RD Sharma 2020 2021 Solutions for Class 6 Maths Chapter 6 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among class 6 students for Maths Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2020 2021 Book of class 6 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2020 2021 Solutions. All RD Sharma 2020 2021 Solutions for class 6 Maths are prepared by experts and are 100% accurate.

Page No 6.11:

Question 1:

Convert each of the following into a mixed fraction:

(i) 289
(ii) 22615
(iii) 1459
(iv) 1285

Answer:

(i) 289Quotient=3, Remainder=1, Denominator=9289=319(ii) 22615Quotient=15, Remainder=1, Denominator=1522615=15115(iii) 1459Quotient=16, Remainder=1, Denominator=91459=1619(iv) 1285Quotient=25, Remainder=3, Denominator=51285=2535

Page No 6.11:

Question 2:

Convert each of the following into an improper fraction:

(i) 714
(ii) 857
(iii) 5310
(iv) 1257

Answer:

(i) 714=7×4+14=294(ii) 857=8×7+57=617(iii) 5310=5×10+310=5310(iv) 12715=12×15+715=18715
 



Page No 6.15:

Question 1:

Write the fractions and check whether they are equivalent or no

Answer:

(i)

Yes, they are equivalent.


(ii)

 
Yes, they are equivalent.

Page No 6.15:

Question 2:

Write the fractions and match fractions in Column I with the equivalent fractions in Column II.

Answer:



Page No 6.16:

Question 3:

Replace     in each of the following by the correct number:

(i) 27=6   
(ii) 58=10   
(iii) 45=   20
(iv) 4560=15   
(v) 1824=   4

Answer:

(i)
27=6As 2×3=6, we will multiply both the numerator & denominator by 3.27×33=621

(ii)
58=10As 5×2=10, we will multiply both the numerator & denominator by 2. 58×22=1016

(iii)
45=20As 5×4=20, we will multiply both the numerator & denominator by 4.45×44=1620

(iv)
4560=15As 45÷3=15, we will multiply both the numerator & denominator by 3.453603=1520

(v)
1824=4As 24÷6=4, we will multiply both the numerator & denominator by 6.186246=34

Page No 6.16:

Question 4:

Find the equivalent fraction of 35, having:
(i) numerator 9
(ii) denominator 30
(iii) numerator 21
(iv) denominator 40

Answer:

(i)
35=9Consider the numerators.As 3×3=9, we will multiply both the numerator & denominator by 3.  35×33=915
(ii)
35=30Consider the denominators. As 5×6=30, we will multiply both the numerator & denominator by 6.35×66=1830

(iii)
35=21Consider the numerators.As 3×7=21, we will multiply both the numerator & denominator by 7.35×77=2135

(iv)
35=40Consider the denominators.As 5×8=40, we will multiply both the numerator & denominator by 8.35×88=2440

Page No 6.16:

Question 5:

Find fraction equivalent of 4560, having:

(i) numerator 15
(ii) denominator 4
(iii) denominator 240
(iv) numerator 135

Answer:

(i)
4560=15DenominatorWe will consider the numerators.As 45÷3=15, we will divide both the numerator & denominator by 3.453603=1520

(ii)
4560=Numerator4We will consider the denominators.As 60÷15=4, we will multiply both the numerator & denominator by 15.45156015=34

(iii)
4560=Numerator240We will consider the denominators.As 60×4=240, we will multiply both the numerator & denominator by 4.4560×44=180240

(iv)
4560=135DenominatorWe will consider the numerators.As 45×3=135, we will multiply both the numerator & denominator by 3.4560×33=135180

Page No 6.16:

Question 6:

Find the fraction equivalent of 3542, having:

(i) numerator 15
(ii) denominator 18
(iii) denominator 30
(iv) numerator 30

Answer:

Firstly, we will reduce 3542 into the lowest term.Now, we will divide both the numerator & denominator by the HCFs of 35 & 42.35÷742÷7=56
(i) 56=15DenominatorWe will consider the numerators. As 5×3=15, we will multiply both the numerator & denominator by 3.5×36×3=1518

(ii)56=Numerator18We will consider the denominators. As 6×3=18, we will multiply both the numerator & denominator by 3.       5×36×3=1518    


(iii) 56=numerator30We will consider the denominators ,Since,6×5=30,therefore multiplying both the numerator & denominator by 5     5×56×5   =2530


(iv) 56=30DenominatorWe will consider the numerators. As 5×6=30, we will multiply both the numerator & denominator by 6. 5×66×6= 3036

Page No 6.16:

Question 7:

Check whether the given fraction are equivalent:

(i) 59,3054
(ii) 27,1642
(iii) 713,511
(iv) 411,3288
(v) 310,1250
(vi) 927,2575.

Answer:

(i) 
59×66=3054 
So, the given fractions are equivalent.

(ii) 
27×88=1656
27  is not equivalent to 1642.

(iii)
 713×55=3565  
  511×77=3577
713 is not equivalent to 511.

(iv)
411×88=3288411 is equivalent to 3288.

(v)
310×44=1240  310is not equivalent to 1250.

(vi)

927=132575=13927 is equivalent to 2575.



Page No 6.17:

Question 8:

Match the equivalent fractions and write another 2 form each:

(i) 250400          (a) 23
(ii) 180200          (b) 25
(iii) 660990          (c) 12
(iv) 180360           (d) 58
(v) 220550          (e) 910

Answer:

(i)250400Dividing both the numerator & denominator by the HCFs of 250 & 400, we get:=2505040050=58
(ii) 180200Dividing both the numerator & denominator by the HCFs of 180 & 200, we get:=1802020020=910
(iii) 660990Dividing both the numerator & denominator by the HCFs of 660 & 990, we get:=6603099030=22113311=23

(iv)180360Dividing both the numerator & denominator by the HCFs of 180 & 360, we get:=180180360180=12

(v) 220550Dividing both the numerator & denominator by the HCFs of 220 & 550, we get:=2201155011=2050=20105010=25

(i) - (d)
(ii) - (e)
(iii) - (a)
(iv) -(c)
(v)- (b)

Page No 6.17:

Question 9:

Write some equivalent fractions which contain all digits from 1 to 9 once only.

Answer:

24=36=7915839=26=58174

Page No 6.17:

Question 10:

Ravish had 20 pencils, Sikha had 50 pencils and Priya had 80 pencils. After 4 months, Ravish used up 10 pencils. Shikha used up 25 pencils and Priya used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of their pencils?

Answer:

Total pencils Ravish had = 20
Pencils used by Ravish = 10
Fraction of pencils used by ravish = 10÷1020÷10=12     (Dividing both the numerator & denominator by the HCFs of 10 & 20)

Total pencils Shikha had = 50
Pencils used by Shikha = 25
Fraction of pencils used by Shikha = 25÷2550÷25=12      (Dividing both the numerator & denominator by the HCFs of 25 & 50)

Total pencils Priya had = 80
Pencils used by Priya = 40
Fraction of pencils used by Priya = 40÷4080÷40=12       (Dividing both the numerator & denominator by the HCFs of 40 & 80)   

Yes, each of them has utilised an equal fraction of pencils.



Page No 6.19:

Question 1:

Reduce each of the following fractions to its lowest term (simplest form):

(i) 4075
(ii) 4228
(iii) 1252
(iv) 4072
(v) 8024
(vi) 8456

Answer:


(i) 4075Factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40.Factors of 75 are 1, 3, 5, 15 and 75.Common factors of 40 & 75 are 1 & 5.HCF = 5Divide both the numerator & denominator by 5.40÷575÷5=815

(ii) 42 28Factors of 42are1,2,3,6,7,14,21,42 Factors of 28 are1,2,4,7,14,28Common factors of 42 & 28 are 1,2,7,14HCF=14Divide both the numerator & denominator by 1442÷14 28÷14=32

(iii) 1252Factors of 12 are 1, 2, 3, 4, 6 and 12.Factors of 52 are 1, 2, 4, 13, 26 and 52.Common factors of 12 & 52 are 1, 2, and 4.HCF= 4Divide both the numerator & denominator by 4.12÷452÷4=313
(iv)4072Factors of 40 are 1,2,4,5,8,10,20,40Factors of 72 are 1,2,3,4,6,8,9,12,18,24,36,72Common factors of 40 & 72 are 1,2,4,8HCF=8Divide both the numerator & denominator by 840÷872÷8=59
(v) 8024Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.Common factors of 80 & 24 are 1, 2, 4 and 8. HCF=8Divide both the numerator & denominator by 4.80÷824÷8=103
(vi) 8456Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56.Common factors of 84 & 56 are 1, 2, 4, 7, 14 and 28.HCF=28Divide both the numerator & denominator by 28.84÷2856÷25=32

Page No 6.19:

Question 2:

Simplify each of the following to his lowest term:

(i) 7580
(ii) 5276
(iii) 8498
(iv) 6817
(v) 15050
(vi) 162108

Answer:

(i) 7580Factors of 75 are 1, 3, 5, 15, 25 and 75.Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.Common factors of 75 & 80 are 1 & 5.HCF of 75 & 80 is 5.Dividing both the numerator & denominator by 5, we get:75÷580÷5=1516

(ii) 5276Factors of 52 are 1, 2, 4, 13, 26 and 52.Factors of 76 are 1, 2, 4, 19, 38 and 76.Common factors of 52 & 76 are 1, 2 and 4. HCF of 52  and 76 is 4.Dividing both the numerator and denominator by 4, we get:52÷476÷4=1319

(iii) 8498Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.Factors of 98 are 1, 2, 7, 14, 49 and 98.Common factors of 84 and 98 are 1, 2, 7 and 14.HCF of 84 and 98 is 14.Dividing both the numerator and denominator by 14, we get:84÷1498÷14=67

(iv) 6817Factors of 68 are 1, 2, 4, 17, 34 and 68.Factors of 17 are 1 and 17.Common factors of 68 and 17 are 1 and 17.HCF of 68 and 17 is 17.Dividing both the numerator and denominator by 17, we get:68÷1717÷17=41

(v) 15050Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 50 and 150.Factors of 50 are 1, 2, 5, 10, 25 and 50.Common factors of 150 and 50 are 1, 2, 5, 10, 25 and 50.Dividing both the numerator and denominator by 50, we get:150÷5050÷50=31

(vi) 162108Factors of 162 are 1, 2, 3, 6, 9, 18, 27, 54, 81 and 162.Factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27 and 54.Common factors of 162 and 108 are 1, 2, 3, 6, 9, 18, 27 and 54.HCF of 162 and 108 is 54.Dividing both the numerator and denominator by 54, we get:162÷54108÷54=32



Page No 6.24:

Question 1:

Write each fraction. Arrange them in ascending and descending order using correct sign '<',
'=', '>' between the fractions:

Answer:

(i) Ascending order



Descending order


(ii) Ascending order


Descending order


(iii) Ascending order


Descending order

Page No 6.24:

Question 2:

Mark 26,46,86 and 66 on the number line and put appropriate signs between fractions given below:

(i) 56......26
(ii) 36......06
(iii) 16......66
(iv) 86......56

Answer:



(i) 56>26 because 5 > 2 and the denominator is the same.
(ii) 36>06 because 3 > 0 and the denominator is the same.
(iii) 16<66 because 6 > 1 and the denominator is the same.
(iv) 86>56 because 8 > 5 and the denominator is the same.

Page No 6.24:

Question 3:

Compare the following fractions and put an appropriate sign:

(i) 36......56
(ii) 45......05
(iii) 320......420
(iv) 17......14

Answer:

(i) 36<56 because 3 < 5 and the denominator is the same.
(ii) 45>05 because 4 > 0 and the denominator is the same.
(iii) 320<420 because 3 < 4 and the denominator is the same.
(iv) 17<14 because 7 > 4; if the numerator is the same, then the fraction that has smaller denominator is greater.

Page No 6.24:

Question 4:

Compare the following fractions using the symbol > ro <:

(i)  67and 611
(ii) 37and 57
(iii) 23and 812
(iv) 15and 415
(v) 83and 813
(vi) 49and 158

Answer:

(i) 67>611 because if the numerator is the same, then the fraction with smaller denominator is greater.
(ii) 37<57 because 3 < 5 and the denominator is the same.

(iii)
812=2×2×22×2×3=23 so 23=812

(iv)
15=15×33=315  315<415 (Because 3 < 4 and the denominator is the same. Therefore, 15<415.)

(v)
83>813  because if the numerator is the same, then the fraction with smaller denominator is greater. 

(vi)
 49=49×88=3272  158=158×99=13572    3272<13572 (Because 135 > 32 and the denominator is the same)

49<158

Page No 6.24:

Question 5:

The following fractions represent just three different numbers. Separate them in to three groups of equal fractions by changing each one to its simplest form:

(i) 212
(ii) 315
(iii) 850
(iv) 16100
(v) 1060
(vi) 1575
(vii) 1260
(viii) 1696
(ix) 1275
(x) 1272
(xi) 318
(xii) 425

Answer:

(i) 212HCF of 2 & 12 is 2.Divide both the numerator & denominator by the HCF of 2 &12. 2÷212÷2=16
(ii) 315HCF of 3 &15 is 3.Divide both the numerator & denominator by the HCF of 3 &15. 3÷315÷3=15

(iii) 850HCF of 8 & 50 is 2.Divide both the numerator & denominator by the HCF of 8 & 50. 8÷250÷2=425

(iv) 16100HCF of 16 &100 is 4.Divide both the numerator &denominator by the HCF of 16 & 100. 16÷4100÷4=425

(v) 1060HCF of 10 & 60 is 10.Divide both the numerator & denominator by the HCF of 10 & 60. 10÷1060÷10=16

(vi) 1575HCF of 15 & 75 is 15.Divide both the numerator & denominator by the HCF of 15 & 75. 15÷1575÷15=15

(vii) 1260HCF of 12 & 60 is 12.Divide both the numerator & denominator by the HCF of 12 & 60.12÷1260÷12=15

(viii) 1696HCF of 16 & 96 is 16.Divide both the numerator & denominator by the HCF of 16 & 96.16÷1696÷16=16

(ix) 1275HCF of 12 & 75 is 3.Divide both the numerator & denominator by the HCF of 12 & 75.12÷375÷3=425

(x) 1272HCF of 12 & 72 is 12.Divide both the numerator & denominator by the HCF of 12 & 72.12÷1272÷12=16

(xi) 318HCF of 3 & 18 is 3.Divide both the numerator & denominator by the HCF of 3 & 18. 3÷318÷3=16

(xii) 425HCF of 4 & 25 is 1.Divide both the numerator & denominator by the HCF of 4 & 25.4÷125÷1=425

Three groups of equal fractions:212,1060,1696,1272,318; 315,1575,1260; 850,16100,1275,425



Page No 6.25:

Question 6:

Isha read 25 pages of a book containing 100 pages. Nagma read 12 of the same book. Who read less?

Answer:

Total pages in the book = 100
Fraction of the book read by Isha = 25÷25100÷25=14            (Dividing numerator & denominator by the HCF of 25 & 100)
Fraction of the book read by Nagma = 12
Now, compare 14 & 12.
LCM of 4 & 2 is 4.

Convert each fraction into equivalent fraction with 4 as its denominator.

1×14×1 &1×22×214  &2414<24
Therefore, Isha read less.




 

Page No 6.25:

Question 7:

Arrange the following fractions in the ascending order:

(i) 29,79,39,49,19,69,59
(ii) 78,725,711,718,710
(iii) 3747,3750,37100,371000,3785,3741
(iv) 35,15,45,25
(v) 25,34,12,35
(vi) 38,312,36,34
(vii) 46,38,612,516

Answer:

(i) 29,79,39,49,19,69,59When denominators are the same & numerators are different, then the fraction with greater numerator has a larger value.(ii) 78,725,711,718,710When numerators are the same & denominators are different, then the fraction with smaller denominator has a smaller value.725<718<711<710<78(iii) 3747,3750,37100,371000,3785,3741When numerators are the same & denominators are different, then the fraction with greater denominator has a smaller value.371000<37100<3785<3750<3747<3741(iv) 35,15,45,25When denominators are the same & numerators are different, then the fraction with greater numerator has a larger value.15<25<35<45

(v)
LCM of 2, 4 and 5 is 20.25=25×44=82034=34×55=152012=12×1010=102035=35×44=1220

25<12<35<34
(vi)38,312,36,34When numerators are the same & denominators are different, then the fraction with smaller denominator has a greater value.312<38<36<34
(vii)
4÷26÷2=23    (Dividing the numerator & denominator by the HCF of 4 & 6)612=12        (Dividing the numerator & denominator by the HCF of 6 & 12)LCM of 2, 8, 3 and 16 is 48.46=23×1616=3248612=12×2424=244838=38×66=1848516=516×33=1548

When denominators are the same & numerators are different, then the fraction with greater numerator has a greater value.

516<38<612<46

Page No 6.25:

Question 8:

Arrange in descending order in each of the following using the symbol >:

(i) 817,89,85,813
(ii) 59,312,13,415
(iii) 27,1135,914,1328

Answer:

When numerators are the same and denominators are different, then the fraction with greater denominator has a smaller value.

When denominators are the same and numerators are different, then the fraction with greater numerator has a larger value.

(i)85>89>813>817
(ii) LCM of 9,12, 3 and 15 is 180.
 59×2020=100180
312×1515=45180
13×6060=60180
415×1212=48180
59>13>415>312
(iii)914>1328>1135>27

Page No 6.25:

Question 9:

Find answers to the following. Write and indicate how you solved them.

(i) Is 59 equal to 45?
(ii) Is 916 equal to 59?
(iii) Is 45 equal to 1620?
(iv) Is 115 equal to 430?

Answer:

Two fractions are equal when:

Numerator of the first fraction × Denominator of the second fraction = Numerator of the second fraction × Denominator of the first fraction
(i)
5×5=254×9=36So, 5×54×959is not equal to 45.
(ii)
9×9=815×16=80So, 9×95×16916 is not equal to 59.

(iii)
4×20=8016×5=80So, 4×20=16×545 is equal to 1620.
(iv)
1×30=304×15=60So, 1×304×15115 is not equal to 430.



Page No 6.29:

Question 1:

Write these fractions appropriately as additions of subtractions:

Answer:

(i) The given fractions are 15 & 25

15+25=351+25=3535=35

(ii) The given fractions are 36 & 26

36+26=563+26=5656=56

Page No 6.29:

Question 2:

Solve:

(i) 512+112
(ii) 315+715
(iii) 322+722
(iv) 14+04
(v) 413+213+113
(vi) 015+215+115
(vii) 731-431+931
(viii) 327+17-237
(ix) 213-123+413
(x) 1-23+73
(xi) 167-57+97

Answer:

(i)512+112=5+112=6÷612÷6=12    (Dividing numerator & denominator by their HCF)

(ii)315+715=3+715=10÷515÷5=23        (Dividing numerator & denominator by their HCF)

(iii)322+722=3+722=10÷222÷2=511     (Dividing numerator  & denominator by their HCF)

(iv)14+04=1+04=14

(v)413+213+113=4+2+113=713

(vi)015+215+115=0+2+115=3÷315÷3=15            (Dividing numerator & denominator by their HCF)

(vii)731-431+931=7-4+931=1231

(viii)327+17-237=237+17-177=23+1-177=77=1
(ix)213-123+413=73-53+133=7-5+133=15÷33÷3=5       (Dividing numerator & denominator by their HCF)

(x)1-23+73=3-2+73=83

(xi)167-57+97=16-5+97=207

Page No 6.29:

Question 3:

Shikha painted 15 of the wall space in her room. Her brother Ravish helped and painted 35 of the wall space. How much did they paint together? How much the room is left unpainted?

Answer:

Shikha painted 15 of the wall space.
Ravish painted 35 of the wall space.
Wall space painted by both of them together  = 15+35=1+35=45
Unpainted part of the room = 1-45=5-45=15

Page No 6.29:

Question 4:

Ramesh bought 212 kg sugar whereas Rohit bought 312 kg or sugar. Find the total amount of sugar bought by both of them.

Answer:

 Quantity of sugar bought by Ramesh = 212kg=(2×2)+12=52kg
 Quantity of sugar bought by Rohit = 312kg=(3×2)+12=72kg
Total amount of sugar bought by them =

52+72=5+72=12÷22÷2=6 kg     (Dividing numerator & denominator by their HCF)

Page No 6.29:

Question 5:

The teacher taught 35 of the book, Vivek revised 15 more on his own. How much does he still have to revise?

Answer:

Fraction of the book taught by the teacher = 35
Fraction of the book revised by Vivek = 15
Fraction of the book still left for revision by Vivek = 35-15=3-15=25 

Page No 6.29:

Question 6:

Amit was given 57 of a bucket of organes. What fraction of oranges was left in the basket?

Answer:

Fraction of oranges given to Amit = 57
Fraction of oranges left in the basket = 1-57=7×1-57=27



Page No 6.30:

Question 7:

Fill in the missing fractions:

(i) 710-   =310
(ii)    -321=521
(iii)    -36=36
(iv)    -527=1227

Answer:

(i)
710-=310=710-310=7-310=410=410

(ii)
-321=521=321+521=3+521=821

(iii)
-36=36=36+36=3+36=66=1

(iv)
-527=1227=1227+527=12+527=1727



Page No 6.35:

Question 1:

Add:

(i) 34and56
(ii) 710and215
(iii) 813and23
(iv) 45and715

Answer:

(i)

34+56LCM of 4 & 6 is 12, so we will convert each fraction into an equivalent fraction with denominator 12.=3×34×3+5×26×2=912+1012=9+1012=1912

(ii)

710+215LCM of 10 & 15 is 30, so we will convert each fraction into an equivalent fraction with denominator 30.=7×310×3+2×215×2=2130+430=21+430=2530

(iii)

813+23LCM of 13 & 3 is 39, so we convert each fraction into an equivalent fraction with denominator 39. =8×313×3+2×133×13=2439+2639=24+2639=5039

(iv)

45+715LCM of 5 & 15 is 15, so we will convert each fraction into an equaivalent fraction with denominator 15.4×35×3+7×115×1=1215+715=12+715=1915

Page No 6.35:

Question 2:

Subtract:

(i) 27from1921
(ii) 2125from1820
(iii) 716from 2
(iv) 415from 215

Answer:

(i)
1921-27LCM of 21 & 7 is 21, so we convert each fraction into an equivalent fraction with denominator 21.=19×121×1-2×37×3=1921-621=19-621=1321

(ii)
1820-2125LCM of 20 & 25 is 100, so we convert each fraction into an equivalent fraction with denominator 100.=18×520×5-21×425×4=90100-84100=6100

(iii)
21-716LCM of 1 & 16 is 16 ,so we convert each fraction into an equivalent fraction with denominator 16.=2×161×16-7×116×1=3216-716=32-716=2516

(iv)
215-415=(5×2)+15-415=115-415LCM of 5 & 15 is 15,so we convert each fraction into an equivalent fraction with denominator 15.11×35×3-4×115×13315-415=33-415=2915

Page No 6.35:

Question 3:

Find the difference of:

(i) 1324and716
(ii) 518and415
(iii) 112and34
(iv) 23and67

Answer:

(i)
1324-716=13×224×2-7×316×3          ( Because LCM of 24 & 16 is 48 )=2648-2148=26-2148=548

(ii)
518-415=5×518×5-4×615×6=2590-2490        (Because LCM of 18 & 15 is 90)=25-2490=190

(iii)
34-112=3×34×3-1×112×1         (Because LCM of 4 & 12 is 12)=912-112=9-112=812

(iv)
67-23=6×37×3-2×73×7          (Because LCM of 7 & 3 is 21)=1821-1421=18-1421=421

Page No 6.35:

Question 4:

Sbutract as indicated:

(i) 83-59
(ii) 425-215
(iii) 567-223
(iv) 434-216

Answer:


(i) 83-59=8×33×3-5×19×1     ( Because LCM of 3 & 9 is 9)=249-59=24-59=199

(ii)  425-215=(5×4)+25-(5×2)+15=225-115=22-115=115

(iii) 567-223=(7×5)+67-(3×2)+23=417-8341×37×3-8×73×7          ( Because LCM of 7 & 3 is 21 )12321-5621=123-5621=6721

(iv) 434-216=(4×4)+34-(6×2)+16=194-13619×34×3-13×26×2           (Because LCM of 4 & 6 is 12)=5721-2621=57-2621=3112

Page No 6.35:

Question 5:

Simplify:

(i) 23+34+12
(ii) 58+25+34
(iii) 310+715+35
(iv) 34+716+58
(v) 423+314+712
(vi) 713+323+516
(vii) 7+74+516
(viii) 56+3+34
(ix) 718+56+1112

Answer:


(i) 23+34+12=2×43×4+3×34×3+1×62×6        (Because LCM of 3,4 &2 is 12 )=812+912+612=8+9+612=2312

 (ii) 58+25+34=5×58×5+2×85×8+3×104×10             (Because LCM of 8,5 & 4 is 40 )=2540+1640+3040=25+16+3040=7140

(iii) 310+715+35=3×310×3+7×215×2+3×65×6          (Because LCM of 10,15 & 5 is 30 )=930+1430+1830=9+14+1830=4130


(iv) 34+716+58=3×44×4+7×116×1+5×28×2            ( Because LCM of 4,16 & 8 is 16 )=1216+716+1016=12+7+1016=2916


 (v) 423+314+712=4×3+23+3×4+14+7×2+12=143+134+152=14×43×4+13×34×3+15×62×6           ( Because LCM of 3,4 & 2 is 12 )=5612+3912+9012=56+39+9012=18512

(vi) 713+324+516=7×3+13+3×4+24+5×6+16=223+144+316=22×43×4+14×34×3+31×26×2            (Because LCM of 3,4 & 6 is 12 )=8812+4212+6212=88+42+6212=192 ÷1212÷12=16                ( HCF of numerator & denominator is 12 )

(vii) 7+74+51671+74+5×6+16=71+74+316=7×121×12+7×34×3+31×26×2            (Because LCM of 1,4 & 6 is 12 )=8412+2112+621284+21+6212=16712

(viii) 56+3+345×26×2+3×121×12+3×34×3              (Because LCM of 6,1 & 4 is 12 )=1012+3612+912=10+36+912=5512

(ix) 718+56+1112718+56+12×1+112=718+56+1312=7×218×2+5×66×6+13×312×3=1436+3036+3936=14+30+3936=8336

Page No 6.35:

Question 6:

Replace     by the correct number:
(i)    -58=14
(ii)    -15=12
(iii) 12-   =16

Answer:


(i) -58=14=14+58=1×24×2+5×18×1          ( Because LCM of 4 & 8 is 8)=28+58=2+58=78=78
 (ii) -15=12=12+15=1×52×5+1×25×2=510+210=5+210=710
(iii) 12-=16=12-16=1×32×3-1×16×1           (Because LCM of 2 & 6 is 6)=36-16=2÷26÷2     (HCF of the numerator & denominator is 2)=13

Page No 6.35:

Question 7:

Savita bought 25 m of ribbon and Kavita 34 m of the ribbon. What was of the total length of the ribbon they bought?

Answer:

Length of the ribbon bought by Savita =25 m
Length of the ribbon bought by Kavita = 34 m
Total length of the ribbon bought by them:

25+34=2×45×4+3×54×5          (Because LCM of 5 & 4 is 20)=820+1520=8+1520=2320 m

Page No 6.35:

Question 8:

Ravish takes 215 minutes to walk across the school ground. Rahul takes 74 minutes to do the same. Who takes less time and by what fraction?

Answer:

Time taken by Ravish = 215=(5×2)+15=115 minutes
Time taken by Rahul = 74 minutes
Comparing 115&74, we get:
11×45×4,7×54×5     (LCM of 4 & 5 is 20, so will we convert each fraction into an equivalent fraction with denominator 20.)4420>3520

Rahul takes less time, i.e., 4420-3520=44-3520=920 minutes.

Page No 6.35:

Question 9:

A piece of a wire 78 metres long broke into two pieces. One piece was 14 meter long. How long is the other piece?

Answer:

Length of the wire = 78m
Length of one piece of wire = 14 m
Let the length of the second piece of wire be x m.

∴ Length of the wire = Length of one piece + Length of the second piece
78=14+xx=78-14x=7×18×1-1×24×2=78-28=7-28x=58 m
Therefore, the length of the second piece is 58m.



Page No 6.36:

Question 10:

Shikha and priya have bookshelves of the same size Shikha's shelf is 56 full of book and Priya's shelf is 25 full. Whose bookshelf is more full? By what fraction?

Answer:

 
Fraction of Shikha's filled bookshelf = 56

Fraction of Priya's filled bookshelf = 25
Comparing 56 & 25, we get:
LCM of 5 & 6 is 30, so we will convert each fraction into an equivalent fraction with denominator 30.

=5×56×5,2×65×62530>1230

Shikha's shelf is more full.

∴​ 2530-1230=25-1230=1330

Page No 6.36:

Question 11:

Ravish's house is 910 Km from his school. He walked some distance and then took a bus for 12 Km up to the school. How far did he walk?

Answer:

Total distance between the house and the school = 910 km
Distance covered in the bus = 12 km
Distance covered by walking + Distance covered in the bus = Total distance between the house and the school
Distance covered by walking = Total distance between the house and the school - Distance covered in the bus
Distance covered by walking:
910-12LCM of 10 and 2 is 10,so we convert each fraction into an equivalent fraction with denominator 10=9×110×1-1×52×5=910-510=9-510=410km=4÷210÷2=25km      (HCF of numerator & denominator is 2)

Page No 6.36:

Question 1:

Which of the following is a proper fraction?

(a) 43
(b) 34
(c) 134
(d) 215

Answer:

(b) 34, because in a proper fraction, the numerator is less than the denominator.

Page No 6.36:

Question 2:

Which of the following is an improper fraction ?

(a) 12
(b) 37
(c) 73
(d) 315

Answer:

(c) 73, because in an improper fraction, the numerator is more than the denominator.

Page No 6.36:

Question 3:

Which of the following is a fraction equivalent of 23?
(a) 45
(b) 86
(c) 1025
(d) 1015

Answer:

(d) 1015=1015=23=10×3=2×1530=30



Page No 6.37:

Question 4:

A fraction equivalent to 35is
(a) 3+25+2
(b) 3-25-2
(c) 3×25×2
(d) None of these

Answer:

(c) 3×25×2
On dividing the numerator & denominator by 2, we get 35.

Page No 6.37:

Question 5:

If 512 is equivalent of x3, then x =

(a) 54
(b) 45
(c) 53
(d) 35

Answer:

(a) 54
512=x3 
On cross-multiplying, we get:
 5×3=12×x
x=5×312=5×34×3=54 

Page No 6.37:

Question 6:

Which of the following are like fractions?

(a) 35,37,311,316
(b) 511,711,1511,211
(c) 23,34,45,67
(d) None of these

Answer:

(b), because like fractions are the fractions with the same denominator.

Page No 6.37:

Question 7:

If 114=77x, then x =

(a) 28

(b) 7728
(c) 44

(d) 308

Answer:

 (a) 28

114=77x
On cross-multiplying, we get:
 11×x=77×4

x=77×411=7×11×411On dividing both the numerator & denominator by 11, we get 28.

Page No 6.37:

Question 8:

1213+1134 is equal to

(a) 714
(b) 1249
(c) 4112
(d) None of these

Answer:

(d) None of these

1213+1134=1(3×2)+13+1(4×1)+34=173+174=37+47=3+47=77=1

Page No 6.37:

Question 9:

If 13+12+1x=4, then x = ?

(a) 518
(b) 619
(c) 185
(d) 2411

Answer:

(b) 619

13+12+1x=41x=4-13-121x=4×61×6-1×23×2-1×32×3=246-26-361x=24-2-361x=196x=619

Page No 6.37:

Question 10:

If 12+1x=2, then x =

(a) 25
(b) 52
(c) 32
(d) 23

Answer:

(d) 23
12+1x=21x=21-121x=2×21×2-1×12×1=42-12=4-12        (Because LCM of 1,2 is 2)1x=32x=23

Page No 6.37:

Question 11:

Which of the following fractions is the smallest?
12,37,35,49

(a) 49
(b) 35
(c) 37
(d) 12

Answer:

(c) 37
The LCM of numerators is 12, so we can convert each fraction into an equivalent fraction with numerator 12.
12=12×1212=122437=37×44=122835=35×44=122049=49×33=1227
When numerator is the same, the fraction with greater denominator is the smallest.
Thus, 37 is the smallest fraction.

Page No 6.37:

Question 12:

Which of the following fractions is the greatest of all?
78,67,45,56

(a) 67
(b) 45
(c) 56
(d) 78

Answer:

(d) 78
The LCM of 8, 7, 6 and 5 is 840, so we can convert each fraction into an equivalent fraction with denominator 840.
78=78×105105=73584067=67×120120=72084045=45×168168=67284056=56×140140=700840

When denominator is the same, the fraction with the largest numerator is the greatest.

Thus, 78 is the greatest fraction among all.

Page No 6.37:

Question 13:

What is the value of a+ba-b, If ab=4?

(a) 35
(b) 53
(c) 45
(d) 54

Answer:

(b) 53
ab=4a=4bOn putting the value of a in a+ba-b, we get:a+ba-b=4b+b4b-b=5b3bOn dividing the numerator & denominator by b, we get 53.



Page No 6.38:

Question 14:

If ab=43, then the value of 6a+4b6a-5b is

(a) −1
(b) 3
(c) 4
(d) 5

Answer:

(c) 4

ab=43a=4b3On putting the value of a=4b3 in 6a+4b6a-5b, we get:6a+4b6a-5b=6(4b3)+4b6(4b3)-5b=24b3+4b24b3-5bLCM of 3 &1 is 3.24b3+4b×31×324b3-5b×31×3=24b3+12b324b3-15b3=24b+12b324b-15b3=36b9bOn dividing the numerator & denominator by the HCF of 36b & 9b, we get:36b÷9b9b÷9b=4

Page No 6.38:

Question 15:

If 15-16=4x, then x =

(a) −120
(b) −100
(c) 100
(d) 120

Answer:

(d) 120

15-16=4x1×65×6-1×56×5=4x630-530=4x6-530=4x130=4xx=4×30=120

Page No 6.38:

Question 16:

The fraction to be added to 6715 to get 815 is equal to

(a) 1115
(b) 1115
(c) 443
(d) 344

Answer:

(b) 1115

Let the fraction to be added is x.
6715+x=81515×6+715+x=8×5+159715+x=415= >x=415-9715=> x =41×35×3-97×115×1=> x =12315-9715==>x=123-9715=> x=2615=>x=15+1115=>x=1515+1115=>x=11115

Page No 6.38:

Question 17:

If 4560 is equivalent to 3x, then x =

(a) 5
(b) 4
(c) 6
(d) 20

Answer:

(b) 4

4560=3xOn cross-multiplying, we get:45×x=3×60x=3×6045x=18045On dividing the numerator & denominator by the HCF of 180& 45, we get:180÷4545÷45=4

Page No 6.38:

Question 18:

A fraction equivalent to 45105 is

(a) 614
(b) 47
(c) 57
(d) 75

Answer:

(a) 614
45105On dividing the numerator & denominator by the HCF of 45 & 105, we get:45÷15105÷15=37

37=37×22=614

Page No 6.38:

Question 19:

58+34-712 is equal to

(a) 1524
(b) 1724
(c) 1924
(d) 2124

Answer:

(c) 1924
58+34-712LCMof 8, 4 and 12 is 24.5×38×3+3×64×6-7×212×21524+1824-142415+18-1424=1924

Page No 6.38:

Question 20:

The correct fraction in the box is -58=14

(a) 68
(b) 78
(c) 12
(d) None of these

Answer:

(b) 78
-58=14=14+58LCM of 4 & 8 is 8.=1×24×2+5×18×1=28+58=2+58=78

Page No 6.38:

Question 1:

A fraction equivalent to 23 is

(a) 2+33+3
(b) 2-13-1
(c) 2×53×5
(d) 2+53+5

Answer:

Fraction equivalent to a given fraction can be obtained by multiplying or dividing its numerator and denominator by a non-zero number.

Therefore, the fraction equivalent to 23 is 2×53×5.

Hence, the correct option is (c).

Page No 6.38:

Question 2:

A fraction equivalent to 812 is

(a) 8+412+4
(b) 8÷412÷4
(c) 8-412-4
(d) None of these

Answer:

Fraction equivalent to a given fraction can be obtained by multiplying or dividing its numerator and denominator by a non-zero number.

Therefore, the fraction equivalent to 812 is 8÷412÷4.

Hence, the correct option is (b).

Page No 6.38:

Question 3:

A fraction equivalent to 3045 is

(a) 34
(b) 32
(c) 23
(d) None of these

Answer:

Fraction equivalent to a given fraction can be obtained by multiplying or dividing its numerator and denominator by a non-zero number.

Therefore, the fraction equivalent to 3045 is 30÷1545÷15=23.

Hence, the correct option is (c).



Page No 6.39:

Question 4:

Which of the following is a proper fraction?

(a) 35
(b) 53
(c) 123
(d) None of these

Answer:

A fraction whose numerator is less than the denominator is called a proper fraction.

Here, 35 is a proper fraction.

Hence, the correct option is (a).

Page No 6.39:

Question 5:

3413 is an example of

(a) a proper fraction
(b) an improper fraction
(c) a mixed fraction
(d) none of these

Answer:

A fraction whose numerator is less than the denominator is called a proper fraction, otherwise it is called an improper fraction.

Here, 3413 is an example of an improper fraction.

Hence, the correct option is (b).

Page No 6.39:

Question 6:

Which of the following fractions is the smallest?
59, 49, 29, 119

(a) 119
(b) 49
(c) 59
(d) 29

Answer:

2 < 4 < 5 < 11
29<49<59<119

∴ the smallest fraction is 29.

Hence, the correct option is (d).

Page No 6.39:

Question 7:

The smallest of the fractions 35, 23, 56, 710 is

(a) 23
(b) 35
(c) 56
(d) 710

Answer:

Fractions can be compared by converting them into like fractions and then arranging them in ascending or descending order.

35=3×65×6=1830,
23=2×103×10=2030,
56=5×56×5=2530,
710=7×310×3=2130.

We know,
18 < 20 < 21 < 25
1830<2030<2130<253035<23<710<56

∴ the smallest fraction is 35.

Hence, the correct option is (b).

Page No 6.39:

Question 8:

If 34 is equivalent to x28, then the value of x is

(a) 6
(b) 21
(c) 8
(d) 9

Answer:

34=3×74×7=2128It is given that 34 is equivalent to x28.2128=x28x=21

Hence, the correct option is (b).

Page No 6.39:

Question 9:

If 4560 is equivalent to 3x, then the value of x is

(a) 3
(b) 6
(c) 4
(d) 9

Answer:

4560=45÷1560÷15=34It is given that 4560 is equivalent to 3x.34=3xx=4

Hence, the correct option is (c).

Page No 6.39:

Question 10:

13+x=3, then x =

(a) 73
(b) 23
(c) 43
(d) 83

Disclaimer: There is a misprint in the question, 1x is written instead of x.

Answer:

13+x=313+x-13=3-13x=3×31×3-13x=93-13x=83

Hence, the correct option is (d).

Page No 6.39:

Question 11:

Aarushi was given 57 of a basket of oranges. What fraction of oranges was left in the basket.

Answer:

Let the total number of oranges in the basket = 1.

Fraction of oranges given to Aarushi = 57.

Fraction of oranges left =
1-57=1×71×7-57=77-57=7-57=27

Thus, 27 fraction of oranges was left in the basket.

Page No 6.39:

Question 12:

Reduce 8498 to its lowest terms.

Answer:

8498=84÷298÷2      =4249      =42÷749÷7      =67

Hence, the lowest term of 8498 is 67.

Page No 6.39:

Question 13:

The cost of a pen is â‚¹623 and that of a pencil is â‚¹416. Which costs more and by how much?623

Answer:

Cost of a pen = â‚¹623 = â‚¹203 = â‚¹406

Cost of pencil = â‚¹416 = â‚¹256

We know, 
25 < 40
256<406416<623

Thus, cost of a pen is more.

Now, 406-256=156=52=212

Hence, a pen costs more than a pencil by â‚¹212.

Page No 6.39:

Question 14:

Simplify: 516-314+313+4.

Answer:

516-314+313+4=316-134+103+4                               =31×26×2-13×34×3+10×43×4+4×121×12                               =6212-3912+4012+4812                               =62+40+48-3912                               =11112                               =374                               =914

Page No 6.39:

Question 15:

Three boxes weigh 1834 kg, 712 kg and 1015 kg respectively. A porter carries all the three boxes. What is the total weight carried by the porter?

Answer:

Since the porter carries all the three boxes, then total weight

 =1834+712+1015=754+152+515=75×54×5+15×102×10+51×45×4=37520+15020+20420=375+150+20420=72920=36920

Hence, the total weight carried by the porter is 36920 kg.



Page No 6.40:

Question 16:

Arrange the following fractions in ascending order: 1318, 815, 1724, 712

Answer:

Fractions can be compared by converting them into like fractions and then arranging them in ascending or descending order.

Now,
1318=13×2018×20=260360
815=8×2415×24=192360
1724=17×1524×15=255360
712=7×3012×30=210360

We know,
192 < 210 < 255 < 260
192360<210360<255360<260360815<712<1724<1318

Hence, the following fractions in ascending order is 815<712<1724<1318.

Page No 6.40:

Question 17:

Subtract the sum of 23 and 12 from the sum of 56, 712 and 415.

Answer:

Sum of 23 and 12 is

23+12=2×23×2+1×32×3            =46+36            =4+36            =76

Sum of 56, 712 and 415 is

56+712+415=5×106×10+7×512×5+4×415×4                        =5060+3560+1660                        =50+35+1660                        =10160

Now,
10160-76=10160-7×106×10               =10160-7060               =101-7060               =3160

Hence, the difference of the sum of 23 and 12 from the sum of 56, 712 and 415 is 3160.

Page No 6.40:

Question 18:

Complete the addition-subtraction box.
 

(a)
 
(b)

 

Answer:

(a) Solving rows first,
23+43=2+43=63=2

13+23=1+23=33=1

Solving the columns,
23-13=2-13=13

43-23=4-23=23

2-1=1

Hence, the complete addition-subtraction box is



(b) Solving rows first,
12+13=1×32×3+1×23×2=36+26=3+26=56

13+14=1×43×4+1×34×3=412+312=4+312=712

Solving the columns,
12-13=1×32×3-1×23×2=36-26=3-26=16

13-14=1×43×4-1×34×3=412-312=4-312=112

56-712=5×26×2-712=1012-712=10-712=312=14

Hence, the complete addition-subtraction box is

Page No 6.40:

Question 19:

Shikha bought 712 litres of milk. Out of this milk, 534 litres was consumed. How much milk is left with her?

Answer:

Total amount of milk bought =  712 = 152 litres

Amount of milk consumed = 534 = 234 litres

Amount of milk left = 152-234=15×22×2-234=304-234=74=134 litres.

Thus, 134 litres of milk is left with Shikha.

Page No 6.40:

Question 20:

Simplify: 3+115+23-715

Answer:

3+115+23-715=31+65+23-715                             =3×151×15+6×35×3+2×53×5-715                             =4515+1815+1015-715                             =45+18+10-715                             =6615                             =225                             =425

Page No 6.40:

Question 21:

Fill in the blanks:

723+.....=9

Answer:

Let the value in the blank be x.

723+x=9233+x=9233+x-233=9-233x=9×31×3-233x=273-233x=27-233x=43x=113

Thus, 723+ 113 =9.

Page No 6.40:

Question 22:

Fill in the blanks:

818-.....=78

Answer:

Let the value in the blank be x.

818-x=78658-x=78658-78=xx=65-78x=588x=728

Thus, 818- 728 =78.

Page No 6.40:

Question 23:

Fill in the blanks:

616-515=.....

Answer:

Let the value in the blank be x.

616-515=xx=376-265x=37×56×5-26×65×6x=18530-15630x=185-15630x=2930

Thus, 616-515= 2930 .

Page No 6.40:

Question 24:

Fill in the blanks:

90108 reduced to simplest form .....

Answer:

Let the value in the blank be x.

90108=90÷3108÷3        =3036        =30÷336÷3        =1012        =10÷212÷2        =56

Thus, 90108 reduced to simplest form  56 .

Page No 6.40:

Question 25:

Fill in the blanks:
923+...=19

Answer:

923+...=19
293+...=19...=19-293...=57-293=283=913
Therefore, 923+913=19



Page No 6.4:

Question 1:

Write the fraction represention the shaded portion:

Answer:

Fraction of the shaded portion = Number of shaded partsTotal number of parts
(i) Total number of parts = 3
   Number of shaded parts = 2
Fraction of the shaded portion = 23
(ii) Total number of parts = 15
    Number of shaded parts = 11
Fraction of the shaded portion = 1115
(iii) Total number of parts = 9
    Number of shaded parts = 8
Fraction of the shaded portion = 89
(iv) Total number of parts = 7
    Number of shaded parts = 3
Fraction of the shaded portion = 37
(v) Total number of parts = 9
    Number of shaded parts = 4
Fraction of the shaded portion = 49
(vi) Total number of parts = 4
    Number of shaded parts = 2
Fraction of the shaded portion = 24=12
(vii) Total number of parts = 2
    Number of shaded parts = 1
Fraction of the shaded portion = 12
(viii) Total number of parts = 5
    Number of shaded parts = 1
Fraction of the shaded portion = 15
(ix) Total number of parts = 4
    Number of shaded parts = 1
Fraction of the shaded portion = 14



Page No 6.5:

Question 2:

Write the fraction represtion the shaded parts:

Answer:

Fraction of the shaded portion = Number of shaded partsTotal number of parts

(i) Total number of parts = 9
    Number of shaded parts = 3
    Fraction of the shaded portion = 39=13

(ii) Total number of parts = 8
    Number of shaded parts = 5
    Fraction of the shaded portion = 48

(iii) Total number of parts = 12
    Number of shaded parts = 3
    Fraction of the shaded portion = 312=14

(iv) Total number of parts = 10
    Number of shaded parts = 5
    Fraction of the shaded portion = 510=12

Page No 6.5:

Question 3:

Write the fraction representing the shaded portion:

Answer:

(i) Total number of portions = 2
   Number of shaded portions = 1
Fraction of the shaded portions = Number of shaded portionsTotal number of portions=12
(ii) Total number of portions = 8
   Number of shaded portions = 4
Fraction of the shaded portions = Number of shaded portionsTotal number of portions=48

Page No 6.5:

Question 4:

Colour the part according to the fraction given:

Answer:

Page No 6.5:

Question 5:

What fraction of an hour is 20 minutes?

Answer:

Minutes in an hour = 60
20 minutes of an hour = 2060 = 13

Page No 6.5:

Question 6:

Write the natural numbers form 2 to 12. What fraction of them are prime numbers?

Answer:

Natural numbers from 2 to 12 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Prime numbers from 2 to 12 are 2, 3, 5, 7 and 11.

Out of 11 numbers, 5 are prime.

Fraction of the prime numbers = 511

Page No 6.5:

Question 7:

Write the natural numbers from 102 to 113. What fraction of them are prime numbers.

Answer:

Natural numbers from 102 to 113 are 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 and 113.

Prime numbers from 102 to 113 are 103, 107, 109 and 113.

Out of 12 natural numbers, 4 are prime.

Fraction of the prime numbers = 412=13



Page No 6.6:

Question 8:

Mukesh has a box of 24 pencils. He gives half of them of Sunita. How many does sunita get?
How many does Mukesh still have?

Answer:

Mukesh has 24 pencils.

Sunita gets half of Mukesh's pencils.

Sunita gets 242 pencils, that is, 12 pencils.
Number of pencils Mukesh still has = 24 − 12 = 12

Page No 6.6:

Question 9:

Kavita has 44 cassettes. She gives 34 of them to Sonia. How many does Sonia get? How many does Kavita keep?

Answer:

Kavita has 44 cassettes.

She gives 34 of the cassettes to Sonia.
For this, Kavita divides 44 cassettes in 4 equal parts and takes 3 parts.

∴ 44 ÷ 4 = 11
It means that Kavita gives 33 cassettes to Sonia.

Number of cassettes Kavita has = 44 - 33 = 11

Page No 6.6:

Question 10:

Shikhas has three frocks that she wears when playing. The material is good, but the colours are faded. Her mother buys some blue dye and uses it on two of the frocks. What fraction of all of the Shikha play frocks did her mother dye?

Answer:

Total frocks Shikha has = 3

Number of frocks dyed by Shikha's mother = 2

Fraction of the dyed frocks = 23
Therefore, Shikha's mother dyed 23 of Shikha's frocks.



Page No 6.7:

Question 1:

Represent 25 on a number line.

Answer:

Page No 6.7:

Question 2:

Represent 010, 110, 510 and 1010 on a number line.

Answer:

Page No 6.7:

Question 3:

Represent 27, 57 and 67 on a number line.

Answer:

Page No 6.7:

Question 4:

How many fractions lie between 0 and 1.

Answer:

Infinite. We can check this by taking numerator less than denominator in a fraction.

Page No 6.7:

Question 5:

Represent 08 and 88 on a number line.

Answer:



Page No 6.8:

Question 1:

Write each of the following division as fraction:

(i) 6 ÷ 3
(ii) 25 ÷ 5
(iii) 125 ÷ 50
(iv) 55 ÷ 11

Answer:

(i) 63
(ii) 255
(iii) 12550
(iv) 5511

Page No 6.8:

Question 2:

Write each of the following fraction as divisions:

(i) 97
(ii) 311
(iii) 9063
(iv) 15

Answer:

(i)  9÷7(ii) 3÷11(iii) 90÷63(iv) 1÷5



View NCERT Solutions for all chapters of Class 6