# Free Fractions Subjective Test 02 Practice Test - 6th grade

### Question 1

Anshu bought a cake from a bakery shop. She has 5 friends to share the cake. In what fraction she should distribute the cake to each of her friends so that every friend gets the same amount of cake? [1 MARK]

#### SOLUTION

Solution :Concept: 1 Mark

The cake has to be divided among 5 people equally, so the fraction of the cake each friend would get will be 15.

### Question 2

Which one of the following is greater? [2 MARKS]

a) 37 and 35

b) 58 and 12

#### SOLUTION

Solution :Each part: 1 Mark

(a) LCM of 7 and 5 is 35

37=3×57×5=1535

35=3×75×7=2135

Clearly, 2135 is greater than 1535

Hence, 35 is greater than 37

(b) LCM of 8 and 2 is 8

58=5×18×1=58

12=1×42×4=48

Clearly, 58 is greater than 48

Hence, 58 is greater than 12

### Question 3

(a) What fraction of the total pizza is left in the given figure below? [2 MARKS]

(b) What is the product of 6729 and its reciprocal?

#### SOLUTION

Solution :Each option: 1 Mark

(a) Given pizza has 4 slicesPizza slices left in total = 3

Fraction of pizza slices left= 34

(b) 6729×7296=1 ( product of a fraction and its reciprocal is always 1)

### Question 4

(a) A piece of wire 78 metre long broke into two pieces. One piece was 14 metre long. How long is the other piece? [3 MARKS]

(b) Sarita bought 25 meters of ribbon and Lalita 34 metre of ribbon. What is the total length of the ribbon they bought?

(c) Simplify: 814−256

#### SOLUTION

Solution :Each option: 1 Mark

(a) Length of the other piece = Total length of the wire - Length of the given piece

= (78−14)m

= (7−28)m

= (58)m

(b) Total length of ribbon = Length of ribbon bought by Sarita + Length of ribbon bought by Lalita

= 25+34

= 8+1520

= 2320

= (1320)m

(c) 814−256 = 334−176

= 198−6824

= 13024

= 6512

= 5512

### Question 5

In a match of 50 overs, the team batting second was given a target of 310.They lost the match by 10 runs. The team batting second i.e. the losing team, scored one-half of their runs in 30 overs and next five overs they scored one-half of the remaining runs. Later in the next 8 overs, they scored 15^{th} of the remaining runs. In last seven overs, they scored the remaining runs. How many runs did they score in last seven overs?

[3 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Answer: 1 Mark

Total runs scored by the team that lost = 300

Now, total no. of overs played by them = 50

That is in 30 overs, this team scored =3002

=150 runs

The balance 300 - 150 = 150 runs were scored by this team in balance 20 overs.

By the given condition,

In the next 5 overs, 12 of 150 runs

= 75 runs were scored.

⇒ 150 runs - 75 runs = 75 runs are still in balance.

Again in the next 8 overs.

15 of 75 = 15 runs were scored.

Now No. of overs left = 20 - (5 + 8)

⇒ 20 - 13 = 7

In seven overs, runs scored will be 75 - 15 = 60.

### Question 6

Mena finished colouring a picture in 715 hour; Vaibhav finished colouring the same picture in 25 hours. Who worked longer? By what fraction was it longer? [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Difference in fractions: 1 Mark

Solution: 1 Mark

Time taken by Mena = 715 hrTime taken by Vaibhav

= 25 hr=2×35×3=615

Clearly, 715>615

∴ Vaibhav worked longer.Time for which Vaibhav has worked more

= 715−25=7−615=115 hrs longer.

### Question 7

(a) Last month, Shakshi and Shashank sold candy to raise money for their debate team. Shashank sold 25 as much candy as Shakshi did. If Shakshi sold 3 boxes of candy, how many boxes of candy did Shashank sell? [4 MARKS]

(b) Naina was given 112 piece of cake and Najma was given 113 piece of cake. Find the total amount of cake that was given to both of them.

#### SOLUTION

Solution :Each part: 2 Marks

(a) Shashank sold 25 times 3 boxes of candy ⇒25×3

Simplify the product 65=115

Shashank sold 115 boxes of candy.

(b) Total mount of cake = Naina's share of cake + Najma's share of cake

= 112+113

= 32+43

= 9+86

= 116

= 156

### Question 8

What are proper fractions, improper fractions and mixed fractions? Give Example for each. [4 MARKS]

#### SOLUTION

Solution :Definitions: 3 Marks

Example: 1 Mark

Saurabh should answer-

Proper Fraction: The fractions in which the numerator is less than the denominator are called proper fractions.

e.g. 13,34,21793

Improper Fraction: The fractions in which the numerator is greater than (or equal to) the denominator are called proper fractions.

e.g. 43,94,229

Mixed Fraction:A whole number and a proper fraction mixed together is called a mixed fraction.

e.g. 213,134,3293

### Question 9

Car A covers 5.5 km in 4 litres of petrol and another car B covers 1800 metres in 1 litre of petrol. Calculate the mileage per litre of petrol, Now calculate which car travels more and by how much distance compared to the other? [4 MARKS]

#### SOLUTION

Solution :Concept: 1 Mark

Application: 3 Marks

Distance covered by car A in 4 litres of petrol is 5500m.Distance covered by car A in 1 litre of petrol =(5500÷4)= 1375m.

Distance covered by car B in 1 litre of petrol = 1800m.

So car B travels more compared to car A

Difference = 1800-1375 = 425m

Car B travels 425m more than car A

### Question 10

The length of the shaded rectangle is (1/10) of the bigger rectangle and its breadth is 0.001 times the bigger rectangle . If the area of the bigger rectangle is 10000 m2. Find the area of the shaded rectangle. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Let the length and breadth of the bigger rectangle be ′a′m and ′b′m respectivelyThen, the length of the shaded rectangle is (0.1 a)m.

Then, the breadth of the shaded rectangle is (0.001 b)m.

Area of the given bigger rectangle = ab metres. = 10000 m2

Area of the shaded rectangle = (0.1a) × (0.001b)

= (0.0001) × (ab)

= (0.0001) × (10000)

= 1m2