Mathematics · Quantitative Aptitude

Ratios and Proportions

309 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 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. $= 40$.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

A sum of $Rs.350$ is to be divided between $A$ and $B$ in the ratio $3:4.$ If the amount distributed in the ratio $4:3$ the amount gained by $A$ is 

  1. $Rs.10$
  2. $Rs.30$
  3. $Rs.50$
  4. $Rs.100$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $A$ got $3x$ and $B$ got $4x$. Then,

$3x+4x=350$

$7x=350$

$x=50$

Therefore, $A$ got $150 Rs.$ and $B$ got  $200 Rs.$.

If the ratio was $4 : 3,$ $A$ got $Rs. 200$ and $B$ got $Rs. 150$.

Therefore, the amount gained by $A = Rs. 50$

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

Instead of dividing $Rs.\,117$ among $P,Q,R$ in the ratio $\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}\colon\displaystyle\frac{1}{4}$, by mistake it was divided in the ratio $2\,\colon\,3\,\colon\,4$. Who gained in the transaction? 

  1. only $P$
  2. only $Q$
  3. only $R$
  4. both $Q$ and $R$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Desired ratio $=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}\colon\displaystyle\frac{1}{4}=\displaystyle\frac{1}{2}\times12\colon\displaystyle\frac{1}{3}\times12\colon\displaystyle\frac{1}{4}\times12$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6\colon4\colon3$
Ratio by mistake $=2\colon3\colon4=6\colon9\colon12$
Hence, it is clear that both $Q$ and $R$ gained in the transaction.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

If Rs. $60$ is divided into two parts in the ratio $2 : 3,$ then the difference between those two parts is ______

  1. Rs. $10$
  2. Rs. $12$
  3. Rs. $5$

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

Rs. $60$ is divided into two parts $2:3$.
Their sum is $2+3=5$.
Thus first part is $60\, \times\, \displaystyle \frac {2}{5} =$ Rs. $24$
and second part will be $60\, \times\, \displaystyle \frac {3}{5}\, =$ Rs. $36$.
Therefore, difference will be $ 36 - 24 =$ Rs. $12$.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

A number $351$ is divided into two parts in the ratio $2 : 7.$ Find the product of the numbers.

  1. $20,294$
  2. $21,294$
  3. $25,295$
  4. $31,294$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the numbers be $2x$ and $7x$.
$2x + 7x = 351$
$x = 39$
Therefore, product of the numbers is $2x \times 7x$ $=$ $14x^2$
$=$ $14\, \times\, (39)^2$
$= 21,294$

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

The ratio of the heights of A and B is $4 : 3$ . If B is $1.2$m tall then the height of A is:

  1. $0.9$ m
  2. $1.8$ m
  3. $1.6$ m

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Height of A $\displaystyle =\frac{4}{3}\times$ height of B
                    

                    $\displaystyle =\dfrac{4}{3} \times 1.2 = 1.6$ m

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

An amount of money is to be divided among $P, Q$ and $R$ in the ratio $4 : 7 : 9.$ If the difference between the shares of $Q$ and $R$ is Rs. $500,$ what will be the difference between the shares of $P$ and $Q$?

  1. Rs. $500$
  2. Rs. $1000$
  3. Rs. $750$
  4. Rs. $850$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the shares of $P, Q$ and $R$ be Rs. $ 4x$, Rs. $7x$ and Rs. $ 9x$ respectively. 

Then, $9x - 7x = 500$
$\Rightarrow 2x = 500$
$ \Rightarrow x = 250$
$\therefore$ required difference $= 7x - 4x = 3x = 3 \times 250 =$ Rs. $750$.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

If Rs. $2,600$ is divided among three persons $A, B$ and $C$ in the ratio $\dfrac {1}{2} : \dfrac {1}{3} : \dfrac {1}{4}$, how much does $A$ get?

  1. Rs. $ 600$
  2. Rs. $ 800$
  3. Rs. $ 1.000$
  4. Rs. $ 1,200$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Among three persons Rs. $2600$ is divided in the ratio as $\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}$.
LCM of $\dfrac {1}{2} : \dfrac {1}{3} : \dfrac {1}{4} $ is $ 6 : 4 : 3$.
Therefore, $A$'s share $= \dfrac {6}{13} \times  2,600 =$ Rs. $1,200$.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

Three numbers are in the ratio $3 : 2 : 5$ and the sum of their squares is $1862$. What are the three numbers?

  1. $18, 12, 30$
  2. $24, 16, 40$
  3. $15, 10, 25$
  4. $21, 14, 35$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the three numbers b $3x, 2x, 5x$

Sum of their squares is $1862$.
Therefore, $(3x)^{2} + (2x)^{2} + (5x)^{2} = 1862$
$\Rightarrow 38x^{2} = 1862$
$\Rightarrow  x^{2} = 1862 \div 38 = 49$
$\Rightarrow x = 7$ 
The numbers are $21, 14, 35$.