0

Using decimals in - class-VII

Description: using decimals in
Number of Questions: 43
Created by:
Tags: how big? how heavy? decimals maths fraction measurement (length) how much does it weigh? length tenths and hundredths weight
Attempted 0/43 Correct 0 Score 0

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 $175cm$

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

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

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

A train $280 m$ long, running with a speed of 63$\mathrm { km } /$ hr will pass a tree in 

  1. $15 sec$

  2. $16 sec$

  3. $18 sec$

  4. $20 sec$


Correct Option: B
Explanation:

$Speed\>of\>train\>=\>63(\dfrac{Km}{hr})\\=(\dfrac{63\cdot100m}{3600s})\\=17.5(\dfrac{m}{s})\\By\>passing\>a\>tree,\>train\>will\>have\>to\>cross\>its\>own\>length\\\therefore\>Time\>taken\>=\>(\dfrac{280}{17.5})=16sec$

Convert the following :
$27$ km $241$ m $319$ mm into $m$

  1. $27241.319\ m$
  2. $27000.319\ m$
  3. $27219.241\ m$
  4. $27319.241\ m$

Correct Option: A
Explanation:

We have,

$27\ km\ 241\ m\ 319\ mm$

We know that
$1\ km =1000\ m$
$1\ mm=10^{-3}\ m$

Therefore,

$\Rightarrow 27\times 10^3\ m+241\ m+ 319\times 10^{-3}\ m$

$\Rightarrow 27000\ m+241\ m+ 0.319\ m$

$\Rightarrow 27241.319\ m$

Hence, this is the answer.

$\displaystyle  \frac { 1 }{2 } of\ 20\ km = $ ............. m

  1. $1000$

  2. $10$

  3. $10,000$

  4. $100$


Correct Option: C
Explanation:

$\displaystyle  \frac { 1 }{ 2 } of\quad 20\ km=10\ km = 10, 000\quad m$

Ankit is $160$ cm tall. What is his height in meters.

  1. $0.16$ meters

  2. $16.0$ meters

  3. $1.6$ meters

  4. $1$ meter


Correct Option: C
Explanation:

we know that

   $1$ meter = $100$ cms
so $160/100$=$1.6$ meters
so Ankit is $1.6$ meters tall

The graphs of $2x+3y-6=0$, $4x-3y-6=0$, $x=2$ and $\displaystyle y=\frac{2}{3}$ intersect in __________.

  1. Four points

  2. One point

  3. In no points

  4. In infinite number of points


Correct Option: B
Explanation:

On solving  2x+3y6=02x+3y−6=0 and 4x3y6=04x−3y−6=0, we get

x=2x=2 and y=23y=23

Hence, from all the four given equations, the!li value of x is 2 and y is 2323

Therefore, the graphs of all the four lines intersect in one point.

A carpenter cuts off $1.3$ inches from a $4$-foot board. What is the new length of the board in inches?

  1. $2.7$ inches

  2. $46.7$ inches

  3. $49.3$ inches

  4. $58.7$ inches


Correct Option: B
Explanation:

Of the two lengths given in the problem, one is inches and the other is in feet. The answer needs to be inches, so convert the length of the board to inches. Since there are $12$ inches in one foot, the 4-foot board is $48$ inches long. Now subtract the length of the piece the carpenter cut off.
$48-  1.3 = 46.7$
Therefore, the new length of the board is $46.7$ inches.

Convert the following measurement into $cm$
$3.44\ m$

  1. $3.44\ cm$

  2. $34.4\ cm$

  3. $344\ cm$

  4. $3440\ cm$


Correct Option: C
Explanation:

We know that


$1$  $meter =100$  $cm$

$1$  $cm =\dfrac1{100}$  $m$

Given That , we have to convert $3.44$  $m$ into $cm$

$3.44$  $m =3.44 \times 100$  $cm$

                $=344$  $cm$

So, Option $C $ is correct 

Convert $1$ feet into meters.

  1. $0.30$ m

  2. $0.3048$ m

  3. $0.3059$ m

  4. $0.3067$ m


Correct Option: B
Explanation:

$1$ Feet $=0.3048$ m

Convert the following into metres and centimetres.
$673\ cm$

  1. $6\ m$ $73\ cm$

  2. $67\ m$ $3\ cm$

  3. $673\ m$ $0\ cm$

  4. $73\ m$ $6\ cm$


Correct Option: A
Explanation:

We know that


$1$  $meter =100$  $cm$

$1$  $cm =\dfrac1{100}$  $m$


$673$  $cm =\dfrac1{100} \times 673$  $m$

                $=\dfrac{673}{100}   $  $m$

                 $= 6.73  meter$

It is equivalent to $6$ $meters$ and $73$ $cms$
So, Option $A$ is correct

Find the value of  $x$?
$x\ meters = 118.1103\ inches$.

  1. $2$

  2. $3$

  3. $4$

  4. $3.5$


Correct Option: B
Explanation:

$\because1 in =0.0254m$

$\therefore 118.1103 in=0.0254\times 118.11.3m=3m$

State whether the following statement is True or False.
$1$ mile is greater than $1$ km.

  1. True

  2. False


Correct Option: A
Explanation:

We know, that $1 \ \text{mile} =1.61 \ \text{km}$

Therefore, $ 1 \ \text{mile}>1 \ \text{km}$
Hence, it is true.

State the following statement is True or False
$10$ m is equal to $32.8084$ foot

  1. True

  2. False


Correct Option: A
Explanation:

$\because 1m=3.28084$ foot

$\therefore10m=32.8084$ foot

$10.5$ litres is equal to how many milliliters

  1. $10500$

  2. $10005$

  3. $10050$

  4. $15000$


Correct Option: A
Explanation:

$10.5$ litres can be written as $10.5\times 1000=10500$ ml.

Thus option A is correct.

Expenditure incurred in cultivating a square field at the rate of Rs. $170$ per hectare is Rs. $680$. What would be the cost of fencing the field at the rate of Rs. $3$ per meter?

  1. Rs. $2400$

  2. Rs. $3600$

  3. Rs. $3000$

  4. Rs. $2000$


Correct Option: A
Explanation:
Total cost of cultivation$=$Rs.$680$
Cost per hectare$=$Rs. $170$
$\therefore $Area of cultivation$=\cfrac { 680 }{ 170 } =4$ hectare$=4\times { 10 }^{ 4 }m^2$
Now, field's shape is square, therefore area $=side \times side$
$\Rightarrow 4\times { 10 }^{ 4 }={ (side) }^{ 2 }$
$\Rightarrow $ side$=\sqrt { 4\times { 10 }^{ 4 } } =200m$
Perimeter of field$=4\times 200=800m$
$\therefore $ Cost of fencing $800m=$ Rs.$\left( 3\times 800 \right) =$ Rs.$2400$

Choose the correct option:

$8$m $5$ cm = .......... cm

  1. $850$

  2. $8005$

  3. $8050$

  4. $805$


Correct Option: D
Explanation:

$1m=100 cm$

$\Rightarrow 8m=800cm$
$\therefore 8m 5cm = 800+5=805 cm$

A person bought $32$L of water for the football game and he divided the water equally into $8$ cooler. Find the quantity of water in each of the coolers by converting it into millilitres.

  1. $40$ ml

  2. $400$ ml

  3. $4000$ ml

  4. $40000$ ml


Correct Option: C
Explanation:
Quantity of water$=32ℓ=32000㎖$
Number of cooler$=8$
$\therefore $Quantity of water in each cooler$=\cfrac { 32000 }{ 8 } =4000㎖$

Rekha started her homework at $1:59$ pm point and finished her homework  $96$ minutes later. Rekha had volleyball practice at $4:00$ pm. How much time(in minutes) did Susan have between finishing her homework and the beginning of volleyball.

  1. $25$ m

  2. $205$ m

  3. $30$ m

  4. $35$ m


Correct Option: A
Explanation:
Starting time of homework$=1.59pm$
Time taken$=96$ minutes $=1$ hour $36$ minutes
i.e., Home work complete by $3:35pm$
Time between finishing homework and beginning of
Volley ball$=4:00pm-3:35pm=25$ minutes.

Capacity of measuring flask is $1$ litre.What it will be in cubic centimeter $?$

  1. $1$ Cubic centimetre

  2. $10$ Cubic centimetre

  3. $100$ Cubic centimetre

  4. $1000$ Cubic centimetre


Correct Option: D
Explanation:

$1000$ cubic centimetre
$1$ litre $=  1000 cm^3$

$\therefore$ The solution is $1000$ cubic centimetre.



State the following statement is True or False
$891$ cm can also be written as $89$ m $3$ cm

  1. True

  2. False


Correct Option: B
Explanation:

$891$ cm $=890$ cm $+$ $1$ cm $=89$ m $1$ cm

Thus $891$ cm cannot be written as $89$ m $3$ cm.

How many meters are there in $1.25$ kilometers?

  1. $10.25$ m

  2. $125$ m

  3. $1250$ m

  4. $12500$ m


Correct Option: C
Explanation:

We know that $1km=1000m$

Suppose $1.25km = x$
$\Rightarrow x={1.25}\times 1000$
$\Rightarrow x=1250$ m
Hence, the answer is $1250$ 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

The total number of donations which can be given from 5 rupee coins and 4 fifty paisa coins, is _____.

  1. 19

  2. 29

  3. 30

  4. 20


Correct Option: A
- Hide questions