Tag: basic operations with same units

Questions Related to basic operations with same units

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