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NCERT Solutions
Class 6
Maths
Chapter 7 Fractions

NCERT Solutions Class 6 Maths Chapter 7 Fractions

NCERT solutions for class 6 - Maths chapter 7 Fractions includes topics like mixed fractions, improper fractions, proper fractions, and how to represent a fraction on a number line. A fraction is a numerical value that denotes a portion of a whole. Even in real life, you need to execute this topic, so you must know about every concept of fractions.  

The NCERT Solutions will be based on the latest CBSE syllabus. Different types of questions will be based on every concept of class 6 Science Chapter 7, so you can easily grasp all principles and fundamentals. It also ensures that students can practice and reinforce their knowledge. 

Therefore, in this article, we have come up with expertly crafted NCERT Solutions for Class 6 of chapter 7, Fractions. Each solution is detailed to help you understand it easily.   

1.0Download NCERT Solutions for Class 6 Maths Chapter 7 : Free PDF

We are providing ALLEN NCERT Solutions a downloadable PDF for Chapter 7 of Class 6 Maths, which covers fractions. The solutions are detailed to make them clearer and easier to understand, helping students solve questions step by step. These solutions are especially important for revising and practising at home, making learning more convenient.

NCERT Solution for Class 6 Maths Chapter 7: Fractions

2.0NCERT Solutions for Class 6 Maths Chapter 7 Fractions: All Exercises

In this section, you will find solutions to all the exercises from Chapter 7 of the Class 6 Maths book. It includes detailed answers for each question, helping students understand how to add, compare, subtract, and simplify fractions. These solutions will help students grasp the concept of fractions and apply them to real-world problems.

Class 6 Maths Chapter 7 Exercise 7.1

7 Questions

Class 6 Maths Chapter 7 Exercise 7.2

8 Questions

Class 6 Maths Chapter 7 Exercise 7.3

8 Questions

Class 6 Maths Chapter 7 Exercise 7.4

4 Questions

3.0NCERT Questions with Solutions Class 6 Maths Chapter 7

7.1 Fractional Units and Equal Shares

Fill in the blanks with fractions.

  • Three guavas together weight 1 kg . If they are roughly of the same size, each guava will roughly weigh kg. Sol. Since three guavas weight 1 kg ., So, one unit will be divided into three part to get weight of each guava. Each guava weight 31​ kg.
  • A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is kg. Sol. Since 1 kg of rice is packed in four packets of equal weight. Weight of each packet is 41​ kg.
  • Four friends ordered 3 glasses of sugarcane juice and shared it equally among themselves. Each one drank glass of sugarcane juice. Sol. Since 3 glasses of sugarcane juice is to be shared equally among four friends, so each one drank 43​ glass of sugarcane juice.
  • The big fish weighs 21​ kg. The small one weighs 41​ kg. Together they weigh kg. Sol. Total weight of both fish =21​+41​ 2×21×2​=42​;4×11×1​=41​ 21​+41​=42​+41​=42+1​=43​ kg.
  • Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: One and a half, three quarters, one and a quarter, half, quarter, two and a half Sol. One and a half =1+21​=121​, half =21​, quarter =41​ One and a quarter =1+41​=141​, two and a half =2+21​=221​

7.2 Fractional Units as Parts of A Whole

A whole chikki The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?
d.
e.
g.
h.
Sol. a. We get the required piece by dividing the whole chikki into 12 equal pieces. So, the required piece is 121​ chikki. b. We get the required piece by dividing the whole chikki into 4 equal pieces. So, the required piece is 41​ chikki.
c. We get the required piece by dividing the whole chikki into 8 equal parts.

So, the required piece is 81​ chikki.

d. We get the required piece by dividing the whole chikki into 6 equal prices. So, the required piece is 61​ chikki. e. We get the required piece by dividing the whole chikki (i.e., 24 pieces) into 8 equal parts, then only we will 3 pieces of chikki in one part. So, the required piece is 81​. f. We get the required piece by dividing the whole chikki into 6 equals. So, the required piece is 61​ chikki.
Note: 6a and 6b makes one piece of chikki. g. We get the required piece by dividing the whole chikki into 24 equal pieces. So, the required piece is 241​ chikki. h. We get the required piece by dividing the whole chikki into 24 equal parts. So, the required piece is 241​ chikki.

7.3 Measuring Using Fractional Units

Let us look at another example.

  • Continue this table of 21​ for 2 more steps. Sol.
  • Can you create a similar table for 41​ ? Sol. Here represents a whole roti.
  • Making 31​ using a paper strip. Can you use this to also make 61​ ? Sol. Making 31​ using a paper strip.
    We can make 61​ by dividing the part 31​ into two equal parts.
  • Draw a picture and write an addition statement as above to show: (a) 5 times 41​ of a roti (b) 9 times 41​ of a roti Sol. (a) 5 times 41​ of a roti.
    ⇒41​+41​+41​+41​+41​=45​ ⇒ Here, one full roti and 41​ part of roti. (b) 9 times 41​ of a roti.
    ​141​+41​+41​+41​​​+141​+41​+41​+41​​​+41​=49​+41​=2+41​​ ⇒ Here, 2 full roti and 41​ part of roti.
  • Match each fractional unit with the correct picture:
    Sol.

7.4 Marking Fraction Lengths on the Number Line

  • On a number line, draw lines of lengths 101​,103​, and 54​. Sol. To represent 101​,103​,54​(=108​) on a number line, divide the distance between ' 0 ' and 1 to 10 equal parts.
    Point P represents 101​ Point Q represents 103​ Point R represents 108​ or 54​
  • Write five more fractions of your choice and mark them on the number line. Sol. Do it yourself.
  • How many fractions lie between 0 and 1 ? Think, discuss with your classmates, and write your answer. Sol. There are infinite number of fractions lie between 0 and 1 . Example : 53​,54​,97​,118​,65​ etc.
  • What is the length of the blue line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is 21​. So, the blue line is 21​ units long. Write the fraction that gives the length of the black line in the box.
    Sol. Length of blue line =21​ Length of black line =21​+21​+21​=23​
  • Write the fraction that gives the lengths of the black lines in the respective boxes.
    Sol.

7.5 Mixed Fractions

  • How many whole units are there in 27​ ? Sol. 27​=3+21​ There are 3 whole units in 27​.
  • How many whole units are there in 34​ and in 37​ ? Sol. 34​=1+31​ and 37​=2+31​ There is 1 whole unit in 34​ and 2 whole units in 37​
  • Figure out the number of whole units in each of the following fractions: (a) 38​ (b) 511​ (c) 49​ Sol. (a) 38​=2+32​=232​ (b) 511​=2+51​=251​ (c) 49​=2+41​=241​
  • Can all fractions greater than 1 be written as such mixed numbers? Sol. Yes, all fractions greater than 1 can be written as mixed numbers/mixed fraction. Ex : −712​=1+75​=175​
  • Write the following fractions as mixed fractions (e.g., 29​=421​ ): (a) 29​ (b) 59​ (c) 1921​ (d) 947​ (e) 1112​ (f) 619​ Sol. (a) 29​=4+21​=421​ (b) 59​=1+54​=154​ (c) 1921​=1+192​=1192​ (d) 947​=5+92​=592​ (e) 1112​=1111​=1111​ (f) 619​=3+61​=361​
  • Write the following mixed numbers as fractions: (a) 341​ (b) 732​ (c) 994​ (d) 361​ (e) 2113​ (f) 3109​ Sol. (a) 341​=3+41​=43×4+1​=413​ (b) 732​=7+32​=37×3+2​=322​ (c) 994​=9+94​=99×9+4​=985​ (d) 361​=3+61​=63×6+1​=619​ (e) 2113​=2+113​=112×11+3​=1125​ (f) 3109​=3+109​=103×10+9​=1039​

7.6 Equivalent Fractions

  • Are 63​,84​,105​ equivalent fractions? Why? Sol. Since, 63​⇒6÷33÷3​=21​ 84​⇒8÷44÷4​=21​ and 105​⇒10÷55÷5​=21​ Since three fractions have same value when simplified, so they are equivalent fractions.
  • Write two equivalent fractions for 32​. Sol. 62​⇒6×22×2​=124​ ⇒6×32×3​=186​
  • 64​=□=□=□=.................(write as many you can) 64​=128​=1812​=2416​= Sol. 64​⇒6×24×2​=128​;6×34×3​=1812​;6×44×4​=2416​; 64​=128​=1812​=2416​=
  • Three rotis are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts. Fraction of roti each child gets is . Division fact: Addition fact: Multiplication fact: Compare your picture and answers
    with your classmates! Sol.
    Division fact ⇒3 wholes divided in 4 parts =3÷4=43​ Addition fact ⇒ Four times 43​ added gives 3 wholes =43​+43​+43​+43​=412​=3 Multiplication fact ⇒4 parts of 43​ make 3 wholes =4×43​=3
  • Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts. Sol. As 2 rotis have to be shared equally by 4 children we divide each roti in 4 parts and give
    (a) 1 part of each roti to each-child as shown below :
    (b) 2 parts to each child as shown below :
    Division fact ⇒2 whole divides in 4 parts 2÷4 or 42​=21​ Addition fact ⇒42​+42​+42​+42​=48​=2 Multiplication fact =4×42​=2
  • Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get? Sol.
    Each cake gets divided into 5 parts and Anil gets one part from each cake i.e. 51​+51​=52​.
  • Find the missing numbers: (a) 5 glasses of juice shared equally among 4 friends is the same as glasses of juice shared equally among 8 friends. So, 45​=8□​ □ (b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in __ bags. So, 34​=□12​ □ (c) 7 rotis divided among 5 children is the same as rotis divided among children. So, 57​=□□​. □ Sol. (a) 45​=8□​ □ Since 4×2=8 So, 5×2=10 ⇒45​=810​ (b) 34​=□12​ Since 4×3=12 So, 3×3=9 34​=912​ (c) 57​=□□​ □ 7×2=14;7×3=21;7×4=28 5×2=10;5×3=15;5×4=20 57​=1014​;57​=1521​;57​=1028​
  • Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. (a) 27​ and 53​ (b) 38​ and 65​ (c) 43​ and 53​ (d) 76​ and 58​ (e) 49​ and 25​ (f) 101​ and 92​ (g) 38​ and 411​ (h) 613​ and 91​ Sol. (a) The smallest common multiple of 2 and 5 is 10. 2×57×5​=1035​;5×23×2​=106​ (b) The smallest common multiple of 3 and 6 is 6 . 3×28×2​=616​=6×15×1​=65​ (c) The smallest common multiple of 4 and 5 is 20 . 4×53×5​=2015​;5×43×4​=2012​ (d) The smallest common multiple of 7 and 5 is 35 . 7×56×5​=3530​;5×78×7​=3542​ (e) The smallest common multiple of 4 and 2 is 4 . 4×19×1​=49​=2×25×2​=410​ (f) The smallest common multiple of 10 and 9 is 90 . 10×91×9​=909​;9×102×10​=9020​ (i) The smallest common multiple of 3 and 4 is 12 . 3×48×4​=1232​=4×311×3​=1233​ (h) The smallest common multiple of 6 and 9 is 18. 6×313×3​=1839​;9×21×2​=182​
  • Express the following fractions in lowest terms: (a) 5117​ (b) 14464​ (c) 147126​ (d) 112525​ Sol. (a) Since, 17 is common factor of 17 and 51, So, 51÷1717÷17​=31​ (b) 14464​⇒144÷264÷2​=7232​⇒72÷232÷2​=3616​ ⇒36÷216÷2​=188​⇒18÷28÷2​=94​ (c) Since 3 is common factor of 126 and 147, 147÷3126÷3​=4942​⇒49÷742÷7​=76​ (d) Since 7 as common multiple of 525 and 112, 112÷7525÷7​=1675​

7.7 Comparing Fractions

  • Compare the following fractions and justify your answers: (a) 38​,25​ (b) 94​,73​ (c) 107​,149​ (d) 512​,58​ (e) 49​,25​ Sol. (a) The smallest common multiple of 3 and 2,4,6. 3×28×2​=616​;2×35×3​=615​ Since 616​>615​ 38​>25​ (b) The smallest common multiple of 9, 7 is 63. 94​×77​=6328​;73​×99​=6327​ ∵6328​>6327​ Now 94​>73​ (c) The smallest common multiple of 10,14 , is 70 . 10×77×7​=7049​;149​×55​=7045​ ∵7049​>7045​ ⇒107​>149​ (d) ∵ Both fraction have same denominators, 512​>58​ (e) The smallest common multiple of 4, 2 is 4 . 49​×21​=49​;2×25×2​=410​ ∵49​<410​ ⇒49​<25​.
  • Write the following fractions in ascending order. (a) 107​,1511​,52​ (b) 2419​,65​,127​ Sol. (a) The smallest common multiple of 10,15,5, is 30 . ∵⇒​10×37×3​=3021​;15×211×2​=3022​;5×62×4​=3012​3012​<3021​<3022​52​<107​<1511​ (ascending order) ​ (b) The smallest common multiple of 24,6,12, is 24 . ⇒​24×119×1​=2419​;6×45×4​=2420​;12×27×2​=2414​127​<2419​<65​ (ascending order) ​
  • Write the following fractions in descending order. (a) 1625​,87​,413​,3217​ (b) 43​,512​,127​,45​ Sol. (a) The smallest common multiple of 16,8,9,32 is 32 ​16×225×2​=3250​;8×47×4​=3228​;4×813×8​=32104​;32×117×1​=3217​∵32104​>3250​>3228​>3217​⇒413​>1625​>87​>3217​ (descending order) ​ (b) The smallest common multiple of 4,5,12, is 60 . 4×153×15​=6045​;5×1212×12​=60144​;12×57×5​=6035​;4×155×15​=6075​ ∵60144​>6075​>6045​>6035​ ⇒512​>45​>43​>127​ (descending order)

7.8 Addition and Subtraction of Fractions

  • Add the following fractions using Brahmagupta's method: (a) 72​+75​+76​ (b) 43​+31​ (c) 32​+65​ (d) 32​+72​ (e) 32​+54​ (g) 54​+32​ (h) 53​+85​ (i) 29​+45​ (j) 38​+72​ (k) 43​+31​+51​ (i) 52​+54​+73​ (m) 29​+45​+67​ Sol. (a) 72​+75​+76​=72+5+6​=713​ (b) 43​×33​=129​;31​×44​=124​ 43​+31​=129​+124​=129+4​=1213​ (c) 32​×22​=64​;6×15×1​=65​ 32​+65​=64​+65​=64+5​=610​=6÷210÷2​=35​ (d) 3×72×7​=2114​;7×32×3​=216​ 32​+72​=2114​+216​=2114+6​=2120​ (e) 43​+31​+51​ The smallest common multiple of 3,4,5 is 60 . 4×153×15​=6045​;3×201×20​=6020​;5×121×12​=6012​ 43​+31​+51​=6045​+6020​+6012​=6045+20+12​=6077​ (f) 3×52×5​=1510​;5×34×3​=1512​ 32​+54​=1510​+1512​=1510+12​=1522​ (g) 5×34×3​=1512​;3×52×5​=1510​ 54​+32​=1512​+1510​=1512+10​=1522​ (h) The smallest common multiple of 5 and 8 is 40 5×83×8​=4024​;8×55×5​=4025​ 4024​+4025​=4024+25​=4049​ (i) 2×29×2​=418​;4×15×1​=45​ 418​+45​=418+5​=423​ (j) 3×78×7​=2156​;7×32×3​=216​ 2156​+216​=2156+6​=2162​ (k) The smallest common multiple of 3,4,5 is 60 . 4×153×15​=6045​;3×201×20​=6020​;8×121×12​=6012​ 6045​+6020​+6012​=6045+20+12​=6077​ (1) The smallest common multiple of 3,5,7 is 105 . 3×352×35​=10570​;5×214×21​=10584​;7×153×15​=10545​ 10570​+10584​+10545​=10570+84+45​=105199​ (m) The smallest common multiple of 2,4,6, is 12 . ​2×69×6​=1254​;4×35×3​=1215​;6×27×2​=1214​1254​+1215​+1214​=1254+15+14​=1283​​
  • Rahim mixes 32​ litres of yellow paint with 43​ litres of blue paint to make green paint. What is the volume of green paint he has made? Sol. Volume of green paint =32​+43​ 3×42×4​=128​;4×33×3​=129​ 32​+43​=128​+129​=128+9​=1217​.
  • Geeta bought 52​ meter of lace and Shamim bought 43​ meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border? Sol. Total length of the lace Geeta and Shamim have bought =52​+43​ The smallest common multiple of 5 and 4 is 20 . 5×42×4​=208​;4×53×5​=2015​ 208​+2015​=2023​ m=1203​ m =1 m+203​×100 cm =1 m15 cm. Since perimeter of border on table cloth as 1 m . Here, length of lack is greater than perimeter, So, the required length of lace will be sufficient to cover the whole border.
  • 85​−83​ Sol. 85​−83​=85−3​=82​=8÷22÷2​=41​
  • 97​−95​ Sol. 97​−95​=97−5​=92​
  • 2710​−271​ Sol. 2710​−271​=2710−1​=279​=27÷99÷9​=31​
  • Carry out the following subtractions using Brahmagupta's method: (a) 158​−153​ (b) 52​−154​ (c) 65​−94​ (d) 32​−21​ Sol. (a) 158​−153​ ⇒158−3​{∵ same denominators } ⇒155​ (b) 52​−154​ Converting given fraction having same denominator into equivalent fraction, 5×32×3​=156​;15×14×1​=154​ ⇒156​−154​=156−4​=152​. (c) 65​−94​ Converting given fraction into equivalent fraction having same denominator, 6×35×3​=1815​; 9×24×2​=188​ ⇒1815​−188​=1815−8​=187​. (d) 32​−21​ Converting given fraction into equivalent fraction having same denominator, 3×22×2​=64​;21​×33​=63​ ⇒64​−63​=64−3​=61​
  • Subtract as indicated: (a) 413​ from 310​ (b) 518​ from 323​ (c) 729​ from 745​ Sol. (a) 413​ from 310​⇒310​−413​ ​3×410×4​=1240​;4×313×3​=1239​⇒1240​−239​=121​​ (b) 518​ from 323​⇒323​−518​ ​3×523×5​=15115​;5×318×3​=1554​⇒15115​−1554​=15115−54​=1561​​ (c) 729​ from 745​⇒745​−729​ 745−29​=716​
  • Solve the following problems: (a) Jaya's school is 107​ km from her home. She takes an auto for 21​ km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school? (b) Jeevika takes 310​ minutes to take a complete round of the park and her friend Namit takes 413​ minutes to do the same. Who takes less time and by how much? Sol. (a)
    Distance between home and school =107​ km. Distance travelled by auto =21​ km. So, remaining distance travelled by walk. ⇒107​−21​ 10×17×1​=107​;21​×55​=105​ ⇒107​−21​=107​−105​=107−5​=102​ =10÷22÷2​=51​ km. (b) Time taken by Jeevika =310​ minutes. Time taken by Namit =413​ minutes. Now, 3×410×4​=1240​;413​×33​=1239​ Comparing 1240​ and 1239​, 1240​>1239​ {For same denominator, fraction having greater numerator is greater} So, 310​>413​ Or we can say Namit takes less time. Also, 310​−413​=1240​−1239​ =1240−39​=121​ minutes Namit takes less time by 121​ minutes.

4.0NCERT Solutions Class 6 Maths Chapter 7 Subtopics

Before understanding the NCERT solutions, it is important to know the subtopics covered under the chapter fractions. This section explains the different subtopics covered in Chapter 7, "Fractions." In the NCERT solutions, you will learn about:

  • Fractional Units and Equal Shares
  • Fractional Units as Parts of a Whole
  • Measuring Using Fractional Units
  • Marking Fraction Lengths on the Number Line
  • Mixed Fractions
  • Equivalent Fractions
  • Addition and Subtraction of Fractions
  • A Pinch of History

NCERT Solutions for Class 6 Maths Other Chapters:-

Chapter 1: Patterns in Mathematics

Chapter 2: Lines and Angles

Chapter 3: Number Play

Chapter 4: Data Handling and Presentation

Chapter 5: Prime Time

Chapter 6: Perimeter and Area

Chapter 7: Fractions

Chapter 8: Playing With Construction

Chapter 9: Symmetry

Chapter 10: The Other Side of Zero


CBSE Notes for Class 6 Maths - All Chapters:-

Class 6 Maths Chapter 1 - Patterns In Mathematics Notes

Class 6 Maths Chapter 2 - Lines And Angles Notes

Class 6 Maths Chapter 3 - Number Play Notes

Class 6 Maths Chapter 4 - Data Handling And PresentationNotes

Class 6 Maths Chapter 5 - Prime Time Notes

Class 6 Maths Chapter 6 - Perimeter And Area Notes

Class 6 Maths Chapter 7 - Fractions Notes

Class 6 Maths Chapter 8 - Playing With Constructions Notes

Class 6 Maths Chapter 9 - Symmetry Notes

Class 6 Maths Chapter 10 - The Other Side Of Zero Notes

Frequently Asked Questions

Practicing NCERT solutions for Class 6 Maths is essential because it enhances conceptual understanding, aids in exam preparation, and develops problem-solving skills. Regular practice helps students become familiar with the types of questions they may encounter in exams, ultimately boosting their confidence.

The most crucial topics in Chapter 7 on fractions include: Understanding Fractions: Basics of fractions and their components. Types of Fractions: Proper, improper, and mixed fractions. Equivalent Fractions: Identifying and creating fractions that represent the same value. Simplest Form: Reducing fractions to their simplest form. Comparing Fractions: Techniques for comparing different fractions. These topics are foundational for mastering fractions and are vital for future mathematical studies.

Fractions are used in NCERT Math. Class 6, Chapter 7, to illustrate parts of a whole. A fraction is expressed as 𝑎/𝑏, where the denominator (the total number of equal parts that make up the whole) is shown by 𝑏, the numerator, and 𝑎, the numerator, indicates how many parts are considered.

Chapter 7 of Maths Class 6, Chapter 7 defines equivalent fractions as fractions with the same value but different numerators and denominators. As long as the number is not zero, you can find equivalent fractions by multiplying or dividing the numerator and denominator by the same number.

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