Free Fractions Subjective Test 02 Practice Test - 6th grade 

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 $\frac{1}{5}$.

Each part: 1 Mark

(a) LCM of 7 and 5 is 35 

$\frac{3}{7}=\frac{3\times 5}{7\times 5}=\frac{15}{35}$

$\frac{3}{5}=\frac{3\times 7}{5\times 7}=\frac{21}{35}$

Clearly,  $\frac{21}{35}$ is greater than $\frac{15}{35}$

Hence, $\frac{3}{5}$ is greater than $\frac{3}{7}$


(b) LCM of 8 and 2 is 8 

$\frac{5}{8}=\frac{5\times 1}{8\times 1}=\frac{5}{8}$

$\frac{1}{2}=\frac{1\times 4}{2\times 4}=\frac{4}{8}$

Clearly, $\frac{5}{8}$ is greater than $\frac{4}{8}$

Hence, $\frac{5}{8}$ is greater than $\frac{1}{2}$

Each option: 1 Mark

(a) Given pizza has 4 slices

Pizza slices left in total = 3

Fraction of pizza slices left= $\frac{3}{4}$

(b) $\frac{6}{729}\times \frac{729}{6}=1$ ( product of a fraction and its reciprocal is always 1)

Each option: 1 Mark

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

= $(\frac{7}{8}-\frac{1}{4})m$ 

= $\left(\frac{7-2}{8}\right)m$

= $\left(\frac{5}{8}\right)m$


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


= $\frac{2}{5}+\frac{3}{4}$


= $\frac{8+15}{20}$


= $\frac{23}{20}$


=  $\left(1\frac{3}{20}\right)m$


(c)  $8\frac{1}{4}-2\frac{5}{6}$ =  $\frac{33}{4}-\frac{17}{6}$

=  $\frac{198-68}{24}$

=  $\frac{130}{24}$

=  $\frac{65}{12}$

=  $5\frac{5}{12}$

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 $=\frac{300}{2}$

=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, $\frac{1}{2}$ of 150 runs

= 75 runs were scored.

$\Rightarrow$ 150 runs - 75 runs = 75 runs are still in balance.

Again in the next 8 overs.

$\frac{1}{5}$ of 75 = 15 runs were scored.

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

$\Rightarrow$ 20 - 13 = 7

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

Steps: 1 Mark
Difference in fractions: 1 Mark
Solution: 1 Mark

Time taken by Mena = $\frac{7}{15}hr$

Time taken by Vaibhav
= $\frac{2}{5}hr=\frac{2\times 3}{5\times 3}=\frac{6}{15}$

Clearly, $\frac{7}{15}>\frac{6}{15}$

$\therefore$ Vaibhav worked longer.

Time for which Vaibhav has worked more
= $\frac{7}{15}-\frac{2}{5}=\frac{7-6}{15}=\frac{1}{15}$ hrs longer.

Each part: 2 Marks

(a) Shashank sold $\frac{2}{5}$ times 3 boxes of candy  $\Rightarrow \frac{2}{5}\times 3$

Simplify the product $\frac{6}{5}=1\frac{1}{5}$

Shashank sold $1\frac{1}{5}$ boxes of candy.


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

=  $1\frac{1}{2}+1\frac{1}{3}$

=  $\frac{3}{2}+\frac{4}{3}$

=  $\frac{9+8}{6}$

= $\frac{11}{6}$

=  $1\frac{5}{6}$

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. $\frac{1}{3},\frac{3}{4},\frac{21}{793}$


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

                                   e.g. $\frac{4}{3},\frac{9}{4},\frac{22}{9}$


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

                                   e.g. $2\frac{1}{3},1\frac{3}{4},3\frac{2}{93}$

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

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Let the length and breadth of the bigger rectangle be ${}^{\prime }{a}^{\prime }$m and ${}^{\prime }{b}^{\prime }$m respectively

Then, 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 $m^2$

Area of the shaded rectangle = (0.1a) $\times$ (0.001b)

= (0.0001) $\times$ (ab)

= (0.0001) $\times$ (10000)

= $1m^2$