Answer \( \color{orange} (d) \displaystyle \ \color{red} \frac{x}{y} = \text{constant} \)

Solution

Let \( x \) be the number of sheets that weigh 2600 grams. \[ \begin{array}{|c|c|c|} \hline \text{Number of sheets} & 15 & x \\ \hline \text{Weight (g)} & 40 & 2600 \\ \hline \end{array} \]

It is a case of direct variation. As the weight increases, the number of sheets also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{15}{40} &= \frac{x}{2600} \\ \\ 15 \times 2600 &= 40 \times x \\ \\ x &= \frac{\cancel{2600}^{325} \times \cancel{15}^3}{\cancel{40}_{\cancel8_1}} \\ \\ x &= 325 \times 3 \\ \color{green} x &= \color{green} 975 \end{align*} \]

Answer \( \color{red} 975 \, \color{orange} (c) \)

Solution

Let \( x \) be the time taken to travel 480 km. \[ \begin{array}{|c|c|c|} \hline \text{Distance (km)} & 540 & 480 \\ \hline \text{Time (hrs)} & 18 & x \\ \hline \end{array} \]

As the distance decreases, the time taken also decreases. It is a case of direct variation.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{540}{18} &= \frac{480}{x} \\ \\ 540 \times x &= 18 \times 480 \\ \\ x &= \frac{\cancel{18}^1 \times \cancel{480}^{16}}{\cancel{540}_{\cancel{30}_1}} \\ \\ \color{green} x &= \color{green} 16 \text{ hours} \end{align*} \]

Answer \( \color{red} 16 \, hours \, \color{orange} (b) \)

Solution

Total work \( = n \) days \( \text{Work done in 1 day} = \displaystyle \color{green} \frac{1}{n} \)

Answer \( \color{red} \displaystyle \frac{1}{n} \ \color{orange} (a) \)

Solution

Let \( x \) be the number of days required by 20 labourers to build the wall. \[ \begin{array}{|c|c|c|} \hline \text{No. of Labourers} & 25 & 20 \\ \hline \text{No. of Days} & 4 & x \\ \hline \end{array} \]

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

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 25 \times 4 &= 20 \times x \\ \\ x &= \frac{\cancel{25}^5 \times \cancel{4}^1}{\cancel{20}_{\cancel5_1}} \\ \\ x &= \color{green} 5 \text{ days} \end{align*} \]

Answer \( \color{red} 5 \, \text{days} \, \color{orange} (b) \)

Solution

Let \( x \) be the time taken at 45 km/hr to cover the same distance. \[ \begin{array}{|c|c|c|} \hline \text{Speed (km/hr)} & 72 & 45 \\ \hline \text{Time (hrs)} & 2.5 & x \\ \hline \end{array} \]

It is a case of inverse variation. As the speed decreases, time increases.

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 72 \times 2.5 &= 45 \times x \\ \\ x &= \frac{\cancel{72}^8 \times 2.5}{\cancel{45}_5} \\ \\ x &= \frac{8 \times 2.5}{5} \\ \\ x &= \frac{20}{5} \\ \\ \color{green} x &= \color{green} 4 \text{ hours} \end{align*} \]

Answer \( \color{red} 4 \, \text{hours} \, \color{orange} (c) \)

Solution

Let \( x \) be the number of cartons required to pack 576 books. \[ \begin{array}{|c|c|c|} \hline \text{Books} & 96 & 576 \\ \hline \text{Cartons} & 3 & x \\ \hline \end{array} \]

This is a case of direct variation. As the number of books increases, the number of cartons also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{96}{3} &= \frac{576}{x} \\ \\ 96 \times x &= 3 \times 576 \\ \\ x &= \frac{3 \times \cancel{576}^{6}}{\cancel{96}_1} \\ \\ \color{green} x &= \color{green} 18 \end{align*} \]

Answer \( \color{red} 18 \, \text{cartons} \, \color{orange} (b) \)

Solution

Let \( x \) be the time taken to fill the tank using 5 pipes.

\( 1 \text{ hour } 20 \text{ minutes } = 60 + 20 \implies 80 \text{ minutes} \) \[ \begin{array}{|c|c|c|} \hline \text{Number of pipes} & 6 & 5 \\ \hline \text{Time (min)} & 80 & x \\ \hline \end{array} \]

This is a case of inverse variation. As the number of pipes decreases, time increases.

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 6 \times 80 &= 5 \times x \\ \\ x &= \frac{6 \times \cancel{80}^{16}}{\cancel5_1} \\ \\ x &= 6 \times 16 \\ \color{green} x &= \color{green} 96 \text{ minutes} \end{align*} \]

Answer \( \color{red} 96 \, \text{minutes} \, \color{orange} (a) \)

Solution

\[ \begin{array}{|c|c|c|} \hline x & 5 & x \\ \hline y & 30 & 10 \\ \hline \end{array} \]

Since it's inverse variation: \( x \times y = \text{constant} \)

\[ \begin{align*} 5 \times 30 &= x \times 10 \\ \\ x &= \frac{5 \times \cancel{30}^3}{\cancel{10}_1} \\ \\ \color{green} x &= \color{green} 15 \end{align*} \]

Answer \( \color{red} x = 15 \ \color{orange} (c) \)

Solution

Scale = \( 1 : 300 \)
Distance on map = 4 km
\( \text{Actual distance} = 4 \times 300 \implies 1200 \, \text{km} \)

Answer \( \color{red} 1200 \, \text{km} \ \color{orange} (d) \)

Solution

Let \( x \) be the cost of 16 toys. \[ \begin{array}{|c|c|c|} \hline \text{Number of toys} & 9 & 16 \\ \hline \text{Cost (₹)} & 333 & x \\ \hline \end{array} \]

It is a case of direct variation. As the number of toys increases, the cost also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{9}{333} &= \frac{16}{x} \\ \\ 9 \times x &= 333 \times 16 \\ \\ x &= \frac{\cancel{333}^{37} \times 16}{\cancel{9}_1} \\ \\ x &= 37 \times 16 \\ \color{green} x &= \color{green} 592 \end{align*} \]

Answer The cost of 16 toys \( = \color{red} \text{₹ }592 \)

Solution

Let \( x \) be the length (in cm) of the rod with mass 105 g. \[ \begin{array}{|c|c|c|} \hline \text{Length (cm)} & 16 & x \\ \hline \text{Mass (g)} & 192 & 105 \\ \hline \end{array} \]

It is a case of direct variation. As the mass decreases, the length also decreases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{16}{192} &= \frac{x}{105} \\ \\ 16 \times 105 &= 192 \times x \\ \\ x &= \frac{\cancel{16}^1 \times \cancel{105}^{35}}{\cancel{192}_{\cancel{12}_4}} \\ \\ x &= \frac{35}{4} \\ \\ \color{green} x &= \color{green} 8.75 \, \text{cm} \end{align*} \]

Answer The length of the rod \( = \color{red} 8.75 \, \text{cm} \)

Solution

Let \( x \) be the number of cows that can graze the field in 12 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of cows} & 44 & x \\ \hline \text{No. of days} & 9 & 12 \\ \hline \end{array} \]

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

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 44 \times 9 &= x \times 12 \\ \\ x &= \frac{\cancel{44}^{11} \times \cancel{9}^3}{\cancel{12}_{\cancel{4}_1}} \\ \\ x &= 11 \times 3 \\ \color{green} x &= \color{green} 33 \, \text{cows} \end{align*} \]

Answer \( \color{red} 33 \, \text{cows} \)

Solution

Let \( x \) be the cost of 39 oranges.
2 dozen = 24 oranges \[ \begin{array}{|c|c|c|} \hline \text{No. of oranges} & 24 & 39 \\ \hline \text{Cost (₹)} & 84 & x \\ \hline \end{array} \]

It is a case of direct variation. As the number of oranges increases, the cost also increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{24}{84} &= \frac{39}{x} \\ \\ 24 \times x &= 84 \times 39 \\ \\ x &= \frac{\cancel{84}^{7} \times 39}{\cancel{24}_2} \\ \\ x &= \frac{7 \times 39}{2} \\ \\ x &= \frac{273}{2} \\ \\ \color{green} x &= \color{green} \text{₹}136.5 \end{align*} \]

Answer The cost of 39 oranges \( = \color{red} \text{₹}136.5 \)

Solution

Let \( x \) be the distance travelled in 12 minutes. \[ \begin{array}{|c|c|c|} \hline \text{Time (minutes)} & 60 & 12 \\ \hline \text{Distance (km)} & 48 & x \\ \hline \end{array} \]

It is a case of direct variation. As time decreases, distance decreases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{60}{48} &= \frac{12}{x} \\ \\ 60 \times x &= 48 \times 12 \\ \\ x &= \frac{\cancel{48}^{9.6} \times \cancel{12}^1}{\cancel{60}_{\cancel5_1}} \\ \\ \color{green} x &= \color{green} 9.6 \, km \end{align*} \]

Answer The distance travelled in 12 minutes \( = \color{red} 9.6 \, km \)

Solution

Let \( x \) be the area occupied by 120 postal stamps. \[ \begin{array}{|c|c|c|} \hline \text{No. of stamps} & 15 & 120 \\ \hline \text{Area }(cm^2) & 60 & x \\ \hline \end{array} \]

It is a case of direct variation. As the number of stamps increases, the area occupied also increases.

\[ \begin{align*} \frac{x}{y} &= k \, \text{(constant)} \\ \\ \frac{15}{60} &= \frac{120}{x} \\ \\ 15 \times x &= 60 \times 120 \\ \\ x &= \frac{\cancel{60}^4 \times 120}{\cancel{15}_1} \\ \\ x &= 4 \times 120 \\ \color{green} x &= \color{green} 480 \, \text{cm²} \end{align*} \]

Answer The area occupied by 120 stamps \( = \color{red} 480 \, \text{cm²} \).

Solution

Let \( x \) be the number of days the food will last for 25 animals. \[ \begin{array}{|c|c|c|} \hline \text{No. of animals} & 20 & 25 \\ \hline \text{No. of days} & 6 & x \\ \hline \end{array} \]

This is a case of inverse variation. As the number of animals increases, the days the food lasts decreases.

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 20 \times 6 &= 25 \times x \\ \\ x &= \frac{\cancel{20}^4 \times 6}{\cancel{25}_5} \\ \\ x &= \frac{24}{5} \\ \\ \color{green} x &= \color{green} 4.8 \text{ days} \end{align*} \]

Answer The food will last \( \color{red} 4.8 \, \text{days} \).

Solution

Let \( x \) be the number of words she types in 6 minutes. \[ \begin{array}{|c|c|c|} \hline \text{Time (min)} & 30 & 6 \\ \hline \text{Words} & 540 & x \\ \hline \end{array} \]

This is a case of direct variation. As the time decreases, the number of words decreases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{540}{30} &= \frac{x}{6} \\ \\ 540 \times 6 &= 30 \times x \\ \\ x &= \frac{\cancel{540}^{108} \times \cancel6^1}{\cancel{30}_{\cancel5_1}} \\ \\ \color{green} x &= \color{green} 108 \, \text{words} \end{align*} \]

Answer Reema would type \( \color{red} 108 \, \text{words} \) in 6 minutes.

Solution

Let \( x \) be the number of men required to do the work in 14 days. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 52 & x \\ \hline \text{No. of days} & 35 & 14 \\ \hline \end{array} \]

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

\[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 52 \times 35 &= x \times 14 \\ \\ x &= \frac{\cancel{52}^{26} \times \cancel{35}^5}{\cancel{14}_{\cancel2_1}} \\ \\ x &= 26 \times 5 \\ \color{green} x &= \color{green} 130 \, \text{men} \end{align*} \]

Answer \( \color{red} 130 \, \text{men} \) will do the same work in 14 days.

Solution

Let \( x \) be the time taken to cover 221 km. \[ \begin{array}{|c|c|c|} \hline \text{Distance (km)} & 51 & 221 \\ \hline \text{Time (minutes)} & 45 & x \\ \hline \end{array} \]

This is a case of direct variation. As the distance increases, the time increases.

\[ \begin{align*} \frac{x}{y} &= k \text{ (constant)} \\ \\ \frac{51}{45} &= \frac{221}{x} \\ \\ 51 \times x &= 45 \times 221 \\ \\ x &= \frac{\cancel{45}^{15} \times \cancel{221}^{13}}{\cancel{51}_{\cancel{17}_1}} \\ \\ x &= 15 \times 13 \\ \color{green} x &= \color{green} 195 \, \text{minutes} \\ \\ \text{Convert } & \text{minutes to hours}\\ &= \frac{195}{60} \\ \\ &= 3 \ hrs \, 15 \ min \\ \end{align*} \]

Answer It will take \( \color{red} 3 \, \text{hours} \, 15 \, \text{minutes} \) to cover 221 km.

Solution

Let \( x \) be the number of days the food will last for \( 50 + 30 = 80 \) girls. \[ \begin{array}{|c|c|c|} \hline \text{No. of girls} & 50 & 80 \\ \hline \text{No. of days} & 40 & x \\ \hline \end{array} \]

This is a case of inverse variation. As the number of girls increases, the number of days decreases.

\[ \begin{align*} a \times b &= k \, \text{(constant)} \\ \\ 50 \times 40 &= 80 \times x \\ \\ x &= \frac{\cancel{50}^{25} \times \cancel{40}^1}{\cancel{80}_{\cancel{2}_1}} \\ \\ \color{green} x &= \color{green} 25 \, \text{days} \end{align*} \]

Answer The food will last \( \color{red} 25 \, \text{days} \).

Solution

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

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

Solution

Let \( x \) be the number of days the remaining food will last after 30 men die.
The initial provision was for 200 days, but after 5 days, only \(200 - 5 = 195\) days worth of food remains for the original 120 men.
\[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 120 & 90 \\ \hline \text{No. of days (remaining)} & 195 & x \\ \hline \end{array} \]

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

\[ \begin{align*} x \times y &= k \text{ (constant)} \\ \\ 120 \times 195 &= 90 \times x \\ \\ x &= \frac{\cancel{120}^{4} \times \cancel{195}^{65}}{\cancel{90}_{\cancel{3}_1}} \\ \\ x &= 4 \times 65 \\ \color{green} x &= \color{green} 260 \, \text{days} \end{align*} \]

Answer The remaining food will last \( \color{red} 260 \, \text{days} \) for 90 men.

Solution

\[ \begin{align*} \text{Length of the train} &= 400 \, m \\ \text{Speed} &= 72 \, km/hr \\ \\ \text{Convert speed from } & {\color{green} km/hr} \text{ to } \color{green} m/s \\ \\ &= \cancel{72}^4 \times \frac{5}{\cancel{18}_1} \, m/s \\ & = 4 \times 5 \, m/s \\ \color{green} Speed &= \color{green} 20 \, m/s \\ \\ \text{Time} &= \frac{\text{Distance}}{\text{Speed}} \\ \\ &= \frac{400}{20} \\ \\ \color{green} Time &= \color{green} 20 \, \text{seconds} \end{align*} \]

Answer The train takes \( \color{red} 20 \, \text{seconds} \) to cross the telegraph post.

Solution

Let \( x \) be the number of books the shopkeeper can buy if he gets a discount of ₹25 on each book. The new price per book is \( \color{green} \text{₹}125 - \text{₹}25 = \text{₹}100\). \[ \begin{array}{|c|c|c|} \hline \text{Cost per book (₹)} & 125 & 100 \\ \hline \text{No. of books} & 40 & x \\ \hline \end{array} \]

It is a case of inverse variation. As the cost per book decreases, the number of books he can buy increases.

\[ \begin{align*} \text{Since } a \times b &= k \text{ (constant)} \\ \\ 125 \times 40 &= 100 \times x \\ \\ x &= \frac{\cancel{125}^{25} \times \cancel{40}^{2}}{\cancel{100}_{\cancel{20}_1}} \\ \\ x &= 25 \times 2 \\ \color{green} x &= \color{green} 50 \text{ books} \end{align*} \]

Answer Shopkeeper can buy \( \color{red} 50 \, \text{books} \) if he gets a discount of ₹25 on each book.

Solution

Let \( x \) be the number of hours he should work per day to finish 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 hours per day increase.

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

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

Solution

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

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

Solution

Let \( x \) be the duration (in minutes) of each period if there are 9 periods. \[ \begin{array}{|c|c|c|} \hline \text{No. of periods} & 8 & 9 \\ \hline \text{Duration (min)} & 45 & x \\ \hline \end{array} \]

The total school time remains the same, so it's a case of inverse variation. As the number of periods increases, the duration per period decreases.

\[ \begin{align*} a \times b &= k \, \text{(constant)} \\ \\ 8 \times 45 &= 9 \times x \\ \\ x &= \frac{8 \times \cancel{45}^{5}}{\cancel{9}_1} \\ \\ x &= 8 \times 5 \\ \color{green} x &= \color{green} 40 \, \text{minutes} \end{align*} \]

Answer Each period would be \( \color{red} 40 \, \text{minutes} \).

Solution

Let \( x \) be the speed of the car to reach office in 75 minutes. \[ \begin{array}{|c|c|} \hline \text{Speed (km/hr)} & 50 & x \\ \hline \text{Time (minutes)} & 90 & 75 \\ \hline \end{array} \]

As the time is reduced, the speed of the car will increase to reach the office. It is a case of inverse variation.

\[ \begin{align*} \implies a \times b &= k \text{ (constant)} \\ \\ 50 \times 90 &= x \times 75 \\ \\ x &= \frac{\cancel{50}^2 \times \cancel{90}^{30}}{\cancel{75}_{\cancel3_1}} \\ \\ x &= 2 \times 30 \\ \color{green} x &= \color{green} 60 \, km/hr \end{align*} \]

To reach the office on time, the speed of the car should be \( \color{red} 60 \, km/hr \)

\[ \begin{align*} \color{green} Distance & = \color{green} Speed \times Time \\ & = 50 \times 1.5 \\ Distance & = 75 \, km \\ \\ \text{Total Daily Distance} & = 75 \, km + 75 \, km \\ & = 150 \, km \end{align*} \]

Answer Speed \( = \color{red} 60 \, km/hr \), Total daily distance \( = \color{red} 150 \, km \)

Solution

Total original provision = 30 days
After 5 days, remaining provision = \( 30 - 5 = 25 \) days (for 120 men)
Now, the number of men becomes \( 120 + 5 = 125 \).

Let \( x \) be the number of days the remaining food will last for 125 men. \[ \begin{array}{|c|c|c|} \hline \text{No. of men} & 120 & 125 \\ \hline \text{No. of days} & 25 & x \\ \hline \end{array} \] It is a case of inverse variation. As the number of men increases, the number of days decreases. \[ \begin{align*} \implies a \times b &= k \text{ (constant)} \\ \\ 120 \times 25 &= 125 \times x \\ \\ x &= \frac{\cancel{120}^{24} \times \cancel{25}^1}{\cancel{125}_{\cancel{5}_1}} \\ \\ \color{green} x &= \color{green} 24 \, \text{days} \end{align*} \]

Answer They can sustain the remaining provision for \( \color{red} 24 \, \text{days} \)

Solution

Let \( x \) be the total number of people now. \[ \begin{array}{|c|c|c|} \hline \text{No. of people} & 3 & x \\ \hline \text{No. of days} & 30 & 18 \\ \hline \end{array} \] It is a case of inverse variation. As the number of people increases, the time the wheat lasts decreases. \[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 3 \times 30 &= x \times 18 \\ \\ x &= \frac{\cancel3^1 \times \cancel{30}^5}{\cancel{18}_{\cancel6_1}} \\ \\ \color{green} x &= \color{green} 5 \, \text{people} \\ \\ \text{New members} &= 5 - 3 \\ &= \color{green} 2 \text{ new members} \end{align*} \]

Answer There are \( \color{red} 2 \, \text{new members} \) in the group.

Solution

Time taken when car travels at \( {\color{green} 30 \ km/hr}: 9 \ am \text{ to } 1 \ pm \color{green} = 4 \ hours \)

Required time: \( 9 \ am \text{ to } 12 \ pm \color{green}= 3 \ hours \)

Let \( x \) be the required speed so that he can reach the place by 12 noon . \[ \begin{array}{|c|c|c|} \hline \text{Time (hrs)} & 4 & 3 \\ \hline \text{Speed (km/hr)} & 30 & x \\ \hline \end{array} \] It is a case of inverse variation. As the time decreases, the speed will increase. \[ \begin{align*} a \times b &= k \text{ (constant)} \\ \\ 4 \times 30 &= x \times 3 \\ \\ x &= \frac{4 \times \cancel{30}^{10}}{\cancel3_1} \\ \\ \color{green} x &= \color{green} 40 \\ \\ \text{Required speed} &= 40 \ km/hr \\ \text{Increase the speed} &= 40 - 30 \implies \color{green} 10 \ km/hr \end{align*} \]

Answer Ranjeet should increase his speed by \( \color{red} 10\, km/hr \) to reach by 12 noon.