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:10 mencan build a wall in5 daysToFind:Solution:Now,5menxTable:[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,inverse proportionWhen,[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]

Henceforth,5mencanbuildthewallin10daysMethod2Weknow,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:Moretoknow:[tex]{ \orange{ \star{ \boxed{ \tt \: {products \: of \: means \: = products \: of \: extremes}}}}}[/tex]

Here,Correct Question :-If

10 mencan build a wall in5 days. Then inhow manydays5 mencan build the same wal?Given:10 mencan build a wall in5 days.Tofind:5 mencan build he same wall?Solution:• Let’s consider

daysRequired for5 mento build the same wall bex.10,5,5&xare given to us.What to do?We need to find

x.Step-by-step explanation :-Here,

10 mencan build a wall in5 days. We have to find outx(how many days5 mencan build the same wall).→

10 : 5 = x : 5→ 10/5 = x/5

→ 5x = 5 × 10

→ 5x = 50

→ x = 50/5

→

x = 10∴ Hence,

5 mencan build the same wall in10 days.