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Grade 6Second quarter at school

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Answer the questions

(1) What is the multiplicative inverse of 870?

(2)Convert the f ollowing to a mixed f raction :

20

6

(3) Solve f or the value of z

6

100 +

3

10 +

9

z + 8 = 8.45

(4) The corner shop sells 188.47 liters of milk on Wednesday and 12.41 liters more than thisamount on Thursday. In total, how many liters of milk did they sell?

(5) When doing operations on f ractions, in which order should the operations be done amongthese f our operations : + x - ÷

(6) What do you get when you subtract 695.23 f rom 783.57?

Choose correct answer(s) from given choice

(7)The f ractions

3

10 and

7

10 are :

a. Like Fractions b. Mixed Fractions

c. Unlike Fractions d. Improper Fractions

(8) Basimah's f ather gave her Dhs174 to spend. She spent Dhs70.36 to buy some toys andDhs26.92 to buy some books. How much does she have lef t?

a. Dhs78.02 b. Dhs76.72

c. Dhs75.12 d. Dhs7.67

(9) What is the dif f erence of nine Tenths and f ive Hundredths?

a. 8.5 b. 0.937

c. 0.85 d. 0.856

(10) Solve the f ollowing

A) _____ + 90.71 = 244.18

a. 153.47 b. 152.84

c. 153.36 d. 154.18

B) 623.09 - _____ = 66.58

a. 556.05 b. 558.2

c. 559.15 d. 556.51

ID : ae-6-Second-quarter-at-school [1]

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C) 675.32 + 32.08 = _____

a. 708.72 b. 707.23

c. 716.45 d. 707.4

D) _____ - 814.74 = 93.54

a. 906.48 b. 912.23

c. 908.28 d. 908.09

Fill in the blanks

(11) Simplif y the f ollowing f ractions:

A) 15

79

133 - 1

6

7 =

B) 27

17

44 - 2

3

4 =

C) 13

31

357 - 4

8

21 =

D) 29

64

195 - 7

6

13 =

(12) There are 34 students in class 8. There are 11 boys in the class. The f raction of girls in the

class is

?

(13)The local community park has space allocated f or jogging, music, and sports.

10

29 of the park

has been allocated f or sports, and out of the sports area 2

10 has been given f or playing

f ootball. The ratio of the park is given f or playing f ootball is

.

(14) Solve the f ollowing:

A)( ÷ 995.69) x 0.7 = 0.7

B)(848.5 ÷ 848.5) x 0.8 =

C)(510.29 ÷ ) x 0.9 = 45.9261

D)(379.51 ÷ ) + 89.02 = 92.8151

Check True/False

(15) 0.31 is lesser than 0.031. True False



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Answers

(1)

1

870

Step 1

Two number f ractions whose product is 1 are called multiplicative inverse of each other.

Step 2

Let us f irst write the whole number 870 into f ractions as 870

1 .

Step 3

To get the multiplicative inverse, let us interchange the numerator and the denominator as

1

870 .

Step 4

Hence, the multiplicative inverse of 870 is 1

870 . A way to verif y our answer is to see that

870 × 1

870 = 1.

(2) 3

2

6



(3) 100

Step 1

The equation that we have been given is:

6

100 +

3

10 +

9

z + 8 = 8.45

⇒ 6

100 +

3

10 +

9

z = 8.45 - 8 [Transpose the term 8 from the L.H.S to the R.H.S]

⇒ 9

z = 0.45 -

6

100 -

3

10 [Transpose the terms

3

10 and

6

100 from the L.H.S to

the R.H.S]

⇒ 9

z =

45

100 -

6

100 -

3

10 [Write 0.45 as

45

100 ]

⇒ 9

z =

45 - 6 - 30

100 [Replace the denominator by their L.C.M and adjust the

numerator accordingly]

⇒ 9

z =

45 - 36

100 [Simplify the numerator]

⇒ 9

z =

9.0000000000001

100 [Simplify the numerator]

By cross multiplying both sides, we get:9 × 100 = 9.0000000000001z

⇒ z = 900

9.0000000000001

⇒ z = 99.999999999999

Step 2

Theref ore, the value of z is 100 .

(4) 389.35

Step 1

Amount of milk sold on Wednesday = 188.47 liters

Step 2

Amount of milk sold on next day (i.e. Thursday) = 188.47 + 12.41 = 200.88 liters

Step 3

Total amount of milk sold in two days = 188.47 + 200.88 = 389.35 liters



(5) ÷ x + -

Step 1

Division is completed bef ore multiplication, multiplication is completed bef ore addition andaddition is completed bef ore subtraction.

Step 2

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

(6) 88.34

Step 1

Decimals with the same number of decimal places are called Like Decimals. The numbersgiven here are theref ore Like Decimals.

Step 2

We need to subtract 695.23 f rom 783.57. Theref ore, 783.57 must be placed on top and695.23 below. The digits of the two numbers must be placed according to their place values.

Step 3

Let us now do the subtraction of the two numbers digit by digit, starting f rom thehundredths and borrowing if needed:

Thousands Hundreds Tens Ones . Tenth Hundredth

7 8 3 . 5 7

- 6 9 5 . 2 3

0 8 8 . 3 4

Step 4

Hence, when we subtract 695.23 f rom 783.57, we get 88.34 .

(7) a. Like Fractions

The f ractions 3

10 and

7

10 have the same denominator. Theref ore, the f ractions

3

10

and 7

10 are Like Fractions.



(8) b. Dhs76.72

Step 1

The amount of money that Basimah's f ather gave her = Dhs174

Step 2

The amount that Basimah spent in buying toys = Dhs70.36

Step 3

The amount that Basimah spent in buying books = Dhs26.92

Step 4

Let us add the amount she spent in buying toys and books to f ind her total expense:

Toys = Dhs70.36

Books = Dhs26.92

Total expense = Dhs97.28

Step 5

In order to f ind the amount lef t with her, we need to subtract the amount Basimah's f athergave and the amount Basimah spent.

Step 6

Bef ore subtracting the two numbers, let us make sure they are written as Like Decimals.Decimals with the same number of decimal places are called Like Decimals. Thus, let uswrite Dhs174 as Dhs174.00, and subtract Dhs97.28 f rom it

Step 7

Let us subtract the amount Basimah spent (Dhs97.28) f rom the amount her f ather gave(Dhs174) digit by digit. The numbers must be placed according to their place values. Thegreater number must be placed on top and the lesser one below:

Thousands Hundreds Tens Ones . Tenth Hundredth

1 7 4 . 0 0

- 9 7 . 2 8

0 7 6 . 7 2

Step 8

Hence, Basimah is lef t with Dhs76.72.



(9) c. 0.85

Step 1

Because nine Tenths = 9

10

and f ive Hundredths = 5

100

Step 2

Theref ore the dif f erence of the numbers nine Tenths and f ive Hundredths = 9

10 -

5

100

= 9 × 10 - 5 × 1

100

= 90 - 5

100

= 85

100

= 0.85

(10) A) a. 153.47

Step 1

Given, _____ + 90.71 = 244.18Let us assume the missing number be x.Theref ore, x + 90.71 = 244.18or x = 244.18 - 90.71

Step 2

Now, we have to subtract 90.71 f rom 244.18 .

2 4 4 . 1 8

- 9 0 . 7 1

1 5 3 . 4 7

Theref ore, x = 153.47

Step 3

Thus, 153.47 + 90.71 = 244.18



B) d. 556.51

Step 1

Given, 623.09 - _____ = 66.58Let us assume the missing number be x.Theref ore, 623.09 - x = 66.58or 623.09 - 66.58 = xor x = 623.09 - 66.58

Step 2

Now, we have to subtract 66.58 f rom 623.09 .

6 2 3 . 0 9

- 6 6 . 5 8

5 5 6 . 5 1


Step 3

Thus, 623.09 - 556.51 = 66.58

C) d. 707.4

Step 1

Given, 675.32 + 32.08 = _____We have to add 675.32 and 32.08 .

6 7 5 . 3 2

+ 3 2 . 0 8

7 0 7 . 4 0

Step 2

Theref ore, 675.32 + 32.08 = 707.4



D) c. 908.28

Step 1

Given, _____ - 814.74 = 93.54Let us assume the missing number be x.Theref ore, x - 814.74 = 93.54or x = 93.54 + 814.74

Step 2

Now, we have to add 814.74 and 93.54 .

9 3 . 5 4

+ 8 1 4 . 7 4

9 0 8 . 2 8


Step 3

Thus, 908.28 - 814.74 = 93.54



(11) A) 15

79

133 - 1

6

7 = 13

14

19

Step 1

Let us f irst convert the given mixed f ractions into improper f ractions. In order to

convert the mixed f raction 1579

133 into an improper f raction, we will need to

multiply the denominator 133 by 15 and add the numerator 79 to the result. Theresulting number will be the numerator of the improper f raction and thedenominator will remain the same:

1579

133 =

15 × 133 + 79

133 =

2074

133

Similarly, the mixed f raction 16

7 can be converted into an improper f raction as:

16

7 =

1 × 7 + 6

7 =

13

7

Step 2

We now need to subtract the f ollowing unlike f ractions: 2074

133 and

13

7 .

Bef ore we subtract, let us f irst convert them to like f ractions.

Step 3

Let us f ind the LCM of the denominators 133 and 7.The LCM of 133 and 7 is 133.

Step 4

What should we multiply with the denominator 133 to get the LCM 133?

It is 133

133 = 1.

So, the f irst f raction can be written as:

2074 × 1

133 × 1 =

2074

133

Step 5

Similarly, the second f raction can be written as:

13 × 19

7 × 19 =

247

133

Step 6

Let us now subtract the two like f ractions:

2074

133 -

247

133



= 2074 - 247

133

= 1827

133

= 1827 ÷ 7

133 ÷ 7 ...[ Dividing both numerator and denominator by their HCF to bring ]

= 261

19

= 1314

19

Step 7

Thus, the simplif ied value of the given f ractions is 1314

19 .



B) 27

17

44 - 2

3

4 = 24

7

11

Step 1





2717

44 =

27 × 44 + 17

44 =

1205

44


4 can be converted into an improper f raction as:

23

4 =

2 × 4 + 3

4 =

11

4

Step 2


44 and

11

4 .


Step 3


Step 4


It is 44

44 = 1.


1205 × 1

44 × 1 =

1205

44

Step 5


11 × 11

4 × 11 =

121

44

Step 6


1205

44 -

121

44



= 1205 - 121

44

= 1084

44

= 1084 ÷ 4

44 ÷ 4 ...[ Dividing both numerator and denominator by their HCF to bring ]

= 271

11

= 247

11

Step 7


11 .



C) 13

31

357 - 4

8

21 = 8

12

17

Step 1





1331

357 =

13 × 357 + 31

357 =

4672

357


21 can be converted into an improper f raction

as:

48

21 =

4 × 21 + 8

21 =

92

21

Step 2


357 and

92

21 .


Step 3


Step 4


It is 357

357 = 1.


4672 × 1

357 × 1 =

4672

357

Step 5


92 × 17

21 × 17 =

1564

357

Step 6


4672

357 -

1564

357



= 4672 - 1564

357

= 3108

357

= 3108 ÷ 21

357 ÷ 21 ...[ Dividing both numerator and denominator by their HCF to bring

]

= 148

17

= 812

17

Step 7


17 .



D) 29

64

195 - 7

6

13 = 21

13

15

Step 1





2964

195 =

29 × 195 + 64

195 =

5719

195


13 can be converted into an improper f raction

as:

76

13 =

7 × 13 + 6

13 =

97

13

Step 2


195 and

97

13 .


Step 3


Step 4


It is 195

195 = 1.


5719 × 1

195 × 1 =

5719

195

Step 5


97 × 15

13 × 15 =

1455

195

Step 6


5719

195 -

1455

195



= 5719 - 1455

195

= 4264

195

= 4264 ÷ 13

195 ÷ 13 ...[ Dividing both numerator and denominator by their HCF to bring

]

= 328

15

= 2113

15

Step 7


15 .



(12) There are 34 students in class 8. There are 11 boys in the class. The f raction of girls in the

class is 23

34

?

Step 1

We have been asked to f ind the f raction of girls in the class 8. Let us f irst f ind the numberof girls in the class.

Step 2

Total number of students in the class = 34Number of boys in the class = 11

Step 3

Number of girls in the class = Total number of students in the class - Number of boys inthe class= 34 - 11= 23

Step 4

Fraction of girls in the class = Number of girls in the class

Total number of students

= 23

34

Step 5

Hence, the f raction of girls in the class 8 is 23

34 .



(13) 2 29

Step 1

: Let the area of the f ull park be 1.

Step 2

The area allocated f or sports = 10

29 of the total park area =

10

29 × 1 =

10

29

Step 3

The area allocated f or playing f ootball out of the sports area = 2

10 of the sports area

= 2

10 ×

10

29

= 2

29

Step 4

Thus the ratio of the park given f or playing f ootball is 2

29 .

(14) A) 995.69

Step 1

Let the unknown number be S. The given equation can then be written as:(S ÷ 995.69) x 0.7 = 0.7

Step 2

Let us solve f or S:(S ÷ 995.69) x 0.7 = 0.7

⇒ S ÷ 995.69 = 0.7

0.7

⇒ S ÷ 995.69 = 1⇒ S = 995.69

Step 3

Theref ore, (995.69 ÷ 995.69) x 0.7 = 0.7 .



B) 0.8

Step 1

Let the unknown number be S. The given equation can then be written as:(848.5 ÷ 848.5) x 0.8 = S

Step 2

Let us solve f or S:(848.5 ÷ 848.5) x 0.8 = S⇒ 1 x 0.8 = S ⇒ S = 0.8

Step 3

Theref ore, (848.5 ÷ 848.5) x 0.8 = 0.8.

C) 10

Step 1

Let the unknown number be S. The given equation can then be written as:(510.29 ÷ S) x 0.9 = 45.9261

Step 2

Let us solve f or S:(510.29 ÷ S) x 0.9 = 45.9261

⇒ 510.29 ÷ S = 45.9261

0.9

⇒ 510.29 ÷ S = 51.029 ⇒ S = 10

Step 3

Theref ore, (510.29 ÷ 10) x 0.9 = 45.9261.



D) 100

Step 1

Let the unknown number be S. The given equation can then be written as:(379.51 ÷ S) + 89.02 = 92.8151

Step 2

Let us solve f or S:(379.51 ÷ S) + 89.02 = 92.8151⇒ 379.51 ÷ S = 92.8151 - 89.02 ⇒ 379.51 ÷ S = 3.7951 ⇒ S = 100

Step 3

Theref ore, (379.51 ÷ 100) + 89.02 = 92.8151.

(15) False



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