Tag: finding the whole when a fraction is given

Questions Related to finding the whole when a fraction is given

$\left ( \frac{\sqrt{625}}{11}\times \frac{14}{\sqrt{25}}\times \frac{11}{\sqrt{196}} \right )$ is equal to:

  1. 5

  2. 6

  3. 8

  4. 11


Correct Option: A
Explanation:

$\frac{25}{11}\times \frac{14}{5}\times \frac{11}{14}=5$.

_____ has no reciprocal

  1. $0$

  2. $1$

  3. $-1$

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


Correct Option: A
Explanation:
Reciprocal of $0$ will be $\dfrac{1}{0}$ and we know that $\dfrac{1}{0}$ is not defined.
Thus, the reciprocal of $0$ is not defined
Hence, $0$ does not have a reciprocal.

Multiply the following. Write the answer as a mixed fraction.
$\cfrac { 2 }{ 9 } \times 5$  

  1. True

  2. False


Correct Option: A
Explanation:

$\dfrac{2}{9} \times 5$


Can be written as


$=\dfrac{10}{9} $

$=1\dfrac{1}{9} $

Multiply the following. Write the answer as a mixed fraction.
$\cfrac { 1 }{ 3 } \times 4$

  1. $1\cfrac{1}{3}$

  2. $\cfrac{4}{3}$

  3. $1\cfrac{2}{3}$

  4. $1\cfrac{4}{3}$


Correct Option: A
Explanation:

$\dfrac{1}{3} \times 4$

Can be written as

$=\dfrac{4}{3} $

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

Find the following product:
$2\cfrac { 1 }{ 3 } \times 3\cfrac { 1 }{ 5 } $

Ans$=7\dfrac{7}{15}$

  1. True

  2. False


Correct Option: A
Explanation:

$2\dfrac{1}{3}\times 3\dfrac{1}{5}$


Can be written as


$\dfrac{7}{3}\times \dfrac{16}{5}$

$=\dfrac{112}{15}$

$=7\dfrac{7}{15}$

Which pair of numbers does not have a product equal to $36$?

  1. ${ -4, -9}$

  2. ${ -3, -12}$

  3. $\left{ \displaystyle\frac{1}{2} , -72\right}$

  4. ${1, 36}$


Correct Option: C
Explanation:

Option A= Product of $-4$ and $-9$=$-4\times -9=36$

Option B=Product of $-3$ and $-12$= $-3\times -12=36$
Option C= Product of $\dfrac{1}{2}$ and $-72$=$\dfrac{1}{2}\times{-72}=-36$
Option D= Product of $1$ and $36$ = $1\times 36=36$
Option C is the correct answer.

Which of the following statements is INCORRECT?

  1. Zero has a reciprocal

  2. The product of two negative rational numbers is always positive

  3. The reciprocal of a positive rational number is always positive

  4. The product of two positive rational numbers is always positive


Correct Option: A
Explanation:

Zero can't have a reciprocal because
$1/0$ is not defined.
The rest statements are
$(-a)\times (-b)=ab$
Reciprocal of $a$ is $1/a$
$a\times b=ab$
where $a$ and $b$ are positive.

Rest all statements are true except for option A.

The value of $\left (-\dfrac {7}{2}\right )^{-1}$ is _________.

  1. $-1$

  2. $\dfrac {7}{2}$

  3. $-\dfrac {2}{7}$

  4. $\dfrac {-7}{2}$


Correct Option: C
Explanation:

To find $\left(-\dfrac{7}{2}\right)^{-1}$


Any fraction raised to negative power yields same result as of its reciprocal with modulus of the power.

$\therefore \left(-\dfrac{7}{2}\right)^{-1} = -\dfrac{2}{7}$

Which of the following statements is true?

  1. Every point on the number line represents a rational number

  2. The product of a rational number and its reciprocal to $0$

  3. $(17\times 12)^{-1}=17^{-1}\times 12$

  4. Reciprocal of $\displaystyle\frac{1}{a}$, $a$ $\neq 0$ is $a$


Correct Option: D
Explanation:

(1) Every point on the line doesn't represent a ration number, it represents the real number,  which includes both rational and irrational numbers.


(2) Let's say $a$ is a rational number,
Its reciprocal will be $\dfrac{1}{a}$
The product will be $ a \times  \dfrac{1}{a} =1$.

(3) $(17 \times  12)^{-1}$
$=$ $ \dfrac{1}{17\times 12}$
$=$ $ \dfrac{1}{17} \times  \dfrac{1}{12}$
$=$ ${17}^{-1} \times  {12}^{-1}$

(4) Reciprocal of $\dfrac{1}{a}$
$= \dfrac{1}{\dfrac{1}{a}}$
$=$ $a$
But $\dfrac{1}{a}$ is defined only if $ a \neq0$

The algebraic expression for the statement "Product of $x$ and reciprocal of $a$, subtracted from the product of $y$ and reciprocal of $b"$ is ___________.

  1. $\dfrac {y}{b} - \dfrac {x}{a}$

  2. $\dfrac {y - x}{a - b}$

  3. $xa - yb$

  4. $\dfrac {1}{yb - xa}$


Correct Option: A
Explanation:

Product of $x$ and reciprocal of $a$ $=$ $x\times \dfrac{1}{a}= \dfrac{x}{a}$
Product of $y$ and reciprocal of $b$ $=$ $y\times \dfrac{1}{b}= \dfrac{y}{b}$
Therefore, the algebraic expression is $\dfrac{y}{b}-\dfrac{x}{a}$.

Hence, option A is correct.