# NCERT solutions for class 8 Maths Chapter 13 Direct And Inverse Proportions

#### Solution for Exercise 13.1

1. Following are the car parking charges near a railway station up to.
4 hours – ₹ 60

8 hours – ₹100

12 hours – ₹140

24 hours – ₹ 180

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

Checking if the given values are in direct proportion by seeing the ratios of time period and the parking time:
Since, we have, $$\frac{4}{60}\neq \frac{8}{100}\neq\frac{12}{140}\neq \frac{24}{180}$$

Therefore, the given charges are not in direct proportion to the parking time

2.A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Let the number to be filled in the blanks be a, b, c and d respectively.
Therefore, $$\frac{1}8=\frac{4}a$$

$$\Rightarrow a=32$$

$$\frac{1}8=\frac{7}b$$

$$\Rightarrow b=56$$

$$\frac{1}8=\frac{12}c$$

$$\Rightarrow c=96$$

$$\frac{1}8=\frac{20}d$$

$$\Rightarrow d=160$$

Hence, we have the required value of bases as:

3.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?

Let the required red pigment be x part
$$\frac{1}x=\frac{75ml}{1800ml}$$

$$\Rightarrow x\times 75=1\times 1800$$

$$\Rightarrow x=\frac{1800}{75}=24\; parts$$

So the required amount of red pigment =24 parts

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

We have 840 bottles filled in six hours. So, let x bottles be filled in five hours.

Since, the two quantities are directly proportional.So we have,

$$\frac{840}x=\frac{6}5$$

$$\Rightarrow 6 \times x=5\times 840$$

Therefore, we have $$x=\frac{5\times 840}{6}$$

$$x=5\times 140=700$$

Hence the required number of bottles filled in five hours=700

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

Original length of bacteria=$$\frac{5}{50,000}cm$$
$$=\frac{1}{10,000}=10^{-4}cm$$

Given:Enlargement of 50,000times attains a length of 5 cm.Similarly for enlargement of 20,000 times let the length attained be x

Since, the two quantities are directly proportional so we have

$$\frac{50,000}5=\frac{20,000}{x}$$

$$\Rightarrow 50,000\times x=5\times 20,000$$

$$\Rightarrow x=\frac{5\times 20,000}{50,000}$$

$$\Rightarrow x=2cm$$
So, the required length is 2cm

6.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?

Given: We have 9cm as the lenght of the model for which 12 cm is the actual length.So, for 28 cm of actual length ,let x be the length of the model

We can see this is the case of direct proportion,So we have

$$\frac{9}{12}=\frac{x}{28}$$

$$\Rightarrow x=28\times \frac{9}{12}$$

$$\Rightarrow x=21 cm$$

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

7.Suppose 2 kg of sugar contains $$9 \times 10^6$$ crystals. How many sugar crystals are there in
(i) 5 kg of sugar?

(ii) 1.2 kg of sugar?

Given: we have 2 kgs of sugar containing $$9 \times 10^6$$ crystals.So let x and y crystals are in 5 kg of sugar and 1.2 kg of sugar.
(i)We can see that this is the case of direct proportion,So we have

$$\frac{9\times 10^6}{2}=\frac{x}{5}$$

$$\Rightarrow x= \frac{5\times 9 \times 10^6}{2}$$

$$\Rightarrow x= \frac{ 9 \times 10^7}{4}$$

$$\Rightarrow x=2.25\times 10^7$$

Hence, for 5kg of sugar we have $$2.25\times 10^7$$ crystals.

(ii) $$\frac{9\times 10^6}{2}=\frac{y}{1.2}$$

$$\Rightarrow x= \frac{1.2\times 9 \times 10^6}{2}$$

$$\Rightarrow x=0.6\times9\times 10^6=5.4\times 10^6$$

Hence, for 1.2kg of sugar, we have $$5.4\times 10^6$$ crystals.

8.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?

Given: Scale of 1 cm represents 18 km.So, let x cm be the distance covered in the map for the drives of 72 km on a road.
We can see that this is the case of direct proportion,So we have

$$\frac{1}{x}=\frac{18}{72}$$

$$\Rightarrow 18x= 1\times72$$

$$\Rightarrow x= \frac{ 1 \times72}{18}$$

$$\Rightarrow x=4cm$$

Hence, for 72km of distance we have 4 cm of scale.

9.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 5 m long.

Given: Scale of 1 cm represents 18 km.So, let x cm be the distance covered in the map for the drives of 72 km on a road.
We can see that this is the case of direct proportion,So we have

$$\frac{1}{x}=\frac{18}{72}$$

$$\Rightarrow 18x= 1\times72$$

$$\Rightarrow x= \frac{ 1 \times72}{18}$$

$$\Rightarrow x=4cm$$

Hence, for 72km of distance we have 4 cm of scale.

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

Given: Truck travels 14 km in 25 minutes.So, let x km be travelled in 5 hours.
And,We can see that this is the case of direct proportion,So we have

$$\frac{14}{25}=\frac{x}{5}$$

$$\Rightarrow x= 14\times\frac{60}{5}=14\times 12$$

$$\Rightarrow x=168$$

Hence, for 5 hours of travelling, times taken= 5 hours

#### Solution for Exercise 13.2

1.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 at 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.

(i) Here in this case we know that more is the number of workers to do a job, less is the time taken to finish the job.So, it is a case of inverse proportion.
(ii) Here,Clearly if more is the time taken, more is the distance travelled in a uniform speed.So, it is a case of direct proportion.

(iii) Here in this case, clearly more is the area cultivated land, more is the crop harvested. So, it is a case of direct proportion.

(iv) Here we know that more is the speed of the vehicle, less is the time to cover a fixed distance.So, it is a case of inverse proportion.

(v) Clearly here we can see that, more is the population, less is the area of land per person in a country.So, it is a case of inverse proportion.

2.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?

Let, the blank spaces be a, b, c, d and e.
So, we see that $$1 ×\times 100,000 = 2 \times 50,000$$

$$\Rightarrow 1,00,000 = 1,00,000$$

Hence, they are inversely proportional.So, we have

$$2 \times 50,000 = 4 \times a$$

∴ $$a=\frac{2\times50,000}4=25,000$$

$$2 \times 50,000 = 5 \times b$$

∴ $$b=\frac{2\times50,000}5=20,000$$

$$2 \times 50,000 = 8 \times c$$

∴ $$c=\frac{2\times50,000}8=12,500$$

$$2 \times 50,000 = 10 \times d$$

∴ $$d=\frac{2\times50,000}{10}=10,000$$

$$2 \times 50,000 = 20 \times e$$

∴ $$e=\frac{2\times50,000}{20}=5,000$$

Hence we have the table as:

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

(i) Are the number of spokes and the angle 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 $$40^o$$?

From the given table we have, $$4\times90^o=6\times60^o$$

$$\Rightarrow 360^o=360^o$$

Thus, we can say that quantities are inversely peoportional

Let the blacks be as a,b and c

So we have, $$4\times90^o=8\times a$$

$$\Rightarrow a=\frac{4\times 90^o}{8}=45^o$$

Similarly, $$4\times90^o=10\times b$$

$$\Rightarrow b=\frac{4\times 90^o}{10}=36^o$$

And, $$4\times90^o=12\times c$$

$$\Rightarrow c=\frac{4\times 90^o}{12}=30^o$$

Hence, we have the table as:

(i) Yes, they are in inverse proportion

(ii) If the number of spokes is 15, then

$$4 \times 90^o = 15 \times x$$

$$x = \frac { 4\times 90 }{ 15 } = 24^o$$

(iii) If the angle between two consecutive spokes is $$40^o$$, then

$$4 \times 90^0 = y \times 40^o$$

$$y = \frac { 4\times 90 }{ 40 } = 9 spokes.$$

Thus the required number of spokes = 9.<\br>

4.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?

Given that 24 children are given 5 sweets each and also the number of the children were reduced by 4 so, their number becomes 24 – 4 = 20.
Let each child get x sweets when their number is 20.

As per given we see that less children will get more sweets. So, it is a case of inverse proportion.

∴$$24 \times 5 = 20 \times x$$

$$\Rightarrow x= \frac { 24\times 5 }{ 20 } = 6$$

Hence, each child will get 6 sweets.

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

Given that a farmer can feed 20 animals in his cattle for 6 days and also let the food will now last for x days.

Clearly we see that, more is the number of animals, less will be the number of days for food to last.So, it is a case of inverse proportion.

∴$$24 \times 6 = 30 \times x$$

$$\Rightarrow x= \frac { 20\times 6 }{ 30 } = 4$$

Hence, the food will now last for 4 days.

6.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?

Given that the 3 persons could rewire the house in 4 days and let x time is taken by 4 persons if used instead of three

Clearly we can see that more is the number of persons, less is the number of days required.

So, it is a case of inverse proportion.

∴ $$4 \times 3 = x \times 4$$

$$\Rightarrow x= \frac { 4\times 3 }{ 3 } = 3$$

Hence, the job will be completed in 3 days.

7.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?

Given that we have 25 boxes with 12 bottles in each box and let x be boxes be filled if 20 bottles are in each box.
We can observe that if the number of bottles is increased then the required number of boxes will be decreased. Thus the two quantities are in inverse proportion.

$$\Rightarrow x_1y_1 = x_2y_2$$

$$25 \times 12 = x \times 20$$

$$x = 15$$

Hence, the required number of boxes = 15.

8.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?

Given that, 42 machines are required to produce a given number of articles in 63 days.
Let the required number of machines be x.

We can see that if the number of machines is increased then less number of days would be required to produce the same number of articles.Thus, the two quantities are in inverse proportion.So we have:

$$\Rightarrow x_1y_1 = x_2y_2$$

$$\Rightarrow 42 \times 63 = x \times 54$$

$$\Rightarrow x = 49$$

Hence, the required number of machines is 49.

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

Given that a car takes 2 hours to reach a destination by traveling at a speed of 60 km/h.And, let the time taken be x hours for travelling at the speed of 80 km/h.

We can see that, on increasing the speed, it will take less time to travel a distance. Thus the two quantities are in inverse proportions.<./br>
So, we have:∴$$x_1y_1 = x_2y_2$$

$$\Rightarrow 60 \times 2 = 80 \times x$$

$$\Rightarrow x = \frac { 3 }{ 2 } hours = 1\frac { 1 }{ 2 }$$ hrs.

Hence, the required time $$x= 1\frac { 1 }{ 2 }$$ hours.

10.Two persons could fit new windows in a house in 3 days.
(i) One of the people 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?

We can see that on increasing the number of persons, less time will be required to complete a job.Thus, the quantities are in inverse proportion.

(i) Let the required number of days be x.

∴$$x_1y_1 = x_2y_2$$

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

$$\Rightarrow x = 6$$

Hence, the required number of days = 6 days

(ii) Let the required number of persons be y.

∴$$x_1y_1 = x_2y_2$$

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

$$\Rightarrow y = 6$$

Hence, the required number of persons = 6.

11.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?

∴$$x_1y_1 = x_2y_2$$
$$\Rightarrow 8 \times 45 = 9 \times x$$
$$\Rightarrow x = 40\; minutes$$