Question

4) The weight of 5 bags of rice is 25 (3/5) Kg. Find the weight of 1 bag ? *

Answer 1

$\sf\huge\bold{Answer:}$

Given :

Weight of 5 bags = $25 \frac{3}{5} kg$

To find :

Weight of 1 bag

Solution :

First of all it can be seen that the weight of 5 bags is in mixed fraction.

To convert

mixed fraction ⇒ improper fraction

Weight of 5 bags = $25 \frac{3}{5} kg = \frac{5 \times 25 + 3}{5} kg = \frac{128}{5} kg$

Now,weight of 1 bag =

$\frac{weight \: of \: 5 \: bags}{number \: of \: initial \: bags} \times number \: of \:final \: bags \:$

Now,on inserting values in the above formula,we get

$⇒\frac{ \frac{128}{5} }{5} \times 1kg$

so, as we know on solving the term further,

we get

$⇒\frac{128}{5} \times \frac{1}{5} \times 1 \\ ⇒ \frac{128}{5 \times 5} \\ ⇒ \frac{128}{25} kg \\ ⇒5 \frac{3}{25} kg$

So,after step by step explanation,it can be easily concluded that the weight of 1 bag is

$⇒5 \frac{3}{25} kgs$

Answer 2

Answer:

\sf\huge\bold{Answer:}Answer:

Given :

Weight of 5 bags = 25 \frac{3}{5} kg25

5

3

kg

To find :

Weight of 1 bag

Solution :

First of all it can be seen that the weight of 5 bags is in mixed fraction.

To convert

mixed fraction ⇒ improper fraction

Weight of 5 bags = 25 \frac{3}{5} kg = \frac{5 \times 25 + 3}{5} kg = \frac{128}{5} kg25

5

3

kg=

5

5×25+3

kg=

5

128

kg

Now,weight of 1 bag =

\frac{weight \: of \: 5 \: bags}{number \: of \: initial \: bags} \times number \: of \:final \: bags \:

numberofinitialbags

weightof5bags

×numberoffinalbags

Now,on inserting values in the above formula,we get

⇒\frac{ \frac{128}{5} }{5} \times 1kg⇒

5

5

128

×1kg

so, as we know on solving the term further,

we get

\begin{gathered} ⇒\frac{128}{5} \times \frac{1}{5} \times 1 \\ ⇒ \frac{128}{5 \times 5} \\ ⇒ \frac{128}{25} kg \\ ⇒5 \frac{3}{25} kg\end{gathered}

⇒

5

128

×

5

1

×1

⇒

5×5

128

⇒

25

128

kg

⇒5

25

3

kg

So,after step by step explanation,it can be easily concluded that the weight of 1 bag is

⇒5 \frac{3}{25} kgs⇒5

25

3

kgs