Let $\dfrac{p}{q}=\dfrac{7}{9}$

Let the number be $n$

$\cfrac{2}{4} \times \cfrac{3}{7} = \cfrac{1}{2} \times \cfrac{3}{7} = \cfrac{3}{14}$

Total number of Animals $=192$

Given $2 \dfrac { 1 } { 2 } \mathrm {  } \text { of } 10 \mathrm { cm }$

$2\cfrac { 1 }{ 4 } \times \cfrac { 5 }{ 12 } +\cfrac { 1 }{ 2 } $

$999\dfrac { 995 }{ 999 } \times 999=\quad 999\times 995\=\dfrac { 999\times 999+995 }{ 999 } \times 999\=998996$

Given expression $\displaystyle \frac{1}{9}\times\frac{1}{6}\times\frac{1}{3}\times56052=346$

$\displaystyle \left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )\left ( 1-\frac{1}{4} \right )......\left ( 1-\frac{1}{n} \right )=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times ......\times \frac{n-2}{n-1}\times \frac{n-1}{n}=\frac{1}{n}$

Let the money with the man at first be Rs 1
$\displaystyle \therefore $ Money spent=$\displaystyle \frac{5}{6}of Rs1=Rs\frac{5}{6}$
Remaining money=$\displaystyle =1-\frac{5}{6}=Rs\frac{1}{6}$
Money earned$\displaystyle =\frac{1}{2}=Rs\frac{1}{6}=Rs \frac{1}{12}$
$\displaystyle \therefore $ Total money with him now=Rs$\displaystyle \frac{1}{6}+Rs.\frac{1}{12}=Rs.\frac{3}{12}=Rs.\frac{1}{4}$
$\displaystyle \therefore \frac{1}{4}th$ part of the money is with him now

$15$ of $\cfrac{1}{5}=15\times\cfrac{1}{5}=3$

The reciprocal of a number is just divide 1 by the number.
$\therefore$ Reciprocal of $3\cfrac 12$ i.e $\cfrac 72$ is $\cfrac 27$
Option B is correct.

Any number multiplied with $0$ gives $0$. 

The given question shows the associative property of multiplication i.e. 

We need to find product of $\displaystyle \frac {12}{24}, \frac {36}{72}$

$3\dfrac12=\dfrac {3\times 2+1}{2}=\dfrac72$

$0.0777\,=\,7\,\times\,0.111$
$=\,7\,\times\, \displaystyle\frac{1}{9}\,=\, \displaystyle\frac{7}{9}$

The product of rational number with its reciprocal is always equal to $1$.

reciprocal of 7/2 is 2/7..

zero multiplied by any number is zero..

zero multiplied by any number is zero

$ \displaystyle \left | x \right | =\left | \frac{-3}{5} \right |  $ and  $ \displaystyle \left | y \right | =\left | \frac{4}{-7} \right |  $ 

(12/24)×(36/72)

To multiply mixed fractions we convert them into improper fractions.

June 2008 has 30 days.
$30\times 3\dfrac {1}{4}=30\times \dfrac {13}{4}=97\dfrac {1}{2} litres$.

$=\dfrac {5}{6}\times \dfrac {2}{3}=\dfrac {10}{18}=\dfrac {5}{9}$

If the product of two numbers is 1, then each number is known as multiplicative inverse or the reciprocal of one another. ... Hence, the product of a fractional number and its multiplicative inverse is 1

Veronica can type words in $1$ minute $=28$.

Let the number of matches lost = x
Number of matches won = x + 4
Total matches played = x + x +4
                                   = 2x + 4
We have,
x + 4 = $\displaystyle\frac{3}{5}\left ( 2x + 4 \right )$
5x + 20 = 6x + 12
x = 8
$\displaystyle\therefore $ total matches played = 2x + 4
$\displaystyle= 2\left ( 8 \right )+ 4= 20$

The equivalent to the given fraction 10/11 having the numerator 40 should have a denominator divisible by 11. The only option with a denominator divisible by 11 is option C. 44 is divisible by 11 and 40 is divisible by 10. So,

Only options A and C have denominators 18 in their options.

$ \displaystyle \frac{1}{5}  $ of 10 km 

When we talk about percents, we should multiply the number with $100$

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

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

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

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

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

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.

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

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

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

Put x = 3 and y = 3 of LHS, we get
5$\dfrac{1}{3}$ $\times$ 3$\dfrac{3}{4}$ = $\dfrac{16}{3}$ $\times$ $\dfrac{15}{4}$ = 4 $\times$ 5 = 20

Let the original fraction be $\dfrac{x}{y}$.

Option A

Part of field in which potatoes are grown $= \dfrac{2}{3}$ 

30% of 40% of 560 = $\dfrac{30}{100}\ \ \times\dfrac{40}{100}\ \ \times 560\ \ = 67.2 $

$\cfrac { m }{ n } =\cfrac { 4 }{ 3 } ,\cfrac { r }{ t } =\cfrac { 9 }{ 14 } \ m=\cfrac { 4n }{ 3 } ,r=\cfrac { 9t }{ 14 } \ mr=\cfrac { 6nt }{ 7 } \ 3mr-nt=\cfrac { 18nt }{ 7 } -nt=\cfrac { 11nt }{ 7 } \ 4nt-7mr=4nt-6nt=-2nt\ \cfrac { 3mr-nt }{ 4nt-7mr } =-\cfrac { 11 }{ 14 } $

Given multiplication of $\dfrac{2}{3}$ with $4$

$\frac{2}{3}$  of $48$
        $=\frac{2}{3}\times48$
        $=2\times16$
        $=32$

$\frac{-39}{3}\times\frac{19}{5}\times\frac{-45}{38}$
$=\frac{-13\times-9}{2}$
$=\frac{117}{2}$

$\frac{-2}{11}\times\frac{-44}{16}$
$=\frac{-1\times-4}{8}$
$=\frac{-1\times-1}{2}=\frac{1}{2}$

$\large{1\frac{2}{7}}$ of $\large{\frac{56}{63}}$
$=\large{\frac{9}{7}} \times \large{\frac{56}{63}}$
$=\large{\frac{8}{7}}$ or $\large{1\frac{1}{7}}$.

$\begin{array}{l} In\, \, L.H.S \ No.=\frac { { 540\times 7 } }{ { 11 } } \times \frac { { 11 } }{ { \left( { 343\times 11+7 } \right)  } }  \ \frac { { 540\times 7 } }{ { 7\left( { { 7^{ 2 } }\times 11+1 } \right)  } }  \ No.=\frac { { 540 } }{ { 539+1 } } =1 \end{array}$

$\sqrt {34-24\sqrt 2}\times (4+3\sqrt 2)$

Let the number of men visiting hotel be $x$

Let the number be x Given $\displaystyle \frac{2}{3}x=\frac{25}{216}\times\frac{1}{x}$
$\displaystyle\Rightarrow x^{2}=\frac{25}{216 _{72}}\times\frac{3^{1}}{2}=\frac{25}{144}$
$\displaystyle\Rightarrow x=\sqrt{\frac{25}{144}}=\frac{5}{12}$

Given expression $\displaystyle =\left ( 999+\frac{999}{1000} \right )\times7$
$\displaystyle=1000-1+1-\frac{1}{1000}\times7$
$\displaystyle=\left ( 1000-\frac{1}{1000} \right )\times7=7000-\frac{7}{1000}$
$\displaystyle=6999+1-\frac{7}{1000}=6999+\frac{993}{1000}$
$\displaystyle=6999\frac{993}{1000}$

Per day consumption $=3\cfrac14$liters.

Let integer = 4 and Rational number = $\displaystyle \frac{2}{1} $
Then product = $\displaystyle 4\times\frac{2}{1}=8 $ (a natural number)
and Quotient = $\displaystyle 4\div \frac{2}{1}=4\times \frac{1}{2}=2 $ (an integer)

Sum of the reciprocals of $\displaystyle \dfrac{3}{5}$ and $\dfrac{7}{3}$
$\displaystyle =\frac{5}{3}+\frac{3}{7}=\frac{35+9}{21}=\frac{44}{21}$
$\displaystyle \therefore $ Required number $\displaystyle =\frac{21}{44} $

We know that the reciprocal of any number $n$ is $\dfrac {1}{n}$. When we multiply a number by its reciprocal, we get $1$ as the answer. For example: $n\times \dfrac { 1 }{ n } =1$ 

The reciprocal of a number is $1$ divided by that number. For example, the reciprocal of $a$ is $\dfrac {1}{a}$

We first rewrite the given mixed fraction $2\dfrac { 1 }{ 3 }$ as $\dfrac {7}{3}$. 

The reciprocal of a number is $1$ divided by that number. For example, the reciprocal of $a$ is $\dfrac {1}{a}$

It is given that Ravi had $\dfrac {5}{6}$ of a cake and from that piece of cake he ate $\dfrac {2}{3}$rd of it which means that:

If the product of two numbers is $1$, then each number is known as multiplicative inverse or the reciprocal of one another. To find the multiplicative inverse of a proper or improper fraction, interchange the numerator and denominator. For example, the multiplicative inverse of the fraction $\dfrac {5}{18}$ is $\dfrac {18}{5}$.

the reciprocal of 5/11 is 11/5..

1/6 of 48 liter

1 rupee=100 paise

$\dfrac{2}{5} $ of $10$ litre $=\dfrac{2}{5}\times 10=4$ litres.

The reciprocal of $6-\sqrt5$ is $\dfrac{1}{(6-\sqrt5)}$
Multiply and divide it by $6+\sqrt5$ , we get $\dfrac{6+\sqrt5}{(6+\sqrt5)(6-\sqrt5)} = \dfrac{6+\sqrt5}{31}$

$\dfrac{2}{3} \div \dfrac{14}{15} = \dfrac{2\times15}{3\times14}$

$\dfrac{6}{7} \times 2$

Rehana works $2\dfrac{1}{2}$ hrs $= 150 $minutes

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

$\boxed{\text{Work(W)}\propto \text{Days(D)}\\dfrac{W _1}{D _1}=\dfrac{W _2}{D _2}}$

given,
$W _1=\dfrac25\quad D _1=1\,\text{day}\W _2=1\,\,\text{(painting the whole house)}$
to find, $D _2=?$

$\dfrac{2/5}{1}=\dfrac{1}{D _2}\Rightarrow D _2=\dfrac52$

$\therefore D _2=2\dfrac12 \,\text{days}$

$1024=2\times2\times2\times2\times2\times2\times2\times2\times2\times2 = 2^{10}$

$\left( 1-\cfrac { 1 }{ 3 }  \right) \left( 1-\cfrac { 1 }{ 4 }  \right) \left( 1-\cfrac { 1 }{ 5 }  \right) ...\left( 1- \cfrac 1n \right) =\cfrac { 2 }{ 3 } .\cfrac { 3 }{ 4 } .\cfrac { 4 }{ 5 } ....\cfrac { n-2 }{ n-1 } .\cfrac { n-1 }{ n } =\cfrac { 2 }{ n } $

Apply BODMAS
$4\frac{4}{5}\div\frac{3}{5}$ of $5+\frac{4}{5}\times\frac{3}{10} -\frac{1}{5}$
$=\frac{24}{5}\div3+\frac{4}{5\times\frac{3}{10}-\frac{1}{5}}$
$=\frac{8}{5}+\frac{6}{25}-\frac{1}{5}$
$=\frac{41}{25}$
$=1\frac{16}{25}$
Option 'A' is the answer

$\quad \left( 1-\cfrac { 1 }{ 3 }  \right) \left( 1-\cfrac { 1 }{ 4 }  \right) \left( 1-\cfrac { 1 }{ 5 }  \right) ...=\cfrac { 2 }{ 3 } .\cfrac { 3 }{ 4 } .\cfrac { 4 }{ 5 } ...\cfrac { n-1 }{ n } =\cfrac { 2 }{ n } $

$\large{\frac{1}{3}} \ of\ \large{4\frac{2}{3}}$ $\div$ $\large{2\frac{1}{3}} of\ \large{1\frac{1}{2}}$

Product of the fraction is $\dfrac{12}{24}$ and $\dfrac{36}{72}$ is 

The product of $\dfrac{11}{12} \times \dfrac{16}{4} \times \dfrac{9}{16} $ is

$\dfrac{1}{9}$ of $45=5$ because $45$ is divided by $9$ in $5$ times.

Let's first convert the mixed fraction into simple fraction $\dfrac{4}{3},$ $\dfrac{13}{4},$ $\dfrac{7}{8}$ .

Reciprocal of $y-1$ is $\dfrac{1}{y-1} = y+1$, so we get ${y}^{2}-1=1$
Which implies $y = \pm \sqrt2$
So, the correct option is $E$.