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 solution 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.
1.0NCERT Solutions for Class 6 Maths Chapter 7 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
Arrange the following in descending order:
(i) 92,32,218
(ii) 51,73,107
Sol. (i) 92,32,218
Changing them to like fractions, we obtain
92=9×72×7=631432=3×212×21=6342218=3×218×3=6324
Since 42>24>14,
∴32>218>92
(ii) 51,73,107
Changing them to like fractions, we obtain
51=5×141×14=701473=7×103×10=7030107=10×77×7=7049
As 49>30>14,
∴107>73>51
In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square ?
114
119
112
113
115
117
118
111
116
Sol. Along the first row, sum
=114+119+112=1115
Along the second row, sum
=113+115+117=1115
Along the third row, sum
=118+111+116=1115
Along the first column, sum
=114+113+118=1115
Along the second column, sum
=119+115+111=1115
Along the third column sum
=112+117+116=1115
Along the first diagonal, sum
=114+115+116=1115
Along the first diagonal, sum
=112+115+118=1115
Since the sum of the numbers in each row, in each column, and along the diagonal is the same, it is a magic square.
A rectangular sheet of paper is 1221cm long and 1032cm wide. Find its perimeter.
Sol. Length =1221cm=225cm
Breadth =1032cm=332cm
Perimeter =2× (Length + Breadth )=2×[225+332]=2×[6(25×3)+(32×2)]=2×[675+64]=2×6139=3139=4631cm
Find the perimeters of (i) △ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Sol. (i) Perimeter of △ABE=AB+BE+EA=(25+411+518)=(25+243+353)=(2×105×10+4×511×5+5×418×4)=2050+55+72=20177=82017cm
(ii) Perimeter of rectangle =2 (Length + Breadth)
Perimeter of rectangle =2[411+67]=2[4×311×3+6×27×2]=2[1233+14]=2×1247=647=765cm
Perimeter of △ABE=20177cm
Changing them to like fractions, we obtain
20177=20×3177×3=60531647=6×1047×10=60470, as 531>470∴20177>647
Perimeter ( △ABE ) > Perimeter (BCDE)
Salil wants to put a picture in a frame. The picture is 753cm wide. To fit in the frame the picture cannot be more than 7103cm wide. How much should the picture be trimmed?
Sol. Width of picture =753=538cm
Required width =7103=1073cm
The picture should be trimmed by
=(538−1073)=(5×238×2−1073)=1076−73=103cm
Ritu ate 53 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much?
Sol. Part of apple eaten by Ritu =53
Part of apple eaten by Somu =1 - Part of apple eaten by Ritu =1−53=52
Therefore, Somu ate 52 part of the apple.
Since 3>2, Ritu had the larger share.
Difference between the 2 shares
Therefore, Ritu's share is larger than the share of Somu by 51.
Michael finished colouring a picture in 127 hour. Vaibhav finished colouring the same picture in 43 hour. Who worked longer? By what fraction was it longer?
Sol. Time taken by Michael =127hr
Time taken by Vaibhav =43hr
Converting these fractions into like fractions, we obtain 43=4×33×3=129 and, 127
Since 9>7, Vaibhav worked longer.
Difference =129−127=122=61 hour
EXERCISE 2.2
Which of the drawings (a) to (d) show :
(i) 2×51
(ii) 2×21
(iii) 3×32
(iv) 3×41
(a)
(b)
(c)
(d)
Sol. (i) 2×51 represents addition of 2 figures, each representing 1 shaded part out of 5 equal parts. Hence, 2×51 is represented by (d).
(ii) 2×21 represents addition of 2 figures, each representing 1 shaded part out of 2 equal parts. Hence, 2×21 is represented by (b)
(iii) 3×32 represents addition of 3 figures, each representing 2 shaded parts out of 3 equal parts. Hence, 3×32 is represented by (a).
(iv) 3×41 represents addition of 3 figures, each representing 1 shaded part out of 4 equal parts. Hence, 3×41 is represented by (c).
Some pictures (a) to (c) are given below. Tell which of them show :
(i) 3×51=53
(ii) 2×31=32
(iii) 3×43=241
(a)
(b)
(c)
Sol. (i) 3×51 represents the addition of 3 figures, each representing 1 shaded part out of 5 equal parts 53 represents 3 shaded parts out of 5 equal parts. Hence, 3×51=53 is represented (c).
(ii) 2×31 represents the addition of 2 figures, each representing 1 shaded part out of 3 equal parts and 32 represents 2 shaded
parts out 3 equal parts. Hence, 2×31=32 is represented by (a)
(iii) 3×43 represents the addition of 3 figures, each representing 3 shaded parts out of 4 equal parts and 241 represents 2 fully shaded figures and one figure having 1 part as shaded out of 4 equal parts. Hence, 3×43=241 is represented by (b)
Multiply and reduce to lowest form and convert into a mixed fraction :
(i) 7×53
(ii) 4×31
(iii) 2×76
(iv) 5×92
(v) 32×4
(vi) 25×6
(vii) 11×74
(viii) 20×54
(ix) 13×31
(x)15×53
Sol.
(i) 7×53=521=451
(ii) 4×31=34=131
(iii) 2×76=712=175
(iv) 5×92=910=191
(v) 32×4=38=232
(vi) 25×6=15
(vii) 11×74=744=672 (viii) 20×54=16
(ix) 13×31=313=431
(x) 15×53=9
Shade:
(i) 21 of the circles in box (a)
(ii) 32 of the triangles in box (b)
(iii) 53 of the squares in box (c)
(a)
(b)
(c)
Sol. (i) It can be observed that there are 12 circles in the given box. We have to shade 21 of the circles in it. As 12×21=6, therefore we will shade any 6 circles of it.
(ii) It can be observed that there are 9 triangles in the given box. We have to shade 32 of the triangles in it. As 9×32=6, therefore, we will shade any 6 triangles of it.
ΔΔΔ
ΔΔΔ
ΔΔΔ
(iii) It can be observed that there are 15 squares in the given box. We have to shade 53 of the squares in it. As 53×15=9, therefore, we will shade any 9 squares of it.
Find :
(a) 21 of
(i) 24
(ii) 4
(b) 32 of
(i) 18
(ii) 27
(c) 43 of
(i) 16
(ii) 36
(d) 54 of
(i) 20
(ii) 35
Sol. (a) (i) 21×24=12
(ii) 21×46=23
(b) (i) 32×18=12
(ii) 32×27=18
(c) (i) 43×16=12
(ii) 43×36=27
(d) (i) 54×20=16
(ii) 54×35=28
Find
(a) 21 of
(i) 243
(ii) 492
(b) 85 of
(i) 365
(ii) 932
Sol. (a) (i) 21×243=21×411=811=183
(ii) 21×492=21×938=919=291
(b) (i) 85×365=85×623=48115=24819
(ii) 85×932=85×329=24145=6241
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 52 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Sol. (i) Water consumed by Vidya =52 of 5 litres =52×5=2 litres
(ii) Water consumed by Pratap =1−52=53 of the total water.
EXERCISE 2.3
Find:
(i) 41 of
(a) 41
(b) 53
(c) 34
(ii) 71 of
(a) 92
(b) 56
(c)103
Sol.
(i) (a) 41×41=161
(b) 41×53=203
(c) 41×34=31
(ii)
(a) 71×92=632
(b) 71×56=356
(c) 71×103=703
Multiply and reduce to lowest from (if possible):
(i) 32×232
(ii) 72×97
(iii) 83×46
(iv) 59×53
(v) 31×815
(vi) 211×103
(vii)54×712
Sol. (i) 32×232=32×38=916=197
(ii) 72×97=92
(iii) 83×46=169
(iv) 59×53=2527=1252
(v) 31×815=85
(vi) 211×103=2033=12013
(vii) 54×712=3548=13513
Multiply the following fractions :
(i) 52×541
(ii) 652×97
(iii) 23×531
(iv) 65×373
(v) 352×74
(vi) 253×3
(vii) 374×53
Sol. (i) 52×541=52×421=1021
This is an improper fraction, and it can be written as a mixed fraction as 2101.
(ii) 652×97=532×97=45224
This is an improper fraction, and it can be written as a mixed fraction as 44544.
(iii) 23×531=23×316=8
This is a whole number.
(iv) 65×273=65×717=4285
This is an improper fraction, and it can be written as a mixed fraction as 2421.
(v) 352×74=517×74=3568
This is an improper fraction, and it can be written as a mixed fraction as 13533.
(vi) 253×3=513×3=539
This is an improper fraction, and it can be written as a mixed fraction as 754.
(vii) 374×53=725×53=715
This is an improper fraction and it can be written as a mixed fraction as 271.
Which is greater :
(i) 72 of 43 or 53 of 85
(ii) 21 of 76 or 32 of 73
Sol. (i) 72×43=14353×85=83
Converting these fractions into like fractions,
143=14×43×4=561283=8×73×7=5621
Since 5621>5612,∴83>143
Therefore, 53 of 85 is greater.
(ii) 21×76=7332×73=72
Since 3>2, ∴73>72
Therefore, 21 of 76 is greater.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 43m. Find the distance between the first and the last sapling? 1st Sol. Sapling 2nd Sapling 3rd Sapling 4th Sapling
From the figure, it can be observed that gaps between 1st and last sapling =3.
Length of 1 gap =43m
Therefore, distance between I and IV sapling =3×43=49=241m
Lipika reads a book for 143 hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Sol. Number of hours Lipika reads the book per day =143=47 hours
Number of days =6
Total number of hours required by her to read the book =47×6=221=1021 hours
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 243 litres of petrol?
Sol. Number of kms a car run per litre petrol = 16 km
Quantity of petrol =243L=411L
Number of kms a car can run for 411 litre
petrol =411×16=44km
It will cover 44 km distance by using 243 litres of petrol.
(a) (i) Provide the number in the box such that 32×□=3010.
(ii) The simplest from the number obtained in □ is -.
(b) (i) Provide the number in the box such that 53×□=7524 ?
(ii) The simplest form of the number obtained in □ is .
Sol. (a) (i) As 32×105=3010, Therefore, the number in the box □, such that 32×□=3010 is 105.
(ii) The simplest form of 105 is 21.
(b) (i) As 53×158=7524, Therefore, the number in the box □, such that 53×□=7524 is 158.
(ii) As 158 cannot be further simplified, therefore, its simplest form is 158
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 73
(ii) 85
(iii) 79
(iv) 56
(v) 712
(vi) 81
(vii)111
Sol. A proper fraction is the fraction which has its denominator greater than its numerator while improper fraction is the fraction which has its numerator greater than its denominator. Whole numbers are a collection of all positive integers including 0 .
(i) Reciprocal of 73=37
Therefore, it is an improper fraction.
(ii) Reciprocal of 85=58
Therefore, it is an improper fraction.
(iii) Reciprocal of 79=97
Therefore, it is a proper fraction.
(iv) Reciprocal of 56=65
Therefore, it is a proper fraction.
(v) Reciprocal of 712=127
Therefore, it is a proper fraction.
(vi) Reciprocal of 81=18
Therefore, it is a whole number.
(vii) Reciprocal of 111=111
Therefore, it is a whole number.
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:-
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.