Tag: ratio and proportion

Quizzes (7) Questions (130)

Questions Related to ratio and proportion

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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. $40$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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. $4:2::3:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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. $12:9::40:13$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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. $\dfrac{d}{a}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given a/b = b/c = c/d = k, then c = dk, b = ck = dk^2, and a = bk = dk^3. Substituting these into the expression: (b^3 + c^3 + d^3) / (a^3 + b^3 + c^3) = (d^3k^6 + d^3k^3 + d^3) / (d^3k^9 + d^3k^6 + d^3k^3) = d^3(k^6 + k^3 + 1) / d^3k^3(k^6 + k^3 + 1) = 1/k^3. Since a/b = k, then b/a = 1/k, so (b/a)^3 = 1/k^3. Alternatively, d/a = d/(dk^3) = 1/k^3.

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 

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

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

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

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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

Solve for $x$:

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

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

  2. <span>$\dfrac{1}{2}$</span>

  3. <span>$\dfrac{3}{4}$</span>

  4. <span>$\dfrac{1}{4}$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$\Rightarrow\dfrac{(2-7x)+(3+7x)}{(2-7x)-(3+7x)}=\dfrac{(1-5x)+(4+5x)}{(1-5x)-(4+5x)}$

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

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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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,

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

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

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

$=k$

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