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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

Edited By Ramraj Saini | Updated on Feb 07, 2024 04:53 PM IST

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals are discussed here. These NCERT solutions are created by the expert team at Careers360 keeping in mind the latest CBSE syllabus 2023. These solutions are simple and comprehensive and cover step by step solutions to each problem. In the solutions of NCERT Class 7 chapter 2 Fractions and Decimals, we will learn questions related to the multiplication and division of fractions and decimals. The study of fractions includes mixed fractions, proper fraction and improper fraction with their addition and subtraction, equivalent fractions, comparison of fractions, ordering of fractions and representation of fractions on a number line.

This Story also Contains
  1. NCERT Solutions for Class 7 Maths Chapter 2 - Important Formulae
  2. NCERT Solutions for Class 7 Maths Chapter 2 - Important Points
  3. NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals
  4. NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (Intext Questions and Exercise)
  5. Fractions and Decimals Class 7 Maths Chapter 2-Topics
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Solutions are the detailed explanation of each and every question of NCERT textbook and this helps to score good marks in the class exams. Some questions related to real-life situations are also explained in the NCERT Solutions for Class 7 . Here you will get solutions to all five exercises of NCERT. Before referring to the solutions, students must complete the NCER Class 7 Maths Syllabus.

Also, read - NCERT Books

NCERT Solutions for Class 7 Maths Chapter 2 - Important Formulae

fraction = p/q, Where p is the numerator and q is the denominator.

Mixed fraction: = p + q/r, Where p is the whole number, q is the numerator and r is the denominator.

Improper fraction to mixed fraction: (p/q) = (p ÷ q) + ( remaining fraction: p % q)/q.

Mixed Fraction to improper fraction: ( p + q/r ) = (( p×r)+ q)/c

(p/q)(a/b) = (pa)/(qb)

(p/q)/(a/b) = (p/q)(b/a)

p/q + a/b = (pb + aq)/(qb)

p/q - a/b = (pb - aq)/qb

NCERT Solutions for Class 7 Maths Chapter 2 - Important Points

Important points for maths class 7 chapter 2 are listed below.

Fractions:

  • A fraction is written as p/q, where p is the numerator and q is the denominator.
  • There are three main types of fractions: proper, improper, and mixed.
  • Proper fractions have the numerator (p) smaller than the denominator (q).
  • Improper fractions have the numerator (p) greater than or equal to the denominator (q).
  • Mixed fractions consist of a whole number and a proper fraction.

Conversions:

  • Improper fractions can be converted to mixed fractions.
  • To convert an improper fraction to a mixed fraction, divide the numerator by the denominator.
  • Mixed fractions can be converted to improper fractions.
  • To convert a mixed fraction to an improper fraction, multiply the whole number by the denominator and add the numerator.

Operations:

  • Multiplication of fractions involves multiplying the numerators and denominators.
  • Division of fractions requires finding the reciprocal of the second fraction and then multiplying.
  • Reciprocal of a fraction is the fraction flipped, swapping the numerator and denominator.

Addition and Subtraction:

  • Adding or subtracting fractions with the same denominator is straightforward.
  • When denominators differ, find the least common multiple (LCM) and then proceed.

Decimals:

  • Decimals represent fractions with denominators of powers of 10.
  • Decimal place value determines the value of each digit in a decimal number.
  • Tenth place, hundredth place, thousandth place, etc., indicate positions after the decimal point.

Operations with Decimals:

  • Multiplying decimals involves ignoring the decimal points, multiplying the numbers, and then placing the decimal point in the product.
  • Dividing decimals requires moving the decimal point in the divisor and dividend to make the divisor a whole number. Then divide as usual.

Comparing Decimals:

  • Start comparing decimals from the left and move towards the right.
  • If digits match up to a certain place value, compare the next digit to determine which decimal is greater.

Free download NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals PDF for CBSE Exam.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (Intext Questions and Exercise)

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.3.1

1.(a) Find:

(a)27×3

If the product is an improper fraction express it as a mixed fraction.

Answer: Given,

27×3=2×37=67

The product is a proper fraction.

1.(b) Find:

(b)97×6

If the product is an improper fraction express it as a mixed fraction.

Answer:

Given the product

97×6

=9×67

=547

This is an improper fraction, so convert it into a mixed fraction.

547=49+57=497+57=7+57=757 .

1.(c) Find:

(c)3×18

If the product is an improper fraction express it as a mixed fraction.

Answer: Given, the product

3×18

=3×18

=38

This is a proper fraction.

1.(d) Find:

(d)1311×6

If the product is an improper fraction express it as a mixed fraction.

Answer: Given the product:

1311×6

=13×611

=7811

This is an improper fraction, so we convert this into a mixed fraction. that is

7811=77+111=7711+111=7+111=7111 .

1.(i) Find:

(i)5×237

Answer: 5×237=5×7×2+37

5×237=5×177

5×237=5×177

5×237=857

5×237=84+17

5×237=847+17

5×237=12+17

5×237=1217

1.(ii) Find :

(ii)149×6

Answer: 149×6=9×1+49×6

149×6=139×6

149×6=13×69

149×6=789

149×6=72+69

149×6=729+69

149×6=8+23

149×6=823 .

1. Can you tell, what is (i)12 of 10? , (ii)14 of 16? , (iii)25 of 25?

Answer: As we know, of means multiply so,

12of10=12×10=5

And

14of16=14×16=4

And

25of25=25×25=10

1. Find: 13×45;23×15

Answer: As we know in the multiplication of a fraction, the numerator of one fraction is multiplied by the numerator of other fraction, and same for denominator, hence,

13×45=4×15×3=415

23×15=2×13×5=215.

1. Find : 83×47;34×23

Answer: As we know in the multiplication of a fraction, the numerator of one fraction is multiplied by the numerator of other fraction, and the same for the denominator, hence,

83×47=8×43×7=322734×23=3×24×3=612=12

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.3.2

1. Fill in these boxes:

(i)12×17=1×12×7= (ii)15×17=

(iii)17×12= (iv)17×15=

Answer: (i)12×17=1×12×7=114

(ii)15×17=135

(iii)17×12=114

(iv)17×15=135

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.4.1

1. Find :

(i)7÷25 (ii)6÷47 (iii)2÷89

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a. So,

(i)7÷25

7÷25=7×25=145

(ii)6÷47

6÷47=6×47=247

(iii)2÷89

2÷89=2×89=169

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.4.2

2. Find:

(i)6÷513 (ii)7÷247

Answer: As we know, The division of a number by a / b is equivalent to the multiplication of that number with b / a. So,

(i)6÷513

6÷513=6÷163=6×316=1816=98=118

(ii)7÷247

7÷247=7÷187=7×718=4918=21318

NCERT Solutions for chapter 2 maths class 7 Fractions And Decimals Topic 2.4.3

3. Find:

(i)35÷12 (ii)12÷35 (iii)212÷35 (iv)516÷92

Answer: As we know, The division of a number by a / b is equivalent to the multiplication of that number with b / a. So,

(i)35÷12

35÷12=35×21=65=115

(ii)12÷35

12÷35=12×53=56

(iii)212÷35

212÷35=52÷35=52×53=256=416

(iv)516÷92

516÷92=316÷92=316×29=6254=1854

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.6

1. Find:

(i)2.7×4 (ii)1.8×1.2 (iii)2.3×4.35

Answer: As we know, The multiplication of the decimal number is just like the multiplication of normal numbers, we just have to put decimal before a digit such that the number of digits after decimal should remain the same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i)2.7×4=10.8

(ii)1.8×1.2=2.16

(iii)2.3×4.35=10.005

2. Arrange the products obtained in descending order.

(i)2.7×4 (ii)1.8×1.2 (iii)2.3×4.35

Answer:(i)2.7×4=10.8

(ii)1.8×1.2=2.16

(iii)2.3×4.35=10.005

The products in descending order are:

10.8>10.005>2.16

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.6.1

1. Find:

(i)0.3×10 (ii)1.2×100 (iii)56.3×1000

Answer: As we know, The multiplication of the decimal number is just like the multiplication of the normal numbers, we just have to put decimal before a digit such that the number of digits after decimal should remain the same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i)0.3×10=3

(ii)1.2×100=120

(iii)56.3×1000=56300

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.7.1

1. Find:

(i) 235.4 ÷ 10 (ii) 235.4 ÷100 (iii) 235.4 ÷ 1000

Answer: As we know, while dividing a number by 10, 100 or 1000, the digits of the number and the quotient are same but the decimal point in the quotient shifts to the left by as many places as there are zeros over 1.

So,

(i) 235.4 ÷ 10 = 23.54

(ii) 235.4 ÷100 = 2.354

(iii) 235.4 ÷ 1000 = 0.2354

NCERT Solutions for Class 7 Chapter 2 Fractions And Decimals Topic 2.7.2

2. ( i) 35.7 ÷ 3 = ?;
(ii) 25.5 ÷ 3 = ?

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 35.7 ÷ 3

35.7÷3=35710÷3=35710×13=12910=12.9

(ii) 25.5 ÷ 3

25.5÷3=25510÷3=25510×13=8510=8.5

3. (i) 43.15 ÷ 5 = ?;
(ii) 82.44 ÷ 6 = ?

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 43.15 ÷ 5

43.15÷5=4315100÷5=4315100×15=863100=8.63

(ii) 82.44 ÷ 6

82.44÷6=8244100÷6=8244100×16=2374100=23.74

4. Find:

(i) 15.5 ÷ 5

(ii) 126.35 ÷ 7

Answer: As we know, dividing by a number is equivalent to multiplying by the reciprocal of that number. So

(i) 15.5 ÷ 5

15.5÷5=15510÷5=15510×15=3110=3.1

(ii) 126.35 ÷ 7

126.35÷7=12635100÷7=12635100×17=1805100=18.05

NCERT Solutions for C lass 7 Chapter 2 Fractions And Decimals Topic 2.7.3

1. Find:

(i)7.750.25 (ii)42.80.02 (iii)5.61.4

Answer: As we know, in the dividing of decimal, we first express the decimal in term of fraction and then divides it, So

(i)7.750.25

7.750.25=77510025100=77525×100100=311=31

(ii)42.80.02

42.80.02=4281002100=4282×100100=214

(iii)5.61.4

5.61.4=56101410=5614×1010=4

NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Exercise 2.1

1. Solve:

(i) 235 (ii) 4+78 (iii) 35+27 (iv) 911415

(v) 710+25+32 (vi) 223+312 (vii) 812358

Answer: As we know we have to make the denominator the same in order to add or subtract the fractions. So,

(i) 235=21×5535=10535=1035=75

(ii)

4+78=41×88+78=328+78=398

(iii)

35+27=35×77+27×55=2135+1035=3135

(iv)

911415=15×911×411×15=13544165=91165


(v)

710+25+32=710+25×22+32×55=710+410+1510=7+4+1510=2610

(vi)

223+312=83+72=166+216=16+216=376=616

(vii)

812358=172298=688298=68298=399=413

2. Arrange the following in descending order:

(i) 29,23,821 (ii) 15,37,710

Answer: (i)

29=23×3×77=143×3×7

23=23×77×33=423×3×7

821=83×7×33=243×3×7

As 14 < 24 < 42

23>821>29

(ii) 15=15×1414=1470

37=37×1010=3070

710=710×77=4970

As 14 < 30 < 49

710>37>15

3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

(Along the first row 411+911+211=1511 ).

Answer: As, we can see that the sum of every row, column or diagonal is 15 / 11 . so yes this is a magic square.

4. A rectangular sheet of paper is 1212 cm long and 1023 cm wide. Find its perimeter.

Answer: Given,

Length of the rectangle :

l=1212cm=12×2+12=252cm

Width of the rectangle :

b=1023cm=10×3+23=323cm

Now, As we know,

Perimeter of the rectangle = 2 x ( length + width )

So,

The perimeter of the given rectangle :

=2×(l+b)

=2×(252+323)

Now let's make the denominator of both fractions equal.

=2×(252×33+323×22)

=2×(756+646)

=2×(1396)

=1393cm

Hence, the perimeter of the rectangle is 1393cm .

5. Find the perimeters of (i)ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

xyz7

Answer: The perimeter of ABE = AB + BE + AE

=52+234+335

=52+4×2+34+3×5+35

=52+114+185

Noe, The LCM of 2,4 and 5 is 20. So let's make the denominator of all fractions, 20.

So,

The perimeter of ABE :

=52×1010+114×55+185×44

=5020+5520+7220

=50+55+7220

=17720cm

Now,

The perimeter of rectangle BCDE = 2 x ( BE + ED )

=2×(234+76)

=2×(4×2+34+76)

=2×(114+76)

The LCM of 4 and 6 is 12. So let's make the denominator of both fractions equal to 12.

=2×(114×33+76×22)

=2×(3312+1412)

=2×(33+1412)

=2×(4712)

=476cm

Hence The perimeter of the Triangle is 177/20cm and the perimeter of the rectangle is 47/6cm .

Now, we have

17720and476

LCM of 20 and 6 is 60, so let's make the denominator of both fractions equal to 60.

So,

17720=17720×33=53160

And

476=476×1010=47060

Now, Since 531 > 470

17720>476.

Area of Triangle > Area of Rectangle.

5. Salil wants to put a picture in a frame. The picture is 735 cm wide. To fit in the frame the picture cannot be more than 7310 cm wide. How much should be trimmed?

Answer: Given, the width of the picture = 735 cm.

The maximum width of the picture which can fit in the frame = 7310 cm.

Hence the length Salil should trim :

7357310

7×5+3510×7+310

3857310

Now LCM of 5 and 10 is 10. So, let's make the denominator of both fractions equal to 10. So,

385×227310×11

76107310

767310

310

Hence Salil should cut 3/10 cm of the picture in order to fit it in the frame.

7. Ritu ate 35 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?

Answer: Given,

Part of Apple eaten by Ritu = 35 .

Part of Apple eaten by Somu:

135

1135

LCM of 1 and 5 is 5. So making the denominator of both fractions equal to 5, we get

11×5535

5535

535

25

Hence, The Part of Apple eaten by Somu is 2/5.

Now, As

35>25

3525=325=15

Hence, Ritu had a larger share of the apple by 15 part.

8. Michael finished colouring a picture in 712 hour. Vaibhav finished colouring the same picture in 34 hour. Who worked longer? By what fraction was it longer?

Answer: Time taken by Michal in colouring = 712 hours.

Time taken by Vaibhav in colouring = 34 hours.

The LCM of 12 and 4 is 12. So making the denominator of both fractions equal to 12, we get,

712and34×33

712and912

As

912>712

For calculating by how much, we do Subtraction

912712=9712=212=16

Vaibhav worked longer by the fraction 16 .

NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Exercise 2.2

1. Which of the drawings (a) to (d) show :

(i)2×15 (ii)2×12 (iii)3×23 (iv)3×14

xyz5

Answer: i)

(d) represents two circles with 1 part shaded out of 5 parts So, it represents

2×15 .

ii) b represents two squares one part out of two of both squares are shaded, So it represents

2×12

(iii)

(a) represents 3 circles 2 parts out of three of all circles are shaded. So it represents

3×23

(iv)

(c) represents 3 squares with one part out of four-part shaded in each square hence it represents

3×14 .

2. Some pictures (a) to (c) are given below. Tell which of them show:

(i)3×15=35 (ii)2×13=23 (iii)3×34=214

xyz3


Answer:(i)3×15=35

As in option (c), In the Left-hand side, there are three figures in which one part out of three-part is shaded and in the right-hand side, three out of five portions are shaded.

Hence this represents

3×15=35 .

(ii)2×13=23

Option (a) represents the equation pictorially.

(iii)3×34=214

Option (b) represents this equation pictorially.

3. Multiply and reduce to lowest form and convert into a mixed fraction:

(i)7×35 (ii)4×13 (iii)2×67 (iv)5×29 (v)23×4

(vi)52×6 (vii)11×47 (viii)20×45 (ix)13×13 (x)15×35

Answer: (i)7×35

On Multiplying, we get

7×35=215=20+15=205+15=4+15=415

(ii)4×13

On Multiplying, we get

4×13=43=3+13=33+13=1+13=113

(iii)2×67

On Multiplying, we get

2×67=127=7+57=77+57=1+57=157

(iv)5×29

On Multiplying, we get

5×29=109=9+19=99+19=1+19=119

(v)23×4

On Multiplying, we get

2×43=83=6+23=63+23=2+23=223

(vi)52×6

On Multiplying, we get

5×62=302=15

(vii)11×47

On Multiplying, we get

11×47=447

Converting this into a mixed fraction, we get

447=42+27=427+27=6+27=627.

(viii)20×45

On Multiplying, we get

20×45=805=16

(ix)13×13

On multiplying, we get

13×13=133

Converting this into a mixed fraction,

133=12+13=123+13=4+13=413 .

(x)15×35

On multiplying, we get,

15×35=455=9 .

4. Shade:

(i)12 of the circles in box (a) (ii)23 of the triangles in box (b)

(iii)35 of the squares in box (c).

xyz2

Answer: 1) In figure a there are 12 circles: half of 12 = 6

2) In figure b there are 9 triangles: 2/3 of 9 = 6

3) In figure c there are 15 triangles: 3/5 of 15 = 9

1643865076786

5. Find:

(a)12of(i)24(ii)46 (b)23of(i)18(ii)27

(c)34of(i)16(ii)36 (d)45of(i)20(ii)35

Answer: (a)(i)12of24

On Multiplying we get,

12of24=12×24=1×242=12

(a)ii)12of46

On multiplying, we get

12of46=12×46=1×462=23.

(b)(i)23of18

On Multiplying, we get

23of18=23×18=2×183=363=12.

(b)(ii)23of27

On multiplying, we get

23of27=23×27=2×273=543=18

(c)(i)34of16

On multiplying, we get

34of16=34×16=3×164=484=12.

(c)(ii)34of36

34of36=34×36=3×364=1084=27

(d)(i)45of20

On multiplying, we get

45of20=45×20=4×205=805=16

(d)(ii)45of35

On Multiplying, we get,

45of35=45×35=4×355=1404=28

6. Multiply and express as a mixed fraction :

(a)3×515 (b)5×634 (c)7×214

(d)4×613 (e)314×6 (f)325×8

Answer:(a)3×515

On Multiplying, we get

3×5×5+15=3×265=3×265=785

Converting This into Mixed Fraction,

785=75+35=755+35=15+35=1535

(b)5×634

On Multiplying, we get

5×6×4+34=5×274=5×274=1354

Converting This into Mixed Fraction,

1354=132+34=1324+34=33+34=3334

(c)7×214

On multiplying, we get

7×214=7×4×2+14=7×94=7×94=634

Converting it into a mixed fraction, we get

634=60+34=604+34=15+34=1534

(d)4×613

On multiplying, we get

4×613=4×3×6+13=4×193=4×193=763

Converting it into a mixed fraction,

763=75+13=753+13=25+13=2513

(e)314×6

Multiplying them, we get

314×6=4×3+14×6=134×6=13×64=784

Now, converting the result fraction we got to mixed fraction,

784=76+24=764+24=19+12=1912

(f)325×8

On multiplying, we get

325×8=5×3+25×8=175×8=17×85=1365

Converting this into a mixed fraction, we get

1365=135+15=1355+15=27+15=2715


7. Find:

(a)12of(i)234(ii)429 (b)58of(i)356(ii)923

Answer: (a)(i)12of234

As we know that of is equivalent to multiply,

12of234=12×234=12×114=118=138

(a)(ii)12of429

As we know that of is equivalent to multiply,

12of429=12×429=12×389=3818=199=219

(b)(i)58of356

As we know that of is equivalent to multiplication, so

58of356=58×356=58×236=11548=21948

(b)(ii)58of923

As we know that of is equivalent to multiplication, so

58of923=58×293=14524=6124

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 25 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?

Answer: Given

Total water = 5 litre.

i) The amount of water vidya consumed :

=25of5liter=25×5=2liter

Hence vidya consumed 2 liters of water from the bottle.

ii) The amount of water Pratap consumed :

=(125)of5liter=35×5=3liter

Hence, Pratap consumed 3 liters of water from the bottle.


NCERT S olutions for fractions and decimals class 7 Exercise 2.3

1. Find:

(i)14of(a)14(b)35(c)43

(ii)17of(a)29(b)65(c)310

Answer: As we know, the term "of " means multiplication. So,

(i)(a)14of14=14×14=116

(i)(b)14of35=14×35=320

(i)(c)14of43=14×43=412=13

(ii)(a)17of29=17×29=263

(ii)(b)17of65=17×65=635

(ii)(c)17of310=17×310=370

2. Multiply and reduce to lowest form (if possible) :

(i)23×223(ii)27×79(iii)38×64

(iv)95×35(v)13×158(vi)112×310(vii)45×127

Answer: As we know, in the multiplication of fraction, the numerator gets multiplied with numerator and denominator gets multiplied by the denominator. So,

(i)23×223=23×83=2×83×3=169=179(ii)27×79=2×77×9=29(iii)38×64=3×68×4=1832=916

(iv)95×35=9×35×5=2725=1225(v)13×158=1×153×8=1524=58(vi)112×310=11×32×10=3320=11320(vii)45×127=4×125×7=4835=11335 .

3. Multiply the following fractions:

(i)25×514(ii)625×79(iii)32×513

(iv)56×237(v)325×47(vi)235×3(vii)347×35

Answer: As we know in the multiplication of fractions, the numerator is multiplied with numerator and denominator is multiplied by the denominator.

So,

(i)25×514=25×214=2×215×4=4220=2110=2110(ii)625×79=325×79=32×75×9=22445=44445(iii)32×513=32×163=3×162×3=486=8

(iv)56×237=56×177=5×176×7=8542=2142(v)325×47=175×47=17×45×7=6835=13335(vi)235×3=135×3=13×35=395=745(vii)347×35=257×35=25×37×5=7535=157=217

4. Which is greater:

(i)27of34or35of58

(ii)12of67or23of37


Answer: (i)27of34or35of58

27× 34or35×58

2×37×4or3×55×8

314or38

Now, As we Know, When the numerator of two fractions is the same the fraction with lesser denominator is the bigger fraction. So,

314<38

Thus,

27of34<35of58 .

(ii)12of67or23of37

12×67or23×37

1×62×7or2×33×7

37or27

As we know, When the denominator of two fractions are the same, the fraction with the bigger numerator is the bigger fraction, so,

37>27

Thus,

12×67>23×37 .

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 34m . Find the distance between the first and the last sapling.

Answer: distance between two adjacent saplings = 34m

distance between the first and the last sapling :

=3×34

=3×34

=94 .

6. Lipika reads a book for 134 hours everyday. She reads the entire book in 6 days.How many hours in all were required by her to read the book?

Answer: Number of time spent in one day = 134 hour

The number of time spent in 6 days :

=6×134=6×74=6×74=424=212=1012hours

Hence 1012hours are required by Lipika to complete the book.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 234 litres of petrol.

Answer: Distance covered in 1 liter petrol = 16 km

The distance that will be covered in 234 liter petrol.

=16×234

=16×114

=11×164

=11×4

=44km

Hence 44 km can be covered using 234 liter petrol.

8. (a) (i) Provide the number in the box xyz8 , such that 23× xyz8 = 1030 .

(ii) The simplest form of the number obtained in xyz8 is _________.

(b) (i) Provide the number in the box xyz8 , such that 35× xyz8 = 2475 .

(ii) The simplest form of the number obtained in xyz8 is _________.

Answer: As we Know, That in the multiplication of fraction, the numerator of both fraction are multiplied to give numerator of new fraction, and denominator of both fractions is multiplied to give the denominator of answer fraction. So,

23× xyz8 = 1030 .

We have to multiply 5 with 2 in the numerator to get numerator equal to 10 and 10 with 3 in the denominator to get 30. So,

23×510=1030

The Simplest Form :

510=12 .

NCERT Solutions for fractions and decimals class 7 Exercise 2.4

1. Find:

(i)12÷34 (ii)14÷56 (iii)8÷73 (iv)4÷83 (v)3÷213 (vi)5÷347

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a. So,

(i)12÷34

12÷34=12×43=483=16

(ii)14÷56

14÷56=14×65=845=1645

(iii)8÷73

8÷73=8×37=247=337

(iv)4÷83

4÷83=4×38=128=32=112

(v)3÷213

3÷213=3÷73=3×37=97=127

(vi)5÷347

5÷347=5÷257=5×725=3525=75=125


2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i)37 (ii)58 (iii)97 (iv)65 (v)127 (vi)18 (vii)111

Answer: As we know, in the reciprocal of any fraction, the numerator and denominator get exchanged. Basically, we flip the number upside down. So

i)Reciprocalof37=73

As numerator is greater than the denominator, it is an improper fraction.

ii)Reciprocalof58=85

As numerator is greater than the denominator, it is an improper fraction.

iii)Reciprocalof97=79

As the Denominator is greater than Numerator, it is a proper fraction.

iv)Reciprocalof65=56

As the Denominator is greater than Numerator, it is a proper fraction.

v)Reciprocalof127=712

As the Denominator is greater than Numerator, it is a proper fraction.

vi)Reciprocalof18=81=8

It is an integer and hence a whole Number.

vii)Reciprocalof111=111=11

It is an integer and hence a whole Number.


3. Find:

(i)73÷2 (ii)49÷5 (iii)613÷7 (iv)413÷3

(v)312÷4 (vi)437÷7


Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a which is also called the reciprocal of a / b. So,

(i)73÷2

73÷2=73×12=76=116

(ii)49÷5

49÷5=49×15=445

(iii)613÷7

613÷7=613×17=691

(iv)413÷3

413÷3=133÷3=133×13=139=149

(v)312÷4

312÷4=72÷4=72×14=78

(vi)437÷7

437÷7=317÷7=317×17=3149


4. Find:

(i)25÷12 (ii)49÷23 (iii)37÷87 (iv)213÷35 (v)312÷83 (vi)25÷112 (vii)315÷123 (viii)215÷115

Answer: As we know, The division of a number by a / b is equivalent to multiplication of that number with b / a which is also called the reciprocal of a / b. So,

(i)25÷12

25÷12=25×21=45

(ii)49÷23

49÷23=49×32=1218=23

(iii)37÷87

37÷87=37×78=38

(iv)213÷35

213÷35=73÷35=73×53=359=389

(v)312÷83

312÷83=72÷83=72×38=2116=1516


(vi)25÷112

25÷112=25÷32=25×23=415

(vii)315÷123

315÷123=165÷53=165×35=4825=12325

(viii)215÷115

215÷115=115÷65=115×56=116=156

NCERT Solutions for fractions and decimals class 7 Exercise 2.5

1. 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.

Answer: As we know, Any decimal is equivalent to the number without decimal divided by 10 x (number of integers after the decimal). in other words,

0.5=510 and0.05=5100

So,

(i) 0.5 > 0.05

(ii) 0.7 > 0.5

(iii) 7 > 0.7

(iv) 1.37 < 1.49

(v) 2.03 < 2.30

(vi) 0.8 < 0.88.

2. Express as rupees using decimals :

(i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise
(iv) 50 paise (v) 235 paise.

Answer: As we know, there are 100 paise in 1 rupee. i.e.

1paise=1100Rupees

So,

i)7paise=7100=0.07Rupees

ii)7Rupees,7paise=7+7100=7+0.07=7.07Rupees

iii)77Rupees,77paise=77+77100=7+0.77=7.77Rupees

iv)50paise=50100=0.5Rupees

v)235paise=235100=200+35100=200100+35100=2+0.35=2.35Rupees


3.(i) Express 5 cm in metre and kilometre

(ii) Express 35 mm in cm, m and km

Answer: As we know,

1 centimeter = 10 millimeter

1 meter = 100 centimeter.

And

1 kilometer = 1000 meter

So,

i) 5 cm

5cm=5100m=0.05m

5cm=5100m=0.05m=0.0051000=0.000005km

ii) 35 mm

35mm=3510cm=3.5cm

35mm=3510cm=3.5cm=3.5100m=0.035m

35mm=3510cm=3.5cm=3.5100m=0.035m=0.0351000km=0.000035km

4. Express in kg:

(i) 200 g (ii) 3470 g (iii) 4 kg 8 g

Answer: As we know,

1 kg = 1000 g.

1g=11000kg

So,

i) 200 g

200g=2001000kg=0.2kg

(ii) 3470 g

3470g=34701000kg=3.47kg

(iii) 4 kg 8 g

4kg,8g=4kg+81000kg=4kg+0.008kg=4.0008kg

5. Write the following decimal numbers in the expanded form:
(i) 20.03 (ii) 2.03 (iii) 200.03 (iv) 2.034

Answer: Decimal in their expanded form are

(i) 20.03

20.03=2×10+0×1+110×0+1100×3

(ii) 2.03

2.03=2×1+110×0+1100×3

(iii) 200.03

200.03=2×100+0×10+0×1+110×0+1100×3

(iv) 2.034

2.034=2×1+110×0+1100×3+4×11000

6. 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.

Answer: (i) 2.56

2 is in one's position.

(ii) 21.37

2 is in ten's position

(iii) 10.25

2 is in one-tenths position

(iv) 9.42

2 is in one-hundredths position.

(v) 63.352.

2 is in one-thousandth's position.

7. 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?

xyz1

Answer: Total distance travelled by Dinesh = AB + BC

= 7.5km + 12.7km

= 20.2 km

Total distance travelled by Ayub = AD + DC

= 9.3km + 11.8 km

= 21.1 km

Hence Ayub travelled More distance than Ayub as 21.1 > 20.2

The difference between path travelled by them = 21.1km - 20.2 km

= 0.9 km.

Hence Ayub travelled 0.9 km more than Dinesh.

8. 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?

Answer: For comparing two quantities, we should make their unit the same first. So,

Fruits bought by Shyama in kg = 5 kg + 300 g + 3 kg + 250 g

= 8 kg + 550 g

= 8 kg + 0.55 kg

= 8.55 kg.

Fruits bought by Sarala in kg = 4 kg + 800 g + 4 kg + 150 g

= 8 kg + 950 kg

= 8 kg + 0.95 kg

= 8.95 kg.

Hence Sarala bought more fruits as 8.95 > 8.55

The difference between the amount of Fruits they bought = 8.95 kg - 8.55 kg

= 0.4 kg

9. How much less is 28 km than 42.6 km?

Answer: Difference = 42.6 km - 28 km

= 14.6 km

Hence 28 is 14.6 km less than 42.6.

NCERT Solutions for maths class 7 chapter 2 Fractions And Decimals Exercise 2.6

1. Find:

(i) 0.2 × 6 (ii) 8 × 4.6 (iii) 2.71 × 5 (iv) 20.1 × 4

(v) 0.05 × 7 (vi) 211.02 × 4 (vii) 2 × 0.86

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i) 0.2 × 6 = 1.2

(ii) 8 × 4.6 = 36.8

(iii) 2.71 × 5 = 13.55

(iv) 20.1 × 4 = 80.4

(v) 0.05 × 7 = 0.35

(vi) 211.02 × 4 = 844.08

(vii) 2 × 0.86 = 1.72

2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.

Answer: Given, Length of rectangle = 5.7 cm.

Width of rectangle = 3 cm

Area of the rectangle = Length x width

= 5.7 x 3

= 17.1 cm2 .

Hence Area of the rectangle is 17.1 cm2 .

3. 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

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(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

4. 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?

Answer: Distance covered by two-wheeler in 1 liter of petrol = 55.3 km.

Distance two-wheeler will cover in 10 liters of petrol = 10 x 55.3 km

= 553 km

Hence two-wheeler will cover a distance of 553 km in 10 liters of petrol.

5. Find:

(i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1 (v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02

(viii) 10.05 × 1.05 (ix) 101.01 × 0.01 (x) 100.01 × 1.1

Answer: As we know, The multiplication of the decimal number is just like multiplication of normal number, we just have to put decimal before a digit such that the number of digits after decimal should remain same.

In other words,

The number of digits after the decimal should be the same before and after the multiplication.

So.

(i) 2.5 × 0.3 = 0.75

(ii) 0.1 × 51.7 = 5.17

(iii) 0.2 × 316.8 = 63.36

(iv) 1.3 × 3.1 = 4.03

(v) 0.5 × 0.05 = 0.025

(vi) 11.2 × 0.15 = 1.680

(vii) 1.07 × 0.02 = 0.0214

(viii) 10.05 × 1.05 = 10.5525

(ix) 101.01 × 0.01 = 1.0101

(x) 100.01 × 1.1 = 110.11

NCERT Solutions for maths class 7 chapter 2 Fractions And Decimals 2.7

1. Find:

(i)0.4÷2 (ii)0.35÷5 (iii)2.48÷4 (iv)65.4÷6

(v)651.2÷4 (vi)14.49÷7 (vii)3.96÷4 (viii)0.80÷5

Answer: As we know, while dividing decimal, we first express the decimal in term of fraction and then divides it,

And

Dividing by a number is equivalent to multiplying by the reciprocal of that number. So

So

(i)0.4÷2

0.4÷2=410÷2=410×12=210=15

(ii)0.35÷5

0.35÷5=35100÷5=35100×15=7100=0.07

(iii)2.48÷4

2.48÷4=248100÷4=248100×14=62100=0.62

(iv)65.4÷6

65.4÷6=654100÷6=654100×16=109100=1.09

(v)651.2÷4

651.2÷4=651210÷4=651210×14=162810=162.8

(vi)14.49÷7

14.49÷7=1449100÷7=1449100×17=207100=2.07

(vii)3.96÷4

3.96÷4=396100÷4=396100×14=99100=0.99

(viii)0.80÷5

0.80÷5=80100÷5=80100×15=16100=0.16

2. 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

Answer: As we know, When we divide a decimal number by 10, the decimal point gets shifted by one digit in the left.

So

(i)4.8÷10

4.8÷10=0.48

(ii)52.5÷10

52.5÷10=5.25

(iii)0.7÷10

0.7÷10=0.07

(iv)33.1÷10

33.1÷10=3.31

(v)272.23÷10

272.23÷10=27.223

(vi)0.56÷10

0.56÷10=0.056

(vii)3.97÷10

3.97÷10=0.397

3. 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

Answer: As we know while dividing a decimal number by 100, the decimal point gets shifted to left by two digits.

So

(i)2.7÷100

2.7÷100=0.027

(ii)0.3÷100

0.3÷100=0.003

(iii)0.78÷100

0.78÷100=0.0078

(iv)432.6÷100

432.6÷100=4.326

(v)23.6÷100

23.6÷100=0.236

(vi)98.53÷100

98.53÷100=0.9853

4. Find:

(i)7.9÷1000 (ii)26.3÷1000 (iii)38.53÷1000 (iv)128.9÷1000

(v)0.5÷1000

Answer: As we know, while dividing a decimal number by 1000 we shift the decimal to left by 3 digits, So

(i)7.9÷1000

7.9÷1000=0.0079

(ii)26.3÷1000

26.3÷1000=0.0263

(iii)38.53÷1000

38.53÷1000=0.03853

(iv)128.9÷1000

128.9÷1000=0.1289

(v)0.5÷1000

5. Find:

(i)7÷3.5 (ii)36÷0.2 (iii)3.25÷0.5 (iv)30.94÷0.7

(v)0.5÷0.25 (vi)7.75÷0.25 (vii)76.5÷0.15 (viii)37.8÷1.4

(ix)2.73÷1.3

Answer: As we know, while dividing decimal, we first express the decimal in term of fraction and then divides it,

And

Dividing by a number is equivalent to multiplying by the reciprocal of that number.

So

(i)7÷3.5

7÷3.5=7÷3510=7×1035=7035=2

(ii)36÷0.2

36÷0.2=36÷210=36×102=180

(iii)3.25÷0.5

3.25÷0.5=325100÷510=325100×105=6510=6.5

(iv)30.94÷0.7

30.94÷0.7=3094100÷710=3094100×107=44210=44.2

(v)0.5÷0.25

0.5÷0.25=510÷25100=510×10025=105=2

(vi)7.75÷0.25

7.75÷0.25=775100÷25100=775100×10025=31

(vii)76.5÷0.15

76.5÷0.15=76510÷15100=76510×10015=510

(viii)37.8÷1.4

37.8÷1.4=37810÷1410=37810×1014=27

(ix)2.73÷1.3

2.73÷1.3=273100÷1310=273100×1013=2110=2.1

6. 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?

Answer: Distance travelled by vehicle in 2.4 litres of petrol = 43.2 km

Distance travelled by vehicle in 1 litre of petrol:

=43.22.4=43224×1010=18

Hence Distance travelled by vehicle in 1 litre is 18 km.

Fractions and Decimals Class 7 Maths Chapter 2-Topics

Here students can find all the topics which are discussed in this chapter, fractions and decimals class 7.

  • How Well Have You Learnt About Fractions?
  • Multiplication Of Fractions
  • Division Of Fractions
  • How Well Have You Learnt About Decimal Numbers
  • Multiplication Of Decimal Numbers
  • Division Of Decimal Numbers

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Integers

Chapter 2

Fractions and Decimals

Chapter 3

Data Handling

Chapter 4

Simple Equations

Chapter 5

Lines and Angles

Chapter 6

The Triangle and its Properties

Chapter 7

Congruence of Triangles

Chapter 8

Chapter 9

Rational Numbers

Chapter 10

Practical Geometry

Chapter 11

Perimeter and Area

Chapter 12

Algebraic Expressions

Chapter 13

Exponents and Powers

Chapter 14

Symmetry

Chapter 15

NCERT Solutions for Class 7 Subject Wise

Importance of NCERT Solutions for Class 7 Chapter 2 Fractions and Decimals

  • Now homework of the chapter becomes an easy breezy thing with CBSE NCERT solutions for Class 7 chapter 2 Fractions and Decimals in hand.
  • Solutions of NCERT Class 7 chapter 2 Fractions and Decimals in hand helps students in self-evaluation
  • The NCERT solutions for Class 7 Maths chapter 2 Fractions and Decimals are helpful in preparation for the exam. A similar type of question can be expected for the class exams.

Also Check NCERT Books and NCERT Syllabus here:

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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