As the men of decrease the no. of days to do the work will increase so they are in inverse proportion
When,
Two quantities are in inverse proportion we direct multiply the means and extreme terms on both the side and transpose numericals to find the value of the variables .
Given :
To Find:
Solution:
Now,
Table :
[tex]\begin{gathered}\begin{gathered}\begin{gathered} \tiny\boxed{\begin{array}{ c |c} \frak{ \pmb{men}}& \rm{ \pmb{days}}\\ \dfrac{\qquad\qquad}{ \sf 10}&\dfrac{\qquad\qquad}{ \sf 5 \: days}& \\ \dfrac{\qquad\qquad}{ \sf 5}& \dfrac{\qquad\qquad}{ \sf x \: days} \end{array}}\end{gathered}& \\ \end{gathered}\end{gathered}[/tex]
Here,
When,
[tex] \longrightarrow \tt \: 10 : 5 \propto 5 : x \\ \\ \\ \longrightarrow \tt \: 50 = 5x \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: x = \cancel \frac{50}{5} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt { \boxed{ \frak{x = 10 \: days}}}[/tex]
Method 2
We know,
So,
[tex] \longrightarrow \tt \: 10 \times 5 \: days \\ \\ \\ \longrightarrow \: { \pink{ \boxed{ \tt{ 50 days}}}}\: \: \: \:[/tex]
Now,
[tex] \longrightarrow \tt \: 50 \div 5 \: days \\ \\ \\ \longrightarrow { \blue{ \boxed{ \tt{10days}} \star}} \: \: \: [/tex]
Hence:
More to know:
[tex]{ \orange{ \star{ \boxed{ \tt \: {products \: of \: means \: = products \: of \: extremes}}}}}[/tex]
Here,
Correct Question :-
If 10 men can build a wall in 5 days. Then in how many days 5 men can build the same wal?
Given:
To find:
Solution:
• Let’s consider days Required for 5 men to build the same wall be x.
10,5,5 & x are given to us.
What to do?
We need to find x.
Step-by-step explanation :-
Here, 10 men can build a wall in 5 days. We have to find out x(how many days 5 men can build the same wall).
→ 10 : 5 = x : 5
→ 10/5 = x/5
→ 5x = 5 × 10
→ 5x = 50
→ x = 50/5
→ x = 10
∴ Hence, 5 men can build the same wall in 10 days.