Tag: composition of ratios

Questions Related to composition of ratios

The reciprocal of $\dfrac {-5}{13}$ is _____

  1. $\dfrac {5}{13}$

  2. $\dfrac {-13}{5}$

  3. $\dfrac {13}{5}$

  4. $\dfrac {-5}{13}$


Correct Option: B
Explanation:
The reciprocal (also known as the multiplicative inverse) is the number we have to multiply to get an answer equal to the multiplicative number with recipocal of it is 1.
Then $\frac{-5}{13}\times \frac{-13}{5}=1$.
So recipocal of $\frac{-5}{13}$ is $\frac{-13}{5}$.
So answer is (B) $\frac{-13}{5}$.

 

The subtriplicate ratio of $a : b$ is ____

  1. $a^{2} : b^{2}$

  2. $a^{3} : b^{3}$

  3. $\sqrt {a} : \sqrt {b}$

  4. $\sqrt [3]{a} : \sqrt [3]{b}$


Correct Option: D
Explanation:

The subtriplicate ratio of $a : b$ is $\sqrt [3]{a} : \sqrt [3]{b} = (a)^{\frac {1}{3}} : (b)^{\frac {1}{3}}$

If $\dfrac {y}{x-z}=\dfrac{y+x}{z}=\dfrac{x}{y}$ then find $x:y:z$

  1. $1:2:3$

  2. $3:2:1$

  3. $4:2:3$

  4. $2:4:7$


Correct Option: C
Explanation:


$ \dfrac{y}{x-z}=\dfrac{y+x}{z}=\dfrac{x}{y} $

 

Now,

$ \dfrac{y}{x-z}=\dfrac{x}{y} $

$ {{y}^{2}}={{x}^{2}}-xz\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(1) $

 

And

$ \dfrac{y+x}{z}=\dfrac{x}{y} $

$ {{y}^{2}}+xy=xz\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(2) $

$ {{x}^{2}}-xz+xy=xz $

$ x-z+y=z $

$ 2z=x+y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(3) $

 

$ And $

$ \dfrac{y}{x-z}=\dfrac{y+x}{z} $

$ yz=xy-yz+{{x}^{2}}-xz $

$ 2yz=xy+{{x}^{2}}-xz $

$ 2yz=x\left( y+x \right)-xz $                    [From equation (3)]

$ 2yz=2xz-xz $

$ 2yz=xz $

$ 2y=x $

$ \dfrac{x}{y}=\dfrac{2}{1}\,\,\,\,\,\,\,\,......\,\,\left( 4 \right) $


Substituting this value in equation (3), we get

$ 2z=2y+y $

$ 2z=3y $

$ \dfrac{y}{z}=\dfrac{2}{3}\,\,\,\,\,......\,\,\left( 5 \right) $


By equation (4) and (5), we get

$ x:y:z=4:2:3 .$


Hence, this is the answer.

If $\left( {p - q} \right)\,:\left( {q - x} \right)\,$ be the duplicate ratio of $p:q$, then : $\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{x}$

  1. True

  2. False


Correct Option: A
Explanation:
$\left(p-x\right):\left(q-x\right)$ is the duplicate ratio of $p:q$

we know,
 if $a^2 : b^2$ is the duplicate  ratio of $a : b$
         now a/c to question,
$(p -x) : (q - x)$ is the duplicate ratio of $p : q$ 
so, from above rule,
$(p -x ) : (q - x ) = p^2 : q^2$


So,$\dfrac{{p}^{2}}{{q}^{2}}=\dfrac{p-x}{q-x}$

$\Rightarrow\,\dfrac{q-x}{{q}^{2}}=\dfrac{p-x}{{p}^{2}}$

$\Rightarrow\,\dfrac{q}{{q}^{2}}-\dfrac{x}{{q}^{2}}=\dfrac{p}{{p}^{2}}-\dfrac{x}{{p}^{2}}$

$\Rightarrow\,\dfrac{1}{q}-\dfrac{x}{{q}^{2}}=\dfrac{1}{p}-\dfrac{x}{{p}^{2}}$

$\Rightarrow\,\dfrac{1}{q}-\dfrac{1}{p}=\dfrac{x}{{q}^{2}}-\dfrac{x}{{p}^{2}}$

$\Rightarrow\,\dfrac{p-q}{pq}=\dfrac{x\left({p}^{2}-{q}^{2}\right)}{{p}^{2}{q}^{2}}$

$\Rightarrow\,p-q=\dfrac{x\left(p-q\right)\left(p+q\right)}{pq}$

$\Rightarrow\,1=\dfrac{x\left(p+q\right)}{pq}$

$\Rightarrow\,\dfrac{1}{x}=\dfrac{\left(p+q\right)}{pq}$

$\Rightarrow\,\dfrac{1}{x}=\dfrac{1}{q}+\dfrac{1}{p}$

$\therefore\,\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{x}$

Hence the given statement is true.

If $2x=3y$ and $4y=5z$, then $x:z=$

  1. $4:3$

  2. $8:15$

  3. $3:4$

  4. $15:8$


Correct Option: D
Explanation:

Given,

$2x=3y$

or, $\dfrac{x}{y}=\dfrac{3}{2}$.....(1).

Again 

$4y=5z$

or, $\dfrac{y}{z}=\dfrac{5}{4}$.....(2).

Now multiplying (4) and (5) we get,

$\dfrac xy \times \dfrac yz=\dfrac 32 \times \dfrac 54$

$\dfrac{x}{z}=\dfrac{15}{8}$

or, $x:z=15:8$

If $\cfrac{a}{2}=\cfrac{b}{3}=\cfrac{c}{4}$, then $a:b:c=$

  1. $2:3:4$

  2. $4:3:2$

  3. $3:2:4$

  4. None of these


Correct Option: A
Explanation:

Given, $\displaystyle \frac{a}{2} = \frac{b}{3} = \frac{c}{4}$


Lets take $\displaystyle \frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k$


So, $\dfrac{a}{2}  = k$

$a = 2k$

$\dfrac{b}{3}  = k$

$b = 3k$

$\dfrac{c}{4} = k$

$c = 4k$

i.e., $a : b: c = 2k : 3k : 4k$

$a : b; c = 2 : 3 : 4$  

If $a:b=3:4$, then $4a:3b=$

  1. $4:3$

  2. $3:4$

  3. $1:1$

  4. None of these


Correct Option: C
Explanation:

Given $a:b=3:4$

or, $\dfrac{a}{b}=\dfrac{3}{4}$
or, $4a={3b}$.....(1).
Now,
$\dfrac{4a}{3b}$
$=\dfrac{3b}{3b}$ [ Using (1)]
$=\dfrac{1}{1}$.
So $4a:3b=1:1$.

What is the compounded ratio of $x : y, y : z$ and $z : w$

  1. $y : w$

  2. $x : w$

  3. $y : z$

  4. $x : z$


Correct Option: B
Explanation:

By the defination of compound ratio these ratio can be expressed as
$\dfrac {x}{y} \times \dfrac {y}{z}\times \dfrac {z}{w} = \dfrac {x}{w}$
Hence $x : w$

If $a:b=5:7$ and $b:c=6:11$, then $a:b:c=$

  1. $35:49:66$

  2. $30:42:77$

  3. $30:42:55$

  4. None of these


Correct Option: B
Explanation:

$\dfrac{a}{b} = \dfrac{5}{7} $       $\dfrac{b}{c} = \dfrac{6}{11}$


$\left.\begin{matrix} a = \dfrac{5}{7} b \end{matrix}\right|\begin{matrix} c = \dfrac{11 b}{6} \end{matrix}$

$a : b : c$

$\dfrac{5}{7} b : b : \dfrac{11b}{6}$

$\dfrac{5}{7} : 1 : \dfrac{11}{6}$

L.C.M of $7, 6$ is $42$

$\dfrac{5}{7} \times 42 : 42 \times 1 : \dfrac{11}{6} \times 42$

$5 \times 6 : 42 : 11 \times 7$

$30 : 42 : 77$

The compounded ratio of (2 : 3), (6 : 11), and (11 : 2) is

  1. 1:2

  2. 2:1

  3. 11:24

  4. 36:121


Correct Option: B
Explanation:

Required compounded ratio $\displaystyle=\frac{2}{3}\times\frac{6}{11}\times\frac{11}{2}=2:1$