Tag: equivalent fractions

Questions Related to equivalent fractions

State whether true or false


$\cfrac { 2 }{ 5 } $ of $\cfrac { 4 }{ 7 } $  is smaller than $\cfrac { 3 }{ 4 } $ of $\cfrac { 1 }{ 2 } $

  1. True

  2. False


Correct Option: A
Explanation:

$\dfrac{2}{5}$ of $\dfrac{4}{7}=\dfrac{2\times 4}{5\times 7}=\dfrac{8}{35}$


$\dfrac{3}{4}$ of $\dfrac{1}{2}=\dfrac{3\times 1}{4\times 2}=\dfrac{3}{8}$


To compare two fractions let us make the denominators same.

LC.M of $8,35=280$

$\therefore \dfrac{8\times 8}{35 \times 8}=\dfrac{64}{280}$

And $\dfrac{3\times 35}{8\times 35}=\dfrac{105}{280}$

We have $\dfrac{105}{280}>\dfrac{64}{280}$

Hence $\dfrac{3}{8}>\dfrac{8}{35}$

Akshit, Rajat, Sanjay and Nikunj each took the same spelling test.
$\bullet$ Akshit spelled $\displaystyle\frac{7}{10}$ of the words correctly.
$\bullet$ Rajat spelled $\displaystyle\frac{3}{4}$ of the words correctly.
$\bullet$ Sanjay spelled $\displaystyle\frac{4}{5}$ of the words correctly.
$\bullet$ Nikunj spelled $\displaystyle\frac{2}{3}$ of the words correctly.
Who spelled the least number of words correctly?

  1. Akshit

  2. Rajat

  3. Sanjay

  4. Nikunj


Correct Option: D
Explanation:

Descending order of the fraction of words spelled correctly is,

$\Rightarrow$  $\dfrac{4}{5}>\dfrac{3}{4}>\dfrac{7}{10}>\dfrac{2}{3}$

$\Rightarrow$  $Sanjay>Rajat>Akshit>Nikunj$

$\therefore$    $Nikunj$ spelled least number of words correctly.

Simplify : $\frac{(-18\frac{1}{3}\times 2\frac{8}{11})}{|\frac{3}{5}+(\frac{-9}{10})| + |-(\frac{-3}{5})|}$

  1. 63$\frac{4}{81}$

  2. -23$\frac{7}{9}$

  3. -67$\frac{7}{9}$

  4. 12$\frac{6}{17}$


Correct Option: C
Explanation:

$\cfrac {\left(-18\cfrac{1}{3}\times 2\cfrac {8}{11}\right)}{\left|\cfrac {3}{5}+\left(\cfrac {-9}{10}\right)\right|+\left|-\left(\cfrac{-3}{5}\right)\right|}$

$=\cfrac{-\cfrac{55}{3}\times \cfrac{30}{11}}{\left|\cfrac {6-9}{10}\right|+\cfrac {3}{5}}$
$=\cfrac {\cfrac {-55\times 30}{33}}{\cfrac {3}{10}+\cfrac {3}{5}}$
$=\cfrac {\cfrac {-55\times 30}{33}}{\cfrac{9}{50}}$
$=\cfrac {-55\times 30\times 50}{33\times 9}$
$=\cfrac {-5\times 10\times 50}{9}$
$=-67\cfrac {7}{9}$

Which of the following options is arranged in descending order?

  1. $\frac{1}{4},\frac{6}{4},\frac{16}{9},\frac{25}{4}$

  2. $\frac{-3}{6},\frac{-4}{3},\frac{-9}{4},\frac{-13}{4}$

  3. $\frac{-5}{8},\frac{-3}{8},\frac{0}{8},\frac{1}{8}$

  4. $\frac{-7}{4},\frac{-3}{4},\frac{5}{4},\frac{8}{3}$


Correct Option: B
Explanation:

To check the order, first make the denominator same of all the fractions, then compare the numerator.

Note- To make the denominator same, multiply both numerator and denominator by HCF of denominator values.
$\cfrac {-3}{6}, \cfrac {-4}{3},\cfrac {-9}{4}, \cfrac {-13}{4}$
Multiply by $12$ both numerator and denominator.
$\cfrac {-36}{12}, \cfrac {-43}{12},\cfrac {-108}{12}, \cfrac {-156}{12}$

The smallest of the fractions given.

  1. $9/10$

  2. $11/12$

  3. $23/28$

  4. $32/33$


Correct Option: C

Compare the given fractions and specify the correct operator

$\dfrac{9}{16}$ ___ $\dfrac{13}{5}$

  1. $\displaystyle = $

  2. $\displaystyle > $

  3. $\displaystyle < $

  4. None of the above


Correct Option: C
Explanation:

To compare fractions, make the denominators equal

LCM of denominators $16$ and $5$ is $80$
$\therefore \dfrac{9}{16} \times \dfrac{5}{5}$ $=\dfrac{45}{80}$ and $\dfrac{13}{5} \times  \dfrac {16}{16}$ $=\dfrac{208}{80}$


By making denominators equal, we find that $208$ is greater than $45$ 
$\therefore \dfrac{9}{16} < \dfrac{13}{5} $.

If a, b, c, are positive $\displaystyle \frac{a+c}{b+c}$ is 

  1. always smaller than $\displaystyle \frac{a}{b}$

  2. always greater than $\displaystyle \frac{a}{b}$

  3. greater than $\displaystyle \frac{a}{b}$ only if a > b

  4. greater than $\displaystyle \frac{a}{b}$ only if a < b


Correct Option: D
Explanation:

Since we need to compare the fraction $\displaystyle \frac{a+c}{b+c}$ with $\displaystyle \frac{a}{b}$, we cross multiply the terms and check since $a,b,c$ are all given to be positive.
We thus get $b(a + c)$ on the L.H.S. and $a(b + c)$ on R.H.S.
Thus, simplifying we are left with $ab + bc$ on the L.H.S. and $ab + ac$ on the R.H.S.
Now, which side is greater depends on $ac$ and $bc$, which in turn depends upon $a$ & $b.$
L.H.S. is greater if $b > a$, which implies $\frac{a + c}{b + c}$ is greater when $b > a.$