Haryana State Board HBSE 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Exercise 13.1

Question 1.

Following are the car parking charges near a railway station upto

4 hours — Rs. 60

8 hours — Rs. 100

12 hours — Rs. 140

24 hours — Rs. 180

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

Solution:

Suppose the parking time of car is x hours and its charges in Rs. is y.

If x and y are direct proportion, then

Question 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.

Solution:

Let the parts of red pigment = x and

part of base = y.

∴ \(\frac{x}{y}\) = \(\frac{1}{8}\)

\(\frac{x}{y}\) = \(\frac{1}{32}\) = \(\frac{4}{32}\) = \(\frac{7}{56}\) = \(\frac{12}{96}\) = \(\frac{20}{160}\)

Question 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 ?

Solution:

Let red pigment = x

∴ \(\frac{1}{75}\) = \(\frac{x}{1800}\)

⇒ x × 75 = 1 × 1800

⇒ x = \(\frac{1800}{75}\)

∴ x = 24

Hence required red pigment = 24.

Question 4.

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

Solution:

Let the no. of bottles = x

Number of time and number of bottles vary directly.

Now, \(\frac{6}{840}\) = \(\frac{5}{x}\)

⇒ 6 × x = 5 × 840

⇒ x = \(\frac{5×840}{6}\)

⇒ x = 700

Hence, required bottles = 700.

Question 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 enlarged length ?

Solution:

Let the length of bacteria = x. Length of bacteria and enlarged times of bacteria vary directly.

Now, \(\frac{50,000}{5}\) = \(\frac{20,000}{x}\)

⇒ 50,000 × x = 5 × 20,000

⇒ x = \(\frac{5×20,000}{50,000}\)

⇒ x = 2

∴ Enlarged length = 2 cm

and Actual length, x_{1} = \(\frac{2}{20,000}\)

= \(\frac{1}{10,000}\) = 0.0001 cm.

Question 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 ?

Solution:

Try Yourself.

Question 7.

Suppose 2 kg of sugar contains 9 × 10^{6} crystals. How many sugar crystals are there in

(i) 5 kg of sugar ?

(ii) 1.2 kg of sugar ?

Solution:

Let 5 kg of sugar contains = x crystals

and 1.2 kg of sugar contains = y crystals.

Sugar and crystals vary directly.

Now, (i) \(\frac{2}{9 \times 10^{6}}\) = \(\frac{5}{x}\)

⇒ 2 × x = 5 × 9 × 10

⇒ x = \(\frac{5 \times 9 \times 10 \times 10^{5}}{2}\)

⇒ x = 225 × 10^{5}

= 2.25 × 10^{7}

∴ 5 kg of sugar contains 2.25 × 10^{7} crystals.

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

⇒ 2 × y = 1.2 × 9 × 10^{6}

⇒ y = \(\frac{1.2 \times 9 \times 10 \times 10^{5}}{2}\)

= 54 × 10^{6}

= 5.4 × 10^{6}

∴ 1.2 kg of sugar contains 5.4 × 10^{6} crystals.

Question 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 ?

Solution:

Let the distance covered in the map = x.

Map with a scale and distance covered in the map vary directly.

Now, \(\frac{1}{18}\) = \(\frac{x}{72}\)

⇒ 18 × x = 72 × 1

⇒ x = \(\frac{72}{18}\) = 4

Hence required distance covered in the map = 4 cm.

Question 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.

Solution:

Let the length of shadow = x

and the height of a pole = y

Length of the shadow and height of pole vary directly.

Now, (i) \(\frac{5.60}{3.20}\) = \(\frac{10.50}{x}\)

⇒ 5.60 × x = 3.20 × 10.50

⇒ x = \(\frac{3.20 \times 10.50}{5.60}\) = 6 m

Hence required length of shadow 6m.

(ii) \(\frac{5.60}{3.20}\) = \(\frac{y}{5}\)

⇒ 3.20 × y = 5.60 × 5

⇒ y = \(\frac{5.60 \times 5}{3.20}\) = 8.75

Hence required height of pole = 8.75 m or 8 m 75 cm.

Question 10.

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

Solution:

60 minutes = 1 hour

∴ 25 minutes = \(\frac{1}{60}\) × 25 = \(\frac{5}{12}\) hour

Let distance of travelling truck = x

Distance of travelling truck and time vary directly.

Now, \(\frac{14}{5 / 12}\) = \(\frac{x}{5}\)

⇒ \(\frac{5}{12}\) × x = 14 × 5

⇒ x = \(\frac{14 \times 5}{5 / 12}\) = \(\frac{14 \times 5 \times 12}{5}\)

⇒ x = 168 km

Hence distance of travelling truck = 168 km.