grade 7 fractions - tests · let us convert the given mixed fractions into improper fractions. 3 1...

$: Grade 7 Fractions - Tests · Let us convert the given mixed fractions into improper fractions. 3 1 6 = 6 × 3 + 1 6 = 19 6 4 1 5 = 5 × 4 + 1 5 = 21 5 Step 3 We find that the two$
Grade 7Fractions

For more such worksheets visit www.edugain.com

Answer the questions

(1) Solve and simplify: ( 32

4 + 2

2

4 ) × ( 3

2

4 - 2

2

4 )

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

A) 4

10 ×

2

3 ÷

2

9 B)

2

8 ×

2

3 ÷

1

6

C) 2

5 ×

3

10 ÷

13

30 D)

1

7 ×

4

5 ÷

5

10

(3) The parking lot of the Shopping Complex has a capacity of 590 cars. On Fridays the ratio of occupiedspots to empty spots in the parking lot was 3:7. How many additional cars could park there?

(4) While doing operations on fractions, in which order should the operations be done among these fouroperations?+ x - ÷

(5) Write the answer in short form (decimal number) :

9000 + 100 + 70 + 6 + 8

10 +

3

1000

(6) Solve the following: 4

5 ×

3

5 - 7

14

.

(7) Ameerah is baking two different types of bread. She needs 31

6 cup of flour for one type of bread and

41

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

bread?

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

A) 14

5 - 1

4

5 ÷ 2

1

5 B) 1

1

6 - 1

5

6 ÷ 4

5

6

C) 14

6 - 3

4

6 ÷ 1

5

6 D) 2

1

5 - 1

1

5 ÷ 2

3

5

ID : ae-7-Fractions [1]

Copyright 2017 www.edugain.com Personal use only. Commercial use is strictly prohibited.

https://www.edugain.com/math/grade-7/Fractions

(9) Aziz covered 32

5 km in the first hour, 3

3

4 km in the second hour, and 4

9

4 km in the third hour.

Find the total distance covered by Aziz in three hours.

(10) Find the value of 3

4 +

( 2

1 -

4

5 ) × 10

1 - 1

2

.

(11) Simplify the following fractions:

A) 17139

195 - 5

13

15 B) 28

59

182 - 2

7

13

C) 17146

255 - 5

12

17 D) 36

99

182 - 7

8

13

(12) There are two containers out of which one can hold 11

3 liters of milk while the other can hold 3

1

2

liters of milk. Suppose there are 9 containers of the first type and 6 containers of the second type. Howmany liters of milk is required to fill them both?

Choose correct answer(s) from the given choices

(13) Add:

+

Figure A Figure B

a. 11

11 b.

11

9

c. 14

9 d.

9

11



(14) Which figure shows fraction 1

3 ?

a.

♣ ♣ ♣ ♣ ♣ ♣♣ ♣ ♣b.

♦ ♦ ♦♦c.

♥ ♥ ♥ ♥♥ ♥ ♥d.

♠ ♠ ♠ ♠ ♠ ♠♠(15) A bear 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 bear reach after walking 7

25

distance on the periphery?

a. D b. C

c. A d. B

© 2017 Edugain (www.edugain.com). All Rights Reserved

Many more such worksheets can begenerated at www.edugain.com



https://www.edugain.com

Answers

(1) 6

Step 1

Let us first convert the given mixed fractions to proper fractions as shown below:

( 32

4 + 2

2

4 ) × ( 3

2

4 - 2

2

4 ) = (

14

4 +

10

4 ) × (

14

4 -

10

4 )

Step 2

Now, the expression involving like fractions, ( 14

4 +

10

4 ) × (

14

4 -

10

4 ), can be simplified as:

( 14

4 +

10

4 ) × (

14

4 -

10

4 )

= ( 14 + 10

4 ) × (

14 - 10

4 )

= ( 24

4 ) × (

4

4 )

= 96

16

Step 3

Let us now find the simplest form of the fraction 96

16 by dividing the numerator 96 and the

denominator 16, by their HCF.The HCF of 96 and 16 = 16.

Thus, the simplest form of the fraction 96

16 =

96/16

16/16 = 6



(2) A) 6

5

Step 1

First, let us simplify the given fraction: 4

10 ×

2

3 ÷

2

9

To simplify it, let us convert the division into multiplication by reciprocating the thirdfraction, as shown below:

4

10 ×

2

3 ×

9

2

= 72

60

Step 2

Now, to reduce 72

60 to the simplest form, we will divide 72 and 60 by their HCF.

The HCF of 72 and 60 = 12

The simplest form of 72

60 =

72

12

60

12

= 6

5



B) 1

Step 1


8 ×

2

3 ÷

1

6


2

8 ×

2

3 ×

6

1

= 24

24

Step 2

Now, to reduce 24




24 =

24

24

24

24

= 1

1



C) 18

65

Step 1


5 ×

3

10 ÷

13

30


2

5 ×

3

10 ×

30

13

= 180

650

Step 2

Now, to reduce 180




650 =

180

10

650

10

= 18

65



D) 8

35

Step 1


7 ×

4

5 ÷

5

10


1

7 ×

4

5 ×

10

5

= 40

175

Step 2

Now, to reduce 40




175 =

40

5

175

5

= 8

35

(3) 413

Step 1

It is given that ratio of occupied spots to empty spots is 3:7, which means if there are 10 ( = 3 + 7)spots, 3 are occupied and 7 are empty

Step 2

But is given that total number of parking spots are 590. Therefore total number of spots are 59times ( = 590/10) of 10

Step 3

Therefore available spots will also be 59 times of 7. Therefore available parking spots,= 7 × 59 = 413



(4) ÷ x + -

Step 1

We know that division is completed before multiplication, multiplication is completed beforeaddition, and addition is completed before subtraction.

Step 2

Therefore, the order of operations should be:÷ x + -

(5) 9176.803

Step 1

We have the place values of all the digits given in the question. To form a number, we’ll need toadd all of them.

Step 2

First let us convert the given fractions into their decimal forms: 8

10 = 0.8,

0

100 = 0,

3

1000 =

0.003.

Step 3

9000 + 100 + 70 + 6 + 0.8 + 0 + 0.003 = 9176.803.

Step 4

Hence the short form of the above given number is 9176.803.



(6) 8

15

Step 1

Let us solve the given equation with the help of the following steps:

4

5 ×

3

5 - 7

14

= 4

5 ×

3

(5 × 14) - 7

14

= 4

5 ×

3

70 - 7

14

= 4

5 ×

3

63

14

= 4

5 ×

3 × 14

63

= 4

5 ×

2

3

Step 2

In order to multiply the fractions, we need to multiply the numerators and the denominators, asshown below

= 4 × 2

5 × 3

= 8

15

= 8

15

(7) 711

30 cups

Step 1

In order to find the total amount of flour needed to bake both types of bread, let us add the 31

6



cup and 41

5 cup.

Step 2

Let us convert the given mixed fractions into improper fractions.

31

6 =

6 × 3 + 1

6 =

19

6

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 6 and 5 = 30

So, 19

6 =

19

6 ×

5

5 =

95

30

and 21

5 =

21

5 ×

6

6 =

126

30

Step 4

Now, let us add the two like fractions.

So, 95

30 +

126

30 =

221

30

Step 5

Converting 221

30 into mixed fraction:

Dividend↴

Divisor→ 30 ) 2 2 1 ( 7 ←Quotient

2 1 0

Remainder← 11

Thus, we have:

221

30 = 7

11

30

Step 6

Thus, Ameerah requires a total of 711

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



(8) A) 54

55

Step 1

Let us first convert the given mixed fractions to proper fractions, as shown below:

21

5 =

2 × 5 + 1

5 =

11

5

14

5 =

1 7 × 5 + 4

5 =

9

5

Step 2

Now, in order to solve the like fractions 9

5 -

9

5 ÷

11

5 , we will reciprocate the third

fraction, as shown below:

9

5 -

9

5 ×

5

11

= 9

5 -

9

11 (Since, the multiplication must be solved before the addition and

subtraction)

= (9 × 11) - (9 × 5)

55

= 99 - 45

55

= 54

55

Step 3

Now, to reduce 54


HCF (54 and 55) = 1

Since, the HCF is 1, this fraction cannot be simplified further and the answer is 54

55



B) 137

174

Step 1


45

6 =

4 × 6 + 5

6 =

29

6

11

6 =

1 7 × 6 + 1

6 =

7

6

15

6 =

1 × 6 + 5

6 =

11

6

Step 2


6 -

11

6 ÷

29

6 , we will reciprocate the

third fraction, as shown below:

7

6 -

11

6 ×

6

29

= 7

6 -

11


subtraction)

= (7 × 29) - (11 × 6)

174

= 203 - 66

174

= 137

174

Step 3

Now, to reduce 137


HCF (137 and 174) = 1


174



C) -1

3

Step 1


15

6 =

1 × 6 + 5

6 =

11

6

14

6 =

1 7 × 6 + 4

6 =

10

6

34

6 =

3 × 6 + 4

6 =

22

6

Step 2


6 -

22

6 ÷

11



10

6 -

22

6 ×

6

11

= 10

6 -

2


subtraction)

= (10 × 1) - (2 × 6)

6

= 10 - 12

6

= -2

6

Step 3

Now, to reduce -2

6 to the simplest form, let us find the HCF of -2 6.

HCF(-2 ,6) = 2

The simplest form of -2

6 =

-2

2

6

2

= -1

3



D) 113

65

Step 1


23

5 =

2 × 5 + 3

5 =

13

5

21

5 =

2 7 × 5 + 1

5 =

11

5

11

5 =

1 × 5 + 1

5 =

6

5

Step 2


5 -

6

5 ÷

13



11

5 -

6

5 ×

5

13

= 11

5 -

6


subtraction)

= (11 × 13) - (6 × 5)

65

= 143 - 30

65

= 113

65

Step 3

Now, to reduce 113


HCF (113 and 65) = 1


65

(9) 132

5 km

Step 1



In order to find the total distance covered by Aziz, we need to add the three distances.

Let us convert the given mixed fractions into improper fractions.

32

5 =

5 × 3 + 2

5 =

17

5

33

4 =

4 × 3 + 3

4 =

15

4

49

4 =

4 × 4 + 9

4 =

25

4

Step 2

We find that all the three 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 5 and 4 = 20

So, 17

5 =

17

5 ×

4

4 =

68

20

15

4 =

15

4 ×

5

5 =

75

20

25

4 =

25

4 ×

5

5 =

125

20

Step 3

Now, let us add the three like fractions.

So, 68

20 +

75

20 +

125

20 =

268

20 =

67

5

Step 4

Converting 67

5 into mixed fraction:

Dividend↴

Divisor→ 5 ) 6 7 ( 13 ←Quotient

5

1 7

1 5

Remainder← 2

We have:



67

5 = 13

2

5

Step 5

Thus, Aziz traveled a total of 132

5 km in three hours.

(10) 99

4

Step 1

Let us first simplify the numerator of the second term:

( 2

1 -

4

5 ) × 10

= (2 × 5) - (4 × 1)

5 × 10

= 6

5 × 10

= 12

Step 2

Similarly, simplify the denominator of the second term:

1 - 1

2

= 1

2

Step 3

Now,

3

4 +

( 2

1 -

4

5 ) × 10

1 - 1

2

= 3

4 +

12

1

2

= 3

4 +

12 × 2

1



= 3

4 +

24

1

= (3 × 1) + (24 × 4)

4

= 3 + 96

4

= 99

4

(11) A) 1111

13

Step 1

Let us first convert the given mixed fractions into improper fractions.

In order to convert the mixed fraction 17139

195 into an improper fraction, we need to

multiply the denominator 195 by 17 and add the numerator 139 to the result.The resulting number is the numerator of the improper fraction and the denominatorremains the same:

17139

195 =

17 × 195 + 139

195 =

3454

195

Similarly, the mixed fraction 513

15 can be converted into an improper fraction as:

513

15 =

5 × 15 + 13

15 =

88

15

Step 2

Now, we need to subtract the following Unlike fractions: 3454

195 and

88

15 .

Before we subtract, let us first convert them to Like fractions.

Step 3

Let us find the LCM of the denominators 195 and 15.The LCM of 195 and 15 is 195.

Step 4

What should we multiply with the denominator 195 to get the LCM 195?

It is 195

195 = 1.

So, the first fraction can be written as:

3454 × 1

195 × 1 =

3454

195



Step 5

Similarly, the second fraction can be written as:

88 × 13

15 × 13 =

1144

195

Step 6

Let us now subtract the two Like fractions:

3454

195 -

1144

195

= 3454 - 1144

195

= 2310

195

= 2310 ÷ 15

195 ÷ 15 ...( Dividing both the numerator and denominator by their HCF )

= 154

13

= 1111

13

Step 7

Hence, the simplified value of the given fractions is 1111

13 .

B) 2511

14

Step 1





2859

182 =

28 × 182 + 59

182 =

5155

182



27

13 =

2 × 13 + 7

13 =

33

13



Step 2


182 and

33

13 .


Step 3


Step 4


It is 182

182 = 1.


5155 × 1

182 × 1 =

5155

182

Step 5


33 × 14

13 × 14 =

462

182

Step 6


5155

182 -

462

182

= 5155 - 462

182

= 4693

182

= 4693 ÷ 13


= 361

14

= 2511

14

Step 7


14 .

C) 1113

15



Step 1





17146

255 =

17 × 255 + 146

255 =

4481

255



512

17 =

5 × 17 + 12

17 =

97

17

Step 2


255 and

97

17 .


Step 3


Step 4


It is 255

255 = 1.


4481 × 1

255 × 1 =

4481

255

Step 5


97 × 15

17 × 15 =

1455

255

Step 6


4481

255 -

1455

255

= 4481 - 1455

255



= 3026

255

= 3026 ÷ 17


= 178

15

= 1113

15

Step 7


15 .

D) 2813

14

Step 1





3699

182 =

36 × 182 + 99

182 =

6651

182



78

13 =

7 × 13 + 8

13 =

99

13

Step 2


182 and

99

13 .


Step 3


Step 4




It is 182

182 = 1.


6651 × 1

182 × 1 =

6651

182

Step 5


99 × 14

13 × 14 =

1386

182

Step 6


6651

182 -

1386

182

= 6651 - 1386

182

= 5265

182

= 5265 ÷ 13


= 405

14

= 2813

14

Step 7


14 .



(12) 33 liters

Step 1

We have been told that the volume of milk the first type of container can hold = 11

3 liters =

4

3

liters

Step 2

Since, there are 9 containers of the first type, the volume of milk they can hold = 4

3 × 9

= 4 × 3= 12 liters.

Step 3

Now, the volume of milk the second type of container can hold = 31

2 liters =

7

2 liters

Step 4

Since, there are 6 containers of the second type, the volume of milk 6 containers of the second

type can hold = 7

2 × 6

= 7 × 3= 21 liters.

Step 5

The volume of milk required to fill both the containers = 12 + 21 = 33 liters.



(13) b. 11

9

Step 1

We see that the 5 out of 9 parts of figure A are shaded.

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

9

Also, 2 out of 3 parts of figure B are shaded.

So, the fraction of shaded part of figure B = 2

3

Step 2

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

L.C.M of 9 and 3 = 9

So, 5

9 =

5

9 ×

1

1 =

5

9

and 2

3 =

2

3 ×

3

3 =

6

9

Step 3

Adding:

5

9 +

2

3 =

5

9 +

6

9 =

11

9

Step 4

Hence, + = 11

9



(14) a.

♣ ♣ ♣ ♣ ♣ ♣♣ ♣ ♣

Step 1

Look at the all of the figures carefully and notice that in figure

♣ ♣ ♣ ♣ ♣ ♣♣ ♣ ♣The number of ♣ in selected box = 3and the total number of ♣ = 9

Step 2

Now the fraction of the number of ♣ in the selected box to the total number of ♣ in given figure =

3

9 =

1

3

Step 3

Therefore the figure

♣ ♣ ♣ ♣ ♣ ♣♣ ♣ ♣shows fraction

1

3 .



(15) d. B

Step 1

If we look at the regular polygon carefully, we notice that there are 5 sides of a regular polygon.

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

5

Step 2

Since, the distance walked by the bear on the periphery, in the clockwise direction = 7

25

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

Distance walked by the bear

Length of a side of the polygon

=

7

25

1

5

= 7

25 ×

5

1

= 7

5

= 1.4

It means that the bear walked on 1 side of the regular polygon and the bear is walking on the 2nd

side of the regular polygon, in the clockwise direction.

Step 3

Since, it started from point s, the bear will be on the side B, after walking 7

25 distance on the

periphery in the clockwise direction.

Step 4

Hence, option d is the correct answer.



grade 7 fractions - tests · let us convert the given mixed fractions into improper fractions. 3 1...

Documents