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If you are in search of **NCERT solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 11 **Direct and Inverse Proportions** Ex 11.1

**Class 8 Maths Chapter** **11 ****Direct and Inverse Proportions** **Exercise 11.1**

**Direct and Inverse Proportions**

**Exercise 11.1**

**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.**

Solution-

Charges per hour:

`\[C1=\frac{60}{4}=₹\ 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C2=\frac{100}{8}=₹\ 12.50\]`

`\[C3=\frac{140}{12}=₹\ 11.67\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C4=\frac{180}{24}=₹\ 7.50\]`

Here, the charges per hour are not the same, i.e. C1 ≠ C2 ≠ C3 ≠ C4

Therefore, the parking 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.**

Parts of red pigment | 1 | 4 | 7 | 12 | 20 |

Parts of Base | 8 | … | … | … | … |

Solution-

`\[Let\ the\ ratio\ of\ parts\ of\ red\ pigment\ and\ parts\ of\ the\ base\ be\ \frac{a}{b}.\]`

1. Here, a1 = 1, b1 = 8

`\[\frac{a1}{b1}=\frac{1}{8}=k\ (say)\]`

2. When a2 = 4, b2 =?

`\[k=\frac{a2}{b2}\ ⟹b2=\frac{a2}{k}=\frac{4}{\left(\frac{1}{8}\right)}=4\times8=32\]`

3. When a3 = 7, b3 =?

`\[k=\frac{a3}{b3}⟹b3=\frac{a3}{k}=\frac{7}{\left(\frac{1}{8}\right)}=7\times8=56\]`

4. When a4 = 12, b4 =?

`\[k=\frac{a4}{b4}⟹b4=\frac{a4}{k}=\frac{12}{\left(\frac{1}{8}\right)}=12\times8=96\]`

5. When a5 = 20, b5 =?

`\[k=\frac{a5}{b5}⟹b5=\frac{a5}{k}=\frac{20}{\left(\frac{1}{8}\right)}=20\times8=160\]`

When combining results for all the cases, we get-

Parts of red pigment | 1 | 4 | 7 | 12 | 20 |

Parts of Base | 8 | 32 | 56 | 96 | 160 |

**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 the parts of red pigment mixed with 1800 mL base be x.

Parts of red pigment | 1 | x |

Parts of Base | 75 | 1800 |

Since it is in direct proportion,

`\[∴\ \frac{1}{75}=\ \frac{x}{1800}\]`

`\[⟹\ 75\times x\ =\ 1\times1800\]`

`\[⟹\ x\ =\ \frac{\left(1\times\ 1800\right)}{75}=\ 24\ parts\]`

Hence, with base 1800 mL, 24 parts red pigment should be mixed.

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

Solution-

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

Hours | 6 | 5 |

Bottles | 840 | x |

Hence, the ratio of hours and bottles are in direct proportion.

`\[∴\ \frac{6}{840}=\ \frac{5}{x}\]`

`\[⟹6\times x=5\times840\]`

`\[⟹\ x\ =\ \frac{\left(5\ \times840\right)}{6}=\ 700\ bottles.\]`

Hence, machine will fill 700 bottles in five hours.

**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?**

Solution-

Let the enlarged length of bacteria be x.

`\[Actual\ length\ of\ bacteria=\frac{5}{50000}=\frac{1}{10000}\ cm=10^{-4}\ cm\]`

Length | 5 | x |

Enlarged Length | 50,000 | 20,000 |

Here, the length and enlarged length of bacteria are in direct proportion.

`\[∴\ \ \frac{5}{50000}=\frac{x}{20000}\]`

`\[⟹x\times50000=5\times20000\]`

`\[⟹x=\frac{\left(5\times20000\right)}{50000}=\ 2\ cm\]`

Hence, the enlarged length of bacteria is 2 cm.

**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-

Let the length of the model ship be x.

Length of actual ship (in m) | 12 | 28 |

Length of model ship (in cm) | 9 | x |

Here, the length of the mast and the actual length of the ship are in direct proportion.

`\[∴\ \frac{12}{9}=\ \frac{28}{x}\]`

`\[⟹x\times12=2\times89\]`

`\[⟹x=\frac{\left(2\times89\right)}{12}=\ 21\ cm\]`

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

**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-

(i) Let sugar crystals be x.

Here, the weight of sugar and the number of crystals are in direct proportion.

`\[∴\ \frac{2}{9\times10^6}=\frac{5}{x}\]`

`\[⟹2\times x=5\times9\times10^6\]`

`\[⟹x=\frac{\left(45\ \times10^6\right)}{2}=\ 22.5\ \times10^6=\ 2.25\times10^7\]`

`\[Hence,\ the\ number\ of\ sugar\ crystals\ is\ 2.25\times10^7.\]`

(ii) Let sugar crystals be x.

Here, the length of the mast and the actual length of the ship are in direct proportion.

`\[∴\ \frac{2}{9\times10^6}=\frac{1.2}{x}\]`

`\[⟹2\times x=1.2\times9\times10^6\]`

`\[⟹x=\frac{\left(1.2\ \times9\times10^6\right)}{2}=\ 0.6\ \times9\times10^6=\ 5.4\times10^6\]`

`\[Hence,\ the\ number\ of\ sugar\ crystals\ is\ 5.4\times10^6.\]`

**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 be x.

Actual Distance (in Km) | 18 | 72 |

Distance covered in Map ( In cm) | 1 | x |

Here, the actual distance and distance covered in the map are in direct proportion.

`\[∴\ \frac{18}{1}=\ \frac{72}{x}\]`

`\[⟹x\times18=1\times72\]`

`\[⟹x=\frac{\left(1\times72\right)}{18}=\ 4\ cm,\]`

Hence, the distance covered in the map is 4 cm.

**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 5m long.**

Solution-

Here, the height of the pole and the length of the shadow are in direct proportion.

And 1 m = 100 cm

5 m 60 cm = 5×100+60 = 560 cm

3 m 20 cm = 3×100+20 = 320 cm

10 m 50 cm = 10×100+50 = 1050 cm

5 m = 5×100 = 500 cm

(i) Let the length of the shadow of another pole be x.

Height of pole (in cm) | 560 | 1050 |

Length of shadow (in cm) | 320 | x |

`\[∴\ \ \frac{560}{320}=\ \frac{1050}{x}\]`

`\[⟹x\times560=320\times1050\]`

`\[⟹x=\ \frac{\left(320\ \times1050\right)}{560}=\ 600\ cm\ =\ 6\ m.\]`

Hence, the length of the shadow of another pole is 6 m.

(ii) Let the height of the pole be x.

Height of pole (in cm) | 560 | x |

Length of shadow (in cm) | 320 | 500 |

`\[∴\ \frac{560}{320}=\ \frac{x}{500}\]`

`\[⟹560\times500=320\times x\]`

`\[⟹x=\ \frac{\left(560\times500\right)}{320}=875\ cm\ =\ 8m\ 75\ cm.\]`

Hence, the height of the pole is 8 m 75 cm.

**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-

Let the distance covered in 5 hours be x km.

1 hour = 60 minutes

Therefore, 5 hours = 5×60 = 300 minutes

Distance in Kilometers | 14 | x |

Time in Minutes | 25 | 300 |

Here, the distance covered and time are in direct proportion.

`\[∴\ \frac{14}{25}=\ \frac{x}{300}\]`

`\[⟹14\times300=x\times25\]`

`\[⟹x=\frac{\left(14\times300\right)}{25}=\ 168\ km.\]`

Hence, the truck can travel 168 km in 5 hours.

**Using Class 8 NCERT solutions offers several key advantages for students:**

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