Class 6 Fractions - TestsLet us first convert them to like fractions. Step 2 Let us find the LCM of...

$: Class 6 Fractions - TestsLet us first convert them to like fractions. Step 2 Let us find the LCM of the denominators 272 and 16. The LCM of 272 and 16 is 272. Step 3 What should we$
Class 6Fractions

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Answer the questions

(1) Solve the following and reduce to the simplest form:

A) ( 11

2 ÷ 2

1

2 ) + 1

1

2 B) ( 2

1

2 ÷ 2

1

2 ) + 1

1

2

C) ( 22

7 ÷ 2

5

7 ) + 3

3

7 D) ( 1

1

2 ÷ 1

1

2 ) + 1

1

2

(2) Simplify the following fractions and reduce them to the simplest form:

A) 716

475 -

14

25 B)

69

40 -

7

8 C)

67

36 -

11

12

D) 229

171 -

4

9 E)

393

272 -

9

16 F)

113

80 -

9

16

(3) Akshiti is reading a book titled 'Creatures of the Deep Sea', and this book has 450 pages. Akshiti

manages to read 2

45 of the book every day. After 2 days, how many pages has she read?

(4) The parking lot of the Shopping Complex has a capacity of 266 cars. On Friday 17

19 of the parking lot

was occupied with cars. How many additional number of cars could be parked there?

(5) Shade the images to show the following fraction addition.

7

12 +

1

12 =

and makes

(6) Shilpa has Rs. 1995 with her. She gives 8

15 of this to her sister. Out of the remaining money, she

gives Rs. 266 to her brother. What fraction of the original amount is left with her?

(7) What will be the result if we divide the sum of 51

5 and 2

2

3 by their difference?

ID : in-6-Fractions [1]

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(8) Sarika is constructing a building. She finishes 1

15 of the construction in 2

1

3 weeks. In how many

days will she finish constructing the building?

(9) Sanjana is baking two different types of bread. She needs 13

4 cup of flour for one type of bread and

41

5 cup of flour for the other type. How many cup of flour will Sanjana need to bake both types of

bread?

Choose correct answer(s) from the given choices

(10) Add:

+

Figure A Figure B

a. 58

40 b.

57

43

c. 40

57 d.

57

40

(11) A cat is walking on the periphery of a regular polygon, as shown below.

If it starts from point s, in the clockwise direction, which side will the cat reach after walking 28

30

distance on the periphery?

a. B b. F

c. A d. E



(12) What is the mixed fraction that represents the shaded part in the figure below?

a. 511

5 b. 5

8

11

c. 5

11 d. 4

5

11

(13) The fractions 18

53 and

22

53 are :

a. Mixed Fractions b. Unlike Fractions

c. Like Fractions d. Improper Fractions

Fill in the blanks

(14) Fill in the blank to make the two fractions equivalent:

A) 6

13 =

117 B)

5 =

24

40

C) 6

= 30

55 D)

11 =

40

110

(15) Add the following fractions and reduce them to the simplest form:

A) 5

1 +

6

7 = B)

2

6 +

6

5 = C)

7

1 +

6

2 =

D) 8

3 +

2

8 = E)

3

8 +

8

6 = F)

7

7 +

5

4 =

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Answers

(1) A) 21

10

Step 1

( 11

2 ÷ 2

1

2 ) + 1

1

2 can be written as:

( 3

2 ÷

5

2 ) +

3

2 ...[ Converting mixed fractions into improper fractions ]

= ( 3

2 ×

2

5 ) +

3

2 ...[ Dividing by a fraction is same as multiplying with its

reciprocal]

= 3

5 +

3

2

= 6 + 15

10

= 21

10

Step 2

Thus, the simplest form of ( 11

2 ÷ 2

1

2 ) + 1

1

2 is

21

10 .



B) 5

2

Step 1

( 21

2 ÷ 2

1

2 ) + 1

1


( 5

2 ÷

5

2 ) +

3


= ( 5

2 ×

2

5 ) +

3


reciprocal]

= 5

5 +

3

2

= 10 + 15

10

= 25

10

= 5

2

Step 2


2 ÷ 2

1

2 ) + 1

1

2 is

5

2 .



C) 568

133

Step 1

( 22

7 ÷ 2

5

7 ) + 3

3


( 16

7 ÷

19

7 ) +

24


= ( 16

7 ×

7

19 ) +

24


reciprocal]

= 16

19 +

24

7

= 112 + 456

133

= 568

133

Step 2


7 ÷ 2

5

7 ) + 3

3

7 is

568

133 .



D) 5

2

Step 1

( 11

2 ÷ 1

1

2 ) + 1

1


( 3

2 ÷

3

2 ) +

3


= ( 3

2 ×

2

3 ) +

3


reciprocal]

= 3

3 +

3

2

= 6 + 9

6

= 15

6

= 5

2

Step 2


2 ÷ 1

1

2 ) + 1

1

2 is

5

2 .

(2) A) 18

19

Step 1

We need to subtract the following unlike fractions:

716

475 and

14

25 .

Let us first convert them to like fractions.

Step 2

Let us find the LCM of the denominators 475 and 25.The LCM of 475 and 25 is 475.

Step 3

What should we multiply with the denominator 475 to get the LCM 475?

It is 475

475 = 1.



So, the first fraction can be written as:

716 × 1

475 =

716

475

Step 4

Similarly, the second fraction can be written as:

14 × 19

475 =

266

475

Step 5

Let us now subtract the like fractions:

716

475 -

266

475

= 716 - 266

475

= 450

475

Step 6

Let us now find the simplest form of the fraction 450

475 , by dividing the numerator 450

and denominator 475, by their HCF.The HCF of 450 and 475 = 25.

Thus, the simplest form of the fraction 450

475 =

450

25

475

25

= 18

19 .

B) 17

20

Step 1


69

40 and

7

8 .


Step 2


Step 3




It is 40

40 = 1.


69 × 1

40 =

69

40

Step 4


7 × 5

40 =

35

40

Step 5


69

40 -

35

40

= 69 - 35

40

= 34

40

Step 6





40 =

34

2

40

2

= 17

20 .

C) 17

18

Step 1


67

36 and

11

12 .


Step 2




Step 3


It is 36

36 = 1.


67 × 1

36 =

67

36

Step 4


11 × 3

36 =

33

36

Step 5


67

36 -

33

36

= 67 - 33

36

= 34

36

Step 6





36 =

34

2

36

2

= 17

18 .

D) 17

19

Step 1


229

171 and

4

9 .


Step 2




Step 3


It is 171

171 = 1.


229 × 1

171 =

229

171

Step 4


4 × 19

171 =

76

171

Step 5


229

171 -

76

171

= 229 - 76

171

= 153

171

Step 6





171 =

153

9

171

9

= 17

19 .

E) 15

17

Step 1


393

272 and

9

16 .




Step 2


Step 3


It is 272

272 = 1.


393 × 1

272 =

393

272

Step 4


9 × 17

272 =

153

272

Step 5


393

272 -

153

272

= 393 - 153

272

= 240

272

Step 6





272 =

240

16

272

16

= 15

17 .

F) 17

20

Step 1




113

80 and

9

16 .


Step 2


Step 3


It is 80

80 = 1.


113 × 1

80 =

113

80

Step 4


9 × 5

80 =

45

80

Step 5


113

80 -

45

80

= 113 - 45

80

= 68

80

Step 6





80 =

68

4

80

4

= 17

20 .



(3) 40

Step 1

Number of pages in the book Akshiti is reading = 450

Step 2

Number of pages read by Akshiti in one day = 2

45 of the book

= 2

45 × 450

= 900

45

= 20

Step 3

Number of pages read by Akshiti in 2 days = 20 × 2 = 40

Step 4

Therefore, after the 2 days, she reads 40 pages.

(4) 28

Step 1

Total number of cars that can be parked = 266

Step 2

Number of cars already parked = 17

19 × 266

= 238

Step 3

Number of cars that can be parked in vacant space = Total - Occupied= 266 - 238= 28

(5) 7

12 +

1

12 =

+ =



(6) 1

3

Step 1

The initial amount of money Shilpa had = Rs. 1995

Step 2

The amount of money Shilpa gives to her sister = 8

15 of the initial money she had

= 8

15 of Rs. 1995

= 8

15 × 1995

= 8 × 133 ... (Dividing 1995 by 15)= Rs. 1064

Step 3

The amount remaining with her after giving money to her sister = Amount Shilpa initially had -Amount she gave to her sister= Rs. 1995 - Rs. 1064= Rs. 931

Step 4

The amount Shilpa gives to her brother = Rs. 266

Step 5

The remaining amount after giving money to her brother = Rs. 931 - Rs. 266 = Rs. 665

Step 6

Now, we know that the final amount of money left with Shilpa is Rs. 665. We need to find whatfraction of Rs. 1995 is Rs. 665. This fraction will be the one with the remaining amount as

numerator and the initial amount as denominator, that is, 665

1995 .

Step 7

Let us now reduce 665

1995 to the simplest form by dividing both the denominator and the

numerator by 665, the HCF of 665 and 1995:

665

1995 =

665 ÷ 665

1995 ÷ 665

= 1

3

Step 8



Hence, the fraction of the original amount that is left with Shilpa is 1

3 .

(7) 59

19

Step 1

We have to divide the sum of 51

5 and 2

2

3 by the difference of 5

1

5 and 2

2

3 .

Step 2

The sum of 51

5 and 2

2

3 = 5

1

5 + 2

2

3

= 26

5 +

8

3

= 26 × 3 + 8 × 5

15

= 78 + 40

15

= 118

15

Step 3

The difference of 51

5 and 2

2

3 = 5

1

5 - 2

2

3

= 26

5 -

8

3

= 26 × 3 - 8 × 5

15

= 78 - 40

15

= 38

15

Step 4

Now, let us divide the sum of 51

5 and 2

2

3 by their difference. We get:



=

118

15

38

15

= 118

15 ×

15

38

= 59

19

Step 5

Hence, we get 59

19 as the result on dividing the sum of 5

1

5 and 2

2

3 by their difference.

(8) 245

Step 1

We know, 1 week = 7 days

Therefore, the number of days in 21

3 weeks = 2

1

3 × 7 days

= 7

3 × 7 days

= 49

3 days

Step 2

1

15 of the construction is finished in

49

3 days.

Therefore, the whole construction will be finished in =

49

3

1

15

days

= 49

3 ×

15

1 days

= 245 days.

(9) 519

20 cups

Step 1



In order to find the total amount of flour needed to bake both types of bread, let us add the 13

4

cup and 41

5 cup.

Step 2

Let us convert the given mixed fractions into improper fractions.

13

4 =

4 × 1 + 3

4 =

7

4

41

5 =

5 × 4 + 1

5 =

21

5

Step 3

We find that the two fractions are unlike fractions.

Let us now take the L.C.M of the denominators to convert them into like fractions.

L.C.M of 4 and 5 = 20

So, 7

4 =

7

4 ×

5

5 =

35

20

and 21

5 =

21

5 ×

4

4 =

84

20

Step 4

Now, let us add the two like fractions.

So, 35

20 +

84

20 =

119

20

Step 5

Converting 119

20 into mixed fraction:

Dividend↴

Divisor→ 20 ) 1 1 9 ( 5 ←Quotient

1 0 0

Remainder← 19

Thus, we have:

119

20 = 5

19

20

Step 6



Thus, Sanjana requires a total of 519

20 cup of flour to bake two different types of bread.

(10) d. 57

40

Step 1

We see that the 5 out of 8 parts of figure A are shaded.

So, the fraction of shaded part of figure A = 5

8

Also, 4 out of 5 parts of figure B are shaded.

So, the fraction of shaded part of figure B = 4

5

Step 2

In order to add the unlike fractions, let us first convert them into like fractions.

L.C.M of 8 and 5 = 40

So, 5

8 =

5

8 ×

5

5 =

25

40

and 4

5 =

4

5 ×

8

8 =

32

40

Step 3

Adding:

5

8 +

4

5 =

25

40 +

32

40 =

57

40

Step 4

Hence, + = 57

40



(11) b. F

Step 1

If we look at the regular polygon carefully, we notice that there are 6 sides of a regular polygon.

Therefore, the length of a side of the regular polygon = 1

6

Step 2

Since, the distance walked by the cat on the periphery, in the clockwise direction = 28

30

Therefore, the number of sides walked by the cat on the regular polygon =

Distance walked by the cat

Length of a side of the polygon

=

28

30

1

6

= 28

30 ×

6

1

= 28

5

= 5.6

It means that the cat walked on 5 sides of the regular polygon and the cat is walking on the 6th

side of the regular polygon, in the clockwise direction.

Step 3

Since, it started from point s, the cat will be on the side F, after walking 28

30 distance on the

periphery in the clockwise direction.

Step 4

Hence, option b is the correct answer.



(12) d. 45

11

Step 1

A fraction represents a particular part of a whole entity.

Step 2

In the given figures we can see that the circles are divided into 11 parts each. The first 4 circles isfully shaded, and in the last circle 5 out of 11 parts are shaded.

Step 3

This means that the first 4 circles will be represented by 4, and the last circle will be represented

by 5

11 .

Step 4

So the mixed fraction that represents the shaded part in the figure is 45

11 .

(13) c. Like Fractions

We find that the fractions 18

53 and

22

53 have the same denominator.

Therefore, the fractions 18

53 and

22

53 are Like Fractions.

(14) A) 54

Step 1

We know that Equivalent fractions are obtained by multiplying or dividing thenumerator and denominator of a fraction by the same number.

Step 2

We can see that we have multiplied the denominator 13 with 9 to get the denominatorof another fraction as 117.

Step 3

Since both the fractions are equivalent, the numerator of the first fraction have to bemultiplied with the same number to get the numerator of the second fraction.

Step 4

This means the new numerator will be equal to 6 × 9 = 54.



B) 3

Step 1


Step 2

We can see that we have divided the denominator 40 by 8 to get the denominator ofanother fraction as 5.

Step 3

Since both the fractions are equivalent, the numerator of the second fraction have tobe divided by the same number to get the numerator of the first fraction.

Step 4

This means the new numerator will be equal to 24

8 = 3.

C) 11

Step 1


Step 2

We can see that we have divided the numerator 30 by 5 to get the numerator ofanother fraction as 6.

Step 3

Since both the fractions are equivalent, the denominator of the second fraction have tobe divided by the same number to get the denominator of the first fraction.

Step 4

This means the new denominator will be equal to 55

5 = 11.



D) 4

Step 1


Step 2

We can see that we have divided the denominator 110 by 10 to get the denominator ofanother fraction as 11.

Step 3

Since both the fractions are equivalent, the numerator of the second fraction have tobe divided by the same number to get the numerator of the first fraction.

Step 4

This means the new numerator will be equal to 40

10 = 4.

(15) A) 5

1 +

6

7 =

41

7

Step 1

The fractions 5

1 and

6

7 are unlike fractions as their denominators are different. We

will first convert the given fractions into equivalent like fractions.

Step 2

Let us first find the LCM of the denominators 7 and 1. The LCM is 7.

Step 3

To write 5

1 as an equivalent fraction which has 7 as denominator, we need to

multiply both the numerator and denominator by 7

1 = 7. So, the equivalent fraction is:

5 × 7

1 × 7 =

35

7

Step 4

Similarly, to write 6

7 as an equivalent fraction which has 7 as denominator, we need

to multiply both the numerator and denominator by 7

7 = 1. So, the equivalent fraction

is:



6 × 1

7 × 1 =

6

7

Step 5

Now, we can add the equivalent like fractions by adding the numerators together andkeeping the denominator same:

35

7 +

6

7 =

35 + 6

7 =

41

7

Step 6

In order to convert the fraction 41

7 in the simplest/lowest form, let us divide both the

numerator and denominator by their HCF.

Step 7

The HCF of 41 and 7 is 1.

Step 8

Hence, the simplest/lowest form of 41

7 is

41

1

7

1

= 41

7

B) 2

6 +

6

5 =

23

15

Step 1

The fractions 2

6 and

6



Step 2


Step 3

To write 2




is:

2 × 5

6 × 5 =

10

30



Step 4




5 = 6. So, the equivalent

fraction is:

6 × 6

5 × 6 =

36

30

Step 5


10

30 +

36

30 =

10 + 36

30 =

46

30

Step 6




Step 7


Step 8


30 is

46

2

30

2

= 23

15

C) 7

1 +

6

2 =

10

1

Step 1

The fractions 7

1 and

6



Step 2


Step 3

To write 7





1 = 2. So, the equivalent fraction is:

7 × 2

1 × 2 =

14

2

Step 4





is:

6 × 1

2 × 1 =

6

2

Step 5


14

2 +

6

2 =

14 + 6

2 =

20

2

Step 6




Step 7


Step 8


2 is

20

2

2

2

= 10

1

D) 8

3 +

2

8 =

35

12

Step 1

The fractions 8

3 and

2





Step 2


Step 3

To write 8




is:

8 × 8

3 × 8 =

64

24

Step 4





fraction is:

2 × 3

8 × 3 =

6

24

Step 5


64

24 +

6

24 =

64 + 6

24 =

70

24

Step 6




Step 7


Step 8


24 is

70

2

24

2

= 35

12



E) 3

8 +

8

6 =

41

24

Step 1

The fractions 3

8 and

8



Step 2


Step 3

To write 3




is:

3 × 3

8 × 3 =

9

24

Step 4





fraction is:

8 × 4

6 × 4 =

32

24

Step 5


9

24 +

32

24 =

9 + 32

24 =

41

24

Step 6




Step 7


Step 8




24 is

41

1

24

1

= 41

24

F) 7

7 +

5

4 =

9

4

Step 1

The fractions 7

7 and

5



Step 2


Step 3

To write 7




is:

7 × 4

7 × 4 =

28

28

Step 4





fraction is:

5 × 7

4 × 7 =

35

28

Step 5


28

28 +

35

28 =

28 + 35

28 =

63

28

Step 6






Step 7


Step 8


28 is

63

7

28

7

= 9

4



Class 6 Fractions - TestsLet us first convert them to like fractions. Step 2 Let us find the LCM of...

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