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NCERT Solutions for Class 7 Maths Chapter 8 Working With Fractions

NCERT Solutions for Class 7 Maths Chapter 8 Working With Fractions

Edited By Komal Miglani | Updated on Jun 26, 2025 09:20 AM IST

We use fractions in our daily lives all the time. Suppose you cut an apple into 4 pieces and gave one piece to your friend, so your friend has 14 of the apple. Again, you take 30 minutes to complete a work, so you have taken 12 an hour to do it. In the chapter 'Working With Fractions', we get to learn how to add, subtract, multiply, divide, and compare fractions easily. There are many occasions where using factions becomes an essential thing for us. Therefore, it is essential to understand this chapter with a proper understanding and clarity, and this is where the NCERT Solutions for Class 7 Maths come into play.

This Story also Contains
  1. NCERT Solutions for Class 7 Maths Chapter 8: Working With Fractions (Exercise)
  2. Working With Fractions Class 7 Maths Chapter 8: Topics
  3. NCERT Solutions for Class 7 Maths Chapter Wise
NCERT Solutions for Class 7 Maths Chapter 8 Working With Fractions
NCERT Solutions for Class 7 Maths Chapter 8 Working With Fractions

These NCERT Solutions provide detailed and step-by-step explanations for the students. With the help of these solutions, they can practice the problems, verify their answer and strengthen their weak areas that need improvement. This will also clear their doubts and give them a proper idea about the concepts of the chapter. Students can refer to the NCERT Solutions for Class 7 to access the subject-wise solutions of Class 7.

NCERT Solutions for Class 7 Maths Chapter 8: Working With Fractions (Exercise)

Page number: 176
Number of Questions: 5

Figure it Out

Question 1: Tenzin drinks 12 glass of milk every day. How many glasses of milk does he drink in a week? How many glasses of milk did he drink in the month of January?

Solution:

Glass of milk drank every day =12

Glasses of milk drank in a week =12×7=72

Days in month of January = 31

Glasses of milk drank in month of January =31×12=312=1512.

Question 2: A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make ___ km of the water canal. If they work 5 days a week, they can make ___ km of the water canal in a week.

Solution:

The length of the canal made by workers in 8 days =1 km

The length of canal made by workers in 1 day =18 km

By working 5 days a week, the length of the canal made by workers =5×18=58 km.

Question 3: Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks?

Solution:

Amount of oil shared by 3 families in a week = 5 litres

Amount of oil one family got in a week =53 litres

Amount of oil one family got in 4 weeks =4×53=203=623 litres.

Question 4: Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets 56 hours later than the previous day. How many hours after 10 pm will the moon set on Thursday?

Solution:

Number of days from Monday to Thursday =3

Delay in moon setting time per day =56 hour

Delay in moon setting time over 3 days =3×56 hour =52=2.5 hours

Thus, the Moon will set 2.5 hours or 2 hours and 30 minutes after 10 PM on Thursday.

Question 5: Multiply and then convert it into a mixed fraction:

(a) 7×35 (b) 4×13 (c) 97×6 (d) 1311×6

Solution:

(a) 7×35=215=415

(b) 4×13=43=113

(c) 97×6=547=757

(d) 1311×6=7811=7111

Page number: 180
Number of Questions: 2

Figure it Out

Question 1: Find the following products. Use a unit square as a whole for representing the fractions:

(a) 13×15 (b) 14×13 (c) 15×12 (d) 16×15

Solution:

(a) 13×15

Unit square divided into 3 rows and 5 columns = 15 parts.

1 part shaded =115. Thus, 13×15=115

1750865343803

(b) 14×13

A unit square divided into 4 rows and 3 columns = 12 parts.

1 part shaded =112. Thus, 14×13=112

1750865343641

(c) 15×12

A unit square divided into 5 rows and 2 columns = 10 parts.

1 part shaded =110. Thus, 15×12=110

1750865343742

(d) 16×15

A unit square divided into 6 rows and 5 columns = 30 parts.

1 part shaded =130. Thus, 16×15=130

1750865344228

Question 2: Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.

(a) 23×45 (b) 14×23 (c) 35×12 (d) 46×35

Solution:

(a) 23×45

First, the whole is divided into 5 rows and 3 columns, creating (5 × 3) = 15 equal parts.

The value we get by dividing 45 into 3 equal parts is 43×5.

Thus, we multiply this result by 2 to get the product. This is 2×43×5.

So, 23×45=2×43×5=815.

1750865345715

(b) 14×23

First, the whole is divided into 3 rows and 4 columns, creating (3 × 4) = 12 equal parts.

The value we get by dividing 23 into 4 equal parts is 24×3.

So, 14×23=1×24×3=212.

1750865344333

(c) 35×12

First, the whole is divided into 2 rows and 5 columns, creating (2 × 5) = 10 equal parts. The value we get by dividing 12 into 5 equal parts is 15×2.

Thus, we multiply this result by 3 to get the product. This is 3×15×2.

So, 35×12=3×15×2=310.

1750865345621

(d) 46×35

First, the whole is divided into 5 rows and 6 columns, creating (5 × 6) = 30 equal parts.

The value we get by dividing 35 into 6 equal parts is 36×5.

Thus, we multiply this result by 4 to get the product. This is 4×36×5.

So, 46×35=4×36×5=1230.

1750865345750

Page number: 183
Number of Questions: 5

Figure it Out

Question 1: A water tank is filled from a tap. If the tap is open for 1 hour, 710 of the tank gets filled. How much of the tank is filled if the tap is open for

(a) 13 hour

(b) 23 hour

(c) 34 hour

(d) 710 hour

(e) For the tank to be full, how long should the tap be running?

Solution:

Part of the tank filled in 1 hour =710

(a) Part of the tank filled in 13 hour =13×710=730

(b) Part of the tank filled in 23 hour =23×710=1430=715

(c) Part of the tank filled in 34 hour =34×710=2140

(d) Part of the tank filled in 710 hour =710×710=49100

(e) Part of the tank filled in 1 hour =710

⇒ Time required to fill 710 of a tank =1 hour

Time required to fill 1 tank =1÷710=1×107=107 hours =137 hours.

Question 2: The government has taken 16 of Somu’s land to build a road. What part of the land remains with Somu now? She gives half of the remaining part of the land to her daughter, Krishna and 13 of it to her son, Bora. After giving them their shares, she kept the remaining land for herself.

(a) What part of the original land did Krishna get?

(b) What part of the original land did Bora get?

(c) What part of the original land did Somu keep for herself?

Solution:

Part of Somu's land acquired by the government =16

Somu's original land =116=616=56

(a) Part of the land given to Krishna =12 of original land =12×56=512

(b) Part of the land given to Bora =13 of original land =13×56=518

(c) Part of the land Somu kept for herself =56(512+518)

=56(3×5+5×236)

=56(15+1036)

=56(2536)

=6×52536

=302536=536

Question 3: Find the area of a rectangle of sides 334ft and 935ft.

Solution:

Area of rectangle =(334×935)ft=(154×485)ft=72020ft=36 ft

Question 4: Tsewang plants four saplings in a row in his garden. The distance between two saplings is 34 m. Find the distance between the first and last sapling.

Solution:

Number of saplings in a row in the garden =4

Distance between two saplings =34

Distance between first and last sapling =34+34+34=3+3+34=94 m=214 m.

Question 5: Which is heavier: 1215 of 500 grams or 320 of 4 kg?

Solution:

1215×500 g=45×500 g=400 grams.

320×4 kg=320×4000 g=600 grams.

Since 600 g is heavier than 400 g

Therefore, 320 of 4 kg is heavier than 1215 of 500 grams.

Page number: 196-198

Number of Questions: 12

Question 1: Evaluate the following:

3÷79

144÷2

23÷23

146÷73

43÷34

74÷17

82÷415

15÷19

16÷1112

323÷138

Solution:

Expression

Calculation

Final Value

3÷79

3÷79
=3×97
=277

277

144÷2

144÷2

=142×12

=72

72

23÷23

23÷23

=23×32

=1

1

146÷73

146÷73

=146×37

=1

1

43÷34

43÷34

=43×43

=169

169

74÷17

74÷17

=74×71

=494

494

82÷415

82÷415

=82×154

=15

15

15÷19

15÷19

=15×91

=95

95

16÷1112

16÷1112

=16×1211

=211

211

323÷138

323÷138
=113÷118
=113×811
=83

83

Question 2: For each of the questions below, choose the expression that describes the solution. Then simplify it.

(a) Maria bought 8 m of lace to decorate the bags she made for school. She used 14 m for each bag and finished the lace. How many bags did she decorate?

(i) 8×14

(ii) 18×14

(iii) 8÷14

(iv) 14÷8

Solution:

Maria bought 8 m of lace to decorate the bags she made for school.
So, total lace = 8 m

One bag took 14 m of lace.
So, total bags she decorated =8÷14=814=8×4=32

Hence, the correct answer is option (iii).

Question 2: For each of the questions below, choose the expression that describes the solution. Then simplify it.

(b) 12 meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

(i) 8×12

(ii) 12÷18

(iii) 8÷12

(iv) 12÷8

Solution:

12 meter of ribbon is used to make 8 badges.
So, for each bandage, the amount of ribbon used
=12÷8=12×18=116

Hence, the correct answer is option (iv).

Question 2: For each of the questions below, choose the expression that describes the solution. Then simplify it.

(c) A baker needs 16 kg of flour to make one loaf of bread. He has 5 kg of flour. How many loaves of bread can he make?

(i) 5×16

(ii) 16÷5

(iii) 5÷16

(iv) 5×6

Solution:

A baker needs 16 kg of flour to make one loaf of bread.

Total flour = 5 kg
So, the number of bread made =5÷16=5×6=30

Hence, the correct answer is option(iii).

Question 3: If 14 kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

Solution:

12 rotis takes 14 kg of flour to be made.
So, 6 rotis will take half the amount, as the number of rotis is also halved.
So, required amount of flour =14÷2=14×12=18

Hence, the correct answer is 18.

Question 4: Pạtiganita, a book written by Sridharacharya in the 9th century CE, mentions this problem: "Friend, after thinking, what sum will be obtained by adding together 1÷16,1÷110,1÷113,1÷19, and 1÷12. What should the friend say?

Solution:

1÷16=1×6=6
1÷110=1×10=10
1÷113=1×13=13
1÷19=1×9=9
1÷12=1×2=2

So, total = 6 + 10 + 13 + 9 + 2 = 40

Hence, the correct answer is 40.

Question 5: Mira is reading a novel that has 400 pages. She read 15 of the pages yesterday and 310 of the pages today. How many more pages does she need to read to finish the novel?

Solution:

Total pages of the novel = 400

15 of the pages =15×400=80 pages
310 of the pages =310×400=120 pages

Pages left to read =40080120=200 pages

Hence, the correct answer is 200 pages.

Question 6: A car runs 16 km using 1 litre of petrol. How far will it go using 234 litres of petrol?

Solution:

Using 1 litre of petrol, a car runs 16 km.
So, using 234=114 litres of petrol, the car will run
=114×16=44 km

Hence, the correct answer is 44 km.

Question 7: Amritpal decides on a destination for his vacation. If he takes a train, it will take him 516 hours to get there. If he takes a plane, it will take him 12 hour. How many hours does the plane save?

Solution:

The train takes 516 hours to get to the destination.
But the plane will take him 12 hour.

Hours saved =51612=31612=286=143 hours

Hence, the correct answer is 143 hours.

Question 8: Mariam's grandmother baked a cake. Mariam and her cousins finished 45 of the cake. The remaining cake was shared equally by Mariam's three friends. How much of the cake did each friend get?

Solution:

Let the total cake be 1.
Mariam and her cousins finished 45 of the cake.
So, remaining cake =115=45
The remaining cake was shared equally by Mariam's three friends.
So, share of every friend =45÷3=45×13=415th of the cake

Hence, the correct answer is 415.

Question 9: Choose the option(s) describing the product of (565465×707676) :

(a) >565465

(b) <565465

(c) >707676

(d) <707676

(e) >1

(f) <1

Solution:

Here, 565 > 465
So, 565465>1
Similarly, 707 > 676
So, 707676>1
When you multiply two fractions that are both slightly greater than 1, the result will be greater than 1.
So, (565465×707676)>1
Since both 565465 and 707676 are greater than 1, their product will be greater than either fraction individually.
Therefore, the product is:

  • Greater than 565465 (a),
  • Greater than 707676 (c),
  • Greater than 1 (e).

Hence, the correct answer is options (a), (c), and (e).

Question 10: What fraction of the whole square is shaded?

Solution:

Here, four small squares of equal size.
Now, consider the small square.
If you divide it into 8 equal triangles, the shaded part will occupy 3 triangles.
So, the shaded part is 38th of the small square.

the shaded part is 38×14=332 part of the big square

Hence, the correct answer is 332 part.

Question 11: A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?


Solution:

Let the total ants at the start be 1.
At the first point, ants split into two ways.
So, 12 the ants go either way.
In the next intersection, the fraction of ants on each way =12÷2=14
At the third point, ants split in half again.
So, fraction of ants at each way =14÷4=116
At the next point, ants split into 2 ways.
So fraction of ants at each way =116÷2=132

Hence, a fraction of ants at the mango tree is:
12+14+116+116+132=2932

The fraction of ants near the sugarcane field is:
132+116=332

Question 12: What is
112?
(112)×(113)?(112)×(113)×(114)×(115)?(112)×(113)×(114)×(115)×(116)×(117)×(118)×(119)×(1110)?

Make a general statement and explain.

Solution:

Expression 1:
112=2212=12

Expression 2:
(112)×(113)
=12×23
=13

Expression 3:
(112)×(113)×(114)
=12×23×34
=14

Expression 4:
(112)×(113)×(114)×(115)×(116)
×(117)×(118)×(119)×(1110)
=12×23×34×45×56×67×78×89×910
=110

The general statement is that multiplying the terms (11k) for k=2,3,4,,n results in 1n.

Working With Fractions Class 7 Maths Chapter 8: Topics

The important topics that are covered in NCERT Class 7 Maths Chapter 8, Working With Fractions, are:

  • Multiplication of Fractions
  • Connection between the Area of a Rectangle and Fraction Multiplication
  • Simplifying to Lowest Form
  • Order of Multiplication
  • Division of Fractions
  • Some Problems Involving Fractions

NCERT Solutions for Class 7 Maths Chapter Wise

Given below are the chapter-wise NCERT solutions of class 7 mathematics:

NCERT Solutions for Class 7 Subject-Wise

The NCERT Solutions for Class 7, Subject-wise, can be downloaded using the links below.

Students can also check the NCERT Books and the 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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