Tag: composition of ratios

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Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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

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

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

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

  4. $\dfrac {-5}{13}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $\frac{-13}{5}$.

 
Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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. $2:4:7$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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

  1. $4:3$

  2. $8:15$

  3. $3:4$

  4. $15:8$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $x:z=15:8$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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., $a : b: c = 2k : 3k : 4k$

$a : b; c = 2 : 3 : 4$  

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $42$

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

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

$30 : 42 : 77$