Mathematics · Quantitative Aptitude

Ratios and Proportions

302 Questions

Ratios and proportions deal with comparing two or more quantities and finding their relationships. The questions involve calculating compound ratios, duplicate ratios, and solving proportional equations. This topic is a crucial part of the mathematics and quantitative aptitude sections in competitive exams.

compound ratiosduplicate ratiosproportion equationssimple ratio calculationscombining multiple ratios

Ratios and Proportions Questions

Multiple choice maths ratio, proportion and unitary method more on proportion terms related to proportion proportion

Find out
(i) the fourth proportional to 4, 9, 12
(ii) the third proportional to 16 and 36
(iii) the mean proportional between 0.08 and 0.18

  1. $27,81,0.12$
  2. $27,81,1.2$
  3. $9,27,0.12$
  4. $27,9,0.12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(i) Let the fourth proportional to $4, 9, 12$ be $x$

Then $4 : 9 : : 12 : x$
$\displaystyle \Rightarrow 4\times x=9\times 12$
$\Rightarrow x=\dfrac{9\times 12}{4}=27$
Therefore, fourth proportional to $4, 9, 12$ is $27$.
(ii) Let the third proportional to $16$ and $36$ is $x$ 
Then $16 : 36 : : 36 : x$
$\Rightarrow$ $16 \times  X = 36 \times  36$
$\displaystyle \Rightarrow  x=\frac{36\times 36}{16}=81$
Therefore, third proportional to $16$ and $36$ is $81$.
(iii) Mean proportional between $0.08$ and $0.18$
$\displaystyle =\sqrt{0.08\times 0.18}=\sqrt{\frac{8}{100}\times \frac{18}{100}}$
$\displaystyle =\sqrt{\frac{144}{100\times 100}}=\frac{12}{100}=0.12$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

In a GP, first term is $1$. If $4T _2 + 5T _3$ is minimum,then its common ratio is.

  1. $\dfrac {2}{5}$
  2. $-\dfrac {2}{5}$
  3. $\dfrac {3}{5}$
  4. $-\dfrac {3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given first term of the G.P is $1.$


Let the G.P be, $1,r,r^2,r^3,.....$

Thus $4T _2+5T _3= 4r+5r^2 = y$ (say)

For minimum value of $y$

$\cfrac{dy}{dr}=0\Rightarrow 4+10r=0\Rightarrow r=-\cfrac{2}{5}$
Hence, option 'B' is correct.

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

Mark the correct alternative of the following.
Two cylindrical jars have their diameters in the ratio $3:1$, but height $1:3$. Then the ratio of their volumes is?

  1. $1:4$
  2. $1:3$
  3. $3:1$
  4. $2:5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $V _1$ and $V _2$ be the volume of the two cylinders with radius $r _1$ and height $h _1$, and radius $r _2$ and height $h _2.$

$\dfrac{2r _1}{2r _2}=\dfrac{3}{1}$ and $\dfrac{h _1}{h _2}=\dfrac{1}{3}$             [ Given ]
So,
$V _1=\pi r _1^2h _1$           ----- ( 1 )
Now,
$V _2=\pi r _2^2h _2$            ---- ( 2 )
From equation ( 1 ) and ( 2 ), we get
$\dfrac{V _1}{V _2}=\left(\dfrac{r _1}{r _2}\right)^2\left(\dfrac{h _1}{h _2}\right)$

$\Rightarrow$  $\dfrac{V _1}{V _2}=\left(\dfrac{2r _1}{2r _2}\right)^2\left(\dfrac{h _1}{h _2}\right)$

$\Rightarrow$  $\dfrac{V _1}{V _2}=(3)^2\left(\dfrac{1}{3}\right)$

$\Rightarrow$  $\dfrac{V _1}{V _2}=\dfrac{3}{1}$

Multiple choice elements of accounts ratio analysis liquidity ratios accounting ratio's accounting ratios

Quick ratio is calculated by using the following formula ___________________.

  1. $\cfrac { Cash+near cash+debtors - Inventories }{ Current\quad liabilities } $
  2. $\cfrac { Cash+debtors }{ Current\quad liabilities } $
  3. $\cfrac { Cash }{ Current\quad liabilities } $
  4. $\cfrac { Cash+near\quad cash+debtors }{ Current\quad assets } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quick ratio is calculated by dividing liquid current assets by total current liabilities. Liquid current assets include cash, marketable securities and receivables. Cash includes cash in hand and cash at bank.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If $y=x^n$, then the ratio of relative errors in $y$ and $x$ is

  1. $1:1$
  2. $2:1$
  3. $1:n$
  4. $n:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$y=x^{n}$
$\Rightarrow \displaystyle \dfrac{dy}{dx}=nx^{n-1}$
Approximate error in y is $\displaystyle dy=\left (\dfrac{dy}{dx}\right) \Delta x$
                                      $=nx^{n-1} \Delta x$
Relative error in y is $\displaystyle \dfrac{dy}{y}=\dfrac{n}{x}\Delta x$
Approximate error in x is $\displaystyle dx=\left (\dfrac{dx}{dy}\right) \Delta y$
                                     $\displaystyle=\dfrac{1}{nx^{n-1}} \Delta y$
Relative error in x is $\displaystyle \dfrac{dx}{x}=\dfrac{1}{nx^{n}}\Delta y$
Required ratio $\displaystyle = \dfrac{\dfrac{n}{x}\Delta x}{\dfrac{1}{nx^{n}}\Delta y}$
                               $\displaystyle =n^{2}x^{n-1} \dfrac{\Delta x}{\Delta y}$
                               $\displaystyle =\dfrac{n}{1}$
So, the ratio of relative errors in y and x is $ n:1$.
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ m }{ P } _{ { m }^{ - } }\quad (n-2)!$
  2. $^{ m }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (n-2)!$
  3. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (m-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P (distinct) = (n * (n-1)) * m! * (n-2)!. Q (alike) = (n-2)!. The ratio P/Q is n! * m! / (n-2)! which simplifies based on the specific arrangement constraints.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ 2^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$
  2. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$
  3. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 459936. The logic remains the same.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$
  2. $^{ m }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$
  3. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(m-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 459936 and 459940.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

If the ratio of the corresponding sides of two similar triangles is 2 : 3, then the ratio of their corresponding altitude is :

  1. 3 : 2

  2. 16 : 81

  3. 4 : 9

  4. 2 : 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In two similar triangles, if the corresponding sides are in a particular ratio, then altitudes will also be in the same ratio.
Hence the ratio of the altitudes will be 2 : 3.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

Ratio of areas of two similar triangles is equal to :

  1. ratio of squares of the corresponding altitudes

  2. ratio of squares of corresponding medians.

  3. Either (A) or (B)

  4. (A) and (B) both
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ratio of areas of two similar triangles is equal to ratio of squares of the corresponding altitudes and ratio of squares of corresponding medians. This means that if the ratio of either altitude or median is given and asked to find ratio of areas , then it will be the ratio of squares of the corresponding altitudes or ratio of squares of corresponding medians.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If the sum of an infinite GP is 20 and sum of their square is 100 then common ratio will be 

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{3}{5}$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $a,ar,a{r}^{2},...$ to $\infty$
Now, sum of infinite G.P$=20$
$\Rightarrow \dfrac{a}{1-r}=20$          .........$\left(1\right)$
where each term of the above G.P is squared, then the progression becomes
${a}^{2},{a}^{2}{r}^{2},{a}^{2}{r}^{4},.....$
Now, first term $A={a}^{2}$ and common ratio$R={r}^{2}$
Sum of above G.P$=100$
$\Rightarrow \dfrac{A}{1-R}=100$
$\Rightarrow \dfrac{{a}^{2}}{1-{r}^{2}}=100$     ....$\left(2\right)$
Squaring $\left(1\right)$ we get
$\dfrac{{a}^{2}}{{\left(1-r\right)}^{2}}=400$    ....$\left(3\right)$
Dividing eqn$\left(2\right)$ by $\left(3\right)$ we get
$\dfrac{{\left(1-r\right)}^{2}}{1-{r}^{2}}=\dfrac{400}{100}=4$
$\Rightarrow \dfrac{{\left(1-r\right)}^{2}}{\left(1-r\right)\left(1+r\right)}=4$
$\Rightarrow \dfrac{1-r}{1+r}=4$
$\Rightarrow 4-4r=1+r$
$\Rightarrow 5r=3$
$\therefore r=\dfrac{3}{5}$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If the sum of an infinite $GP$ is $20$ and sum of their square is $100$ then common ration will be=

  1. $1/2$
  2. $1/4$
  3. $3/5$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $G.P$ is $a, ar, ar^2,....\infty$
sum $=\dfrac {a}{1-r}=20---(1)$
If terms are required the $G.P.$
becomes : $a^2, a^2r^2, a^2 r^4,.....\infty$
Sum $=\dfrac {a^2}{1-r^2}=100---(2)$
Name dividing square of equation $(1)$ by equation $(2)$
$\dfrac {\dfrac {a^2}{1-r^2}}{\dfrac {a^2}{(1-r)^2}}=\dfrac {100}{400}$
$\Rightarrow \ \dfrac {(1-r)}{(1-r)(1+r)}=\dfrac {1}{4}$
$\Rightarrow \ \dfrac {1-r}{1+r}=\dfrac {1}{4}$
$\Rightarrow \ 4-4r=1+r$
$\Rightarrow \ 4-1=4r+r$
$\Rightarrow \ 5r=3$
$\Rightarrow \ r=\dfrac {3}{5}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

In a geometric progression with common ratio 'q', the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. If the sum of the first nine terms of the progression is $\displaystyle \frac {\lambda}{q^{100}}$ then the value of $ \lambda $ equals to

  1. $10$
  2. $14$
  3. $12$
  4. $22$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$r=q$ (common ratio)
${ S } _{ n }=\cfrac { a({ r }^{ n }-1) }{ (r-1) } \\ { S } _{ 109 }={ S } _{ 100 }+12\\ \cfrac { a({ q }^{ 109 }-1) }{ (q-1) } =\cfrac { a({ q }^{ 100 }-1) }{ (q-1) } +12\quad \quad (1)\\ \cfrac { a({ q }^{ 9 }-1) }{ (q-1) } =\cfrac { \lambda  }{ { q }^{ 100 } } \\ \lambda =\cfrac { a({ q }^{ 109 }-{ q }^{ 100 }) }{ (q-1) } \quad \quad \quad (2)$
From $(1)$ and $(2)$
$\lambda =12$
Multiple choice physics motion and measurement physical quantities units - definitions and systems physical quantities like mass and weight

If kg $m^2s^2$ represents 'x' and g $cm^2s^2$ represents 'y', then find the ratio of x to y.

  1. 10$^7$
  2. 10$^5$
  3. 10$^9$
  4. 10$^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that, 

$1 kg=1000g$
$1m=100cm$

Now, converting the unit $kg\ m^2\ s^2$ into $g\ m^2\ s^2$

$1\ kg\ m^2\ s^2=1000\ g\times(100\ m)^2\ s^2$

$=1000\times10^4\ g\ m^2\ s^2$
$=10^7 g\ cm^2\ s^2$

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

Three number are in the ratio of $3 : 4 : 5$ and their L.C.M. is $2400$. Their H.C.F. is:

  1. $40$
  2. $80$
  3. $120$
  4. $200$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers be $3x$, $4x$ and $5x$.
Then, their L.C.M. $= 60x$.
So, $60x = 2400$ or $x = 40$.
$\therefore $ The numbers are $\left( 3\times 40 \right) $, $\left( 4\times 40 \right) $ and $\left( 5\times 40 \right) $.
Hence, required H.C.F. $= 40$.