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Important Questions Class 8 Mathematics Chapter 13 – Direct and Indirect Proportions
Mathematics is one of the most engaging subjects, which requires students of all classes to engross themselves in the concepts and hypotheses relevant to the subject. Class 8 Mathematics Chapter 13 is about Direct and Indirect Proportion. It is one of the most crucial topics in Mathematics. Following are the topics covered in this chapter:
Quick Links
Toggle Introduction
 Direct Proportion
 Inverse Proportion
The official website of Extramarks is here to provide students with a clear and elaborate understanding of concepts, hypotheses, significant questions, NCERT book solutions and many other critical examples.
Sedulous practice on a regular basis is the only way to score excellent marks. Extramarks is a very trustworthy platform that has guided and supported lakhs of students in their exam preparation.. Our team understands the value of regularly solving complex problems when it comes to Mathematics. After carefully researching and analysing CBSE past year question papers, NCERT books, and exemplars, our highly qualified and experienced professionals have come up with the question bank Important Questions Class 8 Mathematics Chapter 13 for the students. Students can take help from our question bank Important Questions Class 8 Mathematics Chapter 13 to practise some fundamental questions relevant to the chapter Direct and Indirect Proportion so they can practice well and maximise their potential and excel in their examinations.
Along with providing Important Questions Class 8 Chapter 13, our Extramarks website also provides the students with the CBSE revision notes, the CBSE sample papers, and CBSE past year question papers that strictly adhere to the latest CBSE syllabus.
Get Access to CBSE Class 8 Maths Important Questions 202223 with ChapterWise Solutions
You can also find CBSE Class 8 Maths ChapterbyChapter Important Questions here:
CBSE Class 8 Maths Important Questions  
Sr No.  Chapters  Chapters Name 
1  Chapter 1  Rational Numbers 
2  Chapter 2  Linear Equations in One Variable 
3  Chapter 3  Understanding Quadrilaterals 
4  Chapter 4  Practical Geometry 
5  Chapter 5  Data Handling 
6  Chapter 6  Squares and Square Roots 
7  Chapter 7  Cubes and Cube Roots 
8  Chapter 8  Comparing Quantities 
9  Chapter 9  Algebraic Expressions and Identities 
10  Chapter 10  Visualising Solid Shapes 
11  Chapter 11  Mensuration 
12  Chapter 12  Exponents and Powers 
13  Chapter 13  Direct and Inverse Proportions 
14  Chapter 14  Factorisation 
15  Chapter 15  Introduction to Graphs 
16  Chapter 16  Playing with Numbers 
Direct and Inverse Proportion Class 8 Extra क़ुएस्तिओन्स – With Solutions
Given below are some of the important questions and their solutions covered in our Chapter 13 Class 8 Mathematics Important Questions:
Question 1: A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer 1: Let the number of bottles filled in 5 hours = x
We can see that here the ratio of hours and bottles are in direct proportion.
6/840 = 5/x
=⟩ 6x = 840 × 5
=⟩ x = (840 × 5) / 6
=⟩ x = 700
Therefore, the machine will fill up 700 bottles in 5 hours.
Question 2: Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in
(i) 5kg of sugar?
(ii) 1.2 kg of sugar?
Answer 2: (i) Let the number of sugar crystals = x
Weight of sugar (in kg)  2  5 
No. of crystals  9 × 106  x 
We can see that the weight of sugar and the number of crystals are directly proportional.
2/(9 × 106) = 5/x
=⟩ 2x = 9 × 106 × 5
=⟩ x = (9 × 106 × 5)/2
=⟩ x = 2.25 × 107
Therefore, there are 2.25 × 107 crystals.
(ii) Let the number of sugar crystals = x
Now, we can see that the weight of sugar and the number of crystals are directly proportional.
Weight of sugar (in kg)  2  1.2 
No. of crystals  9 × 106  x 
2/(9 × 106 ) = 1.2/x
=⟩ 2x = (1.2 × 9 × 106)
=⟩ x = (1.2 × 9 × 106 ) / 2 = 5.4 × 106
Therefore, the number of sugar crystals in 1.2 kg sugar is 5.4 × 106
Question 3: A 5m 60cm high vertical pole casts a shadow which is 3m 20cm long. Find at the same time
(i) length of the shadow cast by a different pole which is 10m 50cm high.
(ii) height of a pole which casts a shadow 5m long.
Answer 3: Here, we can see that the pole’s height and the shadow’s length are directly proportional.
And we know that 1m = 100cm
Hence, 5m 60cm = 5 × 100 + 60 = 560cm
3m 20cm = 3 × 100 + 20 = 320cm
10m 50cm = 10 × 100 + 50 = 1050cm
5m = 5 × 100 = 500cm
(i) Let the length of the shadow of another pole = x
Height of pole (in cm)  560  1050 
Length of shadow (in cm)  320  x 
560/320 = 1050/x
=⟩ x × 560 = 1050 × 320
=⟩ x = (1050 × 320)/560
=⟩ x = 600cm = 6m
Therefore, the length of the shadow of another pole is 6m.
(ii) Let the height of the pole = x
Height of the pole (in cm)  560  x 
Length of the shadow (in cm)  320  500 
560/320 = x/500
=⟩ x × 320 = 560 × 500
=⟩ x = (560 × 500)/320
=⟩ x = 875cm = 8.75m
Therefore, the height of the pole is 8.75m
Question 4: In a Television game show, total prize money of Rs. 1,00,000 has to be divided equally among the winners. Complete the below table and find out whether the prize money given to one individual winner is directly or inversely proportional to the total number of winners:
No. of winners  1  2  4  5  8  10  20 
Prize for each winner (in Rs.)  1,00,000  50,000  …  …  …  …  … 
Answer 4: Here, we can see that the number of winners and prize money are inversely proportional because winners are increasing, prize money is decreasing.
Therefore,
When the number of winners is 4, the winner will receive = 100000/4 = Rs. 25000
When the number of winners is 5, the winner will receive = 100000/5 = Rs. 20000
When the number of winners is 8, the winner will receive = 100000/8 = Rs. 12500
When the number of winners is 10, the winner will receive = 100000/10 = Rs. 10000
When the number of winners is 20, the winner will receive = 100000/20 = Rs. 5000
Question 5: Rehman is making a wheel using spokes. He wants to fix equal spokes so that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.
Number of spokes  4  6  8  10  12 
Angle between a pair of consecutive spokes  90°  60°  …  …  … 
(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes inversely proportional?
(ii) Find out the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) If the angle between a pair of consecutive spokes is 40 degrees, how many spokes would be needed?
Answer 5: Here, we can see that the number of spokes is inversely proportional to the angle between the pair of consecutive spokes as the number of spokes increases and the angle between the pair of consecutive spokes decreases.
And we know that the centre of a circle is 360°.
Therefore,
When the number of spokes is 8, then the angle between the pair of consecutive spokes = 360/8 = 45°
And when the number of spokes is 10, then the angle between a pair of consecutive spokes = 360/10 = 36°
And when the number of spokes is 12, then the angle between a pair of consecutive spokes = 360/12 = 30°
Number of spokes  4  6  8  10  12 
Angle between a pair of consecutive spokes  90°  60°  45°  36°  30° 
(i) Yes. The number of spokes is inversely proportional to the angle between a pair of consecutive spokes.
(ii) When the number of spokes is 15, the angle between the pair of consecutive spokes = 360/15 = 24°
(iii) Number of spokes needed if the angle between the pair of consecutive spokes is 40° = 360/40 = 9
Question 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?
Solution: Let the time it takes to complete the job = x
Persons  3  4 
Days  4  x 
Here, we can see that the number of persons and days are inversely proportional.
Therefore, 3/4 = x/4
=⟩ 3 × 4 = 4x
=⟩ 4x = 12
=⟩ x = 12/4 = 3
Hence, 4 persons will require 3 days to complete the work.
Question 7: 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?
Answer 7: Let the number of required machines = x
Days  63  54 
Machines  42  x 
Here, the number of machines are inversely proportional to the number of days.
Therefore, 63/54 = x/42
=⟩ 63 × 42 = 54x
=⟩ x = (63 × 42)/54
=⟩ x = 49
Hence, 49 machines are required for 54 days.
Question 8: Two people could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many people would be needed to fit the windows in one day?
Answer 8: Let the number of days = x
Persons  2  1 
Days  3  x 
Here, the number of persons is inversely proportional to the number of days.
2/1 = x/3
=⟩ x = 6
Therefore, 6 days are required by 1 person to complete the job.
(ii) Let the number of persons = x
Persons  2  1 
Days  3  x 
Here, the number of persons is inversely proportional to the number of days.
2/x = 1 /3
x = 6
Therefore, 6 persons are required to complete the work in 1 day.
Question 9: Under the condition that the temperature remains constant, the volume of gas is inversely proportional to its pressure. If the volume of gas is 630 cubic centimetres at a pressure of 360 mm of mercury, then what will be the pressure of the gas if its volume is 720 cubic centimetres at the same temperature?
Answer 9: Given that, at a constant temperature, the pressure and volume of a gas are inversely proportional.
Let the pressure = x
Volume  630  720 
Pressure  360  x 
Therefore,
630/720 = x/360
=⟩ x = (630 × 360)/720
=⟩ x = 315 mm of mercury
Hence, 315 mm of mercury will be the pressure of 720 cc gas.
Question 10: A contractor undertook a contract to complete a part of a stadium in 9 months with a team of 560 persons. Later on, it was required to complete the job in 5 months. How many extra persons should be employed to complete the work?
Answer 10: Here, we can see that the number of persons is inversely proportional to the number of months.
Hence, Let the number of persons required = x
Persons  560  x 
Months  9  5 
Therefore, 560/x = 5/9
=⟩ 5x = 560 × 9
=⟩ x = 5040/5
=⟩ x =1008
We know that 1008 persons are required to complete the work in 5 months.
Therefore, the number of extra persons required = 1008 – 560 = 448 persons
Hence, the 448 extra persons are required to complete the work in 5 months.
Benefits of Solving Important Questions Class 8 Mathematics Chapter 13
When it comes to Mathematics, practice is mandatory for a student to gain mastery in this subject. Only thorough practice can help students get a good hold of the concepts and feel confident in Mathematics. By practising the Important Questions Class 8 Mathematics Chapter 13 provided on the website of Extramarks, students prepare themselves properly for their examinations.
Below are some benefits of solving Class 8 Mathematics Chapter 13 Important Questions:
 Important Questions Class 8 Mathematics Chapter 13 is a set of carefully selected questions by experienced Mathematics faculty members. Our experts have prepared these questions after considering the past years’ question papers and examining several CBSE textbooks.
 Every question and answer provided by Mathematics Class 8 Chapter 13 Important Questions strictly adheres to the latest CBSE syllabus and follows the latest CBSE guidelines so that the students can completely rely on the solutions provided by the website.
 All the Chapter 13 Class 8 Mathematics Important Questions have been solved and supported with the required formulas and explanations for the students to understand the concept well and overall clarification.
 Students can trust and enjoy the process of learning help to speed up their learning and improve their academic performance, through revisions and make them aware of their mistakes through guided practice and help to get the best results
Along with Important Questions, Class 8 Mathematics Chapter 13, the online learning platform Extramarks also provides the following:
 NCERT books
 CBSE revision notes
 CBSE sample paper
 Important Formulas
 CBSE past years’ question papers
 CBSE extra questions
Q.1 If 56 men can do a piece of work in 42 days, how many men are required to complete it in 14 days
Marks:3
Ans
Let x no. of men are required to complete the work in 14 days.
No. of men  56  x 
Number of days  42  14 
$\begin{array}{l}56\text{}\times \text{42 = x}\times \text{14}\\ \\ \text{x =}\frac{56\text{}\times \text{42}}{14}\text{= 168}\end{array}$
Therefore, 168 men are required to complete the work in 14 days.
Q.2 A garrison of 500 men had provision for 24 days. However, a reinforcement of 300 men arrived. How long the food will last
Marks:2
Ans
Let the number of days the food last for 500 + 300 = 800 men be x.
Number of men  500  800 
Days  24  x 
$\begin{array}{l}500\text{}\times \text{24 = 800}\times \text{x}\\ \\ \text{x =}\frac{500\text{}\times \text{24}}{800}\text{= 15 days.}\end{array}$
Therefore, the food will last for 15 days.
Q.3 The scale of a map is given as 1 : 40000000. Two cities are 4 cm apart on the map. Find the actual distance between them.
Marks:3
Ans
Let the map distance be x cm and the actual distance be y cm. Then,
1 : 40000000 = x : y
$\begin{array}{l}\frac{1}{4\u2014{10}^{7}}=\frac{\mathrm{x}}{\mathrm{y}}\\ \frac{1}{4\u2014{10}^{7}}=\frac{4}{\mathrm{y}}\\ \mathrm{y}=\text{}16\u2014{10}^{7}\text{}\mathrm{cm}\\ \mathrm{or}\text{}\mathrm{y}\text{}=\text{}1600\text{}\mathrm{km}\end{array}$
Two cities which are 4 cm apart on the map are actually 1600 km away from each other.
Q.4 If 9 kg of rice costs 166.50, how much rice can be purchased for 259
Marks:1
Ans
Let x kg rice can be purchased for 259. Then,
Cost of rice ()  166.50  259 
Amount of rice (kg)  9  x 
$\begin{array}{l}\frac{166.50}{9}\text{=}\frac{259}{\mathrm{x}}\\ \\ \mathrm{x}\text{=}\frac{9\text{}\times \text{259}}{166.5}\text{= 14 kg.}\\ \\ \text{14kg}\mathrm{rice}\text{}\mathrm{can}\text{}\mathrm{be}\text{}\mathrm{purchased}\text{}\mathrm{for}\text{259.}\end{array}$
Q.5 A car can finish a journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance
Marks:2
Ans
Let the speed be x km/hr for covering the same distance in 8 hours.
Time Taken in hours  10  8 
Speed in km/hr  48  x 
$\begin{array}{l}48\text{}\u2014\text{10 = 8}\u2014\text{x}\\ \\ \text{x=}\frac{48\text{}\u2014\text{10}}{8}\text{= 60 km/hours}\end{array}$
Therefore, the speed needs to be increased by 12 km/hr.
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CBSE Class 8 Maths Important Questions
FAQs (Frequently Asked Questions)
1. What are the topics covered in Important Questions Class 8 Mathematics Chapter 13?
Our expert Mathematics faculty members have carefully chosen some significant question bank Important Questions Class 8 Mathematics Chapter 13. The questions have been picked after carefully analysing and discussing all the important CBSE past year question papers, NCERT books and exemplars. The questions in our question bank cover the following topics:
 Introduction
 Direct Proportion
 Inverse Proportion.
2. How will Important Questions Class 8 Mathematics Chapter 13 help the students score well in their examinations?
Mathematics requires students to practice on a regular basis for them to get a good command over the concepts. Deciphering problems and doing selfassessment about their strong and weak areas is an effective approach to gain mastery in Mathematics.
Students can completely trust and rely on the Extramarks question bank, such as Important Questions Class 8 Mathematics Chapter 13, to get an idea about examrelated questions and their solutions. Our professionals have picked up some constructive questions after thoroughly analysing and considering the CBSE past year’s question papers. Students solving these questions on a regular basis can get excellent results in their exams.
3. Which are the chapters covered in CBSE Class 8 Mathematics syllabus?
Many important chapters that form the fundamentals of Class 10 Mathematics are there in Class 8 Mathematics syllabus. Given below is a complete list of these 16 chapters:
 Chapter 1 Rational Numbers
 Chapter 2 – Linear Equations in One Variable
 Chapter 3 – Understanding Quadrilaterals
 Chapter 4 – Practical Geometry
 Chapter 5 – Data Handling
 Chapter 6 – Square and Square Roots
 Chapter 7 – Cube and Cube Roots
 Chapter 8 – Comparing Quantities
 Chapter 9 – Algebraic Expressions and Identities
 Chapter 10 – Visualising Solid Shapes
 Chapter 11 Mensuration
 Chapter 12 – Exponents and Powers
 Chapter 13 – Direct and Inverse Proportions
 Chapter 14 – Factorisation
 Chapter 15 – Introduction to Graphs
 Chapter 16 – Playing with Numbers