Tag: decimal fractions in the units of currency, length, weight, capacity

Questions Related to decimal fractions in the units of currency, length, weight, capacity

Prerna has Rs. $5000$. She bought $2$ boxes of chocolates for Rs. $52.98$ each, $3$ packets of chips for Rs. $12.98$ each and $2$ books for Rs. $300.05$ each. How much money left with her?

  1. Rs. $4255$

  2. Rs. $4750$

  3. Rs. $4000$

  4. Rs. $3875$


Correct Option: A
Explanation:

Total expenditure $=Rs. ((2\times 52.98)+(3\times 12.98)+(2\times 300.05))$
$=Rs. (105.96+38.94+600.10)=Rs. 745$
Amount of money left with her$=Rs. (5000-745)=Rs. 4255$.

Rahul purchased $4$kg $500$g rice, $5$kg $650$g sugar and $12$kg $600$g atta. The total weight of his purchases was __________.

  1. $10.150$kg

  2. $22.750$kg

  3. $25.350$kg

  4. $18.250$kg


Correct Option: B
Explanation:

Weight of rice $=4$kg $500$g $=4.500$kg
Weight of sugar $=5$kg $650$g $=5.650$ kg
Weight of atta $=12$kg $600$g $=12.600$ kg
$\therefore$ Total weight of all items $=(4.500+5.650+12.600)$kg
$=22.750$kg.

Fill in the banks: 
$88\, cm =...................... meter$

  1. $0.88$

  2. $0.66$

  3. $0.55$

  4. $0.99$


Correct Option: A
Explanation:

$1$ meter $= 100$ cm


So, $88 \,  cm =$ $\dfrac{88}{100}  \, meter$

$= 0.88$ meter

A road is $4$ km $ 800$ m long. If trees are planted both its sides at intervals of $9.6$ m, how many trees were planted?

  1. $400$

  2. $500$

  3. $700$

  4. $1000$


Correct Option: B
Explanation:

Total length $=4.8$kms $=4800$m


Interval $=9.6$ m

Number of trees planted $=\dfrac{4800}{9.6}=500$

Abhishek has Rs. $817.45$. He bought toys for Rs. $538.30$. The balance amount left with Abhishek is ___________.

  1. Rs. $279.30$

  2. Rs. $265.45$

  3. Rs. $279.15$

  4. Rs. $315.82$


Correct Option: C
Explanation:

Amount of money Abhishek had $=Rs. 817.45$
Amount of money spent by him on toys $=Rs. 538.30$
$\therefore$ Amount of money left with him $=Rs. (817.45-538.30)=Rs. 279.15$.

Mohan bought $500$ g of cabbage, $250$ g of onions, $750$ g of potatoes, $600$ g of apples.
Calculate total weight in kilograms.

  1. $2100$kg

  2. $21.00$kg

  3. $2.100$kg

  4. $0.2100$kg


Correct Option: C
Explanation:
Weight of cabbage is $500$ gm

Weight of onions is $250$ gm

Weight of potatoes is $750$ gm

Weight of apples is $600$ gm

Total weight of fruits and vegetables in grams is:

$500+250+750+600=2100$ gm

We know that $1$ kg $=1000$ gm

Therefore, $2100$ gm can be converted in kgs as follows:

$\dfrac {2100}{1000}=2.100$ kg

Hence, the total weight is $2.1$ kg.

$1\ km\ 250\  m =....................km$
Fill in the blank by chosing correct option from the following options

  1. $10.360$

  2. $10250$

  3. $1.250$

  4. $10.625$


Correct Option: C
Explanation:

We know that $1\ m$=$1000\ km$.


Now, let us convert $250\ m$ in kms by dividing it by $1000$ as follows:

$\dfrac {250}{1000}=0.25$

Therefore, $250\ m$=$0.25\ km$

Thus we have:

$1\ km$ $250\ m$=$(1+0.25)\ km$=$1.25\ km$

Hence, $1\ km$ $250\ m$=$1.25\ km$

What is the ratio of 1 mm to 1 cm?

  1. 1: 100

  2. 10: 1

  3. 1: 10

  4. 100:1


Correct Option: C
Explanation:

As we know that

$1\ cm =10\ mm$
Ratio of $1\ mm $ to $1\ cm$  is $1\ mm:1\ cm=1\ mm :10\ mm=1:10$

The cost of 1kg of sugar is Rs.18.What is cost of 250gms of sugar?

  1. $Rs.9$

  2. $Rs.4.5$

  3. $Rs.8$

  4. $Rs.5.5$


Correct Option: B
Explanation:

We know that $1$kg=$1000$gm.


It is given that the cost of $1$kg sugar is Rs.$18$ which means that the cost of $1000$g sugar is Rs.$18$ then cost of one gram sugar will be:


$\dfrac {18}{1000}=0.018$

Therefore, $1$gm sugar costs Rs.$0.018$.

Now, the cost of $250$gms of sugar will be as follows:

Cost$\times$number of grams$=0.018\times 250=4.5$

Hence, the cost of $250$gms of sugar is Rs.$4.5$.

How much pure alcohol should be added to $400 ml$ of strength $15\%$ to make its strength $32\%$?

  1. $50 ml$

  2. $75 ml$

  3. $100 ml$

  4. $150 ml$


Correct Option: C
Explanation:

The amount of alcohol in $400ml$ solution of $15\%$ strength

$=400\times \cfrac { 15 }{ 100 } ml=60ml$

Let the required amount of alcohol to be added $=x$ ml.
It will increase the amount of the solution by $x$ ml.

$\therefore $ The amount of alcohol now $=(60+x)ml$
and the amount of the solution$=(400+x)ml$.
Then, the strength $=32\%=\cfrac { 60+x }{ 400+x } =\cfrac { 32 }{ 100 }$

$\Rightarrow 25x+1500=3200+8x\ \Rightarrow 17x=1700\ \Rightarrow x=100$.
So, the required amount of pure alcohol to be added $=100 ml$.