These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1

Question 1.

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.

Answer:

Parking time | Parking charges | Ratio |

4 hours | ₹ 60 | \(\frac { 60 }{ 4 }\) = 15 |

8 hours | ₹ 100 | \(\frac{100}{8}=\frac{25}{2}\) |

12 hours | ₹ 140 | \(\frac{140}{12}=\frac{35}{3}\) |

24 hours | ₹ 180 | \(\frac{180}{24}=\frac{15}{2}\) |

Since \(\frac{15}{1} \neq \frac{25}{2} \neq \frac{35}{3} \neq \frac{15}{2}\)

∴ The parking charges are not in direct proportion with the parking time.

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.

Answer:

Let the number of parts of red pigment be x_{1}, x_{2}, x_{3}, x_{4} , x_{5} and the number of parts of

the base be y_{1}, y_{2}, y_{3}, y_{4}, y_{5}

As the number of parts of red pigment increases, number of parts of the base also increases in the same ratio It is a direct proportion

The required parts of bases are 32, 56, 96 and 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?

Answer:

We have \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{1}{8}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)

Here x_{1} = 1, y_{1} = 75 and y_{2} = 1800

∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)

x_{2} = \(\frac{1 \times 1800}{75}\) = 24

Thus, the required no. of red pigments = 24.

Question 4.

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

Answer:

Number of bottles filled | Number of hours |

840 | 6 |

X | 5 |

More the number of hours, more the number of bottles that would be filled. Thus, given quantities vary directly.

∴ \(\frac{840}{x}=\frac{6}{5}\)

x = \(\frac{840 \times 5}{6}\) = 140 × 5 = 700

∴ The required number of 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 its enlarged length?

Answer:

Actual length = \(\frac{5}{50000}=\frac{1}{10000}\) = 10^{-4} cm

Number of times photograph enlarged | Length (in cm) |

50,000 | 5 |

20,000 | x |

The length increases with an increment in the number of times the photograph enlarged.

∴ It is a case of direct proportion.

\(\frac{50000}{20000}=\frac{5}{x}\)

50,000 × x = 5 × 20,000

x = \(\frac{5 \times 20,000}{50,000}\) = 2

∴The enlarged length = 2 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?

Answer:

Let the required length of the model ship be ‘x’ cm

Length of the ship | Height of the mast |

28 | 12 |

X | 9 |

Since, more the length of the ship, more would be the length of its mast.

∴ It is a direct variation \(\frac{28}{x}=\frac{12}{9}\)

12 × x = 28 × 9

x = \(\frac{28 \times 9}{12}=\frac{7 \times 9}{3}\) = 7 × 3 = 21

The required length of the model = 21 cm.

Question 7.

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

Answer:

(i) Let the required number of sugar crystals be Y.

Since more the amount of sugar, more would be the number of sugar crystals.

∴ It is a direct variation

\(\frac{2}{5}=\frac{9 \times 10^{6}}{\mathrm{x}}\)

2x = 5 × 9 × 10^{6}

x = \(\frac{5 \times 9 \times 10^{6}}{2}=\frac{45 \times 10^{6}}{2}\) = 22.5 × 10^{6}

= 2.25 × 10 × 10^{6}

= 2.25 × 10^{7}

∴ Required number of sugar crystals = 2.25 x 10^{7}

(ii) Let the number of sugar crystals in a sugar be ‘y’

It is a direct variation

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

2y = 9 × 10^{6} × 1.2

y = \(\frac{9 \times 10^{6} \times 1.2}{2}\) = 9 × 10^{6} × 0.6

= 5.4 × 10^{6}

The required number of sugar crystals = 5.4 × 10^{6}

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?

Answer:

Let the required distance covered in the map be ‘x’ cm.

Distance covered on the Road (in km) | Distance represented on the map (in cm) |

18 | 1 |

72 | x |

It is a direct variation

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

18 × x = 72 × 1

x = \(\frac{72 \times 1}{18}\) = 4 × 1 = 4

∴ The required distance on the map is 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.

Answer:

Let the required length of shadow be ‘x’ cm

Height of the pole | Length of the shadow |

5 m 60 cm = 560 cm | 3 m 20 cm = 320 cm |

10 m 50 cm = 1050 cm | cm |

As the height of the pole increases, the length of the shadow also increases in the same ratio

It is a direct variation.

\(\frac{560}{1050}=\frac{320}{\mathrm{x}}\)

560 × x = 1050 × 320

x = \(\frac{1050 \times 320}{560}\) = 600 cm

∴ Required length of the shadow = 600 cm (6 m)

(ii) Let the required height of the pole be y’ cm

Height of the pole | Length of the shadow |

560 cm | 320 cm |

y | 5 m = 500 cm |

It is a direct variation.

\(\frac{560}{y}=\frac{320}{500}\)

y × 320 = 560 × 500

y = \(\frac{560 \times 500}{320}\) = 125 × 7 = 875

7 320

∴ The height of the pole = 875 cm (or) 8 m 75 cm.

Question 10.

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

Answer:

Let the distance travelled in 5 hours be ‘x’ km.

Distance (km) | Time (minutes) |

14 | 25 |

x | 300 |

If the time increases, the distance also increases.

∴ It is a direct variation.

\(\frac{14}{x}=\frac{25}{300}\)

25 × x = 300 × 14

x = \(\frac{300 \times 14}{25}\) = 12 × 14 = 168

∴ The required distance =168 km.