $200$ g then who bought more fruits in terms of weight? 

  1. Sudha bought more $400$ g

  2. Sara bought more $250$ g

  3. Sudha bought more $300$ g

  4. Sara bought more $200$ g


Correct Option: D
Explanation:
Weight bought by Sara$=10\times 250ℊ+5\times 200ℊ=2500ℊ+1000ℊ=3500ℊ$
Weight bought by Sudha$=6\times 250ℊ+9\times 200ℊ=1500ℊ+1800ℊ=3300ℊ$
Sara bought more weight than Sudha by $\left( 3500-3300 \right) 200$ gm

Nehal purchased $7$ kg $200$ g of sugar, $9$ kg $395$ g of rice. What is the total weight which Nehal carried?

  1. $10$ kg $375$ g

  2. $16$ kg $595$ g

  3. $20$ kg $495$ g

  4. $24$ kg $765$ g


Correct Option: B
Explanation:
Total weight carried$=\left( 7.200+9.395 \right) ㎏=16.595㎏$
$=16㎏$ $595 gm$

Chris sold 112 kg 342 g of newspapers and 195 kg of magazines. Find the total quantity of articles sold.

  1. $200$ kg $143$ g

  2. $256$ kg $246$ g

  3. $307$ kg $342$ g

  4. $384$ kg $342$ g


Correct Option: C
Explanation:
Total quantity of articles sold$=\left( 112.342+195 \right) ㎏=307.342㎏$
$=307$ ㎏ $342$ gm

How many spherical bullets can be made out of a solid cube of lead whose edge measures $44\space cm$, each bullet being $4\space cm$ in diameter.

  1. $2451$

  2. $2541$

  3. $2304$

  4. $2536$


Correct Option: B
Explanation:

Number of spherical bullets formed $ = \dfrac {Volume  of 

cube}{Volume   of   one   spherical  bullet} $





Volume of a cube of edge a $ = {a}^{3} $





 Volume of a sphere of radius 'r' $ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $





As the diameter of the sphere is $4$ cm, its radius r $ = 2$ cm





Hence, number of spherical bullets formed $ = \dfrac {44 \times 44 \times 44}{\dfrac {4}{3}

\times \dfrac {22}{7} \times 2 \times 2 \times 2} = 2541 $

A hemispherical tank of radius $1\displaystyle\frac{3}{4}m$ is full of water. It is connected with a pipe which empties it at the rate of $7\space litres$ per second. How much time will it take to empty the tank completely?

  1. $26.74\space min.$

  2. $26.54\space min.$

  3. $26.4\space min.$

  4. $26\space min.$


Correct Option: A
Explanation:

Suppose the pipe takes $x$ seconds to empty the tank. 

Then, 
The volume of the water that flows out of the tank in $x$ seconds =Volume of the hemispherical tank

The volume of the water that flows out of the tank $x$ in seconds= Volume of the hemispherical shell of radius $175cm$

$\Rightarrow 7000x=\cfrac { 2 }{ 3 } \times \cfrac { 22 }{ 7 } \times 175\times 175\times 175$

$\Rightarrow x=\cfrac { 2 }{ 3 } \times \cfrac { 22 }{ 7 } \times \cfrac { 175\times 175\times 175 }{ 7000 } =1604.16seconds$

$\Rightarrow x=\cfrac { 1604.16 }{ 60 } =26.74\quad minutes$

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