Tag: problems on ratios

Questions Related to problems on ratios

Divide $Rs.\,715$ in the ratio $\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}\colon\displaystyle\frac{1}{4}$.

  1. $\;Rs.\,350,\,Rs.\,250,\,Rs.\,115$

  2. $\;Rs.\,300,\,Rs.\,250,\,Rs.\,165$

  3. $\;Rs.\,330,\,Rs.\,220,\,Rs.\,165$

  4. $\;Rs.\,400,\,Rs.\,260,\,Rs.\,55$


Correct Option: C
Explanation:

The ratio in which Rs $715$ is to be divided is $6:4:3$.

So, we have $13x = 715$ or $x = 55$
So, $6x$ will be Rs. $330$
$3x$ will be Rs $165$ and $4x$ is Rs $220$.

So, the answer is option $C$

An amount of money is to be distributed among $A,B\,$and$\,C$ in the ratio $3\,\colon\,1\,\colon\,5$. The difference between $B's$ share and $C's$ share is $Rs.\,3600$. What is the total of $A's$ share and $B's$ share?

  1. $\;Rs.\,5400$

  2. $\;Rs.\,3600$

  3. $\;Rs.\,2700$

  4. $\;Rs.\,1800$


Correct Option: B
Explanation:

Let the shares of $A,B\,$and$\,C$ be $3x,x\,$and$\,5x$ respectively.
$\;\;\;\;\;\;\;$ Given, $5x-x=3600\;\;\;\;\;\Rightarrow\;4x=3600$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow\,x=900$
$\;\;\;\;\;\;\;\;\therefore\;A's\;share+B's\;share=Rs.\,3\times900+Rs.\,900$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=Rs.\,2700+Rs.\,900$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\,Rs.\,3600$

Three numbers are in the ratio $\displaystyle\frac{1}{2}\colon\displaystyle\frac{2}{3}\colon\displaystyle\frac{3}{4}$. The difference between the greatest and the smallest numbers is $36$. The numbers are

  1. $\;72,84,108$

  2. $\;60,72,96$

  3. $\;72,84,96$

  4. $\;72,96,108$


Correct Option: D
Explanation:

The three numbers are in the ratio $\displaystyle\frac{1}{2}\colon\displaystyle\frac{2}{3}\colon\displaystyle\frac{3}{4}$,
$\;\;\;\;\;\;\;\;i.e.,\displaystyle\frac{1}{2}\times12\colon\displaystyle\frac{2}{3}\times12\colon\displaystyle\frac{3}{4}\times12,i.e.,\,6\,\colon8\,\colon\,9$.
$\;\;\;\;\;\;\;\;\;$ Let the number be $6x,8x$ and $9x$.
$\;\;\;\;\;\;\;\;\;$ Given, $9x-6x=36$
$\;\;\;\;\;\;\;\;\;\Rightarrow\,3x=36\;\Rightarrow\;x=12$
$\therefore$The numbers are $72,96$ and $108$.

A certain sum of money is divided between $P$ and $Q$ in the ratio $3\displaystyle\frac{1}{2}\,\colon\,5\displaystyle\frac{1}{2}$. If $P$ gets $Rs.\,180$ less than $Q$, then what is the share of $Q$?

  1. $\;Rs.\,315$

  2. $\;Rs.\,495$

  3. $\;Rs.\,630$

  4. $\;Rs.\,810$


Correct Option: B
Explanation:

Let Q's share be Q, and P's share be $(Q - 180)$


P : Q is $7 : 11$

    $\dfrac { Q -  180 }{ Q } =\dfrac { 7 }{ 11 } $
$11Q - 1980=\quad 7Q$
               $ 4Q= 1980$
                 $Q = 495$

So, Q is Rs $495$.

The ratio of the monthly income to the saving of a family is $9\,\colon\,2$. If the monthly income of the family is $Rs.\,2700$, then its monthly expenditure will be

  1. $\;Rs.\,1800$

  2. $\;Rs.\,2209.09$

  3. $\;Rs.\,2100$

  4. $\;Rs.\,1600$


Correct Option: C
Explanation:

Let the monthly savings of the family be $Rs.\,x$.
Then,$\displaystyle\frac{9}{2}=\displaystyle\frac{2700}{x}\;\Rightarrow\;\;\;x=600$
$\therefore$ Monthly expenditure$=Rs.\,2700-Rs.\,600$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=Rs.\,2100$.

In a particular type of fertilizer, the ratio of two chemicals $A$ and $B$ is $2\,\colon\,5$. In $21:kg$ of this fertilizer, if $3:kg$ of $A$ is added. What will be the ratio of $A$ to $B$ in the new fertilizer?

  1. $\;1\,\colon\,1$

  2. $\;2\,\colon\,3$

  3. $\;3\,\colon\,5$

  4. $\;4\,\colon\,5$


Correct Option: C
Explanation:

Quantity of $A=\displaystyle\frac{2}{7}\times21:kg=6:kg$,
$\;\;\;\;\;\;\;$ Quantity of $B=\displaystyle\frac{5}{7}\times21:kg=15:kg$
$\;\;\;\;\;\;\;$ After adding $3:kg$ of $A$ in the given fertilizer,


$\;\;\;\;\;\;\;$ Required ratio $=\displaystyle\frac{Quantity\;of\;A}{Quantity\;of\;B}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle\frac{(6+3):kg}{15:kg}=\displaystyle\frac{9}{15}=3\,\colon\,5$. 

A company makes a profit of $Rs.\,9,00,000$. $20\%$ of which is paid as taxes. If the rest is divided among the partners $P,Q\,and\,R$ in the ratio $1\,\colon\,1\displaystyle\frac{1}{2}\colon2$, then the share of $P,Q$ and $R$ are respectively.

  1. $\;2,40,000;\,3,20,000;\,1,60,000$

  2. $\;3,20,000;\,2,40,000;\,1,60,000$

  3. $\;1,60,000;\,3,20,000;\,2,40,000$

  4. $\;1,60,000;\,2,40,000;\,3,20,000$


Correct Option: D
Explanation:

Amount to be paid in taxes $=20\%$ of $Rs.\,9,00,000=Rs.\,1,80,000$
$\;\;\;\;\;\;\;$ Amount to be distributed among $P,Q\,and\,R=9,00,000-1,80,000=7,20,000$.
$\;\;\;\;\;\;\;\therefore\;Let\;the\;shares\;of\;P,Q\,and\,R\;be\,x,\displaystyle\frac{3x}{2}$ and $2x$ respectively. 

Then,
$\;\;\;\;\;\;\;\;x+\displaystyle\frac{3x}{2}+2x=7,20,000\;\Rightarrow\;\;x=Rs.\,1,60,000$.
$\;\;\;\;\;\;\;\;The\;share\;of\;P,Q\,and\,R\;are\;Rs.\,1,60,000,Rs.\,2,40,000\,and\,Rs.\,3,20,000$.

If $378$ coins consist of one rupee, $50$ paise and $25$ paise coins whose values are in the ratio $13 : 11 : 7,$ then the number of $50$ paise coins will be

  1. $132$

  2. $128$

  3. $136$

  4. $133$


Correct Option: A
Explanation:

Let, Number of one-rupee coins = x,
Number of 50 paise coins = y,
Number of 25 paise coins = z,
Given, x + y + z = 378              ....... (1)
Also, $1 \times x : \displaystyle \dfrac{50}{100} \times y : \dfrac{25}{100} \times z = 13 : 11 : 7$
$\Rightarrow x : y/2 : z/4 = 13 : 11 : 7$
$\Rightarrow \displaystyle \dfrac{x}{13} = \dfrac{y/2}{11} = \dfrac{z/4}{7} =k $  (say)
$\Rightarrow x = 13k, y = 22 k, z = 28 k.$
Then, $13 k + 22k + 28 k = 378 $        (From (1))
$\Rightarrow 63 k = 378$  hence $ k = 6$
$\therefore $ Number of 50 paise coins $= 22 \times 6 = 132$

The ratio of the number of ladies to that of gents at a party was $3 : 2.$ When $20$ more gents joined the party, the ratio was reversed. The number of ladies present at the party was

  1. $36$

  2. $32$

  3. $24$

  4. $16$


Correct Option: C
Explanation:

Let the number of ladies and gents in the party be 3x and 2x respectively. Then,
$\displaystyle \frac{3x}{2x + 20} = \frac{2}{3} \Rightarrow 9x = 4x + 40 \Rightarrow 5x = 40$
$\Rightarrow x = 8$
$\therefore$ Number of ladies in the party $= 3x = 3 \times 8 = 24$

Two numbers are in the ratio $2\,\colon\,3$. If $2$ is subtracted form the first and $2$ is added to the second, the ratio becomes $1\,\colon\,2$. What is the sum of the number?

  1. $\;30$

  2. $\;28$

  3. $\;24$

  4. $\;10$


Correct Option: A
Explanation:

Let the two numbers be $2x$ and $3x$. Then,
$\;\;\;\;\;\;\;\;\displaystyle\frac{2x-2}{3x+2}=\displaystyle\frac{1}{2}\;\Rightarrow\;4x-4=3x+2\;\Rightarrow\,x=6$
$\therefore$ The numbers are $12$ and $18$.
Required sum$=12+18=30$.