NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals
Fractions and Decimals is the second chapter of Class 7 NCERT Maths. This chapter comprehensively covers essential topics such as Multiplication of Fractions, Division of Fractions, Multiplication of Decimal Numbers, and Division of Decimal Numbers. By mastering these topics, students will strengthen their foundational knowledge of mathematics. The NCERT Solutions for Class 7 Maths Chapter 2 provides clear, step-by-step guidance, helping students tackle complex problems with ease and improve their overall mathematical skills.
1.0Download Class 7 Maths Chapter 2 NCERT Solutions PDF Online
This article provides NCERT Class 7 solutions for Fractions and Decimals, designed to help students build a solid mathematical foundation and improve their problem-solving skills. By practising these solutions, students can gain a deeper understanding of the concepts and enhance their performance in exams, leading to better scores. For comprehensive guidance, students can download the NCERT Class 7 Maths Chapter 2 PDF, meticulously curated by ALLEN’s experts, from the link below.
NCERT Solutions Class 7 Maths Chapter 2: Fractions and Decimals
2.0Class 7 Maths Chapter 2 Fractions and Decimals Overview
Before discussing the specifics of NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals, let's quickly review the key topics and subtopics included in this chapter of the NCERT Class 7 Maths book.
Topics covered in this chapter
Multiplication of Fractions
Division of Fractions
Multiplication of Decimal Numbers
Division of Decimal Numbers
3.0NCERT Questions with Solutions for Class 7 Maths Chapter 2 Fractions Part 1 - Detailed Solutions
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 Questions with Solutions for Class 7 Maths Chapter 2 Decimals Part 2 - Detailed Solutions
Exercise 2.5
Which is greater?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88
Sol. (i) 0.5 or 0.05
Converting these decimal numbers into equivalent fractions,
0.5=105=10×105×10=10050 and 0.05=1005
It can be observed that both fractions have the same denominator.
As 50>5, Therefore, 0.5>0.05
(ii) 0.7 or 0.5
Converting these decimal numbers into equivalent fractions,
0.7=107 and 0.5=105
It can be observed that both fractions have the same denominator.
As 7>5, Therefore, 0.7>0.5
(iii) 7 or 0.7
Converting these decimal numbers into
7=17=1×107×10=1070 and 0.7=107
It can be observed that both fractions have the same denominator.
As 70>7, Therefore, 7>0.7
(iv) 1.37 or 1.49
Converting these decimal numbers into equivalent fractions,
1.37=100137 and 1.49=100149
It can be observed that both fractions have the same denominator.
As 137<149, Therefore, 1.37<1.49
(v) 2.03 or 2.30
Converting these decimal numbers into equivalent fractions,
2.03=100203 and 2.30=100230
It can be observed that both fractions have the same denominator.
As 203 < 230,
Therefore, 2.03<2.30
(vi) 0.8 or 0.88
Converting these decimal numbers into equivalent fractions,
0.8=108=10×108×10=10080 and 0.88=10088
It can be observed that both fractions have the same denominator.
As 80<88, Therefore, 0.8<0.88
Express as rupees using decimals :
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise
Sol. There are 100 paise in 1 rupee. Therefore, if we want to convert paise into rupees, then we have to divide paise by 100 .
(i) 7 paise =₹1007=₹0.07
(ii) 7 rupees 7 paise =₹7+₹1007=₹7.07
(iii) 77 rupees 77 paise =₹77+10077=₹77.77
(iv) 50 paise =₹10050=₹0.50
(v) 235 paise =100235 rupees =₹2.35
(i) Express 5 cm in metre and kilometre
(ii) Express 35 mm in cm,m and km
Sol. (i) 5 cm
5cm=1005m=0.05m5cm=1000005km=0.00005km
(ii) 35 mm
35mm=1035cm=3.5cm35mm=100035m=0.035m35mm=100000035km=0.000035km
Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8 g
Sol. (i) 200g=1000200kg=0.2kg
(ii) 3470g=10003470=3.470kg
(iii) 4kg8g=4kg+10008kg=4.008kg
Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Sol. (i) 20.03=2×10+0×1+0×101+3×1001
(ii) 2.03=2×1+0×101+3×1001
(iii) 200.03=2×100+0×10+0×1+0×101+3×1001
(iv) 2.034=2×1+0×101+3×1001+4×10001
Write the place value of 2 in the following decimal numbers:
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352
Sol. (i) 2.56 - Ones
(ii) 21.37 - Tens
(iii) 10.25 - Tenths
(iv) 9.42 - Hundredths
(v) 63.352 - Thousandths
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
Sol. Distance travelled by Dinesh =AB+BC=(7.5+12.7)km7.5+12.7=20.2
Therefore, Dinesh travelled 20.2 km .
Distance travelled by Ayub =AD+DC=(9.3+11.8)km9.3+11.8=21.1
Therefore, Ayub travelled 21.1 km .
Hence, Ayub travelled more distance.
Difference =(21.1−20.2)km21.1−20.2=0.9
Therefore, Ayub travelled 0.9 km more than Dinesh.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Sol. Total fruits bought by Shyama =5kg300g+3kg250g=8kg550g=(8+1000550)kg=8.550kg
Total fruits bought by Sarala =4kg800g+4kg150g=8kg950g=(8+1000950)kg=8.950kg∴ Sarala bought more fruits
How much less is 28 km than 42.6 km ?
42.6
Sol. 14.6−28.0
Therefore, 28 km is 14.6 km less than 42.6 km.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm .
Sol. Length =5.7cm
Breadth =3cm
Area = Length × Breadth =5.7×3=17.1cm2
Find:
(i) 1.3×10
(ii) 36.8×10
(iii) 153.7×10
(iv) 168.07×10
(v) 31.1×100
(vi) 156.1×100
(vii) 3.62×100
(viii) 43.07×100
(ix) 0.5×10
(x) 0.08×10
(xi) 0.9×100
(xii) 0.03×1000
Sol. We know that when a decimal number is multiplied by 10,100,1000, the decimal point in the product is shifted to the right by as many places as there are zeroes. Therefore, these products can be calculated as
(i) 1.3×10=13
(ii) 36.8×10=368
(iii) 153.7×10=1537
(iv) 168.07×10=1680.7
(v) 31.1×100=3110
(vi) 156.1×100=15610
(vii) 3.62×100=362
(viii) 43.07×100=4307
(ix) 0.5×10=5
(x) 0.08×10=0.8
(xi) 0.9×100=90
(xii) 0.03×1000=30
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Sol. Distance covered in 1 litre of petrol =55.3 km
Distance covered in 10 litre of petrol = 10 ×55.3=553km Therefore, it will cover 553 km distance in 10 litre petrol
Find:
(i) 4.8÷10
(ii) 52.5÷10
(iii) 0.7÷10
(iv) 33.1÷10
(v) 272.23÷10
(vi) 0.56÷10
(vii) 3.97÷10
Sol. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10 , 100,1000 , etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 10 , the decimal will shift to the left by 1 place.
(i) 4.8÷10=0.48
(ii) 52.5÷10=5.25
(iii) 0.7÷10=0.07
(iv) 33.1÷10=3.31
(v) 272.23÷10=27.223
(vi) 0.56÷10=0.056
(vii) 3.97÷10=0.397
Find:
(i) 2.7÷100
(ii) 0.3÷100
(iii) 0.78÷100
(iv) 432.6÷100
(v) 23.6÷100
(vi) 98.53÷100
Sol. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10 , 100,1000 , etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 100, the decimal will shift to the left by 2 places.
(i) 2.7÷100=0.027
(ii) 0.3÷100=0.003
(iii) 0.78÷100=0.0078
(iv) 432.6÷100=4.326
(v) 23.6÷100=0.236
(vi) 98.53÷100=0.9853
Find:
(i) 7.9÷1000
(ii) 26.3÷1000
(iii) 38.53÷1000
(iv) 128.9÷1000
(v) 0.5÷1000
Sol. We know that when a decimal number is divided by a multiple of 10 only (i.e., 10 , 100,1000 , etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 1000 , the decimal will shift to the left by 3 places.
(i) 7.9÷1000=0.0079
(ii) 26.3÷1000=0.0263
(iii) 38.53÷1000=0.03853
(iv) 128.9÷1000=0.1289
(v) 0.5÷1000=0.0005
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Sol. Distance covered in 2.4 litres of petrol = 43.2 km
∴ Distance covered in 1 litre of petrol =43.2÷2.4=10432÷1024=10432×2410=18
Therefore, the vehicle will cover 18 km in 1 litre petrol.
5.0Benefits of Studying this Chapter
Studying NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals offers several benefits:
Concept Clarity: It helps students understand complex topics like multiplication and division of fractions and decimals with step-by-step solutions.
Exam Preparation: Practicing these solutions boosts problem-solving skills, improving accuracy and speed in exams.
Strong Foundation: It strengthens the mathematical foundation, making it easier to tackle advanced topics in higher classes.
NCERT Solutions for Class 7 Maths Other Chapters:-
Yes, ALLEN provides the NCERT Solutions for Class 7 Maths Chapter 2 in PDF format, which students can download. The PDFs created by the experts are of high quality, and the free download links allow students to access them.
The topics covered in the NCERT Solutions for Class 7 Maths Chapter 2 are: Multiplication of Fractions Division of Fractions Multiplication of Decimal Numbers Division of Decimal Numbers
Chapter 2, Fractions and Decimals, in Class 7 Maths can be challenging for some students, but it is generally manageable with regular practice. The concepts build on basic arithmetic, and understanding them clearly is key to mastering the topics. Using NCERT Solutions for Class 7 Maths Chapter 2 helps simplify the learning process.