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Properties of proportion - class-XII

Description: properties of proportion
Number of Questions: 35
Created by:
Tags: ratio and proportion maths ratio and proportions
Attempted 0/35 Correct 0 Score 0

If two ratios are equal then their inverse ratios are equal. This property is known as

  1. Dividendo property

  2. Invertendo property

  3. Componendo property

  4. Alternendo property


Correct Option: B
Explanation:

If $a : b :: c : d$ then $b : a :: d : c \Rightarrow $ invertendo property

If $\dfrac {x}{y}=\dfrac {3}{4}$ and $\dfrac {x}{2z}=\dfrac {3}{2}$, then $\dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )$ will be equivalent to

  1. 8

  2. 9

  3. 11

  4. 12


Correct Option: A
Explanation:

$\because \dfrac {x}{y}=\dfrac {3}{4}\Rightarrow \dfrac {y-x}{y+x}=\dfrac{4-3}{4+3}=\dfrac {1}{7}$
$\because \dfrac {x}{2z}=\dfrac {3}{2}\Rightarrow 2x=6z$ or, $x=3z$
$\therefore \dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )=\dfrac {6z+z}{3z-2z}+\left (\dfrac {6}{7}+\dfrac {1}{7}\right )$
$=\dfrac {7z}{z}+\dfrac {7}{7}=7+1=8$.

If $a=2+\sqrt 3$ then the value of $a+\frac {1}{a}$ is

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: D
Explanation:

$\because a=2+\sqrt 3$
$\therefore \frac {1}{a}=\frac {1}{2+\sqrt 3}\times \frac {2-\sqrt 3}{2-\sqrt 3}=\frac {2-\sqrt 3}{4-3}=2-\sqrt 3$
$a+\frac {1}{a}=2+\sqrt 3+2-\sqrt 3=4$

After applying invertendo to $3:7::2:9$ we get

  1. $3:7::2:9$

  2. $3:2::7:9$

  3. $9:7::2:3$

  4. $7:3::9:2$


Correct Option: D
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $3:7::2:9$
After applying invertendo we get
$7:3::9:2$
Option D is correct

If $x=7-4\sqrt 3$, the value of $x^2+\displaystyle\frac{1}{x^2}$ will be

  1. $146$

  2. $148$

  3. $194$

  4. $196$


Correct Option: C
Explanation:

Given,
$x=7-4\sqrt 3$
$\therefore \displaystyle\frac{1}{x}=\displaystyle\frac{1}{(7-4\sqrt 3)}\times \displaystyle\frac{7+4\sqrt 3}{7+4\sqrt 3}$
$=\displaystyle\frac{7+4\sqrt 3}{(49-48)}$
$=7+4\sqrt 3$
Now,
$x+\displaystyle\frac{1}{x}=(7-4\sqrt 3)+(7+4\sqrt 3)$
$=14$
$=> \left( x+\displaystyle\frac{1}{x}\right)^2=(14)^2$
Using $(a+b)^2=a^2+b^2+2ab$ we get,
$=> x^{ 2 }+\frac { 1 }{ x^{ 2 } } +2(x)(\frac { 1 }{ x } )=196$
$=> x^2+\displaystyle\frac{1}{x^2}=196-2$
$\therefore x^2+\displaystyle\frac{1}{x^2}=194$

If $x=3+\sqrt 8$ then the value of $x^2+\frac {1}{x^2}$ is

  1. 30

  2. 32

  3. 34

  4. 36


Correct Option: C
Explanation:

$x=3+\sqrt 8\Rightarrow \frac {1}{x}=\frac {1}{3+\sqrt 8}\times \frac {3-\sqrt 8}{3-\sqrt 8}=\frac {3-\sqrt 8}{9-8}$
$=3-\sqrt 8$
$x^2+\frac {1}{x^2}=\left (x+\frac {1}{x}\right )^2-2=36-2=34$

If $a : b :: c : d$ then $b : a :: d : c$. This property is known as :

  1. Invertendo property

  2. Alternendo property

  3. Componendo property

  4. Dividendo property


Correct Option: A
Explanation:

If $a : b :: c : d$ then $b : a :: d : c\Rightarrow $invertendo property.

proof
$\dfrac{a}{b}=\dfrac{c}{d}$

inverting both sides, we get
$\Rightarrow \dfrac{b}{a}=\dfrac{d}{c}$

After applying invertendo to $31:15::18:19$  we get $15:a::b:18$ 
What is the value of $a+b$

  1. $40$

  2. $50$

  3. $30$

  4. $60$


Correct Option: B
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $31:15::18:19$
After applying invertendo we get
$15:31::19:18\equiv  15:a::b:18\ a+b=31+19=50$
Option B is correct

Find the value of $a$ and $b$ respectively
After applying invertendo to $2:5::3:9$  we get $a:2::9:b$ 

  1. $9$ and $2$

  2. $2$ and $9$

  3. $3$ and $5$

  4. $5$ and $3$


Correct Option: D
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $2:5::3:9$
After applying invertendo we get
$5:2::9:3\equiv a:2::9:b$
$\Rightarrow a=5,b=3$
Option D is correct

Here 'x' in the following is : $\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=b$

  1. $\dfrac{2ab}{(b^2+1)}$

  2. $\dfrac{2ab}{a+b}$

  3. $\dfrac{a+b}{2ab}$

  4. $\dfrac{b^2+1}{2ab}$


Correct Option: A
Explanation:

$\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-{\sqrt{a-x}}}=\dfrac{b}{1}$

$\Rightarrow \dfrac{\sqrt{a+x}}{\sqrt{a-x}}=\dfrac{b+1}{b-1}$

$\Rightarrow \dfrac{a+x}{a-x}=\dfrac{(b+1)^2}{(b-1)^2}$

$\Rightarrow \dfrac{(a+x)+(a-x)}{(a+x)-(a-x)}=\dfrac{(b+1)^2+(b-1)^2}{(b+1)^2-(b-1)^2}$

$\Rightarrow \dfrac{2a}{2x}=\dfrac{2(b^2+1)}{4b}$

$\Rightarrow x=\dfrac{2ab}{b^2+1}$

After applying invertendo to $30:50::80:20$  we get $50:a::b:80$ 
Find the value of $a-b$

  1. $10$

  2. $20$

  3. $30$

  4. $40$


Correct Option: A
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $30:50::80:20$
After applying invertendo we get
$50:30::20:80\equiv 50:a::b:20\ a-b=30-20=10$
Option A is correct

After applying invertendo to $1:2::3:4$ we get

  1. $1:2::3:4$

  2. $2:1::4:3$

  3. $1:3::2:4$

  4. $4:2::3:1$


Correct Option: B
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $1:2::3:4$
After applying invertendo we get
$2:1::4:3$
Option B is correct

After applying invertendo to $9:12::13:40$ we get

  1. $9:13::12:40$

  2. $40:12::13:9$

  3. $9:12::13:40$

  4. $12:9::40:13$


Correct Option: D
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $9:12::13:40$
After applying invertendo we get
$12:9::40:13$
Option D is correct

If $\dfrac { a }{ b } =\dfrac { b }{ c } =\dfrac { c }{ d }$ then $\dfrac { { b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 } }{ { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }$ will be equal to

  1. $\dfrac{a}{b}$

  2. $\dfrac{b}{c}$

  3. $\dfrac{c}{d}$

  4. $\dfrac{d}{a}$


Correct Option: D

If $\cfrac{{x}^{3}+{x}^{2}+x+1}{{x}^{3}-{x}^{2}+x-1}=\cfrac{{x}^{2}+x+1}{{x}^{2}-x+1}$, then the number of real-value of $x$ satisfying are

  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: A
Explanation:

Given that


$\dfrac{x^3+x^2+x+1}{x^3-x^2+x+1}=\dfrac{x^2+x+1}{x^2-x+1}$

Performing Componendo Dividendo, we get

$\Rightarrow \dfrac{x^3+x^2+x+1+x^3-x^2+x+1}{x^3+x^2+x+1-x^3+x^2-x+1}=\dfrac{x^2+x+1+x^2-x+1}{x^2+x+1-x^2+x-1}$

$\Rightarrow \dfrac{2x^3+2x}{2x^2+2}=\dfrac{2x^2+2}{2x} $

$\Rightarrow \dfrac{x^3+x}{x^2+1}=\dfrac{x^2+1}{x} $

$\Rightarrow x(x^3+x)=(x^2+1)^2$

$\Rightarrow x^4+x^2=x^4+1+2x^2$

$\Rightarrow x^2=-1$

Hence no real values of $x$ satisfy this equation.

If $\cfrac{a+3d}{a+9d}=\cfrac{a+d}{a+5d}=k$, then $k$ is equal to $(a,d> 0)$

  1. $\dfrac{1}{2}$

  2. $2$

  3. $6$

  4. $0$


Correct Option: A
Explanation:

$\dfrac{a+3d}{a+9d}=\dfrac{a+d}{a+5d}$


Applying Componendo-Dividendo

$\Rightarrow \dfrac{a+3d+a+9d}{a+3d-a-9d}=\dfrac{a+d+a+5d}{a+d-a-5d}$ 

$\Rightarrow \dfrac{2a+12d}{-6d}=\dfrac{2a+6d}{-4d}$

$\Rightarrow \dfrac{a+6d}{-3}=\dfrac{a+3d}{-2}$

$\Rightarrow -2a-12d=-3a-9d$

$\Rightarrow a=3d$

Now, given that 

$\Rightarrow k=\dfrac{a+3d}{a+9d}$

$\Rightarrow k=\dfrac{3d+3d}{3d+9d}$

$\Rightarrow k=\dfrac{6d}{12d}$

$\Rightarrow k=\dfrac{1}{2}$

Solve for $x$:

$\dfrac{{2 - 7x}}{{1 - 5x}} = \dfrac{{3 + 7x}}{{4 + 5x}}$

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

  2. $\dfrac{1}{2}$

  3. $\dfrac{3}{4}$

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


Correct Option: B
Explanation:

$\dfrac{2-7x}{1-5x}=\dfrac{3+7x}{4+5x}$


Applying Alternendo,

$\Rightarrow \dfrac{2-7x}{3+7x}=\dfrac{1-5x}{4+5x}$

Now applying Componendo and Dividendo,
$\Rightarrow\dfrac{(2-7x)+(3+7x)}{(2-7x)-(3+7x)}=\dfrac{(1-5x)+(4+5x)}{(1-5x)-(4+5x)}$

$\Rightarrow\dfrac{5}{-1-14x}=\dfrac{5}{-3-10x}$

$\Rightarrow-1-14x=-3-10x\\$
$\Rightarrow 14x-10x=3-1\\$
$\Rightarrow 4x=2\\$
$\Rightarrow x=\dfrac{2}{4}=\dfrac{1}{2}$

A naughty student breaks the pencil in such a way that the ratio of two broken parts is same as that of the original length of the pencil to one of the larger part of the pencil, The ratio of the other part to the original length of pencil is:

  1. $1 :2 \sqrt{5}$

  2. $2 : (3+\sqrt{5})$

  3. $2 : \sqrt{5}$

  4. $can't\ be\ determined $


Correct Option: B
Explanation:

Suppose that,

The length of larger part be$=x$

And the smaller part be $=y$


So, their ratio is $x$ is to

$ \dfrac{x}{y}=\dfrac{kx+ky}{kx} $

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

$ \Rightarrow {{x}^{2}}=xy+{{y}^{2}} $

$ \Rightarrow {{x}^{2}}-{{y}^{2}}=xy $


Let $y=1$

$ {{x}^{2}}-{{1}^{2}}=x $

$ \Rightarrow {{x}^{2}}-x-1=0 $

$ \Rightarrow x=\dfrac{1\pm \sqrt{5}}{2}\,\,\,\,\left( \text{On}\,\text{solving}\,\text{by}\,\text{second}\,\text{degree}\,\text{equation}\,\text{rule} \right) $


Negative value cannot be considered

So,

$x=\dfrac{1+\sqrt{5}}{2}$ so, the ratio

$ \dfrac{x}{y}=\dfrac{\dfrac{1+\sqrt{5}}{2}}{1} $

$ \Rightarrow x:y=\left( 1+\sqrt{5} \right):2 $


Therefore,

$ \dfrac{y}{x+y}=\dfrac{2}{\left( 1+\sqrt{5}+2 \right)} $

$ \Rightarrow \dfrac{y}{x+y}=\dfrac{2}{\left( 3+\sqrt{5} \right)} $


Hence, this is the answer.

$\dfrac{7a-3b}{7c-3d}=1, then\ \dfrac{a}{b}=\dfrac{d}{c}$

  1. True

  2. False


Correct Option: B

If  a/b = x/y = p/q , then $\dfrac{6a + 9x + 2p}{6b + 9y + 2q}$ = _________. 

  1. a/b

  2. x/y

  3. p/q

  4. All of these


Correct Option: D
Explanation:

All of these.


Given,

$\dfrac{a}{b}=\dfrac{x}{y}=\dfrac{p}{q}$

Let, $\dfrac{a}{b}=\dfrac{x}{y}=\dfrac{p}{q}=k$

$\Rightarrow a=bk, x=yk, p=qk$

$\Rightarrow 6a=(6b)k, 9x=(9y)k, 2p=(2q)k$

Now,

$\dfrac{6a+9x+2p}{6b+9y+2q}$

$=\dfrac{(6b)k+(9y)k+(2q)k}{6b+9y+2q}$

$=k\dfrac{6b+9y+2q}{6b+9y+2q}$

$=k$

$=\dfrac{a}{b}=\dfrac{x}{y}=\dfrac{p}{q}$


Three numbers $A,B$ and $C$ are in the ratio $12\colon15\colon25$. If the sum of these numbers is $312$, find the ratio between the difference of $A$ and $B$ and the difference of $C$ and $B$.

  1. $\;3\colon7$

  2. $\;10\colon3$

  3. $\;3\colon10$

  4. $\;7\colon3$


Correct Option: C
Explanation:

The ratio of $A,B,C =12:15:25$
Then, total of the ratio is $12+15+25=52$
Then, $ A=\displaystyle \frac{12}{52}\times 312=72$

$B=\displaystyle \frac{15}{52}\times 312=90$

$C=\displaystyle \frac{25}{52}\times 312=150$
Then, $B-A=18$ and $C-B=60$
Hence, the required ratio $=18:60\Rightarrow 3:10$

Mark the correct alternative of the following.
If $x : y =1 : 1$, then $\dfrac{3x+4y}{5x+6y}=?$

  1. $\dfrac{7}{11}$

  2. $\dfrac{17}{11}$

  3. $\dfrac{17}{23}$

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


Correct Option: A
Explanation:

Given, $x:y=1:1$

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

Now,
$\dfrac{3x+4y}{5x+6y}$
$=\dfrac{7x}{11x}$ [ Using (1)]
$=\dfrac{7}{11}$.

The given property $a : b :: c : d$ then $(a - b) : b :: (c - d) : d.$ is known as

  1. Alternendo property

  2. Dividendo property

  3. Componendo property

  4. Invertendo Property


Correct Option: B
Explanation:

If $\dfrac{a}{b}=\dfrac{c}{d},$ then by dividendo property

$\dfrac{a-b}{b}=\dfrac{c-d}{d}\Rightarrow (a-b):b::(c-d):d$

After applying invertendo to $1:2::8:9$ we get:

  1. $2:1::9:8$

  2. $1:8::2:9$

  3. $8:9::1:2$

  4. $1:9::8:2$


Correct Option: A
Explanation:

If $a : b :: c : d$ then $b : a :: d : c$ is invertendo property of ratios

We have $1:2::8:9$
After applying invertendo we get
$2:1::9:8$
Option A is correct

The given property $a : b :: c : d$ then $a : c :: b : d$ is known as:

  1. Componendo property

  2. Dividendo property

  3. Invertendo property

  4. Alternendo property


Correct Option: D
Explanation:
If $\dfrac{a}{b}=\dfrac{c}{d}$ then, by alternedo

$\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow a:c::b:d$

If $\displaystyle \frac{x}{y}=\frac{6}{5}$ then $\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}$ is:

  1. $\displaystyle \frac{36}{25}$

  2. $\displaystyle \frac{25}{36}$

  3. $\displaystyle \frac{61}{11}$

  4. $\displaystyle \frac{11}{61}$


Correct Option: C
Explanation:

$\displaystyle \frac{x}{y}=\frac{6}{5}$

$x = 6k, y = 5k$

$\displaystyle \frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{\left ( 6k \right )^{2}+\left ( 5k \right )^{2}}{\left ( 6k \right )^{2}-\left ( 5k \right )^{2}}=\frac{61k^{2}}{11k^{2}}=\frac{61}{11}$

If $x=\cfrac { 4ab }{ a+b } $ then value of $\cfrac { x+2a }{ x-2a } +\cfrac { x+2b }{ x-2b } $

  1. $a$

  2. $b$

  3. $0$

  4. $2$


Correct Option: D
Explanation:

Given, $x = 4ab/(a+b)$

=> $x = 2a\times 2b / (a+b)$
=> $x/2a = 2b / (a+b)$
Using componendo and dividendo 
=> $x+2a / x-2a = 2b+a+b / 2b-a-b$
=> $x+2a / x-2a = 3b+a / b-a$

Similarly, $x+2b / x-2b = 3a+b / a-b$

Now, adding both,
$[x+2a / x-2a] + [x+2b / x-2b ] = [3b+a / b-a ] + [ 3a +b / a-b ]$
RHS => $[3b+a / b-a ] - [ 3a+b / b-a ]$
=> $3b+a-3a-b / b-a$
=> $2b-2a / b-a$
=> $2(b-a)/b-a$
=> $2$

Hence, LHS $= 2$ 

If $a : b = c : d = e : f$, then the value of each ratio is $(a + c + e) : (b + d + f)$
This property is called as 

  1. Componendo property

  2. Convertendo property

  3. Addendo property

  4. Dividendo property


Correct Option: C
Explanation:

$\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=k$

$a=bk,c=dk,e=fk\ \Rightarrow \dfrac{a+c+e}{b+d+f}=\dfrac{bk+dk+fk}{b+d+f}=k\ \Rightarrow a:b=c:d=e:f=(a+c+e):(b+d+f) $
addendo property

 $\dfrac{a+be^y}{a-be^y} = \dfrac{b+ce^y}{b-ce^y}  =  \dfrac{c+de^y}{c-de^y}$, then $a,b,c,d$  are  in

  1. $A.P.$

  2. $G.P.$

  3. $H.P.$

  4. $A.G.P.$


Correct Option: B

If $\cfrac { { a }^{ 3 }+3a{ b }^{ 2 } }{ 3{ a }^{ 2 }b+{ b }^{ 3 } } =\cfrac { { x }^{ 3 }+3x{ y }^{ 2 } }{ 3{ x }^{ 2 }y+{ y }^{ 3 } } $ then

  1. $bx=ay$

  2. $by=ax$

  3. ${ b }^{ 2 }y={ a }^{ 2 }x$

  4. ${ b }^{ 2 }x={ a }^{ 2 }y$


Correct Option: A
Explanation:

$\dfrac{{{a^3} + 3a{b^2}}}{{3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2}}}{{3{x^2}y + {y^3}}}$


Use of compodendo and Devidendo

$\dfrac{{{a^3} + 3a{b^2} + 3{a^2}b + {b^3}}}{{3{a^2}b + {b^3} - 3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2} + 3{x^2}y + {y^3}}}{{3{x^2}y + {y^3} - 3{x^2}y + {y^3}}}$

$ \Rightarrow \dfrac{{{{\left( {a + b} \right)}^3}}}{{{{(a - b)}^3}}} = \dfrac{{{{\left( {x + y} \right)}^3}}}{{{{(x - y)}^3}}}$

$ \Rightarrow {\left( {\dfrac{{a + b}}{{a + b}}} \right)^3} = {\left( {\dfrac{{x + y}}{{x - y}}} \right)^3}$

$ \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{x + y}}{{x - y}}$

$ \Rightarrow \left( {a + b} \right)\left( {x + y} \right) = \left( {x + y} \right)\left( {a - b} \right)$

$ \Rightarrow ax{\text{ }} - {\text{ }}ay{\text{ }} + {\text{ }}bx{\text{ }} - {\text{ }}by{\text{ }} = {\text{ }}xa{\text{ }} - {\text{ }}xb{\text{ }} + {\text{ }}ay{\text{ }} - {\text{ }}yb$

$ \Rightarrow bx{\text{ }} + {\text{ }}xb{\text{ }} = {\text{ }}ay{\text{ }} + {\text{ }}ay$

$ \Rightarrow 2bx{\text{ }} = {\text{ }}2ay$

$ \Rightarrow bx{\text{ }} = {\text{ }}ay$


If $x=\cfrac { 2\sqrt { 5 }  }{ \sqrt { 3 } +\sqrt { 5 }  } $, then what is the value of $\cfrac { x+\sqrt { 5 }  }{ x-\sqrt { 5 }  } +\cfrac { x+\sqrt { 3 }  }{ x-\sqrt { 3 }  } $

  1. $\sqrt {5}$

  2. $\sqrt {3}$

  3. $\sqrt {15}$

  4. $2$


Correct Option: D
Explanation:

$x=\cfrac { 2\sqrt { 5 }  }{ \sqrt { 3 } +\sqrt { 5 }  } \Rightarrow \cfrac { x }{ \sqrt { 3 }  } =\cfrac { 2\sqrt { 5 }  }{ \sqrt { 3 } +\sqrt { 5 }  } $
and $\cfrac { x }{ \sqrt { 5 }  } =\cfrac { 2\sqrt { 3 }  }{ \sqrt { 3 } +\sqrt { 5 }  } $
Applying components and dividendo, we get
$\cfrac { x+\sqrt { 5 }  }{ x-\sqrt { 5 }  } =-\left( 7+2\sqrt { 15 }  \right) $
and
$\cfrac { x+\sqrt { 3 }  }{ x-\sqrt { 3 }  } =9+2\sqrt { 15 } $
$\Rightarrow \cfrac { x+\sqrt { 5 }  }{ x-\sqrt { 5 }  } +\cfrac { x+\sqrt { 3 }  }{ x-\sqrt { 3 }  } =2$

Which of the following ratios is equal to $13:4$ in its simplest form?

  1. $18:8$

  2. $105:36$

  3. $91:28$

  4. $144:250$


Correct Option: C
Explanation:

A.  $18 : 8 = \dfrac{18}{8}$

      Cancelling both numerator and denominator by $2$, the ratio becomes $9 : 4$
      Hence this option is wrong.

B. $105 : 36 = \dfrac{105}{36}$
    Cancelling both numerator and denominator by $3$, the ratio becomes $35 : 12$
     Hence this option is wrong.

C.  $91 : 28 = \dfrac{91}{28}$
      Cancelling both numerator and denominator by $7$, the ratio becomes $13 : 4$
      Therefore this is correct option. 

D.  $144 : 250 = \dfrac{144}{250}$
      Cancelling both numerator and denominator by $2$, the ratio becomes $72 : 125$
      Hence this option is also wrong. 

If $a : b = 3 : 5$  then $ a - b : a + b =$

  1. $\displaystyle \frac{-1}{4}$

  2. $\displaystyle \frac{1}{4}$

  3. $-4$

  4. $4$


Correct Option: A
Explanation:

Let $a = 3x$ and $b = 5x$.
$\therefore \displaystyle \frac{a-b}{a+b}=\frac{3x-5x}{3x+5x}=\frac{-2x}{8x}=-\frac{1}{4}$

If $\left( {{p^2} + {q^2}} \right)/\left( {{r^2} + {s^2}} \right) = \left( {pq} \right)/\left( {rs} \right)$, then what is the value of $\left( {p - q} \right)/\left( {p + q} \right)$ in terms of $r$ and $s$?

  1. $\left( {r + s} \right)/\left( {r - s} \right)$

  2. $\left( {r - s} \right)/\left( {r + s} \right)$

  3. $\left( {r + s} \right)/\left( {r s} \right)$

  4. $\left( {r s} \right)/\left( {r - s} \right)$


Correct Option: A

If $ \displaystyle \frac {1}{x} : \frac {1}{y} : \frac {1}{z} = 2:3:5, $ then $x:y:z =?$

  1. $2:3:5$

  2. $15:10:6$

  3. $5:3:2$

  4. $6:10:15$


Correct Option: B
Explanation:

$\dfrac { 1 }{ x } :\dfrac { 1 }{ y } :\dfrac { 1 }{ z } =\quad 2:3:5\ \dfrac { yz:xz:xy }{ xyz } =\quad 2:3:5\ yz:xz:xy\quad =\quad 2xyz:3xyz:5xyz\ 1:1:1=\quad 2x:3y:5z\ x:y:z=\quad \dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 } :\dfrac { 1 }{ 5 } =\dfrac { 15:10:6 }{ 30 } \ So,\quad x:y:z\quad is\quad 15:10:6$

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