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How much does it weigh? - class-VI

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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 $175cm$

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

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

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

Simplify : $4 \times 4 - 2 \times 3 + 16 + 4 =$

  1. $20$

  2. $13/2$

  3. $30$

  4. None of these


Correct Option: C
Explanation:
$ 4\times 4-2\times 3+16+4 $
Using BODMAS Rule (BBracket, O of D Diviion,
M Multiplication, A Addition, S Subtraction) we get 
$ = 16-6+16+4 $
$ = 16-6+20 $
$ = 36-6 $
$ = 30 $ 

Divide $4\ kg\ 64\ g$ by $4$.

  1. $1.16\ kg$

  2. $10.16\ kg$

  3. $1.016\ kg$

  4. $101.6\ kg$


Correct Option: C
Explanation:

$1\ \ kg=1000g$

$\Rightarrow 1\ \ g=\dfrac{1}{1000}\ \ kg$
$4\ \ kg\ \ 64\ \ g=4+\dfrac{64}{1000}=4.064\ \ kg$
$\dfrac{4.064}{4}=1.016\ \ kg$
Option $C$ is correct.

Vineet packed $\dfrac {3}{4} kg$ of sugar each into $42$ plastic bags. Find the total weight of sugar Vineet packed.

  1. $34.5\ kg$

  2. $3.45\ kg$

  3. $31.5\ kg$

  4. $345\ kg$


Correct Option: C
Explanation:

Weight of sugar in $1\ bag = \dfrac {3}{4} kg$
$\therefore$ Weight of sugar in $42$ bags
$= \left (\dfrac {3}{4}\times 42\right ) kg = \dfrac {63}{2} kg = 31.5\ kg$.

What is $2:15+2:00$ in minutes and hours?

  1. $2$ hrs $15$ minutes

  2. $2$ hrs $45$ minutes

  3. $4$ hrs $45$ minutes

  4. $4$ hrs $15$ minutes


Correct Option: D
Explanation:

$2:15+2:00$

$=4:15$
So it is $4$ hours and $15$ minutes.

What is $3:13 A.M.+1:15 A.M. =$ ?

  1. $4:28 A.M.$

  2. $3:28 A.M.$

  3. $5:08A.M.$

  4. $1:15A.M.$


Correct Option: A
Explanation:

$3:13 A.M.+1:15A.M.$

$=4:28 A.M.$
Option $A$ is correct.

Calculate the time $1:30+4:45$.

  1. $5:15$

  2. $5:45$

  3. $6:15$

  4. $6:45$


Correct Option: C
Explanation:

$1:30+4:45$

Adding hours $=1+4=5$ hours
Adding minutes $30+45=75$ minutes
Now $60$ minutes $=1 $ hour
$\therefore 75$ minutes $=1$ hour $+15$ minutes
So total time is $(5+1=6)$ hours and  $15$ minutes 
$=6:15$

If Tina weighs $\dfrac{2}{3}$times than Reema and Reema weighs $45$ kg, then Tina's weight is

  1. $45$

  2. $30$

  3. $15$

  4. $50$


Correct Option: B
Explanation:

Weight of Reema $=w=45\ \ Kg$

Weight of Tina $=\dfrac{2}{3}w=\dfrac{2}{3}\times{45}=30\ \ Kg$
Option $B$ is correct.

What is $3:50-2:45$ ? 

  1. $1:00$

  2. $1:05$

  3. $1:15$

  4. $1:25$


Correct Option: B
Explanation:

$3:50-2:45$

Subtracting hour from hour and minutes from minutes 
$\Rightarrow 1:05$
So option $B$ is correct.

What is $4:15-3:30$ ?

  1. $00:45$

  2. $-1:15$

  3. $-00:15$

  4. $1:15$


Correct Option: A
Explanation:

$4:15$

$=4\times 60+15$ minutes
$=255$ minutes
$3:30$
$=3\times 60+30$ minutes
$210$ minutes
$\therefore 4:15-3:30=255-210=45$ minutes
So time is $00:45$

Gabriel watched $3$ old movies on videotape. The first movie was $62$ minutes long. The
second was $1$ hour $34$ minutes long. The third was $1$ hour $25$ minutes long. He started
watching at $3:15$ P.M. At what time did the last movie end?

  1. $7:00$ P.M.

  2. $7:08$ P.M.

  3. $7:12$ P.M.

  4. $7:16$ P.M.


Correct Option: D
Explanation:

First movie $62$ min.

Second movie $94$ min
Third movie $85$ min
so total time of movie will be $241$ minutes
which makes $4$ hours and 1 minute
So he should finish $3:15 + 4:01$ = $7:16 $ P.M.

The ratio at which the point $(5,4)$ divides the line $(3,2)$ and $(8,7)$

  1. $\dfrac 23$

  2. $\dfrac{-3}{2}$

  3. $\dfrac{1}{2}$

  4. $\dfrac{1}{9}$


Correct Option: A
Explanation:
Given points $(3,2);(8,7)$
Let the ratio be $m:n$
The dividing point is given as $\dfrac{8m+3n}{m+n}=5\\8m+3n=5m+5n\\3m=2n\\\dfrac mn=\dfrac 23$

Mr Sahoo attended a 1-day workshop from 09:15 a.m. to half five in the evening. The workshop included a $1\frac{1}{4}$ hour lunch break, two 15 minutes tea breaks and 13 activities, each of equal duration. Calculate the duration of each activity.

  1. $30$ minutes

  2. $20$ minutes

  3. $25$ minutes

  4. $40$ minutes


Correct Option: A
Explanation:

Total time of the workshop =$8$ hours and $15$ min

$=8 \times 60+15=495 $ min

Total time of tea break = $30$ min

lunch break time = $75$ min

hence time available for 13 activities 
$=495-75-30$ $=6$ hours and $30$ min
$=390 $ min

So, time devoted to each activity $=\dfrac{390}{13}=30$ min

The greatest length which may be used to measure exactly $\displaystyle13\frac{3}{4}:ft$., $\displaystyle17\frac{1}{2}$., 20 ft. and $\displaystyle21\frac{1}{4}:ft$. is given by

  1. 3"

  2. 15"

  3. 18"

  4. 21"


Correct Option: B
Explanation:

$\displaystyle13\frac{3}{4}:ft=165:inches$

$\displaystyle17\frac{1}{2}:ft=210:inches$ 

$20:ft=240:inches$ 

and $\displaystyle21\frac{1}{4}:ft=255:inches$ 

In order to find the greatest length we have to find the G. C. M. of 165", 210", 240" and 255" and the G. C. M. of these four numbers is 15".

One inch is equivalent to $2.54$ cm. How many centimeters are in two feet? Round your answer to two decimal places.

  1. $5.08$ cm

  2. $12.62$ cm

  3. $60.96$ cm

  4. $30.48$ cm


Correct Option: C
Explanation:

First convert 2 feet to inches and then convert the result to centimeters. One foot is equivalent to 12 inches, so there are 24 inches in 2 feet. Since there are 2.54 cm in a foot, multiply 24 by 2.54 to convert feet to centimeters.
$24$ $\times$ $2.54 = 60.96$
Therefore, there are $60.96$ cm in $2$ feet.

Subtract $778\ m\ 78\ cm$ from $2\ km\ 768\ cm$

  1. $1228.9\ m$

  2. $571.1\ m$

  3. $1221.988\ m$

  4. $189.22\ m$


Correct Option: A
Explanation:

First we have to convert both the quantities in $m$

$1\ \ m=100\ \ cm$
$\Rightarrow 1\ \ cm=.01 \ m$
$778 \ \ m\ \ 78\ \ cm =778+.01\times 78=778.78\ \ m$
$1\ \ km=1000\ \ m$
$2\ \ km\ \ 768\ \ cm =2\times1000+.01\times768=2007.68\ \ m$
Now subtracting both 
$2007.86-778.8=1128.9\ \ m$
Option $A$ is correct.

Add $1.25\ cm$ and $13.45\ cm$

  1. $2.595$

  2. $1.470$

  3. $14.70$

  4. $25.95$


Correct Option: C
Explanation:

$1.25\ \ cm+13.45\ \ cm=14.70\ \ cm$

So option $C$ is correct.

Subtract :
$23.67\ m$ from $1.4\ km$

  1. $116.33$

  2. $1376.33\ m$

  3. $13976.33\ m$

  4. $22.27\ m$


Correct Option: B
Explanation:

First we have to convert $km$ into $m$

$1\ \ km=1000\ \ m$
$\Rightarrow 1.4\ \ km=1.4\times1000=1400\ \ m$
Now subtracting both
$1400 \ \ m-23.67\ \ m=1376.33\ \ m$
So option $C$ is correct.

Add $23.456\ m$ and $26.5\ cm$ .

  1. $499.56\ cm$

  2. $49.956\ cm$

  3. $237.16\ cm$

  4. $2372.6\ cm$


Correct Option: D
Explanation:
First we have to convert $m$ into $cm$
$1\ \ m=100\ \ cm$
$\Rightarrow 23.456\ \ m=23.456\times100=2345.6\ \ cm$
Now adding both
$2345.6\ \ cm+26.5\ \ cm=2372.6\ \ cm$
None of the options are correct

A pole is painted yellow and black. The yellow part is $1.8\ m$ long and the black is three times longer than yellow part. Find the length of pole.

  1. $5.4\ m$

  2. $7.2\ m$

  3. $3.6\ m$

  4. $none\ of\ these$


Correct Option: B
Explanation:

Length of yellow Part $=Y=1.8m$

Length of Black Part $=3Y=5.4m=B$
So, Length of Pole $=L=Y+B=1.8+5.4=7.2m$

Rita had $\displaystyle 38\frac { 1 }{ 4 } $ m long rope. She cut it into 5 equal parts. Then the length of each piece will be- 

  1. $\displaystyle 1\frac { 1 }{ 4 } $ m

  2. $\displaystyle 2\frac { 3 }{ 4 } $ m

  3. $\displaystyle 3\frac { 1 }{ 4 } $ m

  4. $\displaystyle 1\frac { 3 }{ 4 } $ m


Correct Option: D
Explanation:

Length of the rope $\displaystyle= 8\dfrac { 3 }{ 4 } m\ Number\quad of\quad peices\quad cut\quad =5\ Length\quad of\quad each\quad peice\quad =8\dfrac { 3 }{ 4 } m\div 5=\dfrac { 35 }{ 4 } m\times \dfrac { 1 }{ 5 } \ =\left( \dfrac { 35 }{ 4 } \times \dfrac { 1 }{ 5 }  \right) m=\dfrac { 7 }{ 4 } m=1\dfrac { 3 }{ 4 } m\ \therefore The\quad length\quad of\quad each\quad piece\quad of\quad the\quad rope\quad =1\dfrac { 3 }{ 4 } m.$

If Harry runs $23\ m\ 5\ cm$ and Joy runs $14.37\ m$ from the same starting point, then how far is Joy from Harry.

  1. $8.68\ m$

  2. $9.13\ m$

  3. $86.8\ m$

  4. $91.3\ m$


Correct Option: A
Explanation:

We low that $1 m=100 cm$

Hence distance run by Harry $=23 m + 5 cm\=23 m+ \dfrac{5}{100} m\=23.05 m$
Distance run by Joy $=14.37 m$
Distance between them $=23.05 m -14.37 m\=8.68 m$
                                                 

${\rm{1}}\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ is equal to____________.

  1. ${\rm{1}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$

  2. ${\rm{1}}{{\rm{0}}^3}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$

  3. ${\rm{1}}{{\rm{0}}^{ - 2}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$

  4. ${\rm{1}}{{\rm{0}}^{ - 3}}\,{\rm{kg/c}}{{\rm{m}}^{\rm{3}}}$


Correct Option: D
Explanation:

We have,

$1\ g/cm^3$

Since,
$1000\ g=1\ kg$
$1\ g=10^{-3} \ kg$

So,

$1\ g/cm^3 =10^{-3}\ kg/cm^3$

Hence, this is the answer.

A spherical bowl  with diameter  $24$ m. How much it weigh  (approx. in kg)  (if $1$ gm $= 1$ cubic cm).

  1. $7216473.6 kg$

  2. $216473.6 kg$

  3. $721473.6 kg$

  4. None of these


Correct Option: A

Weight of $1$ mango is $250$ gms. Then identify the weight of $16$ mangos. (Write your answer in kilograms)

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: D
Explanation:
the weight of $1$ mango $=250 $  $gram$

weight of $16$ mangoes $=250 \times 16 = 4000$  $gram$

We know that

$1$  $kg =1000$  $gram$

$1$  $gram =\dfrac1{1000}$  $kg$

Given That, we have to convert $4000$  $gram$  to   $kg$

$4000$  $gm =\dfrac{1}{1000} \times 4000$  $kg$

                    $ =\dfrac{4000}{1000} $  $kg$

                   $=4$  $kg$

So, option $D$ is correct

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