Direct And Inverse Proportions - Class 8 Maths - Extra Questions

The number of men doing a work and the number of days taken are in inverse proportion.
Let the number of men doing the work in $$ 16 $$ days be $$ a $$
Then, $$ 24:a :: $$ inverse ratio of $$ 40:16 $$

$$ 24 \times 16 = a \times 40 $$

Applying the rule, product of extremes $$ = $$ product of means

$$ 24 \times 40 = a \times 16 $$

$$  a = \dfrac{24 \times 40}{16} $$

$$ a = 60 $$

The number of men doing a work and the number of days taken are in inverse proportion. 

Let the number of days taken by $$ 20 $$ men to reap the field be $$ a $$.
Then, $$ 30:20 :: $$ inverse ratio of $$ 14:a $$

$$ => 30:20::a:14 $$
Applying the rule, product of extremes $$ = $$ product of means

$$ 30 \times 14 = 20 \times a $$

$$ a = \displaystyle \frac{30 \times 14}{20} $$

$$ a = 21 $$

Hence, $$ 20 $$ men can reap the field in $$ 21 $$ days.

$$ 18 \times 18 = a \times 27 $$

$$ a = \dfrac{18 \times 18}{27} $$


$$ a = 12 $$

Hence, $$ 12 $$ men can do the work in $$ 27 $$ days.

$${\textbf{Step - 1: Form the proportion table from condition}}{\text{.}}$$

                   $${\text{Given that 2 kg sugar contains 9106 crystals}}{\text{.}}$$

                   $${\text{Let's assume 5 kg sugar contains x crystals}}{\text{.}}$$

                   $${\text{Let's assume 1}}{\text{.2 kg sugar contains y crystals}}{\text{.}}$$

                   $${\text{Clearly it is a direct proportion}}{\text{.}}$$

                   $$ \Rightarrow {\text{with increasing of quantity of sugar , number of crystal will also increase}}{\text{.}}$$

                   $${\text{so , }}\dfrac{2}{{9106}} = \dfrac{5}{{\text{x}}}$$

                   $${\text{so , }}\dfrac{2}{{9106}} = \dfrac{{1.2}}{{\text{y}}}$$

 $${\textbf{Step - 2: Solve the equation and find value of x}}{\text{.}}$$

                    $${\text{Now , }}\dfrac{2}{{9106}} = \dfrac{5}{{\text{x}}}$$

                    $$ \Rightarrow {\text{x}} = \dfrac{{5 \times 9106}}{2}$$

                    $$ \Rightarrow {\text{x}} = \dfrac{{45530}}{2}$$

                    $${\text{x}} = 22765$$ 

  $${\textbf{Step -  3: Solve the equation and find value of y}}{\text{.}}$$

                      $${\text{Now , }}\dfrac{2}{{9106}} = \dfrac{{1.2}}{y}$$

                      $$ \Rightarrow {\text{y}} = \dfrac{{1.2 \times 9106}}{2}$$

                      $$ \Rightarrow y = \dfrac{{10927.2}}{2}$$

                      $$ \Rightarrow y = 5463.6{\text{ }}$$                        

$${\textbf{Thus, there are total 22765 sugar crystals in 5 kg of sugar and 5463}}{\textbf{.6 crystals in 1}}{\textbf{.2 kg of sugar}}{\textbf{.}}$$

$${\textbf{Step-1: Finding number of days required when one person fell ill.}}$$

                 $${\text{Given, Let the required number of days be x and the required number of persons be y.}}$$

                 $${\text{2 persons can fit a new window in a house in 3 days.}}$$

               $$\Rightarrow$$               $${x_1}{y_1} = {x_2}{y_2}$$

                $$\Rightarrow$$             $$2 \times 3 = 1 \times x$$

                $$\Rightarrow$$                    $$x = 6.$$

 $${\textbf{Step-2: Finding number of persons required to fit the windows in one day.}}$$

                   $$\Rightarrow$$               $${x_1}{y_1} = {x_2}{y_2}$$

                    $$\Rightarrow$$             $$2 \times 3 = y \times 1$$

                    $$\Rightarrow$$                    $$y = 6.$$

$${\textbf{Hence, (1) One of the persons got ill, job would take 6 days.}}$$

$${\textbf{(2)To fit the window in 1 day 6 persons would be required.}}$$

$${\textbf{Step -1: Assume number of bacteria and use given condition.}}$$

                  $$\text{Let }y\text{ be the number of bacteria at time }t.$$

                  $${\text{It is given that,}}$$

                  $$\dfrac{dy}{dT}\propto y$$

                  $$\Rightarrow\dfrac{dy}{dt}=\mu y$$

                  $$\Rightarrow\dfrac{dy}{y}=\mu {dt}$$

$${\textbf{Step -2: Integrate both sides.}}$$

                  $${\text{Integrating both sides,}}$$

                  $$\Large{\int}$$$${\dfrac{dy}{y}}=\Large{\int}$$$$\mu dt$$ 

                  $$\Rightarrow\log y= \mu t+c......(1)$$

                  $${\text{when}}$$ $$ t =0, y = 1,00,000$$

                  $$\therefore \log (1,00,000) =\mu\times0+c$$

                  $$c = \log 1,00,000$$

                  $${\text{Now putting}}$$ $$c = \log 1,00,000$$ $${\text{in (1), we have}}$$

                  $$\log y = \mu+\log{(1,00,000)}\ldots(2)$$

                  $${\text{Now, when}}$$ $$t = 2$$ $${\text{hours}}$$

                  $$y = 1,00,000 + \dfrac{10}{100}\times1,00,000 = 1,10,000$$

                  $$\therefore \log 1,10,000 = 2\mu+\log1,00,000$$

                  $$\Rightarrow \log \left(\dfrac{1,10,000}{1,00,000}\right)=2\mu\ $$     $$ \mathbf{\left[\because log{ \ n- log{ \ m=log}}\dfrac{n}{m}\right]}$$

                  $$\Rightarrow\mu=\dfrac{1}{2}\log{\left(\dfrac{11}{10}\right)}$$

$$\textbf{Step -3: Finding t when bacteria is 2,00,000.}$$

                  $$ \Rightarrow\log 2,00,000 =\dfrac{1}{2}\log{\left(\dfrac{11}{10}\right)}t+\log{\left(\ 1,00,000\right)}$$              

                  $$\Rightarrow\log \left(\dfrac{2,00,000}{1,00,000}\right) =  \dfrac{1}{2}\  \log{\left(\dfrac{11}{10}\right)}t$$

                  $$\Rightarrow \log 2 =  \dfrac{1}{2}\  \log{\left(\dfrac{11}{10}\right)t}$$

                  $$\Rightarrow t = \dfrac{2\log{2}}{\ \log{\left(\dfrac{11}{10}\right)}}$$

                  $$  = 14.54$$$${\text{ hours}}$$

$${\textbf{Hence, the count will reach 2,00,000 in 14.54 hours.}}$$