Solution

Let \( x \) be the number of hours Ramit should work per day to finish the work in 20 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of days} & 25 & 20 \\ \hline \text{Hours per day} & 8 & x \\ \hline \end{array} \]

It is a case of inverse variation. As the number of days decreases, the number of hours per day increases.

\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 25 \times 8 &= 20 \times x \\ \\ x &= \frac{\cancel{25}^{5} \times \cancel8^2}{\cancel{20}_{\cancel4_1}} \\ &= 5 \times 2 \\ \color{green} x &= \color{green} 10 \text{ hours/day} \end{align*} \]

Answer Ramit should work \( \color{red} 10 \, \text{hours/day} \) to finish the same work in 20 days.

Solution

Let \( x \) be the amount of work completed in 8 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of days} & 10 & 8 \\ \hline \text{Amount of work} & 1 & x \\ \hline \end{array} \] \[ \begin{align*} \text{This is a case } & \text{ of direct variation} \\ \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{10}{1} & = \frac{8}{x} \\ \\ 10 \times x &= 8 \times 1 \\ \\ x &= \frac{\cancel8^4}{\cancel{10}_5} \\ \\ \color{green} x &= \color{green} \left(\frac{4}{5} \right)^{th} work \end{align*} \]

Answer Amount of work completed in 8 days \( \begin{align*} = \color{red} \left(\frac{4}{5} \right)^{th} work \end{align*} \)

Solution

Let \( x \) be the number of days required for 12 men to build the same wall. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 20 & 12 \\ \hline \text{No. of days} & 9 & x \\ \hline \end{array} \]

It is a case of inverse variation. As the number of men decreases, the number of days increases.

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 20 \times 9 &= 12 \times x \\ \\ x &= \frac{\cancel{20}^5 \times \cancel9^3}{\cancel{12}_{\cancel4_1}} \\ \\ x &= 5 \times 3 \\ \color{green} x &= \color{green} 15 \text{ days} \end{align*} \]

Answer 12 men can build the wall in \( \color{red} 15 \, \text{days} \).

Solution

Let \( x \) be the number of days required for Geetika to weave 120 baskets. \[ \begin{array}{|c|c|c|} \hline \text{No. of baskets} & 20 & 120 \\ \hline \text{No. of days} & 30 & x \\ \hline \end{array} \]

It is a case of direct variation. As the number of baskets increases, the number of days will also increase.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{20}{30} &= \frac{120}{x} \\ \\ 20 \times x &= 30 \times 120 \\ \\ x &= \frac{30 \times \cancel{120}^6}{\cancel{20}_1} \\ \\ x &= 30 \times 6 \\ \color{green} x &= \color{green} 180 \text{ days} \end{align*} \]

Answer \( \color{red} 180 \, \text{days} \) required to weave 120 baskets

Solution

\[ \begin{align*} \text{Length of the train} &= 280 \, m \\ \text{Speed} &= 42 \, km/hr \\ \\ \text{Convert speed from } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ \\ &= \cancel{42}^7 \times\frac{5}{\cancel{18}_3} \text{ m/s} \\ \\ & = \frac{7 \times 5}{3} \text{ m/s} \\ \\ \color{green} Speed &= \color{green} \frac{35}{3} \text{ m/s} \\ \\ Distance &= 280 \, m \\ \\ \color{red} Time &= \color{red} \frac{Distance}{Speed} \\ \\ & = \frac{280}{\left(\frac{35}{3}\right)} \\ \\ & = \cancel{280}^8 \times \frac{3}{\cancel{35}_1} \\ \\ & = 8 \times 3 \\ \color{green} Time &= \color{green} 24 \,seconds \end{align*} \]

Answer The train will cross the man standing on the platform in \( \color{red} 24 \text{ seconds} \)

Solution

\[ \begin{align*} \text{Distance} &= 350 \, \text{m} \\ \text{Time} &= 28 \, \text{seconds} \\ \\ \color{red} Speed &= \color{red} \frac{Distance}{Time} \\ \\ &= \frac{\cancel{350}^{\cancel{175}^{25}}}{\cancel{28}_{\cancel{14}_2}} \text{m/s} \\ \\ \color{green} Speed &= \color{green}\frac{25}{2} \text{m/s} \\ \\ \text{Convert speed from } & { \color{green} m/s} \text{ to } {\color{green} km/hr} \\ \\ \text{Speed} &= \frac{\cancel{25}^5}{\cancel2_1} \times \frac{\cancel{18}^9}{\cancel5_1} \, \text{km/hr} \\ \\ &= 9 \times 5 \text{ km/hr} \\ \color{green} Speed &= \color{green} 45 \text{ km/hr} \end{align*} \]

Answer Speed of the train \( = \color{red} 45 \, \text{km/hr} \)

Solution

\[ \begin{align*} \text{Speed of the train} &= 72 \, \text{km/hr} \\ \text{Convert } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ \\ &= \cancel{72}^4 \times \frac{5}{\cancel{18}_1} \text{ m/s} \\ \\ \color{green} Speed &= \color{green} 20 \, \text{m/s} \\ \\ \text{Time taken to cross the bridge} &= 13 \, \text{seconds} \\ \text{Length of the train} &= 150 \, \text{m} \\ \text{Length of bridge} &= x \\ \text{Distance covered} &= \text{Length of train} + \text{Length of bridge} \\ & = 150 + x \\ \\ \text{Distance} &= \text{Speed} \times \text{Time} \\ 150 + x&= 20 \times 13 \\ 150 + x&= 260 \\ x & = 260 -150 \\ \color{green} x &= \color{green} 110 \, \text{m} \\ \end{align*} \]

Answer The length of the bridge \( = \color{red} 110 \, \text{m} \)

Solution

\[ \begin{align*} \text{Speed of the train} &= 50 \, \text{km/hr} \\ \text{Convert } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ &= \cancel{50}^{25} \times \frac{5}{\cancel{18}_9} \text{ m/s} \\ \\ & = \frac{25 \times 5}{9} \ m/s \\ \\ \color{green} Speed &= \color{green} \frac{125}{9} \, \text{m/s} \\ \\ \text{Length of the train} &= 120 \, \text{m} \\ \text{Length of the platform} &= 130 \, \text{m} \\ \text{Total distance to cover} &= \text{Length of train} + \text{Length of platform} \\ &= 120 \, \text{m} + 130 \, \text{m} \\ \text{Distance} &= 250 \, \text{m} \\ \\ \color{red} \text{Time} &= \color{red} \frac{\text{Distance}}{\text{Speed}} \\ \\ &= \frac{250}{\left(\frac{125}{9}\right)} \\ \\ &= \cancel{250}^{2} \times \frac{9}{\cancel{125}_1} \\ \\ & = 2 \times 9 \\ \color{green} \text{Time} & = \color{green} 18 \, \text{seconds} \\ \end{align*} \]

Answer The train will take \( \color{red} 18 \, \text{seconds} \) to clear the platform.

Solution

\[ \begin{align*} \text{Length of the train} &= 210 \, \text{m} \\ \text{Length of the tunnel} &= 90 \, \text{m} \\ \text{Total distance to cover} &= \text{Length of train} + \text{Length of tunnel} \\ &= 210 + 90 \, \text{m} \\ \text{Distance} &= 300 \, \text{m} \\ \text{Time} &= 12 \, \text{seconds} \\ \\ \color{red} \text{Speed} &= \color{red} \frac{\text{Distance}}{\text{Time}} \\ \\ &= \frac{\cancel{300}^{25}}{\cancel{12}_1} \, \text{m/s} \\ \\ \color{green} \text{Speed} &= \color{green} 25 \, \text{m/s} \\ \\ \text{Convert speed from } & {\color{green} m/s} \text{ to } \color{green} km/hr \\ \\ &= \cancel{25}^5 \times \frac{18}{\cancel5_1} \, \text{km/hr} \\ \\ &= 5 \times 18 \, \text{km/hr} \\ \color{green} \text{Speed} &= \color{green} 90 \, \text{km/hr} \end{align*} \]

Answer The speed of the train is \( \color{red} 90 \, \text{km/hr} \)

Solution

\[ \begin{align*} \text{Speed of the train} &= 80 \, \text{km/hr} \\ \text{Convert } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ \\ &= \cancel{80}^{40} \times \frac{5}{\cancel{18}_9} \, \text{m/s} \\ \\ \color{green} Speed &= \color{green} \frac{200}{9} \, \text{m/s} \\ \\ \text{Length of the train} &= 270 \, \text{m} \\ \text{Length of the platform} &= 130 \, \text{m} \\ \text{Total distance to cover} &= \text{Length of train} + \text{Length of platform} \\ &= 270 + 130 \, \text{m} \\ \text{Distance} &= 400 \, \text{m} \\ \\ \color{red} \text{Time} &= \color{red} \frac{\text{Distance}}{\text{Speed}} \\ \\ &= \frac{400}{(\frac{200}{9})} \\ \\ &= \cancel{400}^2 \times \frac{9}{\cancel{200}_1} \\ \\ \color{green} \text{Time} & = \color{green} 18 \, \text{seconds} \end{align*} \]

Answer The train will take \( \color{red} 18 \, \text{seconds} \) to cross the platform.