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# NCERT Solutions for Class 8 Maths Chapter 11: Direct and Inverse Proportions - Exercise 11.1

Last updated date: 09th Sep 2024
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## NCERT Solutions for Class 8 Maths Chapter 11 (EX 11.1)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 8 Subject: Class 8 Maths Chapter Name: Chapter 11 - Direct and Inverse Proportions Exercise: Exercise - 11.1 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

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## Access NCERT Solutions for Class 8 Chapter 11 - Direct and Inverse Proportions

Exercise 11.1

Refer to page 1-8 for Exercise 11.1 in the PDF

1. Following are the car parking charges near a railway station up to:

4 hours Rs 60

8 hours Rs 100

12 hours Rs 140

24 hours Rs 180

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

Ans: Tabulate the given information.

 Number of hours 4 8 12 24 Parking charges (in Rs) 60 100 140 180

Now, find the ratio of parking charges to the respective number of hours (Rs/hour).

$\dfrac{{60}}{4} = 15$

$\dfrac{{100}}{8} = \dfrac{{25}}{2}$

$\dfrac{{140}}{{12}} = \dfrac{{35}}{3}$

$\dfrac{{180}}{{24}} = \dfrac{{15}}{2}$

We can see that all the ratios are different so 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 pigment with 8 parts of base. In the Following Table,find parts base that need to be added.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 - - - -

Ans: The mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. This implies that for more parts of red pigments, the part of the base will also be more. Thus, the parts of the base are in direct proportion.

Assume the parts of base as ${x_1}$, ${x_2}$,${x_3}$ and ${x_4}$.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 ${x_1}$ ${x_2}$ ${x_3}$ ${x_4}$

Since the given problem follows direct proportion then, all the ratios will be equal. Thus, the first ratio of part of base to red pigment  is $\dfrac{8}{1}$.

$\dfrac{{{x_1}}}{4} = \dfrac{8}{1}$

Multiply both sides by 4 to get the value of ${x_1}$.

${x_1} = 4 \times 8$

${x_1} = 32$

Now, we will find the value of ${x_2}$.

$\dfrac{{{x_2}}}{7} = \dfrac{8}{1}$

Multiply both sides by 7 to get the value of ${x_2}$.

${x_2} = 7 \times 8$

${x_2} = 56$

Now, we will find the value of ${x_3}$.

$\dfrac{{{x_3}}}{{12}} = \dfrac{8}{1}$

Multiply both sides by 12 to get the value of ${x_3}$.

${x_3} = 12 \times 8$

${x_3} = 96$

Now, we will find the value of ${x_4}$.

$\dfrac{{{x_4}}}{{20}} = \dfrac{8}{1}$

Multiply both sides by 20 to get the value of ${x_4}$.

${x_4} = 20 \times 8$

${x_4} = 160$

Now, tabulate all the values obtained.

 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 required 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Ans: Let the amount of red pigment required to mix with 1800 mL of base be $x$.

Tabulate the given information as follows.

 Parts of red pigment 1 $x$ Parts of base (in mL) 75 1800

Since the given problem follows direct proportion then, all the ratios will be equal. Thus, the first ratio of part of base to red pigment  is $\dfrac{1}{{75}}$.

$\dfrac{1}{{75}} = \dfrac{x}{{1800}}$

Multiply both sides by 1800.

$\dfrac{{1800}}{{75}} = \dfrac{x}{1}$

$24 = x$

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

4. 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 $x$.

Tabulate the given information as follows.

 Number of bottles 840 $x$ Time taken (in hours) 6 5

The number of bottles and time taken to fill the bottles are in direct proportion.

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

Multiply both sides by 5.

$\dfrac{x}{5} \times 5 = \dfrac{{840}}{6} \times 5$

$x = \dfrac{{840}}{6} \times 5$

$x = 700$

Therefore, 700 bottles will be filled in 5 hours.

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

Ans: Let us assume that the actual length of the bacteria is $x$ cm.

Let the enlarged length of the bacteria be $y$ cm.

Tabulate the information as follows.

 Length of bacteria (in cm) 5 $x$ $y$ Number of times the bacteria was enlarged 50000 1 20000

The parameters, number of times then bacteria is enlarged and the length of bacteria are in direct proportion.

The fixed ratio is $\dfrac{5}{{50000}}$.

Since all the ratios are in  direct proportion, then , they will be equal.

$\dfrac{5}{{50000}} = \dfrac{x}{1}$

On simplifying we get,

$x = \dfrac{1}{{10000}}$

$x = {10^{ - 4}}$

Hence, the actual length of the bacteria is ${10^{ - 4}}$ cm.

Now, we will calculate $y$.

On equating the ratio involving $y$with $\dfrac{5}{{50000}}$, we get,

$\dfrac{y}{{20000}} = \dfrac{5}{{50000}}$

Multiply both sides by 20000.

$y = \dfrac{{20000 \times 5}}{{50000}}$

$y = 2$

Hence, the enlarged length of the given 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?

Ans: Assume the length of the mast of the model ship as $x$ cm.

Tabulate the given information.

 Height of the mast Length of the ship Model ship 9 cm $x$ Actual ship 12 m 28 m

We know that the dimension of the actual ship and the model ship are directly proportional to each other.

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

Cross multiply all the terms.

$12x = 28 \times 9$

Divide both sides by 12.

$x = \dfrac{{28 \times 9}}{{12}}$

$x = 21$

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 the amount of sugar given below?

i. 5 kg of sugar

Ans: Assume the number of sugar crystals in 5 kg of sugar as $x$.

Tabulate the data as follows.

 Amount of sugar ( in kg) 2 5 Number of crystals $9 \times {10^6}$ $x$

Here, the amount of sugar and the number of crystals have a direct proportion.

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

Cross multiply both sides.

$5 \times \left( {9 \times {{10}^6}} \right) = 2x$

Divide both sides by 2.

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

$x = 2.25 \times {10^7}$

Hence, the number of crystals in 5 kg of sugar are $2.25 \times {10^7}$.

ii. $1.2$ kg of sugar

Ans: Assume the number of sugar crystals in $1.2$ kg of sugar as $x$.

Tabulate the data as follows.

 Amount of sugar ( in kg) 2 $1.2$ Number of crystals $9 \times {10^6}$ $y$

Here, the amount of sugar and the number of crystals have a direct proportion.

$\dfrac{{1.2}}{y} = \dfrac{2}{{9 \times {{10}^6}}}$

Cross multiply both sides.

$5 \times \left( {9 \times {{10}^6}} \right) = 2y$

Divide both sides by 2.

$y = \dfrac{{1.2 \times 9 \times {{10}^6}}}{2}$

$x = 5.4 \times {10^6}$

Hence, the number of crystals in $1.2$ kg of sugar is $5.4 \times {10^6}$.

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?

Ans: Assume the distance that is represented on the map as $x$ cm.

Tabulate the data as follows.

 Distance covered on road (in km) 18 72 Distance represented on map (in cm) 1 $x$

The distance covered on road and represented on map are directly proportional to each other.

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

Cross-multiply.

$18x = 72$

Divide both sides by 18.

$x = \dfrac{{72}}{{18}}$

$x = 4$

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

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find the following at the same time.

i. The length of the shadow cast by another pole 10 m 50 cm high.

Ans: First, we will convert the values from centimetres to meters.

We know that 1 metre is equal to 100 cm.

5 m 60 cm will be equal to $5.60$ m.

3 m 20 cm will be equal to $3.20$ m.

10 m 50 cm will be equal to $10.50$ m

Let us assume the length of the shadow of the pole as $x$ m.

 Height of pole (in m) $5.60$ $10.50$ Length of shadow (in m) $3.20$ $x$

The Higher the height of an object, the more will be the length of its shadow.

The height of the pole and length of the shadow are directly proportional.

$\dfrac{{5.60}}{{3.20}} = \dfrac{{10.50}}{x}$

Cross multiply the terms.

$5.60x = 10.50 \times 3.20$

Divide both sides by $5.60$.

$x = \dfrac{{10.50 \times 3.20}}{{5.60}}$

$x = 6$

Thus, the length of the shadow is 6 m.

ii. The height of a pole which casts a shadow 5 m long.

Ans: Let us assume the length of the shadow of the pole as $y$ m.

 Height of pole (in m) $5.60$ $y$ Length of shadow (in m) $3.20$ $5$

The Higher the height of an object, the more will be the length of its shadow.

The height of the pole and length of the shadow are directly proportional.

$\dfrac{{5.60}}{{3.20}} = \dfrac{y}{5}$

Cross multiply the terms.

$5.60 \times y = 3.20 \times 5$

Divide both sides by $5.60$.

$x = \dfrac{{5.60 \times 5}}{{3.20}}$

$x = 8.75$

Thus, the length of the shadow is $8.75$ m.

10. 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.

1 hour is equal to 60 minutes.

So, convert 5 hours to minutes.

$5 \times 60 = 300\,{\text{minutes}}$

Now, tabulate the information as follows.

 Distance travelled (in km) 14 $x$ Time (in minute) 25 300

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

$\dfrac{{14}}{{25}} = \dfrac{x}{{300}}$

Multiply both sides by 300.

$\dfrac{{14}}{{25}} \times 300 = \dfrac{x}{{300}} \times 300$

$14 \times 12 = x$

$168 = x$

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

## NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions (Ex 11.1) Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 11 Exercise 11.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 11 Exercise 11.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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## FAQs on NCERT Solutions for Class 8 Maths Chapter 11: Direct and Inverse Proportions - Exercise 11.1

1. What are direct and Inverse proportions?

Indirect proportion if one quantity is risen or declined then the other quantity increases or decreases respectively. But in the case of indirect or inverse proportion 1 quantity increases then other quantity decreases and vice versa.

2. Is the example of direct and Inverse proportion?

For instance, when you buy more apples then you will have to pay more money. Similarly, if we rise the speed of a vehicle the time that it takes to cover some distance goes down. The initial is an example of direct proportionality and the second is an example of quantities that are inversely proportional.

3. What are the differences between direct and inverse proportion?

Why increases by X increases in the case of director relationship. A direct relationship always has a positive slope on a graph. An Inverse relationship means that the variable changes in the opposite direction or one increases while the other decreases and vice versa.

4. Is chapter 11 direct and Inverse proportion in NCERT class 8 math important?

Absolutely yes chapter 11 direct and Inverse proportion is a very crucial chapter in class 8 math. This will give you a base for the upcoming class and also be useful for many competitive exams. So, students should give this chapter extra focus and understand the concepts. You can refer to vedantu, where you can get details and step solutions to chapter 11 direct and Inverse proportion in the free PDF format. Also, you will be able to build a better understanding level and learn a different approach to solving the problem in proportion.

5. NCERT solution for class 8 math chapter 11 direct and Inverse proportion exercise 11.1 for board exams?

NCERT solution for class 8 math chapter 11 direct and Inverse proportion exercise 11.1 provides answers with detailed description as per the syllabus prescribed by the CBSE board. For the student to finish the assignment on time solving the solution would be fantastic practice. The NCERT solution for class 8 math chapter 11 direct and Inverse proportion (EX 11.1) exercise 11.1 are obviously necessary to achieve excellent exam scores. Students can practice writing exams and become more comfortable with this process.