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NCERT Solutions for Class 6 Maths Chapter 7 - Fractions

NCERT Solutions for Class 6 Maths Chapter 7 - Fractions

Edited By Komal Miglani | Updated on Jul 08, 2025 09:31 AM IST

Have you ever tried to share a pizza with your friends? Congratulations, you’ve already used fractions! In the world of numbers, fractions are the art of splitting and still staying connected. We are surrounded by fractions on all our sides. Fractions are an integral part of our lives. Fractions are a part of a whole. If we share a single chocolate among two people, each person gets half of the chocolate, which is a part of a whole. This NCERT Solutions for class 6 article contains the answers to all the questions in this chapter in a comprehensive, step-by-step manner.

This Story also Contains
  1. NCERT Solutions for Class 6 Maths Chapter 7 Exercise
  2. Fractions Class 6 Maths Chapter 7 - Topics
  3. NCERT Solutions for Class 6 Maths Chapter 7 Fractions - Notes
  4. NCERT Solutions for Class 6 Maths Chapter 7 Fractions - Points to Remember
  5. NCERT Solutions for Class 6 Maths Chapter Wise
NCERT Solutions for Class 6 Maths Chapter 7 - Fractions
NCERT Solutions for Class 6 Maths Chapter 7 - Fractions

Fractions help us realise that even small parts can add up to something whole. These NCERT Solutions for Class 6 Maths are highly reliable resources, as they were created by experienced subject matter experts from Careers360. Students can use these solutions to practice more questions and check their answers during exam preparation to identify their weaknesses and areas for improvement later. Explore the full NCERT curriculum, along with revision notes and PDFs, at this link: NCERT.

NCERT Solutions for Class 6 Maths Chapter 7 Exercise

Page number: 152-153
Number of Questions: 4

Fill in the blanks with fractions.

Question 1: Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____ kg.

Answer:

Given, weigh of 3 guavas =1 kg,
then weigh of each of the guava = total weight 3=13 kg

Question 2: A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is ___ kg.

Answer:

1 kg of rice is packed in 4 packets of equal weight, then the weight of each packet is

= Total weight  No. of packets =14 kg.

Question 3: Four friends ordered 3 glasses of sugarcane juice and shared them equally among themselves. Each one drank ____ glass of sugarcane juice.

Answer:

The Amount of sugarcane consumed by four friends = 3 glasses.
Amount of sugarcane consumed by each friend =34 glass.
Hence, each one drank 34 glass of sugarcane juice.

Question 4: The big fish weighs 12 kg. The small one weighs 14 kg. Together they weigh _______ kg.

Answer:

Given the weighs of big fish =12 kg and the weighs of small fish =14 kg
Total weight of both fish =12±14=2+14 kg =34 kg

Question 5: Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: One and a half, three-quarters, one and a quarter, half, quarter, two and a half.

Answer:

One and a half means 1+12=112.
Three-quarters means 34.
One and a quarter means 1+14=114.
Half means 12.
Quarter means 14
Two and a half mean 2+12=212.

Now, arranging the fraction words in order of size from the smallest to the biggest parts: Quarter < Half < Three Quarters < One and a quarter < One and a half < Two and a half.

Page number: 55
Number of Questions: 1

Question 1: The figures below show different fractional units of a whole chikki. How much of a whole chikki is each piece?

Screenshot%202025-06-10%20212851

Answer:

a) We get this piece by breaking the chikki into 12 equal parts.
Hence, it is 112

b) We get this piece by breaking the chikki into 4 equal parts.
Hence, it is 14

c) We get this piece by breaking the chikki into 8 equal parts.
Hence, it is 18

d) We get this piece by breaking the chikki into 6 equal parts.
Hence, it is 16

e) We get this piece by breaking the chikki into 8 equal parts.
Hence, it is 18

f) We get this part by breaking the chikki into 6 equal pieces.
Hence, it is 16

g) We get this part by breaking the chikki into 24 equal pieces.
Hence, it is 124

h) We get this part by breaking the chikki into 24 equal pieces.
Hence, it is 124mfrac

Page number: 158
Number of Questions: 5

Question 1: Continue this table of 12 for 2 more steps.
Screenshot%202025-06-10%20213903

Answer:

Here, 1749705283042represents a full roti (whole)

Step 1:

1749705354032

=12=1 times half

Step 2:

1749705382517

=12+12=2 times half

Step 3:

1749705404700

=12+12+12=3 times half

Step 4:

1749705448388

=12+12+12+12=4 times half

Step 5:

1749705467112

=12+12+12+12+12=5 times half

Step 6:

1749705490672

=12+12+12+12+12+12=6 times half =6×12=62=3

Step 7:

1749705540444

=12+12+12+12+12+12+12=7 times half =7×12=72

Question 2: Can you create a similar table for 14?

Answer:

Yes, we can create a similar table for 14.
Here

1749705616790

represents a full roti (whole).

1749705659137

14=1 times quarter

1749705684043

24=2 times quarter

1749705713545

34=3 times quarter

1749705733315

44=4 times quarter

1749705777083

54=5 times quarter

1749705799001

64=6 time quarter

1749705843875

74=7 times quarter

1749705862744

84=8 times quarter

Question 3: Make 13 using a paper strip. Can you use this to also make 16?

Answer: Yes, we can use this 13 to make 16.

Take a strip of paper.

Fold the strip into three equal parts and then open it up.

Screenshot%202025-06-10%20214549

When we divide each part of this 13 it is equal to Half of 13=13×12=16mfrac

Screenshot%202025-06-10%20214640

Question 4: Draw a picture and write an addition statement as above to show:

a. 5 times 14 of a roti
b. 9 times 14 of a roti

Answer:

a)

5 times 14 of a roti

=14+14+14+14+14=54
Which means 1 full and 14 roti.

Screenshot%202025-06-10%20214737

b)

9 times 14 of a roti

=14+14+14+14+14+14+14+14+14=94
Which means 2 full and 14 of roti.

Screenshot%202025-06-10%20215009

Question 5: Match each fractional unit with the correct picture:

13151816

Screenshot%202025-06-10%20213107

Answer:

Screenshot%202025-06-10%20215129

Page number: 160
Number of Questions: 5

Question 1: On a number line, draw lines of lengths 110,310, and 45.

Answer:

Step 1. Draw a line I. Mark a point O on it.

Step 2. Mark point A at a distance of 1 unit from 0.

Step 3. Divide OA into 10 equal parts.

Here, OP represents 1102,OQ represents 310, and OR represents 810=45mfrac

1749706163134

Question 2: Write five more fractions of your choice and mark them on the number line.

Answer:

Step 1. Let a number line OJ be divided into 10 equal parts.

Step 2. Now mark points A, B, C, D, ...., J on it.

Step 3. Here

OF represents 610=35
OG represents 710
OB represents 210=15
OH represents 810=45
Ol represents 910mfrac

1749706188830

Question 3: How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.

Answer: There are an infinite number of fractions between 0 and 1.

Example: 35,45,710,12 etc.

Question 4: What is the length of the blue line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is 12. So the blue line is 12 units long. Write the fraction that gives the length of the black line in the box.

Answer:
Length of black line is 12.
Length of black line is 12+12+12
Fraction that gives length of black line =32mfrac

1749706256173

Question 5: Write the fraction that gives the lengths of the black lines in the respective boxes.

1749706295189

Answer:

1749706312225

Page number: 162
Number of Questions: 2

Question 1: How many whole units are there in 72 ?

Answer:

72=7 times 12
=12+12+12+12+12+12+12
=1+12+1+12+1+12+12
=1+1+1+12
=3+12
=312
Hence, 3 whole units are there in 72.

Question 2: How many whole units are there in 43 and in 73 ?

Answer:
Here 43=4 times 13=13+13+13+13

=1+1+13+13=1+13=113
Hence, 1 whole unit is there in 43.

73=7 times 13
=13+13+13+13+13+13+13
=1+1+13+1+1+13+13
=1+1+13
=2+13
=213

Hence, 2 whole units are therein 73.

Page number: 162
Number of Questions: 3

Question 1: Figure out the number of whole units in each of the following fractions:
a. 83
b. 115
C. 94

Answer:
a) Numerator 8 is split/divided into as many 3 's, i.e., denominator number:
8=3+3+2=1+1+23=223

No. of whole units in 83=2 whole units.

b)
115=55+55+15[ Numerator 11=5+5+1]
=1+1+15
=215

Number of whole units =2

94=44+44+14

c) =1+14[ Numerator 9=4+4+1]

=214

Number of whole units =2

Question 2: Can all fractions greater than 1 be written as mixed numbers?

Answer: Yes, all fractions greater than 1 can be written as mixed fractions/numbers.

Question 3: Write the following fractions as mixed fractions (e.g., 92=412 ):
a. 92
b. 95
C. 2119
d. 479
e. 1211
f. 196

Answer:

(a)
92=22+22+22+22+12
=1+1+1+1+12
=412

(b)
95=55+45
=1+45
=145

(c)
2119
=1919+219
=1+219
=1219

(d)
479=99+99+99+99+99+29
=1+1+1+1+1+29
=5+29
=529

(e)
1211
=1111+111
=1+111
=1111

(f)
196
=66+66+66+16
=1+1+1+16
=3+16
=316

Page number: 163
Number of Questions: 1

Question 1: Write the following mixed numbers as fractions:
a. 314
b. 723
c. 949
d. 316
e. 2311
f. 3910

Answer:
a) Here 314=3+14
=1+1+1+14=44+44+44+14=134

b) Here 723=7+23
=1+1+1+1+1+1+1+23=33+33+33+33+33+33+33+23=233

c) Here 949=9+49
=9times99+49=819+49=859

d) Here 316=3+16
=3times66+16=186+16=196

e) We have 2311=2+311
=2times1111+311=2211+311=2511

(f)
3910=3+910=3times1010+910=3010+910=3910

Page number: 165
Number of Questions: 3

Question 1: Are 36,48,510 equivalent fractions? Why?

Answer: Here, simplest form of 36=3÷36÷3=12 [HCF of 3 and 6 is 3 ]
Simplest form of 48 is 4÷48÷4=12 [HCF of 4 and 8 is 4 ]
Simplest form of 510 is 5÷510÷5=12 [HCF of 5 and 10 is 5] Hence, 36,48,510 are equivalent fractions.

Question 2: Write two equivalent fractions for 26.

Answer: Here it is given 262×26×2,2×36×3,2×46×4
Hence 412,618,824 are equivalent fractions of 26.

Question 3: 46=== (Write as many as you can)

Answer:

46=4×26×2=4×36×3=4×46×4=812=1218=1624

Page number: 166
​​​​​​​Number of Questions: 3

Question 1: Three rotis are shared equally by four children. Show the division in the picture and write a fraction for how much each child gets. Also, write the corresponding division facts, addition facts, and multiplication facts.

The fraction of roti each child gets is________.

Division fact:

Additional fact:

Multiplication fact:

Compare your picture and answers with your classmates!

1749706577296

Answer:

1749706597504

Division fact:

3 wholes divided into 4 parts
3÷4=34

Addition fact:
Four times 34 added gives 3 wholes

=34+34+34+34=124=3

Multiplication fact:
4 parts of 34 make 3 wholes

4×34=3

Question 2: Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.

Answer:

As 2 rotis have to be shared equally by 4 children, we divide each roti into 4 parts and give

(a) 1 part of each roti to each child as shown below:

1749706648259

(b) 2 parts to each child as shown below

1749706668068

Question 3: Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?

1749706695718

Answer: Anil is in a group where 2 cakes were divided equally among 5 children.

1749706727668

Each cake gets divided into 5 parts, and Anil gets one part from each cake, i.e., 15+15=25 of a cake.

Page number: 168
​​​​​​​Number of Questions: 1

Question: Find the missing numbers:
a) 5 glasses of juice shared equally among 4 friends is the same as _____ glasses of juice shared equally among 8 friends.

Answer:  number of glasses  number of friends =54
Now to determine how many glasses of juice would be needed to give each of the 8 friends the same amount =8×54 =10 glasses

So, 10 glasses of juice shared equally among 8 friends is the same as 5 glasses of juice shared equally among 4 friends.

54=108

b) Find the missing numbers: 4 kg of potatoes divided equally in 3 bags is the same as 12 kg of potatoes divided equally in _______ bags.

Answer: Let x be the number of bags for 12 kg of potatoes, where each bag has the same amount of potatoes, then

12 kgx bags =43 kg per bag 12×3=4×x36=4xx=364x=943=129

c) Find the missing numbers: 7 rotis divided among 5 children is the same as _______ rotis divided among ________ children.

Answer: Dividing 7 rotis among 4 children gives 7 each child =75 of a roti. We can find an equivalent fraction by multiplying both the numerator and the denominator by the same number. For example, multiplying both by 2.

7×25×2=1410
So, 7 rotis divided among 5 children is the same as 14 rotis divided among 10 children.

75=1410mfrac

Page number: 173
​​​​​​​Number of Questions: 1

Question: Express the following fractions in their lowest terms:
a. 1751
b. 64144
C. 126147
d. 525112

Answer:
a) 1751=1717×3=13

b)

64144=64÷16144÷16=49

c) Here, 126147

=126÷7147÷7

[ HCF of 126 and 147 is 7×3=21 ]

=18÷321÷3=67

d) Here, 525 and 112 are both multiples of 7 ; we divide both by 7 .

=525÷7112÷7=7516mfrac

Page number: 174
​​​​​​​Number of Questions: 3

Question 1: Compare the following fractions and justify your answers:
a. 83,52
b. 49,37
c. 710,914
d. 125,85
e. 94,52

Answer:
a. Given fractions are 83 and 52.

Here LCM of the denominators 3 and 2 is 6.
Then, multiplying and dividing 83 by 2 and 52 by 3 , then

8×23×2=166,5×32×3=156 Hence, 166156166>15683>52

b. Given fractions are 49 and 37.

Here LCM of the denominators 9 and 7 is 63 .
then multiplying and dividing 49 by 7 and 37 by 9,
then
Here, 4×79×7=2863

3×97×9=27632863>276349>37

c. Given fractions are 710 and 914.

Here LCM of the denominators 10 and 14 is 70.
then multiplying and dividing 710 by 7 and 914 by 5,
then

7×710×7=4970,9×514×5=45704970>4570710>914

d. Given fractions are 125,85.
Here, denominator is same so 125>85

e. Given fractions are 94,52.

Here LCM of the denominators 4 and 2 is 4 . Then, multiplying and dividing 52 by 2 and Equivalent fractions with 4 are 5×22×2

104>9452>94


Question 2: Write the following fractions in ascending order.
a. 710,1115,25
b. 1924,56,712

Answer:
a)
LCM of 10,15 and 5=2×3×5=30
Now let us make the denominator of each fraction as LCM
7×310×3,11×215×2,2×65×62130,2230,1230
Clearly 1230<2130<2230

25<710<115

b) The given fractions are 1924,56,712

Here LCM of 24,6,12 is 24.

19×124×1,5×46×4,7×212×2

Thus 1924<2024>1424
On arranging in ascending Order, we get

1424,1924,2024712,1924,56

Question 3: Write the following fractions in descending order.
a. 2516,78,134,1732
b. 34,125,712,54

Answer:
a) LCM of 16,8,4,32=2×2×2×2×2=32

Now let us make the denominator of each fraction as LCM, thus

25×216×2,7×48×4,13×84×8,17×132×1=5032,2832,10432,1732
On arranging in descending order, we get

10432>5032>2832>1732134>2516>78>1732
Hence, given fractions in descending order are 134,2516,78 and 1732

b) Here LCM of 4,5,12,4 is 60.
Now let us make the denominator of each fraction as LCM, then
3×154×15,12×125×12,7×512×5,5×154×15=4560,14460,3560,756014460>7560>4560>3560
Hence, fractions in descending order are
125>54>34>712mfrac

Page number: 179
​​​​​​​Number of Questions: 3

Question 1: Add the following fractions using Brahmagupta's method:

a. 27+57+67
b. 34+13
c. 23+56
d. 23+27
e. 34+13+15
f. 23+45
g. 45+23
h. 35+58
i. 92+54
j. 83+27
k. 34+13+15
l. 23+45+37
m. 92+54+76

Answer:

(a)
Given: 2+5+67
=137
=7+67
=77+67
=1+67
=167

(b)
Here 34+13

Here, the LCM of the denominators 4 and 3 is 12
Equivalent fraction of 34 with denominators 12 is 912 and equivalent fraction of 13 with denominators 12 is 412

Hence,
912+412
=9+412=1312
=12+112
=1+112
=1112

(c)
23+56=2×23×2+5×16×1=46+56=96=6+36=1+36=1+12[ Bring to lowest term 36]=112

(d)
Expressing as equivalent fractions with denominators 21, we get

2×73×7+3×37×3=1421+921=2321=2121+221=1+221=1221

(e)
3×154×15+1×203×20+1×125×12=4560+2060+1260=45+20+1260=7760=60+1760=11760

(f)
Expressed as equivalent fractions with denominators 15, we get

2×53×5+4×35×3
=1015+1215
=2215
=15+715
=1515+715
=1+715
=1715

(g)
4×35×3+2×53×5
=1215+1015
=12+1015
=2215
=1515+715
=1+715
=1715

(h)
3×85×8+5×58×5
=2440+2540
=24+2540
=4940
=4040+940
=1+940
=1940

(i)
9×22×2+5×14×1=184+54=18+54=234=20+34=204+34=5×4+34=534

(j)
Expressing as equivalent fractions with denominators 21(LCM), we get

8×73×7+2×37×3
=5621+621
=6221
=42+2021
=21×221+2021
=21×221+2021
=2+2021
=22021

(k)
Expressing as equivalent fractions with denominators 60(LCM), we get

=3×154×15+1×203×20+1×125×12
=4560+2060+1260
=45+20+1260
=7760
=60+1760
=6060+1760
=1+1760
=11760

(l)

I) Expressing as equivalent fractions with denominators 105(LCM), we get

2×353×35+4×215×21+3×157×15
=70105+84105+45105
=70+84+45105
=199105
=105+94105
=105105+94105
=194105

(m)
Expressing as equivalent fractions with denominators 12(LCM), we get

9×62×6+5×34×3+7×26×2
=5412+1512+1412
=54+15+1412=8312
=12+12+12+12+12+12+1112
=12×6+1112
=12×612+1112
=61112

Question 2: Rahim mixes 23 litres of yellow paint with 34 litres of blue paint to make green paint. What is the volume of green paint he has made?

Answer:
Given quantity of yellow paint =23 litres
and quality of blue paint =34 litres
Volume of green paint made

=23 litres +34 litres 
Here LCM of 3 and 4 is 12.
Now expressed as equivalent fractions with a denominator of 12, we get

2×43×4+3×34×3=8+912=1712=12+512=1212+512 litres =1512 litres 

Question 3: Geeta bought 25 meter of lace and Shamim bought 34 meter of the same lace to put a complete border on a tablecloth whose perimeter is 1 meter long. Find the total length of the lace they have both bought. Will the lace be sufficient to cover the whole border?

Answer:
Given length of lace bought by Geeta =25 meter
and length of lace bought by Shamim =34 metre
Total length of lace =25+34 metre
LCM of 5 and 4 is 20
Now expressed as equivalent fractions with a denominator of 20, we get,

=2×45×4+3×54×5=8+1520=2320=2020+320=1+320 m
Total lace required = Perimeter =1 m
Hence, the total lace Geeta and Shamim have together is sufficient to cover the whole border.

Page number: 181
​​​​​​​Number of Questions: 3

Question 1: 5838

Answer:
As fractional unit is same i.e., 18 we shall simply subtract numerators keeping fractional unit as 18
Then 5838=538=28=14 (representing in simplest form)

Question 2: 7959

Answer:
As fractional unit is same i.e., 19 we shall simply subtract numerators keeping fractional unit as 19

7959=759=29

Question 3:
1027127

Answer: Here 1027127

=10127=927=13

Page number: 182
​​​​​​​Number of Questions: 3

Question 1: Carry out the following subtractions using Brahmagupta's method:
a. 815315
b. 25415
c. 5649
d. 2312

Answer:
a) Given 815315

Fractional unit for both fractions is 115 then

815315=8315=515=13

b) Given 25415mfrac

Here LCM of 5 and 15 is 15. Fractional unit for both fractions should be 115 then

25415=2×35×34×115×1=615415=6415=215

c) Given 5649

Hence LCM of 6 and 9 is 18. Fractional unit for both fractions should be 118 then
Given 5649

=5×36×34×29×2=1518818=15818=718

d) Given 2312

Here LCM of 3 and 2 is 6 Fractional unit for both fractions should be 16

So, 2312

=2×23×21×32×3=4636=436=16

Question 2: Subtract as indicated:
a. 134 from 103
b. 185 from 233
C. 297 from 457

Answer:

(a)
103134
=10×43×413×34×3
=40123912
=403912
=112

(b)
233185
=23×53×518×35×3
=115155415
=1155415
=6115
=4115

(c)

457297
=457297=45297=167=7×2+27=7×27+27=2+27=227

Question 3: Solve the following problems:

a. Jaya's school is 710 km from her home. She takes an auto for 12 km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?
b. Jeevika takes 103 minutes to take a complete round of the park, and her friend Namit takes 134 minutes to do the same. Who takes less time, and by how much?

Answer:
a) Given distance between Jaya's school and home is 710 km and distance covered by Jaya in auto is 12 km.
Distance Jaya covered by walking =710 km12 km
LCM of 10 and 2 is 10.

=[7101×52×5]km=[7510]km=210 km=15 km( lowest form )

b) Time taken by Jeevika to cover 1 round of the park =103 mm Time taken by Namit to cover 1 round of the 13 park =134 min

To find who takes less time we need to compare 103 and 134 Here, LCM of 3 and 4 is 12 .

So, the time taken by Jevika and Namit will be 10×43×4=4012 minutes and 13×34×3=3912 minutes respectively.
Clearly 4012>3912
Hence, Namit takes less time by 40123912=403912=112 minutes.

Fractions Class 6 Maths Chapter 7 - Topics

The topics discussed in the NCERT Solutions for class 6, chapter 7, Fractions, are:

  • Fractional units and equal shares
  • Fractional units as a part of a whole
  • Measuring using fractional units
  • Marking fractions on number lines
  • Mixed fractions
  • Equivalent fractions
  • Comparing fractions
  • Addition and Subtraction of Fractions

NCERT Solutions for Class 6 Maths Chapter 7 Fractions - Notes

Understanding fractions is learning to appreciate the beauty of parts within a whole. Here are some important notes that will help you to understand fractions better.

Fractions

Fractions can be defined as a part of a whole. When one unit is divided into several equal parts, each part is called a fractional unit. The fractional units are 12,13,14,15,16,,110,,150,,1100, etc. We also sometimes refer to fractional units as unit fractions.

Fractions Chart Grade 2-8 - LRC

We usually read the fraction 34 as 'three quarters' or 'three upon four', but reading it as ' 3 times 14 ', helps us to understand the size of the fraction because it shows what the fractional unit is (14) and how many such fractional units (3) there are.

Fraction on Number Line

1749486664869

We can easily mark fractions on a number line. Here we mark numbers as 0 and 1 and divide them into 7 equal parts. So, starting from 17,27 and till we write 1 as 77.

Mixed Fractions

These fractions are represented as a mix of 3 numbers instead of 2. They are improper fractions in which the numerator is bigger than the denominator, unlike the normal ones.
For example,
32=12+12+12=1+12
52=12+12+12+12+12=2+12mfrac

Equivalent Fractions

'Equivalent fractions' are those that denote the same length, but they are expressed in terms of different fractional units. For eg, 12,24, etc.

Addition and Subtraction of Fractions

Fractions can be added and subtracted provided their denominator are the same. If not so, we take the LCM to make the denominators the same and then perform operations.

Image result for addition of fractions

In the above case, the denominator is the same, so we have directly added the numerators. We must remember that neither in addition nor subtraction is the denominator added or subtracted.

Example:
25+13=2×35×3+1×53×5=615+515=1115
12+13=1×32×3+1×23×2=36+26=56mfrac

In the above example, we notice that the denominators of both fractions are not the same. Hence, we take the LCM first and then multiply the numerator and denominator by the desired factor to finally add them.

Subtraction of Fractions:

7636=46

In the above case, the denominator is the same, so we have directly subtracted the numerators.

If the denominators of both fractions are not the same. We take the LCM first and then multiply the numerator and denominator by the desired factor to finally subtract them.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions - Points to Remember

Fractions: Fractions are numbers that are used to represent a part of the whole. For example, 12 is a fraction representing half of the whole.

Improper fractions: Improper fractions are fractions in which the numerator is greater than the denominator. Eg. 72

Mixed fractions: Mixed fractions are another form of improper fractions in the form abc where bc is a proper fraction. Eg. 312

Equivalent Fractions: Equivalent fractions are fractions with the same value but different representations. Example: 12=36=48

Addition or Subtraction of Fractions: Addition and subtraction of fractions are possible only when the denominators are the same. For example, 15+35=45

If the denominators of the fractions to be added or subtracted are different, the LCM should be taken to make the denominators the same to add or subtract the fractions.
13+3626+36=56mfrac

NCERT Solutions for Class 6 Maths Chapter Wise

We at Careers360 compiled all the NCERT class 6 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Solutions for Class 6 Subject Wise

The NCERT Solutions provide the step-by-step solutions for all the questions in each chapter of the NCERT Book. To access subject-wise NCERT Solutions for Class 6, students can refer to the links below.

Students can also check the NCERT Books and the NCERT Syllabus for Class 6 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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