Reduce the following fractions to their lowest forms. a. $$\dfrac{36}{144}$$ b. $$\dfrac{65}{117}$$ c. $$\dfrac{180}{120}$$

$$a=\dfrac{1}{4}, b=\dfrac{5}{9}, c=\dfrac{3}{2}$$

$$a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$$

$$a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{3}{2}$$

$$a=\dfrac{3}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$$

MCQ Questions for Class 6 Maths Fractions Quiz 6

Example of proper fraction is _________

0%
$\displaystyle \frac{5}{7}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{16}{15}$
0%
$\displaystyle \frac{22}{21}$

Explanation

When the numerator is less than the denominator, it is called a proper fraction.

So only

$\dfrac{5}{7}$ is a proper fraction.

Hence, the answer is

$\dfrac{5}{7}$ .

$\displaystyle 1\frac{3}{4}$ is a _______ fraction.

0%
Proper
0%
improper
0%
mixed
0%
None of the above

Explanation

$1\dfrac{3}{4}$ is a mixed fraction because it has both integer and a proper fraction.

Hence, the answer is 'mixed'.

_____

$\displaystyle -\dfrac{3}{7}=\dfrac{3}{7}$

0%
$0$
0%
$1$
0%
$\displaystyle \frac{3}{7}$
0%
$\displaystyle \frac{6}{7}$

Explanation

What should be subtracted from

$\dfrac{3}{7}$ so that we would get

$\dfrac{3}{7}$

So if we subtract

$\dfrac{6}{7}$ from

$\dfrac{3}{7}$ , then we will get

$\dfrac{3}{7}$ .

Hence, the answer is

$\dfrac{6}{7}$ .

$\displaystyle \frac { 3 }{17}+ ......... = \frac { 3 }{ 17 }$ , find the missing value.

0%
$1$
0%
$0$
0%
$\displaystyle \frac { 3 }{17}$
0%
$\displaystyle \frac { 2 }{17}$

Explanation

Mising value is

$=\displaystyle \frac { 3 }{17}-\frac { 3 }{ 17 }$

$=0$

Mixed fraction for

$\displaystyle \frac{39}{12}$ is

0%
$\displaystyle 3\tfrac{1}{12}$
0%
$\displaystyle 3\tfrac{2}{12}$
0%
$\displaystyle 3\tfrac{3}{12}$
0%
$\displaystyle 2\tfrac{14}{12}$

Explanation

To convert an improper fraction to a mixed fraction, we divide the numerator by the denominator, then write down the whole number answer.

Finally we write down any remainder above the denominator.

$39÷12=3$ leaving remainder

$3$

Therefore, the answer will be,

$3$ whole

$\dfrac {3}{12}$

Hence, the mixed fraction of

$\dfrac {39}{12}$ is

$3\dfrac {3}{12}$ .

Difference of

$\displaystyle \frac{19}{25}$ and

$\displaystyle \frac{27}{25}$ is

0%
$\displaystyle \frac{9}{25}$
0%
$\displaystyle \frac{3}{25}$
0%
$\displaystyle \frac{8}{25}$
0%
$\displaystyle \frac{6}{25}$

Explanation

Difference of the fractions

$\dfrac{19}{25}$ and

$\dfrac{27}{25}$ is

$\dfrac{27-19}{25} =\dfrac{8}{25}$

Hence, the answer is

$\dfrac{8}{25}$ .

Subtract

$\displaystyle\frac{11}{35}$ from

$\displaystyle\frac{12}{15}$

0%
$\displaystyle\frac{84}{35}$
0%
$\displaystyle\frac{33}{35}$
0%
$\displaystyle\frac{11}{15}$
0%
$\displaystyle\frac{17}{35}$

Explanation

The subtraction of the fraction

$\dfrac{11}{35}$ from

$\dfrac{12}{15}$ can be written as

$\dfrac{12}{15}-\dfrac{11}{35}$

LCM of

$35$ and

$15$ is

$105$

$\therefore \dfrac{12}{15}-\dfrac{11}{35}$

$=\dfrac{12\times7-11\times3}{105} = \dfrac{51}{105}$

After simplifying it can be written as

$\dfrac{17}{35}$ .

Hence, the answer is

$\dfrac{17}{35}$ .

Sum of

$\displaystyle \frac{7}{12},\frac{12}{12},\frac{8}{12}$ is

0%
$\displaystyle \frac{26}{12}$
0%
$\displaystyle \frac{19}{12}$
0%
$\displaystyle \frac{27}{12}$
0%
$\displaystyle \frac{23}{12}$

Explanation

In all the three fractions

$\displaystyle \frac{7}{12},\frac{12}{12},\frac{8}{12}$ , denominators are same

So, the sum of the fraction is

$\dfrac{7+12+8}{12}= \dfrac{27}{12}$ .

Hence, the answer is

$\dfrac{27}{12}$ .

Peggy needs to ship two packages. The first package weighs pounds

$16 \dfrac14$ , and the second package weighs

$23\dfrac38$ pounds. What is the total weight of the packages that Peggy needs to ship?

0%
$38 \dfrac89$ pounds
0%
$39 \dfrac58$ pounds
0%
40 pounds
0%
$40 \dfrac58$ pounds

What should be subtracted from

$\dfrac {-7}{8}$ so as to get

$\dfrac {5}{12}$ ?

0%
$\dfrac {31}{24}$
0%
$\dfrac {8}{5}$
0%
$\dfrac {-31}{24}$
0%
$\dfrac {-6}{8}$

Explanation

Let the number be

$x$ to be subtracted from

$\dfrac {-7}{8}$ to get

$\dfrac {5}{12}$ . Then it can be solved as follows:

$\dfrac { -7 }{ 8 } -x=\dfrac { 5 }{ 12 }$

$\Rightarrow -\dfrac { 7 }{ 8 } -x=\dfrac { 5 }{ 12 }$

$\Rightarrow -\dfrac { 7 }{ 8 } -\dfrac { 5 }{ 12 }=x$

$\Rightarrow x=-\dfrac { 7 }{ 8 } -\dfrac { 5 }{ 12 }$

$\Rightarrow x=-\dfrac { 7\times 3 }{ 8\times 3 } -\dfrac { 5\times 2 }{ 12\times 2 }$

$\Rightarrow x=-\dfrac { 21 }{ 24 } -\dfrac { 10 }{ 24 }$

$\Rightarrow x=\dfrac { -21-10 }{ 24 }$

$\Rightarrow x=-\dfrac { 31 }{ 24 }$

Hence,

$x=-\dfrac { 31 }{ 24 }$

Convert

$\dfrac{7}{4}$ into mixed fraction.

0%
$1\dfrac{3}{4}$
0%
$2\dfrac{3}{4}$
0%
$6\dfrac{3}{4}$
0%
None of the above

Explanation

$\dfrac{7}{4}=1\dfrac{3}{4}$

So, option A is correct.

Considering the positions on the number line, which of the following could be a value for x?

0%
$5/3$
0%
$3/5$
0%
$-2/5$
0%
$-5/2$
0%
None

Convert

$\dfrac{11}{4}$ into a mixed fraction.

0%
$\displaystyle 2\frac{3}{4}$
0%
$\displaystyle 3\frac{3}{4}$
0%
$\displaystyle 1\frac{3}{4}$
0%
none of the above

Explanation

Divide

$11$ by

$4$ . The quotient is

$2$ and remainder

$3$ .

$\therefore \displaystyle \frac{11}{4}= 2\frac{3}{4}$

So. option A is correct.

Which of the following statement is true?

0%
Fractions with same numerator are called like fractions
0%
Fractions with same denominator are called unlike fractions
0%
Difference of two like fractions $\displaystyle =\frac{difference\,of\,numerator}{common\,denominator}$
0%
A fraction with numerator greater than or equal to the denominator is called proper fraction

The lowest form of

$\displaystyle \frac{20}{50}$ is ____

0%
$\displaystyle \frac{1}{5}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{2}{5}$
0%
$\displaystyle \frac{10}{25}$

Explanation

For getting the lowest form of the fraction firstly divide the fraction by the same number to the numerator and denominator.

So after dividing both numerator and denominator of

$\dfrac{20}{50}$ by

$10$ we get

$\dfrac{2}{5}$ as the lowest form

Simplifying the fraction

$\displaystyle \frac{6}{5}\times 4\frac{1}{2}$ gives

0%
$\displaystyle \frac{27}{5}$
0%
$\displaystyle \frac{6}{5}$
0%
$\displaystyle \frac{11}{5}$
0%
None of the above

Explanation

$\displaystyle \frac{6}{5} \times 4\frac{1}{2}$

$=\displaystyle \frac{6}{5}\times \frac{9}{2}$

$= \displaystyle \frac{27}{5}$

So, option

$A$ is correct.

Can we say that following figures represent equal fractions?

0%
Yes, 1st and 2nd figure show equal parts.
0%
Yes, all the figures show equal parts.
0%
Yes, 2nd and 3rd figure show equal parts.
0%
No

Explanation

In the given first object, there are a total of

$4$ triangles and

$1$ triangle is shaded amongst them. Therefore, the ratio of the shaded triangle to the total number of triangles is:

$\dfrac {Shaded}{Total}=\dfrac { 1 }{ 4 }$

Now, consider the second circular object, there are a total of

$4$ parts and

$1$ part is shaded amongst them. Therefore, the ratio of shaded part to the total number of parts is:

$\dfrac {Shaded}{Total}=\dfrac {1 }{ 4 }$

Now, in the third object, there are a total of

$4$ rectangles and

$1$ rectangle is shaded amongst them. Therefore, the ratio of the shaded rectangle to the total number of rectangles is:

$\dfrac {Shaded}{Total}=\dfrac { 1 }{ 4 }$

Since the fraction of all of the objects is the same that is

$\dfrac {1}{4}$ .

Hence, all the given figures show equal parts.

$7\dfrac {1}{2} + 8\dfrac {1}{2} = ?$

0%
$16$
0%
$15\dfrac {1}{2}$
0%
$5\dfrac {1}{2}$
0%
$15$

Add the fractions:

0%
$\dfrac{4}{9}$
0%
$\dfrac{2}{9}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{6}{9}$

Explanation

There are 9 squares in the first figure and out of them 2 are filled the fraction of first square is =

$\dfrac{2}{9}$

There are 9 squares in the second figure and out of them 4 are filled the fraction of first square is =

$\dfrac{4}{9}$

By adding them , we get

$\dfrac{2}{9}+\dfrac{4}{9}=\dfrac{6}{9}=\dfrac{2}{3}$

Subtract the fractions:

0%
$\dfrac{1}{4}$
0%
$\dfrac{2}{4}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{4}{4}$

Explanation

Three out of four sectors of the circles are dark the fraction of darken sectors =

$\dfrac{3}{4}$

Two out of four sectors of the circles are dark the fraction of darken sectors =

$\dfrac{2}{4}$

Subtracting them , we get

$\dfrac{3}{4}-\dfrac{2}{4}=\dfrac{1}{4}$

If the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is?

0%
Always greater than the original fraction
0%
Always less than the original fraction
0%
Always equal to the original fraction
0%
None of the above

Explanation

If the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is always greater than the original fraction.

Let us understand and verify it with the help of couple of examples:

$(i)$ Let

$\dfrac{1}{2}$ be an original fraction.

$\dfrac{1}{2}=0.5$

If the numerator and denominator are increased by

$5$ , the fraction becomes:

$\dfrac{1+5}{2+5}=\dfrac{5}{7}=0.71$

$\therefore$

$\dfrac{1}{2}<\dfrac{5}{7}$

$(ii)$ Let

$\dfrac{7}{9}$ be an original fraction.

$\dfrac{7}{9}\approx 0.78$

If the numerator and denominator are increased by

$1$ , the fraction becomes:

$\dfrac{7+1}{9+1}=\dfrac{8}{10}=0.8$

$\therefore$

$\dfrac{7}{9}<\dfrac{8}{10}$

Hence, if the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is always greater than the original fraction.

Therefore, option A is correct.

Ritu painted

$\dfrac{2}{5}$ of the wall space in her room. Her sister Rashmi helped and painted

$\dfrac{3}{5}$ of the wall space. How much did they paint together?

0%
$\dfrac{5}{10}$
0%
$\dfrac{6}{25}$
0%
$\dfrac{1}{5}$
0%
$1$

Explanation

Total fraction of wall painted by Ritu and Rashmi =

$\dfrac{2}{5} + \dfrac{3}{5}$ =

$\dfrac{5}{5}$ = 1

$2+\cfrac { 1 }{ 100 } +\cfrac { 3 }{ 1000 } =\_ \_ \_ \_ \_ \_$

0%
$0.213$
0%
$2.13$
0%
$2.013$
0%
$2.130$

Reduce the following fractions to their lowest forms.
a.

$\dfrac{36}{144}$

$\dfrac{65}{117}$

$\dfrac{180}{120}$

0%
$a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$
0%
$a=\dfrac{1}{4}, b=\dfrac{5}{9}, c=\dfrac{3}{2}$
0%
$a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{3}{2}$
0%
$a=\dfrac{3}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$

Explanation

$\\(a.)(\frac{36}{144})=(\frac{3}{12})=(\frac{1}{4})\\(b.)(\frac{65}{117})=(\frac{13\cdot 5}{13\cdot 9})=(\frac{5}{9})\\(c.)(\frac{180}{120})=(\frac{60\cdot 3}{60\cdot 2})=(\frac{3}{2})$

Compare the following

$\dfrac{1}{8},\ \dfrac{1}{4}$ .

0%
$\dfrac {1}{4}>\dfrac {1}{8}$
0%
$\dfrac {1}{4}<\dfrac {1}{8}$
0%
$\dfrac {1}{4}=\dfrac {1}{8}$
0%
none of these

Solve :

$\dfrac {1}{4}-\left(\dfrac {1}{4}\right)$

0%
$1$
0%
$\dfrac {2}{4}$
0%
$\dfrac {-3}{4}$
0%
$0$

Explanation

$\dfrac{1}{4}-\left(\dfrac{1}{4}\right)$ .

$=\dfrac{1}{4}-\dfrac{1}{4}=0$

A proper fraction with denominator

$3$ , which is less than 1 is

0%
$\frac{4}{3}$
0%
$\frac{2}{3}$
0%
$1 \frac{2}{3}$
0%
None

Explanation

Clearly from question denominator

$=3$

numerator

$=2$

Fraction

$=\dfrac{2}{3}$

A fraction whose numerator is greater than its denominator is fraction.

0%
an improper
0%
a proper
0%
a mixed
0%
None of these

Explanation

That will be an improper fraction

$\dfrac{6}{5}$

The value of

$\dfrac 12+\dfrac 13+\dfrac 14$

0%
$\dfrac {13}{12}$
0%
$\dfrac1 {15}$
0%
$\dfrac 1{12}$
0%
$\dfrac 1{24}$

Explanation

$\dfrac 12+\dfrac 1 3+\dfrac 14$

$=\dfrac{6+4+3}{12}$

$=\dfrac{13}{12}$

$289/391$ when reduced to the lowest terms is

0%
$11 / 23$
0%
$13 / 31$
0%
$17 / 31$
0%
$17 / 23$

Explanation

Dividing both numerator and denominator by the HCF of

$289 / 391$
Hence

$289 / 391 = 17 / 23$

$17 / 23$ is the correct answer
Option (d) is the correct answer
1635259_1777839_ans_ed70506207bc453ab766a0b322c018ee.png

CBSE Questions for Class 6 Maths Fractions Quiz 6 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Fractions Quiz 6 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

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