Free Fractions Subjective Test 01 Practice Test - 6th grade 

Solution: 1 Mark

The amount of cake Saumya ate more than Shreya was ;

$\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$ more cake.

Steps: 1 Mark
Answer: 1 Mark


(i) $2+\frac{3}{5}=\frac{10+3}{5}=\frac{13}{5}$.


(ii) $5+\frac{1}{5}=\frac{25+1}{5}=\frac{26}{5}$.


(iii) $1+\frac{4}{5}=\frac{5+4}{5}=\frac{9}{5}$.

Steps: 1 Mark
Answer: 1 Mark

Shobha spent $\frac{1}{3}$ times of 10 hours watching television.

$\Rightarrow \frac{1}{3}\times 10=\frac{10}{3}or3\frac{1}{3}$

$\therefore$ Shobha spent $\frac{10}{3}$ hours or  $3\frac{1}{3}$ hours (which is 3 hours 20 minutes)  watching  television.

Steps: 1 Mark
Fraction: 1 Mark
Answer: 1 Mark

There are 24 hours in a day.
$\Rightarrow$ 6 hours of a day
$=\frac{6}{24}$ 
$=\frac{1}{4}$ (expressed in the simplest form) 
$\Rightarrow \frac{6}{24}$ and $\frac{1}{4}$ are equivalent fractions.
$\therefore$ 6 hours of a day represents $\frac{1}{4}th$ of a day.

Smallest Fraction: 1 Mark
Greatest Fraction: 1 Mark
Steps: 1 Mark

$\frac{6}{3}=2=\frac{2}{1}$

$\therefore$ LCM for the fractions $\frac{1}{2},\frac{3}{4},\frac{4}{5},\frac{3}{2},\frac{2}{1}$ is 20

$\Rightarrow$$\frac{1}{2}=\frac{10}{20}$

$\Rightarrow$$\frac{3}{4}=\frac{15}{20}$

$\Rightarrow$$\frac{4}{5}=\frac{16}{20}$

$\Rightarrow$$\frac{3}{2}=\frac{30}{20}$

$\Rightarrow$$\frac{2}{1}=\frac{40}{20}$

Now that the denominator is equal, we can compare the fractions. The numerator 40 is the greatest and 10 is the least. Hence, the corresponding original fractions will be the greatest and the least.
$\therefore$ Greatest fraction = $\frac{6}{3}$ = 2
Smallest fraction = $\frac{1}{2}$ = 0.5

Steps: 2 Marks
Answer: 1 Mark

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

$\frac{3}{1}\times \frac{1}{4}=\frac{3}{4}$

$\frac{7}{1}\times \frac{1}{3}=\frac{7}{3}=2\frac{1}{3}$

$\frac{17}{1}\times \frac{1}{5}=\frac{17}{5}=3\frac{2}{5}$

Each option: 1 Mark

There are total of 11 numbers from 0 to 10

1) Number of odd numbers (1,3,5,7,9) = 5. 

$\therefore$ Fraction of odd numbers = $\frac{5}{11}$


2) Number of prime numbers (2,3,5,7) = 4

$\therefore$ Fraction of prime numbers $\frac{4}{11}$


3) Number of even numbers (2,4,6,8,10) = 5

$\therefore$ Fraction of even numbers $\frac{5}{11}$

Each option: 2 Marks

(a) $\frac{4}{4}=1$

     $\frac{2}{4}=\frac{1}{2}$

     $\frac{3}{4}$ is in its simplest form

Representation of the fractions on the number line is shown below

(b) Length of rope = 9.85m

After dividing the rope into 5 equal parts each part will be $\frac{1}{5}th$ of the total part.

$\Rightarrow \frac{1}{5}\times 9.85=\frac{9.85}{5}=1.97m$

(a) Concept: 1 Mark
      Solution: 1 Mark
(b) Solution: 1 Mark
(c) Solution: 1 Mark

(a) Total no of balloons = 55

Balloons left with Nisha = 55 - 46 = 9

Fraction of balloon left = $\frac{9}{55}$

Balloons left with Nisha = 46 - 6 = 40

Fraction of balloon left with Nisha = $\frac{40}{46}=\frac{20}{23}$


(b)The total weight of the fruits = $3\frac{1}{2}+4\frac{3}{4}$

= $\frac{7}{2}+\frac{19}{4}$

=  $\frac{14}{4}+\frac{19}{4}$

=  $\frac{33}{4}$

=  $8\frac{1}{4}$


(c) $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}$

                         $=\frac{8\times 2+7\times 3}{6}$

                        $=\frac{16+21}{6}$

                        $=\frac{37}{6}$

                       $=6\frac{1}{6}$

(a) Solution: 1 Mark
      Name: 1 Mark
(b) Concept: 1 Mark
     Steps: 1 Mark

(a) No of parts Arush has eaten = 4

Total no of parts = 6

So fraction of parts he has eaten = $\frac{4}{6}=\frac{2}{3}$

Fraction of watermelon remaining = $1-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}$

The fractions $\frac{2}{3}$ and $\frac{1}{3}$ are called proper fractions.



(b) Total time of Suman’s study =  $5\frac{2}{3}=\frac{17}{3}$ hours

Time devoted by her for Science and Mathematics =  $2\frac{4}{5}=\frac{14}{5}$ hours

Thus, time devoted by her for other subjects = $\frac{17}{3}-\frac{14}{5}$

= $\frac{17\times 5}{15}-\frac{14\times 3}{15}$

= $\frac{43}{15}$

= $2\frac{13}{15}$