NCERT Solutions Class 8 Maths Chapter 13

Q.1

$\begin{array}{l}\begin{array}{l}\text{A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base.}\\ \text{In the following table, find the parts of base that need to be added.}\end{array}\\ \begin{array}{cccccc}\text{Parts of red pigment}& \text{1}& \text{4}& \text{7}& \text{12}& \text{20}\\ \text{Parts of base}& \text{8}& \mathrm{\dots }& \mathrm{\dots }& \mathrm{\dots }& \mathrm{\dots }\end{array}\end{array}$

Ans

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. Since the ratio of red pigment and the base is same every time, therefore, the parts of red pigments and the parts of base are in direct proportion.

Let us take the unknown values as x₁, x₂, x₃ and x_4.
The given information in the form of a table is as follows.

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& x_1& x_2& x_3& x_4\end{array}$ $\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{8}=\frac{4}{{\mathrm{x}}_1}\Rightarrow {\mathrm{x}}_1=8\times 4\Rightarrow {\mathrm{x}}_1=32\\ \frac{1}{8}=\frac{7}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=8\times 7\Rightarrow {\mathrm{x}}_2=56\\ \frac{1}{8}=\frac{12}{{\mathrm{x}}_3}\Rightarrow {\mathrm{x}}_3=8\times 12\Rightarrow {\mathrm{x}}_3=96\\ \frac{1}{8}=\frac{20}{{\mathrm{x}}_4}\Rightarrow {\mathrm{x}}_4=8\times 20\Rightarrow {\mathrm{x}}_4=160\end{array}$

Therefore, the parts of a base to be added are shown as follows:

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& 32& 56& 96& 160\end{array}$

Q.2 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Ans

Let the parts of red pigment required to mix with 1800 mL of base be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Parts of red pigment}& 1& \mathrm{x}\\ \text{Parts of base(in mL)}& 75& 1800\end{array}$

The parts of red pigment and the parts of base are in direct proportion.
Therefore, we obtain

$\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{75}=\frac{\mathrm{x}}{1800}\Rightarrow 75\mathrm{x}=1800\Rightarrow \mathrm{x}=24\end{array}$

Thus, 24 parts of red pigments should be mixed with 1800 mL of base.

Q.3 A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Ans

Let the number of bottles filled by the machine in five hours be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Number of bottles}& 840& \mathrm{x}\\ \text{Time taken(in hours)}& 6& 5\end{array}$

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

$\begin{array}{l}\frac{840}{6}=\frac{\mathrm{x}}{5}\Rightarrow 6\mathrm{x}=840\times 5\\ \Rightarrow 6\mathrm{x}=4200\\ \Rightarrow \mathrm{x}=700\end{array}$

Thus, 700 bottles will be filled in 5 hours.

Q.4 A photograph of a bacterium enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacterium? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Ans

Let the actual length of the bacterium be x cm.
The given information in the form of a table is as follows.

\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& x\\ \begin{array}{l}\text{Number of times photograph of}\\ \text{the bacterium was enlarged}\end{array}& 50000& 1\end{array}

The number of times the photograph of the bacterium was enlarged and the lengths of the bacterium are in direct proportion.
Therefore, we obtain

\begin{array}{l}\frac{5}{50000}=\frac{x}{1}\\ \Rightarrow 50000x=5\\ \Rightarrow x=\frac{1}{10000}\\ \Rightarrow x={10}^{-4}\\ \\ \text{Hence},\text{the actual length of the bacterium is 1}{0}^{-\text{4}}\text{cm}.\end{array}

Now, we have to find its enlarged length if the photograph is enlarged 20,000 times only.

Let the enlarged length of the bacterium be y cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& \mathrm{y}\\ \begin{array}{l}\text{Number of times photograph}\\ \text{of the bacterium was enlarged}\end{array}& 50000& 20000\end{array}$

The number of times the photograph of the bacterium was enlarged and the lengths of bacterium are in direct proportion.

Therefore, we obtain

$\begin{array}{l}\frac{5}{50000}=\frac{\mathrm{y}}{20000}\\ \Rightarrow 100000=50000\mathrm{y}\\ \Rightarrow \mathrm{y}=\frac{100000}{50000}\\ \Rightarrow \mathrm{y}=2\\ \text{Hence},\text{the enlarged length of the bacterium is 2 cm}.\end{array}$

Q.5 In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Ans

Let the length of the mast of the model ship be x cm.
The given information in the form of a table is as follows:

$\begin{array}{ccc}& \text{Height of mast}& \text{Length of ship}\\ \text{Model of ship}& 9& \mathrm{x}\\ \text{Actual ship}& 12& 28\end{array}$

The dimensions of the actual ship and the model ship are directly proportional to each other.

Therefore, we obtain:

$\begin{array}{l}\frac{9}{12}=\frac{\mathrm{x}}{28}\\ \Rightarrow 9\times 28=12\mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{252}{12}\\ \Rightarrow \mathrm{x}=21\\ \end{array}$

Thus, the length of the model ship is 21 cm.

Q.6 Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in
(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Ans

(i) Let the number of sugar crystals in 5 kg of sugar be x.
The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 5\\ \text{Number of crystals}& 9\times {10}^6& \mathrm{x}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{2}{9\times {10}^6}=\frac{5}{\mathrm{x}}\\ \Rightarrow 2\mathrm{x}=5\times 9\times {10}^6\\ \Rightarrow \mathrm{x}=\frac{5\times 9\times {10}^6}{2}\\ \Rightarrow \mathrm{x}=22.5\times {10}^6\\ \Rightarrow \mathrm{x}=2.25\times {10}^7\\ \end{array}$

Hence, the number of sugar crystals is 2.25 × 10⁷.

(ii) Let the number of sugar crystals in 1.2 kg of sugar be y.

The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 1.2\\ \text{Number of crystals}& 9\times {10}^6& \mathrm{y}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other

.

$\begin{array}{l}\frac{2}{9\times {10}^6}=\frac{1.2}{\mathrm{y}}\\ \Rightarrow 2\mathrm{y}=1.2\times 9\times {10}^6\\ \Rightarrow \mathrm{x}=\frac{1.2\times 9\times {10}^6}{2}\\ \Rightarrow \mathrm{x}=5.4\times {10}^6\\ \\ \text{Hence},\text{the number of sugar crystals is}5.4\times {10}^6.\end{array}$

Therefore, we obtain

Q.7 Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Ans

Let the distance represented on the map be x cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance covered on map(in cm)}& 1& \mathrm{x}\\ \text{Distance covered on road(in km)}& 18& 72\end{array}$

The distances covered on road and represented on map are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{1}{18}=\frac{\mathrm{x}}{72}\\ \Rightarrow 72=18\mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{72}{18}\\ \Rightarrow \mathrm{x}=4\end{array}$

Hence, the distance represented on the map is 4 cm.

Q.8 A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5m long.

Ans

Let the length of the shadow of the other pole be x m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& 10.50\\ \text{Length of the shadow(in m)}& 3.20& \mathrm{x}\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{10.50}{\mathrm{x}}\\ \Rightarrow 5.60\mathrm{x}=10.50\times 3.20\\ \Rightarrow \mathrm{x}=\frac{33.6}{5.60}\\ \Rightarrow \mathrm{x}=6\end{array}$

Hence, the length of the shadow will be 6 m.

(ii) Let the height of the pole be y m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& \mathrm{y}\\ \text{Length of the shadow(in m)}& 3.20& 5\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{\mathrm{y}}{5}\\ \Rightarrow 5.60\times 5=3.20\mathrm{y}\\ \Rightarrow \mathrm{y}=\frac{5.60\times 5}{3.20}\\ \Rightarrow \mathrm{y}=8.75\end{array}$

Thus, the height of the pole is 8.75 m or 8 m 75 cm.

Q.9 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Ans

Let the distance travelled by the truck in 5 hours be x km.
We know, 1 hour = 60 minutes
∴ 5 hours = (5 × 60) minutes = 300 minutes
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance travelled(in km)}& 14& \mathrm{x}\\ \text{Time(in min)}& 25& 300\end{array}$

The distance travelled by the truck and the time taken by the truck are directly proportional to each other.

Therefore, we obtain

$\begin{array}{l}\frac{14}{25}=\frac{\mathrm{x}}{300}\\ \Rightarrow 300\times 14=25\mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{4200}{25}\\ \Rightarrow \mathrm{x}=168\end{array}$

Hence, the distance travelled by the truck is 168 km.

Q.10 Which of the following are in inverse proportion?

(i) The number of workers on a job and the time to complete the job.

(ii) The time taken for a journey and the distance travelled in a uniform speed.

(iii) Area of cultivated land and the crop harvested.

(iv) The time taken for a fixed journey and the speed of the vehicle.

(v) The population of a country and the area of land per person.

Ans

(i) It is in inverse proportion because if there are more workers, then it will take less time to complete that job.

(ii) No, these are not in inverse proportion because in more time, we may cover more distance with a uniform speed.

(iii) No, these are not in inverse proportion because in more area, more quantity of crop may be harvested.

(iv) It is in inverse proportion because with more speed, we may complete a certain distance in a lesser time.

(v) It is in inverse proportion because if the population is increasing/decreasing, then the area of the land per person will be decreasing/increasing accordingly.

Q.11 In a Television game show, the prize money of ₹ 1, 00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Ans

Let the unknown prizes be x₁, x₂, x₃, x₄ and x_5.

The given information is represented in a table below:

From the table, we obtain

1 × 1,00,000 = 2 × 50,000 = 1,00,000

Thus, the number of winners and the amount given to each winner are inversely proportional to each other.

Therefore, we obtain

$\begin{array}{l}1\times 1,00,000=4\times {\mathrm{x}}_1\\ \Rightarrow {\mathrm{x}}_1=\frac{1,00,000}{4}=25,000\\ \\ 1\times 1,00,000=5\times {\mathrm{x}}_2\\ \Rightarrow {\mathrm{x}}_2=\frac{1,00,000}{5}=20,000\\ 1\times 1,00,000=8\times {\mathrm{x}}_3\\ \Rightarrow {\mathrm{x}}_3=\frac{1,00,000}{8}=12,500\\ \\ 1\times 1,00,000=10\times {\mathrm{x}}_4\\ \Rightarrow {\mathrm{x}}_4=\frac{1,00,000}{10}=10,000\\ \\ 1\times 1,00,000=20\times {\mathrm{x}}_5\\ \Rightarrow {\mathrm{x}}_5=\frac{1,00,000}{20}=5,000\end{array}$

Q.12 Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.

(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?

(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.

(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?

Ans

Let the unknown angles be x₁, x₂ and x₃.

\begin{array}{cccccc}\text{Number of spokes}& 4& 6& 8& 10& 12\\ \begin{array}{l}\text{Angle between a pair of}\\ \text{consecutive spokes}\end{array}& {90}^{\circ }& {60}^{\circ }& x_1& x_2& x_3\end{array}

$\begin{array}{l}\text{From the given table},\text{we obtain}\\ \text{4}\times \text{9}0\text{°}=\text{36}0\text{°}=\text{6}\times \text{6}0\text{°}\\ \begin{array}{l}\left(\text{i}\right)\text{Thus},\text{the number of spokes and the angle between a pair of consecutive}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}spokes are inversely proportional to each other}.\end{array}\\ \text{Now},{\text{we will find x}}_{\text{1}},{\text{x}}_{\text{2}}{\text{and x}}_{\text{3}}.\\ \\ 4\times 90\text{°}=8\times {\text{x}}_1\\ \Rightarrow {\text{x}}_1=\frac{4\times 90\text{°}}{8}={45}^{\circ }.\\ \\ \text{Similarly},{\text{x}}_2=\frac{4\times 90\text{°}}{10}={36}^{\circ }{\text{and x}}_3=\frac{4\times 90\text{°}}{12}={30}^{\circ }.\\ \text{Thus},\text{the following table is obtained}.\end{array}$

\begin{array}{cccccc}\text{Number of spokes}& 4& 6& 8& 10& 12\\ \begin{array}{l}\text{Angle between a pair of}\\ \text{consecutive spokes}\end{array}& {90}^{\circ }& {60}^{\circ }& {45}^{\circ }& {36}^{\circ }& {30}^{\circ }\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\left(\text{ii}\right)\text{Let the angle between a pair of consecutive spokes on a wheel with 15 spokes be}x.\\ \text{Therefore},\text{4}\times \text{9}0°=\text{15}\times x\\ \Rightarrow x=\frac{4\times 90°}{15}={24}^{\circ }.\\ \text{Hence},\text{the angle between a pair of consecutive spokes of a wheel},\text{which has 15 spokes in it},\text{is 24}°.\end{array}

$\begin{array}{l}\begin{array}{l}\left(\text{iii}\right)\text{Let the number of spokes in a wheel},\text{which has 4}0º\text{angles between}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} a pair of consecutive spokes},\text{be}\mathrm{y}.\end{array}\\ \text{Therefore},\text{4}\times \text{9}0°=\text{y}\times \text{4}0°\\ \Rightarrow \mathrm{y}=\frac{4\times 90°}{{40}^{\circ }}=9\\ \text{Hence},\text{the number of spokes in such a wheel is 9}.\end{array}$

Q.13 If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Ans

Total number of children =24
If the number of the children is reduced by 4, the number of remaining children =24 – 4= 20
Let the number of sweets which each of the 20 students will get be x.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of students}& 24& 20\\ \text{Number of sweets}& 5& \mathrm{x}\end{array}$

If the number of children is reduced, then each student will get more number of sweets. So, this is the case of inverse proportion.

Therefore, we obtain

$\begin{array}{l}24\times 5=20\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{24\times 5}{20}=6\end{array}$

Hence, each student will get 6 sweets.

Q.14 A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if would there were 10 more animals in his cattle?

Ans

Total number of animals is 20.
If there are 10 more animals, then the total number of animals would be 30.

Let the number of days that the food will last if there were 10 more animals in the cattle be x.
The following table is obtained.

$\begin{array}{ccc}\text{Number of animals}& 20& 30\\ \text{Number of days}& 6& \mathrm{x}\end{array}$

The food will last longer if there is less number of animals.
Hence, the number of days the food will last and the number of animals are inversely proportional to each other.

Therefore,

$\begin{array}{l}\text{2}0\times \text{6}=\text{3}0\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{20\times 6}{30}=4\end{array}$

Therefore, the food will last for 4 days.

Q.15 A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?

Ans

Let the number of days required by 4 persons to complete the job be x.

The given information is represented in the following table.

$\begin{array}{ccc}\text{Number of days}& 4& \mathrm{x}\\ \text{Number of persons}& 3& 4\end{array}$

If more persons are employed, it will take lesser time to complete the job.

Hence, the number of days and the number of persons required to complete the job are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 4\times \text{3}=\mathrm{x}\times 4\\ \Rightarrow \mathrm{x}=\frac{4\times 3}{4}=3\end{array}$

Therefore, the number of days required by 4 persons to complete the job is 3.

Q.16 A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Ans

Let the number of boxes filled by using 20 bottles in each box be x.
The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of bottles}& 12& 20\\ \text{Number of boxes}& 25& \mathrm{x}\end{array}$

If more number of bottles are used, there will be less number of boxes.
Hence, the number of bottles and the number of boxes required to pack these are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 12\times \text{25}=20\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{12\times 25}{20}=15\end{array}$

Hence, the number of boxes required to pack 20 bottles is 15.

Q.17 A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Ans

Let the number of machines required to produce the same number of articles in 54 days be x.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of machines}& 42& \mathrm{x}\\ \text{Number of days}& 63& 54\end{array}$

If more number of machines is used, then less number of days it will take to produce the given number of articles.

So, the number of machines and the number of days required to produce the given number of articles are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 42\times \text{63}=54\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{42\times \text{63}}{54}=49\end{array}$

Hence, the required number of machines to produce the given number of articles in 54 days is 49.

Q.18 A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?

Ans

Let the time taken by the car to reach the destination, when it travels at a speed of 80 km/hr, be x hours.
The given information is represented in the following table:

$\begin{array}{ccc}\text{Speed(in km/hr)}& 60& 80\\ \text{Tine taken(in hours)}& 2& \mathrm{x}\end{array}$

If the speed of the car is more, then it will take less time to reach the destination.
Hence, the speed of the car and the time taken by the car are inversely proportional to each other.

\begin{array}{l}\end{array} $\begin{array}{l}\text{Therefore},\\ \text{6}0\times \text{2}=\text{8}0\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{\text{6}0\times \text{2}}{80}=\frac{3}{2}\\ \\ \therefore \text{The time required by the car to reach the given}\\ \text{destination is}\frac{3}{2}\text{or 1}\frac{1}{2}\text{hours}.\end{array}$

Q.19 Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?

Ans

(i) Since one of the persons fell ill before the work started, so we are left with only one person.
Let the number of days required by 1 man to fit all the windows be x.

The following table is obtained by the given information.

$\begin{array}{ccc}\text{Number of persons}& 2& 1\\ \text{Number of days}& 3& \mathrm{x}\end{array}$

If less number of persons are employed, then it will take more number of days to fit all the windows.
Hence, this is a case of inverse proportion.

Therefore,
2 × 3 = 1× x
x = 6

Hence, the number of days taken by 1 man to fit all the windows is 6.

(ii) Let the number of persons required to fit all the windows in one day be y.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of persons}& 2& \mathrm{y}\\ \text{Number of days}& 3& 1\end{array}$

If less number of days is given, then more people would be required to fit all the windows.

Hence, it is a case of inverse proportion.
Therefore, by given information, we get
2 × 3 = y × 1
y = 6

Hence, 6 persons are required to fit all the windows in one day.

Q.20 A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Ans

Let the duration of each period, when there are 9 periods a day in the school, be x minutes.
The following table is obtained from the given information.

$\begin{array}{ccc}\begin{array}{l}\text{Duration of each}\\ \text{period(in minutes)}\end{array}& 45& \mathrm{x}\\ \text{Number of periods}& 8& 9\end{array}$

If there is more number of periods a day in the school, then the duration of each period will be lesser.

Hence, it is case of inverse proportion.

$\begin{array}{l}\text{Therefore},\\ \text{45}\times \text{8}=\mathrm{x}\times \text{9}\\ \Rightarrow \mathrm{x}=\frac{\text{45}\times \text{8}}{9}=40\end{array}$

Hence, the duration of each period, when there are 9 periods a day in the school, will be 40 minutes.

Q.21 Following are the car parking charges near a railway station upto

4 hours ₹ 60

8 hours ₹ 100

12 hours ₹ 140

24 hours ₹ 180

Check if the parking charges are in direct proportion to the parking time.

Ans

The given information is represented in a table.

The ratios of the parking charges to the number of hours are as follows:

$\frac{60}{4}=15,\frac{100}{8}=\frac{25}{2},\frac{140}{12}=\frac{35}{3},\frac{180}{24}=\frac{15}{2}$

Since all the ratios are different from each other, the parking charges are not in a direct proportion to the parking time.